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# Acst

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How long (in years, to 2 decimal places) will it take for \$1240 to accumulate to \$1860 at 5.1% p.a. simple interest? | | | Student Response | Value | Correct Answer | Answer: | 9.80 | 100% | 9.80 | | General Feedback: | | I = | S - P | = | 1860 - 1240 | = | 620 | | | | | I = | Prt | | | 620 = | 1240 × 0.051 × t | | | t = | 620
1240 × 0.051 | | | t = | 9.8 | | Score: | 9/9 | |

2. On 7 April, Mr X borrows \$1300 at 6.2% p.a. and repays on 12 November of the same year. Find the amount of simple interest paid. | | | Student Response | Value | Correct Answer | Answer: | 48.36 | 100% | 48.36 | | General Feedback: | |

| Number of days is 219 | | I | = 1300 × 0.062 × | 219
365 | | I | = 48.36 | |
Score: 9/9
3.

A bank discounts a note which will mature for \$3600 in 4 months. If the simple discount rate is 8.3% p.a. what is the price paid for the note? | | | Student Response | Value | Correct Answer | Answer: | 3500.40 | 100% | 3,500.40 | | General Feedback: | |

P | = S(1 - dt) |

| = 3600(1 - 0.083 × | 4 12 | ) |

| = 3500.4 |
Score: 9/9
4.

A bank discounts a note maturing for \$36,000 in 3 months using a simple discount rate of 9.2% p.a. What simple interest rate p.a. are they obtaining (as a %, to 2 decimal places)? (The rate of simple interest is always greater than the rate of simple discount.) | | | Student Response | Value | Correct Answer | Answer: | 9.42 | 100% | 9.42 | | General Feedback: | |

P = | S(1 - dt) | P = | 36000(1 - 0.092 × 3/12) | = | 35172 | | | I = | 36000 - 35172 | = | 828 | | | I = | Prt | 828 = | 35172 × r × 3/12 | r = | 828
(35172 × 3/12) | = | 0.094166 p.a. | = | 9.42% p.a. | Do not type % as part of your answer. | | | or solve 36000 = | 35172(1 + r × 3/12) |
Score: 9/9
5.

A note maturing for \$100,000 on 20 September is sold for \$96700 on 12 May of the same year. What rate of simple discount p.a. (as a %, to 2 decimal places) was used? | | | Student Response | Value | Correct Answer | Answer: | 8.42 | 0% | 9.19 | | General Feedback: | |

Number of days is 131 | | | | | | | | P | = | S(1 - dt) | 96700 | = | 100000(1 - d × 131/365) | d | = | (1 - | 96700
100000 | ) × | 365
131 | | | = | 0.091947 p.a. | | | | | | | | | = | 9.19% p.a. | | | | | | | | | | | | | | | or D | = | S - P | | = | 100000 - 96700 | | = | 3300 | D | = | Sdt | 3300 | = | 100000 × d × 131/365 | d | = | 3300 × 365
100000 × 131 |
Score: 0/10
6.

A note maturing for \$100,000 on 12 July is sold for \$96700 on 2 March of the same year. What rate of simple interest p.a. (as a %, to 2 decimal places) will the purchaser realise on the investment if it is held to maturity? | | | Student Response | Value | Correct Answer | Answer: | 9.13 | 0% | 9.44 | | General Feedback: | |

Number of days is 132 | | | | | | | | S | = | P(1 + rt) | 100000 | = | 96700(1 + r × 132/365) | r | = | ( | 100000
96700 | - 1) × | 365
132 | | | = | 0.094364 p.a. | | = | 9.44% p.a. | | | | | | | | or I | = | S - P | | = | 100000 - 96700 | | = | 3300 | I | = | Prt | 3300 | = | 96700 × r × 132/365 | r | = | 3300 × 365
96700 × 132 |
Score: 0/10
7.

The following transactions are for an account which had an opening balance of \$250 on 1 July. The rate of simple interest is 5.1% p.a., payable on the minimum monthly balances. Date | Deposit | Withdrawal | Balance | 1 July | - | - | \$250 | 15 August | \$300 | - | \$550 | 13 September | \$300 | - | \$850 | | | | | 25 October | - | \$250 | \$600 | 2 December | - | \$300 | \$300 | 18 March | \$310 | - | \$ 610 |
Calculate the interest due at the end of the financial year assuming no further transactions are made. (The financial year is 1 July to 30 June in the following year. Hence the total interest for the 12 months July, August, ... , June is required. As a check, the total of the minimum monthly balances for July, August and September is \$1,050.) Do not include the \$ sign in your answer. | | | Student Response | Value | Correct Answer | Answer: | 9.11 | 0% | 22.44 | | General Feedback: | |
The minimum monthly balances are
250, 250, 550, 600, 600, 300, 300, 300, 300, 610, 610, 610
The total of the minimum monthly balances is 5280
Interest = Prt where P is the sum of minimum monthly balances and t is 1/12 Interest = 5280 × 0.051 × | 1
12 |

Interest = 22.44 | |
Score: 0/10
8.

The following transactions are for an account which had an opening balance of \$250 on 1 July. The rate of simple interest is 5.4% p.a., payable on the daily balances. Date | Deposit | Withdrawal | Balance | 1 July | - | - | \$250 | 15 August | \$300 | - | \$550 | 13 September | \$300 | - | \$850 | | | | | 25 October | - | \$250 | \$600 | 2 December | - | \$300 | \$300 | 11 March | \$340 | - | \$640 |
Calculate the interest due at the end of the financial year assuming no further transactions are made. (The total interest for 1 July to 30 June in the next year inclusive is required. Make sure that the total number of days add to 365. The number of days for the \$250 balance is 45, for the \$550 balance is 29 etc.) | | | Student Response | Value | Correct Answer | Answer: | 29.45 | 0% | 27.68 | | General Feedback: | |
The number of days interest for the \$250 balance is from 1 July to 14 August including both dates. This is equivalent to the number of days between 1 July and 15 August including only one end point. Except for the final balance the number of days of interest is found in the usual fashion by counting the number of days between the dates given in the question but including only one end point.
For the final balance we must count the number of days between 11 March and 30 June including both these dates. This is equivalent to counting the number of days between 11 March and 1 July counting only one end day. Interest | = Prt for each balance | Interest | = 250 × 0.054 × | 45
365 | + 550 × 0.054 × | 29
365 | + 850 × 0.054 × | 42
365 | | + 600 × 0.054 × | 38
365 | + 300 × 0.054 × | 99
365 | + 640 × 0.054 × | 112
365 | Interest | = 27.68 | | | | | | |
Score: 0/10
9.

A 90-day promissory note will mature for \$10,000 plus simple interest at 6% p.a. for the term of the note. After 30 days it is sold to a bank which uses a simple discount rate of 4% p.a.

1. What is the maturity value of the note?
2. Find the price paid by the bank for the note.
3. What simple rate of interest p.a. did the original owner who paid \$10,000 for the note on the issue date actually earn? (as a %, to 2 decimal places)
4. What simple interest rate p.a. did the bank earn on its investment, assuming that it held the note until maturity? (as a %, to 2 decimal places) | | | Student Response | Value | Correct Answer | 1. | 10147.95 | 25% | Equals 10147.95 (25%)
Equals 10147.94 (25%)
Equals 10147.96 (25%)
Equals 10147.93 (25%)
Equals 10147.97 (25%) | 2. | 527.19 | 0% | Equals 10081.22 (25%)
Equals 10081.21 (25%)
Equals 10081.23 (25%)
Equals 10081.20 (25%)
Equals 10081.2 (25%)
Equals 10081.24 (25%) | 3. | 9950.68 | 0% | Equals 9.88 (25%)
Equals 9.87 (25%)
Equals 9.89 (25%)
Equals 9.86 (25%)
Equals 9.90 (25%)
Equals 9.9 (25%) | 4. | 4.94 | 0% | Equals 4.03 (25%)
Equals 4.02 (25%)
Equals 4.04 (25%)
Equals 4.01 (25%)
Equals 4.05 (25%) | | General Feedback: | |
1. S = P(1 + rt) = 10000(1 + 0.06 × 90/365) = 10147.95
2. If the note is issued at time 0, the bank buys at time 30 and maturity is at time 90. Hence for the bank S = 10147.95, d = 0.04 and t which is the time between P and S is (90 - 30) = 60 days.
P = S(1 - dt) = 10147.95(1 - 0.04 × 60/365) = 10081.22
OO in diagram for 3. and 4. means Original owner Price | 10000 | | 10081.22 | | 10147.95 | | |____ | ... | _____|_____ | ... | ____| | | Time (days) | 0 | ... | 30 | ... | 90 | | | OO buys | | OO sells/
Bank buys | | Maturity | |
3. Original owner pays P = 10000 at time 0 and receives S = 10081.22 at time 30.
Hence I = S - P = 81.22.
I = Prt, 81.22 = 10000 × r × 30/365, r = 81.22 × 365 / (10000 × 30) = 9.88% p.a.
4. For the bank P = 10081.22 at time 30 and S = 10147.95 at time 90.
Hence t = (90 - 30) = 60 days. I = S - P = 66.73
I = Prt, 66.73 = 10081.22 × r × 60/365, r = 66.73 × 365 / (10081.22 × 60) = 4.03% p.a. Score: 6/24

Title: | ACST101 Assignment 2 | Started: | 11 March 2009 2:02 PM | Submitted: | 16 March 2009 1:57 PM | Time spent: | 119:54:43 | Total score: | 82/100 = 82% Total score adjusted by 0.0 Maximum possible score: 100 | 1. | | | Find the future value if the principal is \$8500, the nominal rate of interest is 5.7% p.a. convertible 2 times a year and the term of the investment is 4 years. (The wording means that j2 = 5.7%, hence the half-yearly rate is 5.7%/2.) | | | Student Response | Value | Correct Answer | Answer: | 10642.74 | 100% | 10,642.74 | | General Feedback: | |
Work in half-yearly time units as determined by the interest rate.
The number of time units is 2 × 4 = 8
The interest rate per time unit is 0.057/2 = 0.0285
S = P(1 + i)n = 8500(1.0285)8 = 10642.74 | Score: | 7/7 | |
2.

Find the present value if \$8500 is due at the end of 7 years and the nominal rate of interest is 6.2% p.a. convertible 4 times a year. (Make sure that your answer for the present value is less than 8500.) | | | Student Response | Value | Correct Answer | Answer: | 13075.46 | 0% | 5,525.62 | | General Feedback: | |
Work in quarterly time units as determined by the interest rate.
The number of time units is 4 × 7 = 28
The interest rate per time unit is 0.062/4 = 0.0155
P = S(1 + i)-n = 8500(1.0155)-28 = 5525.62 Score: 0/7
3.

Find the future value of \$1,500 over 8 years at 10% p.a. payable: 1. annually 2. half-yearly 3. quarterly 4. monthly 5. daily (Check that the future value becomes slightly larger as the frequency of the interest payment increases.) | | | Student Response | Value | Correct Answer | 1. | 3215.38 | 20% | Equals 3215.38 (20%)
Equals 3215.37 (20%)
Equals 3215.39 (20%) | 2. | 3274.31 | 20% | Equals 3274.31 (20%)
Equals 3274.30 (20%)
Equals 3274.32 (20%)
Equals 3274.3 (20%) | 3. | 3305.64 | 20% | Equals 3305.64 (20%)
Equals 3305.63 (20%)
Equals 3305.65 (20%) | 4. | 3327.26 | 20% | Equals 3327.26 (20%)
Equals 3327.25 (20%)
Equals 3327.27 (20%) | 5. | 3337.95 | 20% | Equals 3337.95 (20%)
Equals 3337.94 (20%)
Equals 3337.96 (20%) | | General Feedback: | |
1. 1500(1 + 0.1)8 = 3215.38
2. 1500(1 + 0.1/2)2×8 = 3274.31
3. 1500(1 + 0.1/4)4×8 = 3305.64
4. 1500(1 + 0.1/12)12×8 = 3327.26
5. 1500(1 + 0.1/365)365×8 = 3337.95 Score: 20/20
4.

If the interest rate is 5.5% p.a. convertible 4 times a year, find the equivalent effective annual rate of interest. (as a %, to 2 decimal places) (Check that your calculated value of j1 is slightly larger than the j4 value of 5.5%.) | | | Student Response | Value | Correct Answer | Answer: | 5.61 | 100% | 5.61 | | General Feedback: | |

(1 + j1/1)1 | = (1 + j4/4)4 | 1 + j1 | = (1 + 0.055/4)4 | j1 | = (1.01375)4 - 1 | | = 0.056145 | | = 5.61% | Do not type % as part of your answer. |
Score: 7/7
5.

If the rate of interest is 9.4% p.a. effective, find the equivalent nominal rate of interest p.a. if it were payable 2 times a year. (as a %, to 2 decimal places) (j2 is required and not the half-yearly rate.
Check that your calculated value of j2 is slightly smaller than the j1 value of 9.4%.) | | | Student Response | Value | Correct Answer | Answer: | 9.19 | 100% | 9.19 | | General Feedback: | |

(1 + j2/2)2 | = (1 + j1/1)1 | | = 1.094 | j2 | = 2[(1.094)1/2 - 1] | | = 0.091889 | | = 9.19% |
Score: 7/7
6.

If the interest rate is 6.8% p.a. convertible 2 times a year, find the equivalent nominal rate of interest p.a. if it were payable 12 times a year. (as a %, to 2 decimal places) (j12 is required and not the monthly rate.
Check that your calculated value of j12 is slightly smaller than the j2 value of 6.8%.) | | | Student Response | Value | Correct Answer | Answer: | 6.71 | 100% | 6.71 | | General Feedback: | |

(1 + j12/12)12 | = (1 + j2/2)2 | | = (1 + 0.068/2)2 | j12 | = 12[(1.034)2/12 - 1] | | = 0.067056 | | = 6.71% |
Score: 8/8
7.

If the interest rate is 8.9% p.a. convertible daily, find the equivalent nominal rate of interest p.a. if it were payable half-yearly. (as a %, to 2 decimal places) (j2 is required and not the half-yearly rate.
Check that your calculated value of j2 is slightly larger than the j365 value of 8.9%.) | | | Student Response | Value | Correct Answer | Answer: | 9.10 | 100% | 9.10 | | General Feedback: | |

(1 + j2/2)2 | = (1 + j365/365)365 | | = (1 + 0.089/365)365 | j2 | = 2[(1 + 0.089/365)365/2 - 1] | | = 0.090999 | | = 9.1% |
Score: 8/8
8.

How much would Marina have to deposit today in an investment fund which guarantees to pay 6.0% p.a. payable 4 times a year, to have \$2900 in 6.5 years time? | | | Student Response | Value | Correct Answer | Answer: | 1969.16 | 100% | 1,969.16 | | General Feedback: | |
The 2900 payment needs to be moved back 6.5 years or 26 quarters.
The interest rate per quarter is 0.06/4. S | = P(1 + i)n | or P | = S(1 + i)-n | P | = 2900(1.015)-26 | | = 1969.16 |
Score: 9/9
9.

A sum of \$1200 is deposited today into an account earning interest at 7.4% p.a. convertible 2 times a year. What deposit should be made in 1 years time if the account is to accumulate to \$5600 in 4 years from today? (The deposit of \$1200 accumulated for (2×4) time units plus the accumulation of \$X for the appropriate number of time units will equal \$5600. As an example to check your method, suppose that \$1,200 is deposited today into an account earning 10% p.a. convertible 4 times a year. What deposit should be made in 3 years time if the account is to accumulate to \$5,000 in 4 years from today? The answer is \$2,915.89) | | | Student Response | Value | Correct Answer | Answer: | 3212.70 | 100% | 3,212.70 | | General Feedback: | |

Deposits | 1200 | | X | | |____ | ... | _____|_____ | ... | ____| | i = 0.074/2 | Time (½ yrs) | 0 | ... | 2 | ... | 8 | | Accumulation | | | | | 5600 | | | | | | |  | |
The equation at time 8 for the deposit X is 1200(1.037)8 + X(1.037)(8 - 2) | = 5600 | X | = 3212.7 |
Score: 9/9
10.

Find the total value on 1 July 2007 of payments of \$760 on 1 July 1996 and \$750 on 1 July 2011 if the rate of interest is 4.4% p.a. convertible 2 times a year. (You will need to find the future value of the \$760 payment and add the present value of the \$750 payment. As an example to check your method, find the total value on 1 July 2007 of payments of \$1100 on 1 July 1989 and \$800 on 1 July 2018 if the rate of interest is 8% p.a. convertible 12 times a year. The answer is 4953.43) | | | Student Response | Value | Correct Answer | Answer: | 1815.06 | 0% | 1,856.85 | | General Feedback: | |

Payments | 760 | | | | 750 | | |____ | ... | _____|_____ | ... | ____| | i = 0.044/2 | Date | 7/1996 | ... | 7/2007 | ... | 7/2011 | | | | |  | | | |
The 760 payment needs to be moved forward 11 years or 22 half-years.
The 750 payment needs to be moved back 4 years or 8 half-years.
The total value is 760(1.022)22 + 750(1.022)-8 = 1856.85 Score: 0/10
11.

Funds that are borrowed and the repayment of which is a contractual obligation are called | | | Student Response | Value | Correct Answer | Feedback | A. | debt finance | 100% | | Compare with equity finance which is funds that are contributed by the owners of a business. | B. | equity finance | | | | | Score: | 1/1 | |
12.

Government bonds are an example of a debt security. | | | Student Response | Value | Correct Answer | Feedback | A. | True | | | | B. | False | 0% | | The government will make specified payments for a certain length of time. This is a debt security. | | Score: | 0/1 | |
13.

Short-term equity is traded on the money market. | | | Student Response | Value | Correct Answer | Feedback | A. | True | | | | B. | False | 100% | | Equity is long-term not short-term. Question should say "short-term debt". For a list of money market securities see Table 1.1. | | Score: | 1/1 | |
14.

A contract committing two parties to trade a specific security or commodity at a stipulated price at a specified future date is called | | Feedback | | Student Response | Value | Correct Answer | | A. | a futures contract | 100% | | | B. | an option | | | | Score: | 1/1 | |
15.

In the flow of funds in Australia, the household sector is a deficit sector. | | | Student Response | Value | Correct Answer | Feedback | A. | True | | | | B. | False | 100% | | Whilst some households would be borrowers, on average this sector is a net lender. | | Score: | 1/1 | |
16.

For each statement in the left hand column choose the appropriate response from the right hand column. | | Statement | Response | Value | Correct Match | The possibility that financial transactions will not achieve their desired outcome is called | risk | 50.0% | risk | The ability for an investment to be easily converted to cash with little or no loss of capital is called | liquidity | 50.0% | liquidity | | Score: | 1/1 | |
17.

For each type of financial institution the percentage share of total financial institution assets can be calculated. For banks this percentage has been greater than 50% for the last 10 years. | | | Student Response | Value | Correct Answer | Feedback | A. | True | | | | B. | False | 100% | | The percentage has been between 40% and 50%. | | Score: | 1/1 | |
18.

The market which trades in longer-term securities is called the | | | Student Response | Value | Correct Answer | Feedback | A. | Capital market | 100% | | The secondary market trades in existing securities which may be short-term or long-term. | B. | Secondary market | | | | | Score: | 1/1 |

Title: | ACST101 Assignment 3 | Started: | 20 March 2009 1:11 PM | Submitted: | 23 March 2009 3:57 PM | Time spent: | 74:45:51 | Total score: | 82/100 = 82% Total score adjusted by 0.0 Maximum possible score: 100 | 1. | | | Find the nominal annual rate of interest convertible quarterly (as a % to 2 decimal places) given that:

Amount at the beginning of the period is \$2000

Amount at the end of the period is \$3300

Time period is 3 years and 9 months (Work in quarters and calculate the quarterly interest rate. Your final answer however should be j4 and not the quarterly rate.) | | | Student Response | Value | Correct Answer | Answer: | 13.58 | 100% | 13.58 | | General Feedback: | |
Use S = P(1 + i)n and work in quarterly time units at i per quarter The number of time units | = 4×3 + 3 = 15 | 2000(1 + i)15 | = 3300 at i per quarter | (1 + i)15 | = 1.65 | (1 + i) | = 1.651/15 | j4 | = 4i | | = 0.135794 | | = 13.58% | Do not type % as part of your answer | | Score: | 9/9 | |
2.

An obligation of \$1800 falls due at the end of 7 years. Find an equivalent debt at the end of 6 years, if the interest rate is 7% p.a. convertible monthly. (The \$1800 will need to be moved back the appropriate number of months at the monthly rate of 7%/12.) | | | Student Response | Value | Correct Answer | Answer: | 1678.65 | 100% | 1,678.65 | | General Feedback: | |
Work in monthly time units at 0.07/12 per month
The 1800 has to be moved back (7 - 6) years or 12 months using P = S(1 + i)-n Payments | | | X | ← | 1800 | | |____ | ... | _____|_____ | ... | ____| | i = 0.07/12 | Time (mths) | 0 | ... | 72 | ... | 84 | | | | | ↓ | | | |

X | = 1800(1 + 0.07/12)-12 | | = 1678.65 |
Score: 9/9
3.

An obligation of \$3800 falls due at the end of 7 years. Find an equivalent debt at the end of 14 years, if the interest rate is 9% p.a. convertible monthly. | | | Student Response | Value | Correct Answer | Answer: | 7118.17 | 100% | 7,118.17 | | General Feedback: | |
Work in monthly time units at 0.09/12 per month
The 3800 has to be moved forward (14 - 7) years or 84 months using S = P(1 + i)n Payments | | | 3800 | → | X | | |____ | ... | _____|_____ | ... | ____| | i = 0.09/12 | Time (mths) | 0 | ... | 84 | ... | 168 | | | | | | | ↓ | |

X | = 3800(1 + 0.09/12)84 | | = 7118.17 |
Score: 9/9
4.

The following timeline represents a loan of \$1,000 being repaid by payments of \$100 at time 1, \$200 at time 2 and \$R at time 3. The rate of interest is i per time interval. Loan | 1000 | | |____ | ____|____ | ____|____ | ____| | use rate i | Time | 0 | 1 | 2 | 3 | | Repayments | | 100 | 200 | R | | | ↓ | | | | |
If the student chooses the focal date at time 0, a correct equation of value would be: | | | Student Response | Value | Correct Answer | Feedback | A. | 1000 = 100(1+i)1 + 200(1+i)2 + R(1+i)3 | | | | B. | 1000 = 100(1+i)-1 + 200(1+i)-2 + R(1+i)-3 | 100% | | | C. | 1000 + 100(1+i)1 + 200(1+i)2 = R(1+i)3 | | | | D. | 1000 + 100(1+i)-1 + 200(1+i)-2 = R(1+i)-3 | | | | | Score: | 4/4 | |
5.

A person borrows \$2,500 at 10% p.a. convertible quarterly.

If she promises to pay \$600 at the end of one year, \$800 at the end of 2 years and the balance at the end of 3 years, what will the final balance be? (As the interest rate is j4, draw a timeline in quarters and use the quarterly rate 10%/4. The \$2,500 should be on one side of the timeline at time 0. On the other side of the timeline the repayment of \$600 should be at time 4, \$800 at time 8 and \$X at time 12. To find X set up an equation of value with the focal date at time 12.) | | | Student Response | Value | Correct Answer | Answer: | 1816.88 | 0% | 1,748.13 | | General Feedback: | |
Work in quarters at 0.1/4 per quarter = 0.025
Put the focal date at time 12 (quarters) which is the time of the unknown payment Loan | 2500 | | |____ | ____|____ | ____|____ | ____| | i = 0.025 | Time (¼ yrs) | 0 | 4 | 8 | 12 | | Repayments | | 600 | 800 | X | | | | | | ↓ | |

2500(1.025)12 | = 600(1.025)8 + 800(1.025)4 + X | X | = 1748.13 |
Score: 0/9
6.

A person borrows \$2,500 at 11% p.a. convertible quarterly.

If he promises to pay \$X at the end of one year, \$700 at the end of 2 years and \$900 at the end of 3 years, what will the payment of \$X be? (Draw a timeline in a similar way to the previous question. Working in quarters, this time set up an equation of value with focal date at time 4.) | | | Student Response | Value | Correct Answer | Answer: | 1434.12 | 100% | 1,434.12 | | General Feedback: | |
Work in quarters at 0.11/4 per quarter = 0.0275
Put the focal date at time 4 (quarters) which is the time of the unknown payment Loan | 2500 | | |____ | ____|____ | ____|____ | ____| | i = 0.0275 | Time (¼ yrs) | 0 | 4 | 8 | 12 | | Repayments | | X | 700 | 900 | | | | ↓ | | | |

2500(1.0275)4 | = X + 700(1.0275)-4 + 900(1.0275)-8 | X | = 1434.12 |
Score: 9/9
7.

How much will \$6100 accumulate to in 11 years' time if the interest rate is 10.5% p.a. effective for the first 6 years and 5.6% p.a. effective for the next 5 years? | | | Student Response | Value | Correct Answer | Answer: | 14582.20 | 100% | 14,582.20 | | General Feedback: | |
Use S = P(1 + i)n twice.
The S value after the first 6 year accumulation, 6100(1.105)6, becomes the P value for the second 5 year accumulation. Future value after 11 years | = [6100(1.105) 6](1.056)5 | | = 14582.2 |
Score: 9/9
8.

A company wishes to replace the following three debts:

\$25,000 due on 1 July 2005
\$19,000 due on 1 January 2007
\$11,000 due on 1 July 2010

with a single debt of \$Y payable on 1 January 2007. Calculate Y if the interest rate is 4.4% p.a. payable half-yearly up to 1 January 2007 and 6.0% p.a. payable half-yearly after 1 January 2007. (Note that some dates are 1 January and others are 1 July. The \$25,000 payment should be moved forward using the first interest rate, the \$19,000 payment is on the focal date and therefore no interest adjustment is required and the \$11,000 payment is moved back using the second interest rate.) | | | Student Response | Value | Correct Answer | Answer: | 59408.25 | 0% | 54,630.57 | | General Feedback: | |
In the timeline i1 = 0.044/2 and i2 = 0.06/2 where half-yearly time units are used. | |← | i1 = 0.022 | →|← | i2 = 0.03 | →| | Payments | 25000 | | 19000 | | 11000 | | |____ | ... | _____|_____ | ... | ____| | Date | 1/7/2005 | ... | 1/1/2007 | ... | 1/7/2010 | | | | ↓ | | |

Total value | = 25,000(1.022)3 + 19,000 + 11,000(1.03)-7 | | = 54630.57 |
Score: 0/9
9.

How long (in years to 2 decimal places) will it take to increase an investment by 89% if interest is at 7.1% p.a. compounded daily? (If P is say \$100 then S will be \$189. Work in days with a daily rate of 7.1%/365. Convert your final answer to years.) | | | Student Response | Value | Correct Answer | Answer: | 8.97 | 100% | 8.97 | | General Feedback: | |

Let the investment P be 100 and the time n in days.
The future value S will then be 189.
The rate of interest per time unit is 0.071/365. S = | P(1 + i)n | 189 = | 100(1 + 0.071/365)n | n = | log 1.89 log(1 + 0.071/365) | n = | 3272.86 days | = | 8.97 years |
Score: 9/9
10.

Solve the following equation for 'n' (to 2 decimal places) using linear interpolation:

1.1n = 8.58
(1.1 raised to the power of n is equal to 8.58) The starting points for linear interpolation should be the integer values either side of the exact value of n calculated using logarithms e.g. if the exact value of n is say 6.59 then interpolate between n=6 and n=7. The exact value of n using logs and the value of n using linear interpolation may be different. If they are different the answer calculated using logs will be marked as incorrect. | | | Student Response | Value | Correct Answer | Answer: | 22.54 | 100% | 22.54 | | General Feedback: | |

The exact solution for n is | log 8.58 log 1.1 | = 22.55 |

Linear interpolation between n = 22 and n = 23 is required. |

From calculator 1.122 | = | 8.140275 | Given 1.1n | = | 8.58 | From calculator 1.123 | = | 8.954032 | n - 22
23 - 22 | = | 8.580000 - 8.140275
8.954302 - 8.140275 | n | = | 22.54 |

Linear interpolation is an approximation and may not give the exact answer. |
Score: 9/9
11.

At 6% p.a. the present value of a series of payments is \$1,809. At 6.5% p.a. the present value of the same series of payments is \$1,764. Use linear interpolation to estimate the rate of interest (as a % to 2 decimal places) for which these payments have a present value of \$1,786. (Knowing that we are trying to find i which is between 6% and 6.5%, the left hand side of the interpolation expression can be written down. The entries on the right hand side will be the present values corresponding to the interest rate in the same position on the left hand side.) | | | Student Response | Value | Correct Answer | Answer: | 6.26 | 100% | 6.26 | | General Feedback: | |

Present value at 6% | = | 1809 | Present value at i% | = | 1786 | Present value at 6.5% | = | 1764 | The linear interpolation expression for i as a % is | i - 6
6.5 - 6 | = | 1786 - 1809
1764 - 1809 | | i | = | 6.26% | |

The curve which represents a present value decreases as the interest rate increases. |
Score: 9/9
12.

A bank takes a deposit from a customer. This is recorded in the bank's balance sheet as | | | Student Response | Value | Correct Answer | Feedback | A. | an asset | | | | B. | a liability | 100% | | The bank has a liability to repay the deposit. | | Score: | 1/1 | |
13.

An certificate of deposit is similar to | | | Student Response | Value | Correct Answer | Feedback | A. | a debenture | | | | B. | a promissory note | 100% | | A CD is a short-term instrument similar to a P-note. | C. | an ordinary share | | | | | Score: | 1/1 | |
14.

An arrangement in which a bank allows a business to place its operating account into deficit up to an agreed limit is called | | | Student Response | Value | Correct Answer | Feedback | 1. | an overdraft | 100% | | In a fixed term loan the amount borrowed is a fixed amount and the amount of the repayments will be agreed with the bank. | 2. | a fixed term loan | | | | | Score: | 1/1 | |
15.

Which of the following is an example of off-balance-sheet business? | | | Student Response | Value | Correct Answer | Feedback | A. | bill acceptance | | | | B. | letters of credit | 100% | | The bank only has a liability if one of its clients defaults to a third party. | | Score: | 1/1 | |
16.

The two main categories of bank assets are term deposits and current deposits. | | | Student Response | Value | Correct Answer | Feedback | A. | True | | | | B. | False | 100% | | These are liabilities. | | Score: | 1/1 | |
17.

Company A arranges for a bank to accept a bill. The bank charges Comapny A a fee. Company A sells the bill in the market to Company B at a price calculated to yield Company B a rate of simple interest of i p.a. For the overall transaction including the bank fee, the rate of simple interest p.a. being paid by Company A is | | | Student Response | Value | Correct Answer | Feedback | A. | less than i | | | | B. | equal to i | | | | C. | more than i | 100% | | If there was no bank fee Company A would be paying rate i p.a. However the bank fee will increase the amount to be paid back, hence the rate being paid on the borrowed money is greater than i p.a. | | Score: | 1/1 |

Title: | ACST101 Assignment 4 | Started: | 31 March 2009 6:10 PM | Submitted: | 31 March 2009 9:15 PM | Time spent: | 03:04:59 | Total score: | 38/100 = 38% Total score adjusted by 0.0 Maximum possible score: 100 | 1. | | | For n payments of size R at rate i Rs(n, i) finds | | | Student Response | Value | Correct Answer | Feedback | A. | the future value at the time of the last payment | 100% | | | B. | the future value one time unit after the last payment | | | | C. | the present value at the time of the first payment | | | | D. | the present value one time unit before the first payment | | | | | Score: | 2/2 | | | 2. | | | For n payments of size R at rate i Ra(n, i) finds | | | Student Response | Value | Correct Answer | Feedback | A. | the future value at the time of the last payment | | | | B. | the future value one time unit after the last payment | | | | C. | the present value at the time of the first payment | | | | D. | the present value one time unit before the first payment | 100% | | | | Score: | 2/2 | | | 3. | | | A woman deposits \$300 each year into a savings account that pays interest at 4.3% p.a. effective. If she makes her first deposit on 1 July 2000, how much will she have in her account just after she makes her deposit on 1 July 2010? (Make sure you count the number of deposits correctly. If the first deposit was on 1 July 2000 and the last on 1 July 2010 there would be 11 deposits.) | | | Student Response | Value | Correct Answer | Answer: | 3652.34 | 0% | 4,109.39 | | General Feedback: | |
In the assignment solutions the notation used for the 's' function will be s(n,i) where n is the number of payments and i is the interest rate per time unit. Similar notation will be used for the 'a' function. When counting the number of payments, there are 10 years between the date of the first payment and date of the last payment but as there is a payment on both dates the number of payments is (10 + 1) = 11. The focal date is at the time of the last payment so the 's' function is exact. Payment | 300 | 300 | ... | 300 | 300 | | |____ | ____|__ | ... | __|____ | ____| | i = 0.043 | Date | 7/2000 | 7/2001 | ... | 7/2009 | 7/2010 | | | | | | | ↓ | | Future value | = 300 | s(11, 0.043) | | = 300 | (1.043)11 - 1 0.043 | | = 300 × 13.69797 | | = 4109.39 | | Score: | 0/8 | |
4.

A woman deposits \$1600 into an investment account each 1 January, starting in 1997 and continuing until 2008 inclusive. If the fund pays interest at 5% p.a. effective, how much will be in her account on 1 January 2015? | | | Student Response | Value | Correct Answer | Answer: | 35835.19 | 100% | 35,835.19 | | General Feedback: | |
When counting the number of payments, there are 11 years between the date of the first payment and date of the last payment but as there is a payment on both dates the number of payments is (11 + 1) = 12.
As the focal date is after the time of the last payment, the 's' function is not exact. The accumulation at 1 January 2008 has to be moved forward 7 years by multiplying by (1.05)7. Payment | 1600 | 1600 | ... | 1600 | 1600 | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __| | i = 0.05 | Date | 1/1997 | 1/1998 | ... | 1/2007 | 1/2008 | ... | 1/2015 | | | | | | | ↓ | → | ↓ | |

Future value | = 1600s(12, 0.05)(1.05)7 | | = 35835.19 |
The numerical value of s(12, 0.05) is 15.91713 which is greater than 12 as expected. Score: 8/8
5.

A man deposits \$600 at the end of each year for 5 years, and then \$200 at the end of each year for 6 years. Find the future (accumulated) value of these deposits at the end of 11 years if interest is at 5.3% p.a. effective. | | | Student Response | Value | Correct Answer | Answer: | 6998.37 | 0% | 5,917.49 | | General Feedback: | |
Considering the 5 payments of 600, the 's' function will give an exact accumulation to time 5 and as in Question 4 this will have to be multiplied by (1.053)6 to move the accumulation to time 11.
The 's' function will give an exact accumulation for the 6 payments of 200. Payment | | 600 | ... | 600 | 200 | ... | 200 | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __| | i = 0.053 | Time | 0 | 1 | ... | 5 | 6 | ... | 11 | | | | | | | | | ↓ | |

Future value | = 600s(5, 0.053)(1.053)6 + 200s(6, 0.053) | | = 5917.49 |

| | Score: | 0/9 | |
6.

Find the present value of an ordinary annuity of \$4,000 p.a. for 8 years if interest is at 8.7% p.a. effective. | | | Student Response | Value | Correct Answer | Answer: | 22388.29 | 100% | 22,388.29 | | General Feedback: | |
The focal date is one time unit before the first payment so the 'a' function is exact. Payment | | 4000 | 4000 | ... | 4000 | 4000 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.087 | Time (yrs) | 0 | 1 | 2 | ... | 7 | 8 | | | ↓ | | | | | | |

Present value | = 4000 | a(8, 0.087) | | = 4000 | 1 - (1.087)-8 0.087 | | = 4000 × 5.59707 | | = 22388.29 |
Score: 7/7
7.

Repeat the previous question with the interest rate increased to 10.1% p.a. effective.
Note that the present value decreases when the interest rate increases. | | | Student Response | Value | Correct Answer | Answer: | 21262.24 | 100% | 21,262.24 | | General Feedback: | |

Present value | = 4000 | a(8, 0.101) | | = 4000 | 1 - (1.101)-8 0.101 | | = 4000 × 5.31556 | | = 21262.24 |
Score: 7/7
8.

Calculate at j1 = 7.7% the present value at 1 June 2007 of annual payments of \$1400 payable from 1 June 2011 to 1 June 2018 inclusive.
(The 'a' function will value one year before 1 June 2011 and NOT at 1 June 2011.) | | | Student Response | Value | Correct Answer | Answer: | 6514.10 | 100% | 6,514.10 | | General Feedback: | |
When counting the number of payments, there are 7 years between the date of the first payment and date of the last payment but as there is a payment on both dates the number of payments is (7 + 1) = 8.
As the focal date is 4 years before the time of the first payment, the 'a' function is not exact. The 'a' function will find the value of the payments one time unit before the first payment ie at 1 June 2010. The value at 1 June 2010 has to be moved back 3 years by multiplying by (1.077)-3. Payment | | | | | 1400 | ... | 1400 | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __| | i = 0.077 | Date | 6/2007 | 6/2008 | ... | 6/2010 | 6/2011 | ... | 6/2018 | | | ↓ | ← | ↓ | | | | |

Present value | = 1400a(8, 0.077)(1.077)-3 | | = 6514.1 |

The numerical value of a(8, 0.077) is 5.81265 which is less than 8 as expected. |
Score: 9/9
9.

An annuity pays \$4,000 at the end of each year for 3 years and then \$5,000 at the end of each year for the next 9 years. Find the present value of these payments if the effective rate of interest is 8.9% p.a. | | | Student Response | Value | Correct Answer | Answer: | 40241.74 | 0% | 33,448.89 | | General Feedback: | |
Considering the 9 payments of 5000, the 'a' function will give an exact present value at time 3 (one time unit before the first 5000 payment) and this will have to be multiplied by (1.089)-3 to move the value to time 0.
The 'a' function will give an exact present value for the 3 payments of 4000. Payment | | 4000 | ... | 4000 | 5000 | ... | 5000 | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __| | i = 0.089 | Time (yrs) | 0 | 1 | ... | 3 | 4 | ... | 12 | | | ↓ | ← | ↓ | | | | |

Present value | = 4000a(3, 0.089) + 5000a(9, 0.089)(1.089)-3 | | = 33448.89 |
Score: 0/9
10.

It is estimated that a machine will need replacing 16 years from now at a cost of \$80,000. How much must be set aside at the end of each year for 16 years to provide that money if the company's savings earn interest at 7.9% p.a. effective? (Draw a timeline showing the payments of \$R into the fund on one side of your timeline and the payment out of the fund at time 16 on the other side. Choose a focal date and set up your equation of value to solve for R.) | | | Student Response | Value | Correct Answer | Answer: | 20202.05 | 0% | 2,660.45 | | General Feedback: | |

Deposit | | R | R | ... | R | R | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.079 | Time (yrs) | 0 | 1 | 2 | ... | 15 | 16 | | Replacement | | | | | | 80000 | | cost | | | | | | ↓ | |
Writing down an equation of value at time 16 Rs(16, 0.079) | = 80000 | 30.07011R | = 80000 | R | = 2660.45 |
Score: 0/9
11.

A family needs to borrow \$20,000 for some home renovations. The loan is to be repaid with level monthly payments over 7 years.

If they go to a finance company the interest rate will be j12 = 12.7%. Find the monthly payment. | | | Student Response | Value | Correct Answer | Answer: | not answered | 0% | 360.59 | | General Feedback: | |

Loan | 20000 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.127/12 | Time (mths) | 0 | 1 | 2 | ... | 83 | 84 | | Payments | | R | R | ... | R | R | | | ↓ | | | | | | |
Writing down an equation of value at time 0 where the 20000 is borrowed Ra(84, 0.127/12) | = 20000 | 55.46538R | = 20000 | R | = 360.59 |
Score: 0/7
12.

A family needs to borrow \$20,000 for some home renovations. The loan is to be repaid with level monthly payments over 7 years. If they go to a credit union the interest rate will be j12 = 11.6%. Find the monthly payment. (It should be less than the monthly payment of Question 11.) | | | Student Response | Value | Correct Answer | Answer: | not answered | 0% | 348.79 | | General Feedback: | |
Writing down an equation of value at time 0 where the 20000 is borrowed Ra(84, 0.116/12) | = 20000 | 57.34098R | = 20000 | R | = 348.79 |
Score: 0/7
13.

A person has \$4000 in an account which is earning interest at j2 = 4.2%. What level deposit is required at the end of each half-year for the next 5 years if the account balance is to be \$10000 at the end of the 5 years? (The deposits will be \$4000 at time 0 and \$R at times 1 to (2×5) inclusive. With focal date at time (2×5) the accumulation of these deposits will equal \$10000. Don't forget to give interest to the single payment of \$4000.) | | | Student Response | Value | Correct Answer | Answer: | not answered | 0% | 461.46 | | General Feedback: | |

Deposits | 4000 | R | R | ... | R | R | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.042/2 | Time (½ yrs) | 0 | 1 | 2 | ... | 9 | 10 | | Accumulation | | | | | | 10000 | | | | | | | | ↓ | |
With focal date at time 10 in half-years, the single payment of 4000 has to be moved forward 10 time units using S = P(1 + i)n and the level payments of R are moved forward to the time of the last payment using the 's' function. 4000(1.021)10 + Rs(10, 0.021) | = 10000 | 1.231 × 4000 + 10.99991R | = 10000 | R | = 461.46 |
Score: 0/10
14.

In the Capital Adequacy Requirement for banks, capital is | | | Student Response | Value | Correct Answer | Feedback | A. | debt securities issued by the bank | | | | B. | equity securities issued by the bank | 100% | | Capital includes ordinary shares, preference shares issued by the bank. | | Score: | 1/1 | |
15.

Off-balance-sheet business is not covered by the Capital Adequacy Requirements for banks. | | | Student Response | Value | Correct Answer | Feedback | A. | True | 0% | | | B. | False | | | | | Score: | 0/1 | |
16.

In the Capital Adequacy Requirements for banks, allowance is made for | | | Student Response | Value | Correct Answer | Feedback | A. | credit risk only | | | | B. | market risk only | | | | C. | both credit risk and market risk | 100% | | A specified formula is used to determine the amount of capital required for credit risk. Either a specified formula or a formula agreed with APRA can be used for market risk. | | Score: | 1/1 | |
17.

Banks are required to open a special account with the Reserve Bank of Australia to cover losses arising from bad debts. | | | Student Response | Value | Correct Answer | Feedback | A. | True | 0% | | The SRD and non-callable deposits have been abolished. | B. | False | | | | | Score: | 0/1 | |
18.

Banks use a standard formula to determine the minimum amount of liquid assets that must be held. | | | Student Response | Value | Correct Answer | Feedback | A. | True | 0% | | PAR has been abolished. It was replaced by Liquidity Management Guidelines which are agreed between each bank and APRA. | B. | False | | | | | Score: | 0/1 | |
19.

The institution responsible for monetary policy in Australia is | | | Student Response | Value | Correct Answer | Feedback | A. | the Australian Prudential Regulation Authority (APRA). | | | | B. | the Reserve Bank of Australia (RBA). | 100% | | | | Score: | 1/1 |

Title: | ACST101 Assignment 5 | Started: | 1 April 2009 7:53 PM | Submitted: | 8 April 2009 8:26 AM | Time spent: | 157:33:16 | Total score: | 32/100 = 32% Total score adjusted by 0.0 Maximum possible score: 100 | 1. | | | Find the annual rate of compound interest (as a %, 2 decimal places) at which annual deposits of \$500 will accumulate to \$6110 in 8 years.
(Set up an equation of value equating the value of the deposits and the required accumulation. Solve using linear interpolation between 11% and 12%) | | | Student Response | Value | Correct Answer | Answer: | 11.82 | 100% | 11.82 | | General Feedback: | | Deposits | | 500 | 500 | ... | 500 | 500 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i to be found | Time (yrs) | 0 | 1 | 2 | ... | 7 | 8 | | Accumulation | | | | | | 6110 | | | | | | | | ↓ | |
With focal date at time 8 500s(8, i) | = | 6110 | | = | 12.22 | Now s(8, 0.11) | = | 11.85943 from calculator using formula for 's' function | and s(8, 0.12) | = | 12.29969 | i - 11%
12% - 11% | = | 12.22 - 11.85943
12.29969 - 11.85943 | i | = | 11% + 1% × 0.81899 | | = | 11.82% | | Score: | 8/8 | |
2.

An insurance company will pay \$87000 immediately in a lump sum to a beneficiary or \$1000 at the end of each month for 10 years.

What rate j12 (as a %, 2 decimal places) is the insurance company using? (Working in monthly time periods, draw a timeline showing the \$87000 lump sum and the monthly payments of \$1000 in their correct positions. Choose a focal date and write down the equation of value for the monthly rate.
To calculate the monthly rate, use linear interpolation between 0.5% and 0.75%. Your final answer should be j12 and not the monthly rate.) | | | Student Response | Value | Correct Answer | Answer: | 6.99 | 0% | 6.83 | | General Feedback: | |

Lump sum | 87000 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i to be found | Time (mths) | 0 | 1 | 2 | ... | 119 | 120 | | Annuity | | 1000 | 1000 | ... | 1000 | 1000 | | | ↓ | | | | | | |
With focal date at time 0 1000a(120, i) | = | 87000 at i per month | a(120, i) | = | 87 | Now a(120, 0.005) | = | 90.07345 | and a(120, 0.0075) | = | 78.94169 | i - 0.5%
0.75% - 0.5% | = | 87 - 90.07345
78.94169 - 90.07345 | i | = | 0.5% + 0.25% × 0.2761 | j12 | = | 12i | | = | 6.83% |
Score: 0/8
3.

Find the future value of a \$900 17-year ordinary annuity with annual payments if the interest rate is 8.9% p.a. effective for 11 years and 6.1% p.a. effective thereafter. (The future value at time 11 of the first 11 payments should be moved to time 17 at 6.1% and not 8.9%.) | | | Student Response | Value | Correct Answer | Answer: | not answered | 0% | 28,718.84 | | General Feedback: | |
The interest rate for the first 11 years is 0.089.
The interest rate for the last 6 years is 0.061. | |← | i1 = 0.089 | →|← | i2 = 0.061 | →| | Annuity | | 900 | ... | 900 | 900 | ... | 900 | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __| | Time (yrs) | 0 | 1 | ... | 11 | 12 | ... | 17 | | | | | | | | ↓ |
The payments from time 1 to 11 inclusive will be accumulated to time 11 using the 's' function at 8.9% but this accumulation is moved forward for the remaining 6 years at the new interest rate of 6.1%. Future value | = 900s(11, 0.089)(1.061)6 + 900s(6, 0.061) | | = 28718.84 |
Score: 0/8
4.

Deposits of \$200 are made at the beginning of each half-year for 9 years into an account paying j2 = 9.2%.

How much is in the account at the end of the 9 years? (Working in half-years, the first deposit is at time 0 instead of time 1. The last of the 2×9 deposits will be one time unit before the focal date.) | | | Student Response | Value | Correct Answer | Answer: | 5670.37 | 100% | 5,670.37 | | General Feedback: | |
There will be 2×9 = 18 deposits but since the deposits are at the beginning of each time period they will be at times 0 to 17 rather than at times 1 to 18. The focal date will be at the end of 9 years which is time 18. Deposits | 200 | 200 | 200 | ... | 200 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.092/2 | Time (½ yrs) | 0 | 1 | 2 | ... | 17 | 18 | | | | | | | ↓ → ↓ | |
The 's' function will accumulate the 18 payments to the time of the last payment which is time 17. This accumulation then has to be moved forward 1 time unit by multiplying by (1 + i). Amount | = 200s(18, 0.046)(1.046) | | = 5670.37 |
Score: 8/8
5.

The premium for a life insurance policy can be paid either in one annual payment at the beginning of the year or in twelve monthly payments at the beginning of each month. If the annual payment is \$420, what monthly payment would be equivalent at j12 = 9.3%? (You need to set up an equation of value equating the value of the annual payment which is at time 0 to the value of the monthly payments which are at times 0 to 11 and not times 1 to 12. Note the word beginning in questions 4 and 5.) | | | Student Response | Value | Correct Answer | Answer: | not answered | 0% | 36.51 | | General Feedback: | |
The annual premium is at time 0 which is the beginning of the year. The 12 monthly payments are at time 0 to time 11 inclusive since they are at the beginning of each month. Annual prem | 420 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.093/12 | Time (mths) | 0 | 1 | 2 | ... | 11 | 12 | | Mthly prem | R | R | R | ... | R | | | | ↓ | | | | | | |
The 'a' function will value the monthly payments at time -1 which is one time unit before the first payment. To move the focal date to time 0 multiply by (1 + i). Ra(12, 0.093/12)(1 + 0.093/12) | = 420 | R | = 36.51 |
The numerical value of a(12, 0.093/12) is 11.41674 Score: 0/8
6.

Find the value on 1 July 2006 of half-yearly payments of \$480 if interest is 6.7% p.a. payable half-yearly and payments are from 1 July 1995 to 1 July 2002 inclusive. (As there is a payment on each of 1 July 1995 and 1 July 2002 the number of payments equals the number of half-years between these dates plus 1. For example if the first payment was on 1 July 1990 and the last on 1 July 2000 the number of half-yearly payments would be 21.) | | | Student Response | Value | Correct Answer | Answer: | 11922.90 | 100% | 11,922.90 | | General Feedback: | |
There are 2×(2002 - 1995) = 14 half-years between the date of the first payment and the date of the final payment. Since there is a payment on both end dates, the number of half-yearly payments will be (14 + 1) = 15
The number of half-yearly time units between time of last payment and the focal date is 2 × (2006 - 2002) = 8 Payment | 480 | 480 | ... | 480 | 480 | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __| | i = 0.067/2 | Date | 7/1995 | 1/1996 | ... | 1/2002 | 7/2002 | ... | 7/2006 | | | | | | | ↓ | → | ↓ | |

Value | = 480s(15, 0.0335)(1.0335)8 | | = 11922.9 |
Score: 8/8
7.

Find the value on 1 July 2007 of half-yearly payments of \$260 if interest is 10.8% p.a. payable half-yearly and payments are from 1 July 2001 to 1 July 2012 inclusive. (It might be easier to move all the payments to 1 July 2012 using the 's' function and then move this result back to 1 July 2007 using the appropriate (1+i)-n term.) | | | Student Response | Value | Correct Answer | Answer: | 6693.51 | 100% | 6,693.51 | | General Feedback: | |
There are 11 years between the date of the first payment and the date of the final payment. Since there is a payment on both end dates, the number of half-yearly payments will be (2×11 + 1) = 23
Number of half-yearly time units between 1 July 2007 and 1 July 2012 = 10 Payment | 260 | 260 | ... | 260 | 260 | ... | 260 | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __| | i = 0.108/2 | Date | 7/2001 | 1/2002 | ... | 7/2007 | 1/2008 | ... | 7/2012 | | | | | | ↓ | ← | ↓ | |

Value | = 260s(23, 0.054)(1.054)-10 | | = 6693.51 |

An alternative expression would be 260s(13, 0.054) + 260a(10, 0.054) Score: 8/8
8.

Find the value on 1 July 2007 of half-yearly payments of \$120 if interest is 6.2% p.a. payable half-yearly and payments are from 1 July 2009 to 1 January 2016 inclusive. (The last date is 1 January 2016 not 1 July 2016.
If you are using the 'a' function it will bring the payments back to 1 January 2009 which is one time unit before the first annuity payment and not to 1 July 2009.) | | | Student Response | Value | Correct Answer | Answer: | 1155.74 | 0% | 1,228.51 | | General Feedback: | |
There are 6.5 years between the date of the first payment and the date of the final payment (note the last payment is in January not July). Since there is a payment on both end dates, the number of half-yearly payments will be (2×6.5 + 1) = 14 Payment | | | | | 120 | ... | 120 | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __| | i = 0.062/2 | Date | 7/2007 | 1/2008 | ... | 1/2009 | 7/2009 | ... | 1/2016 | | | ↓ | ← | ↓ | | | | |
The 'a' function will value the 14 payments on 1 January 2009 which is one time unit before the first payment. The focal date has then to be moved back 1.5 years or 3 time units to 1 July 2007. Value | = 120a(14, 0.031)(1.031)-3 | | = 1228.51 |
An alternative expression would be 120s(14, 0.031)(1.031)-17 Score: 0/8
9.

Find the present value of a perpetuity paying \$50 a month if j12 = 6.1% and the first payment will be made immediately. (The first payment is at time 0 and not at time 1. Hence 50a(infinity, i) which equals 50/i will value the payments at time -1 and not time 0 as required.) | | | Student Response | Value | Correct Answer | Answer: | not answered | 0% | 9,886.07 | | General Feedback: | |
The first payment is at time 0 since it is payable immediately. Payment | 50 | 50 | 50 | 50 | ... | | |____ | ____|____ | ____|____ | ____|__ | ... | i = 0.061/12 | Time (mths) | 0 | 1 | 2 | 3 | ... | | | ↓ | | | | | |
The 'a' function will value the payments at time -1. This value must then be brought forward for 1 time unit to the focal date at time 0. Value | = 50a(∞, 0.061/12)(1 + 0.061/12) | | = 50 | (1 + 0.061/12) 0.061/12 | since a(∞, i) = | 1 i | | = 9886.07 |
Score: 0/8
10.

\$11,000 is invested today into a fund earning interest at j12 = 6.4%. Level monthly payments are to be withdrawn in perpetuity from the fund, the first payment being 2 years from today. Calculate the size of the monthly withdrawals. (Set up an equation of value equating the value of the payment into the fund with the payments out of the fund. The 'a' function will value the payments at time 23 and not at time 24.) | | | Student Response | Value | Correct Answer | Answer: | not answered | 0% | 66.30 | | General Feedback: | |
The first withdrawal is at time 24 in months. Deposit | 11000 | | |____ | ____|__ | ... | __|____ | ____|____ | ____|__ | ... | i = 0.064/12 | Time (mths) | 0 | 1 | ... | 23 | 24 | 25 | ... | | Withdrawals | | | | | R | R | ... | | | ↓ | ← | ↓ | | | | |
The 'a' function will value the payments at time 23 which is one time unit before the first payment. This value must then be brought back 23 time units to the focal date at time 0. 11,000 | = Ra(∞, 0.064/12)(1 + 0.064/12)-23 | 11,000 | = R | (1 + 0.064/12)-23 0.064/12 | since a(∞, i) = | 1 i | R | = 11,000 × (1 + 0.064/12)23 × (0.064/12) | | = 66.3 |
Score: 0/8
11.

A sum of \$6,000 is required at the end of 11 years. Calculate the level annual deposit which is required at the end of each year for 11 years if interest is at j1 = 5.1% for 3 years followed by j1 = 6.4% for 8 years. (Set up an equation of value equating the payments into the fund to the value of the payment out of the fund.) | | | Student Response | Value | Correct Answer | Answer: | not answered | 0% | 394.11 | | General Feedback: | |
The interest rate for the first 3 years is 0.051.
The interest rate for the last 8 years is 0.064. | |← | i1 = 0.051 | →|← | i2 = 0.064 | →| | Deposits | | R | ... | R | R | ... | R | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __| | Time (yrs) | 0 | 1 | ... | 3 | 4 | ... | 11 | Withdrawal | | | | | | | 6000 | | | | | | | | ↓ |
The deposits from time 1 to 3 inclusive will be accumulated to time 3 using the 's' function at 5.1% but this accumulation is moved forward for the remaining 8 years at the new interest rate of 6.4%. 6,000 | = R[s(3, 0.051)(1.064)8 + s(8, 0.064)] | R | = 394.11 | | | If the focal date is taken at time 3 | 6,000(1.064)-8 | = R[s(3, 0.051) + a(8, 0.064)] |
Score: 0/10
12.

The following timeline shows that deposits of \$4,000 were made into a fund at the end of each year for 29 years. Level annual withdrawals of \$R will then be made at the end of each of the following 17 years. Calculate R if interest is at 0.063 p.a. effective. Deposits | | 4000 | ... | 4000 | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __|____ | ____| | i = 0.063 | Time (yrs) | 0 | 1 | ... | 29 | 30 | ... | 45 | 46 | | Withdrwls | | | | | R | ... | R | R | | | | | Student Response | Value | Correct Answer | Answer: | not answered | 0% | 30,221.42 | | General Feedback: | |

Deposits | | 4000 | ... | 4000 | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __|____ | ____| | i = 0.063 | Time (yrs) | 0 | 1 | ... | 29 | 30 | ... | 45 | 46 | | Withdrwls | | | | | R | ... | R | R | | | | | | ↓ | | | | | |
With focal date at the time of the last deposit (time 29) 4,000s(29, 0.063) | = Ra(17, 0.063) | R | = 30221.42 |

Title: | ACST101 Assignment 6 | Started: | 28 April 2009 12:42 PM | Submitted: | 28 April 2009 10:23 PM | Time spent: | 09:40:38 | Total score: | 47/100 = 47% Total score adjusted by 0.0 Maximum possible score: 100 | 1. | | | \$800 per quarter was deposited from 1 July 1999 to 1 July 2006 inclusive into a fund paying interest at j4 = 7.9%. Calculate the size of the fund at 1 July 2006. (Make sure you count the number of payments carefully. Remember that if there is a payment on both end dates, the number of payments equals the number of time periods + 1.) | | | Student Response | Value | Correct Answer | Answer: | 30917.08 | 100% | 30,917.08 | | General Feedback: | |
There are 7 years between the date of the first payment and the date of the last payment. The number of quarterly payments will be (4×7 + 1) = 29 since there is a payment on both end dates. The 's' function will be exact as the focal date coincides with the last payment. Payment | 800 | 800 | ... | 800 | 800 | | |____ | ____|__ | ... | __|____ | ____| | i = 0.079/4 | Date | 7/1999 | 10/1999 | ... | 4/2006 | 7/2006 | | | | | | | ↓ | | Fund at 1/7/2006 | = 800s(29, 0.01975) | | = 30917.08 | | Score: | 7/7 | |
2.

A fund was of size \$10,000 on 1 July 2006. The fund earns interest at j4 = 7.5%. Equal quarterly withdrawals will be made from 1 July 2008 to 1 January 2013 inclusive so that the balance is zero after the last withdrawal. Calculate the quarterly withdrawal. (Some dates are 1 January others are 1 July. If using the 'a' function it will value the payments at 1 April 2008 and not 1 July 2008.) | | | Student Response | Value | Correct Answer | Answer: | 773.43 | 0% | 718.04 | | General Feedback: | |
There are 4.5 years between the date of the first payment and the date of the final payment so the number of quarterly payments will be (4×4.5 + 1) = 19. Fund in | 10000 | | |____ | ____|__ | ... | __|____ | ____|____ | ____|__ | ... | __| | i = 0.075/4 | Date | 7/2006 | 10/2006 | ... | 4/2008 | 7/2008 | 10/2008 | ... | 1/2013 | | Time (¼ yrs) | 0 | 1 | ... | 7 | 8 | 9 | ... | 26 | | Payts out | | | | | R | R | ... | R | | | ↓ | ← | ↓ | | | | | |
The 'a' function will value the payments on 1/4/2008 which is one time unit (i.e one quarter) before the first payment. There is 1 year and 9 months (7 quarters) back to the focal date on 1 July 2006. The equation of value will be 10,000 | = Ra(19, 0.01875)(1.01875)-7 | R | = 718.04 |
Score: 0/9
3.

Ms Smith purchases furniture for \$3100. She pays \$300 deposit and agrees to pay the balance by monthly payments of \$130 followed by a smaller final payment, the first payment due one month after purchase. If interest is at j12 = 8.1%, how many level payments of \$130 will she make? (Your answer should be an integer e.g. 12 not 12.00 and should not count any final smaller payment.) | | | Student Response | Value | Correct Answer | Answer: | 23 | 100% | 23 | | General Feedback: | |
Let there be n payments of 130. The loan is for 3100 - 300 = 2800. Loan | 2800 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.081/12 | Time (mths) | 0 | 1 | 2 | ... | n-1 | n | | Payments | | 130 | 130 | ... | 130 | 130 | | | ↓ | | | | | | |
With focal date at time 0 the equation of value is 2800 | = 130a(n, 0.081/12) | a(n, 0.081/12) | = 21.53846 | (1 + 0.081/12)-n | = 1 - 21.53846 × 0.081/12 = 0.85462 | n | = - log 0.85462 / log 1.00675 (use unrounded values) | n | = 23.35 |
23 payments of 130 are required followed by a smaller payment at time 24 Score: 9/9
4.

Ms Smith actually purchased the furniture for \$20,300. She paid \$300 deposit and agreed to pay the balance by 47 monthly payments of \$504 followed by a smaller final payment one month after the last \$504 payment. The first \$504 payment is due one month after purchase. If interest is at j12 = 9%, what will be the size of the smaller final payment? (Set up an equation of value equating the value of the amount borrowed with the value of the payments made. Make sure you move all payments to your focal date.) | | | Student Response | Value | Correct Answer | Answer: | 141.67 | 100% | 141.67 | | General Feedback: | |
The loan is for 20,300 - 300 = 20,000 Loan | 20000 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.09/12 | Time (mths) | 0 | 1 | 2 | ... | 47 | 48 | | Payments | | 504 | 504 | ... | 504 | X | | | ↓ | | | | | | |

With focal date at time 0 the equation of value is | 20,000 | = 504a(47, 0.0075) + X(1.0075)-48 | | R | = 141.67 | | | | | With focal date at time 48 the equation of value would be | 20,000(1.0075)48 | = 504s(47, 0.0075)(1.0075) + X | |
Score: 9/9
5.

Paul wants to accumulate \$6000 by depositing \$390 every 3 months into an account paying 8.8% p.a. convertible quarterly. How many full deposits should he make? (Your answer should be an integer e.g. 12 not 12.00 and should not count any final smaller payment which may be required.) | | | Student Response | Value | Correct Answer | Answer: | 13 | 100% | 13 | | General Feedback: | |

Deposits | | 390 | 390 | ... | 390 | 390 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.088/4 | Time (¼ yrs) | 0 | 1 | 2 | ... | n-1 | n | | Accumulation | | | | | | 6000 | | | | | | | | ↓ | |
With focal date at time n the equation of value is 6000 | = 390s(n, 0.022) | s(n, 0.022) | = 15.38462 | (1.022)n | = 1 + 15.38462 × 0.022 = 1.33846 | n | = log 1.33846 / log 1.022 | n | = 13.4 |
13 full deposits of 390 are required. Score: 9/9
6.

Paul actually wants to accumulate \$6000 by depositing \$297 every 3 months into an account paying 6% p.a. convertible quarterly. He will make 17 full deposits followed by a final smaller deposit 3 months after the last full deposit. What will be the size of the concluding deposit? | | | Student Response | Value | Correct Answer | Answer: | 297.20 | 0% | 211.66 | | General Feedback: | |

Deposits | | 297 | 297 | ... | 297 | 297 | X | | |____ | ____|____ | ____|__ | ... | __|____ | ____|____ | ____| | i = 0.06/4 | Time (¼ yrs) | 0 | 1 | 2 | ... | 16 | 17 | 18 | | Accumulation | | | | | | | 6000 | | | | | | | | | ↓ | |
With focal date at time 18 the equation of value is 6,000 | = 297s(17, 0.015)(1.015) + X | X | = 211.66 |
Score: 0/9
7.

Payments of \$1800 are made at the end of each year for 10 years into a bank account where interest is paid at j4 = 7%. Find the future value of the account at the end of 10 years. (Since payments are yearly you should use the equivalent yearly interest rate.) | | | Student Response | Value | Correct Answer | Answer: | 19485.72 | 0% | 25,089.06 | | General Feedback: | |

Payment | | 1800 | 1800 | ... | 1800 | 1800 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.07186... | Time (yrs) | 0 | 1 | 2 | ... | 9 | 10 | | | | | | | | ↓ | |
The payments are annual and the interest rate is quarterly. The equivalent annual interest rate must be found. Store the unrounded annual rate in the memory of your calculator. (1 + j1/1)1 | = (1 + j4/4)4 | | = (1 + 0.07/4)4 | j1 | = (1.0175)4 - 1 | | = 0.07186... | | | Future value at end of 10 years | = 1800s(10, 0.07186...) | | = 1800 × 13.93836... | | = 25089.06 |
Score: 0/8
8.

Deposits of \$171 are made at the end of each quarter to a bank account for 9 years and 6 months. Find the future value of these payments if interest is at j1 = 5%. (To check that you are using the exact quarterly interest rate corresponding to j1 = 5%, if \$100 is invested at the end of each quarter for 11 years and 6 months the future value is 6132.35) | | | Student Response | Value | Correct Answer | Answer: | 8252.96 | 0% | 8,215.96 | | General Feedback: | |

Payment | | 171 | 171 | ... | 171 | 171 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.01227... | Time (¼ yrs) | 0 | 1 | 2 | ... | 37 | 38 | | | | | | | | ↓ | |
The payments are quarterly but the interest rate is yearly. The equivalent quarterly interest rate must be found. (1 + j4/4)4 | = (1 + j1/1)1 | | = (1 + 0.05) | j4/4 | = (1.05)1/4 - 1 | | = 0.01227... | | | Future value at end of 9.5 years | = 171s(38, 0.01227...) | | = 171 × 48.04654... | | = 8215.96 |
Score: 0/8
9.

Deposits of \$111 are made at the end of each quarter to a bank account for 5 years and 3 months. Find the future value of these payments if interest is at j12 = 5%. | | | Student Response | Value | Correct Answer | Answer: | 2648.22 | 100% | 2,648.22 | | General Feedback: | |

Payment | | 111 | 111 | ... | 111 | 111 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.01255... | Time (¼ yrs) | 0 | 1 | 2 | ... | 20 | 21 | | | | | | | | ↓ | |
The payments are quarterly but the interest rate is monthly. The equivalent quarterly interest rate must be found. (1 + j4/4)4 | = (1 + j12/12)12 | | = (1 + 0.05/12)12 | j4/4 | = (1 + 0.05/12)3 - 1 | | = 0.01255... | | | Future value at end of 5.25 years | = 111s(21, 0.01255...) | | = 111 × 23.85786... | | = 2648.22 |
Score: 8/8
10.

Find the future value at the end of 4 years of payments of \$400 made at the beginning of each month for 4 years at j2 = 8.0%. (Note the word "beginning". The last of the 4×12 payments is one month before the focal date.) | | | Student Response | Value | Correct Answer | Answer: | 22479.90 | 0% | 22,627.33 | | General Feedback: | |
The 48 payments are at the beginning of each month and hence are at times 0 to 47 inclusive. The focal date is at time 48. Payment | 400 | 400 | 400 | ... | 400 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.00656... | Time (mths) | 0 | 1 | 2 | ... | 47 | 48 | | | | | | | ↓ → ↓ | |
The payments are monthly but the interest rate is half-yearly. The equivalent monthly interest rate must be found. (1 + j12/12)12 | = (1 + j2/2)2 | | = (1 + 0.08/2)2 | j12/12 | = (1 + 0.04)1/6 - 1 | | = 0.00656... | | | Future value at end of 4 years | = 400s(48, 0.00656...)(1.00656...) | | = 400 × 56.19975... × 1.00656... | | = 22627.33 | since focal date is one time unit after the last monthly payment Score: 0/8
11.

\$12,000 was borrowed on 1 March 2006. The loan will be repaid by monthly instalments commencing on 1 March 2009 and concluding on 1 January 2012. Calculate the monthly instalment if interest is at 6% p.a. convertible quarterly. (Count the number of payments carefully.
Since payments are monthly you should use the equivalent monthly interest rate.
If using the 'a' function it will value the payments at 1 February 2009 and not 1 March 2009.) | | | Student Response | Value | Correct Answer | Answer: | 334.95 | 0% | 445.45 | | General Feedback: | |
The payments are monthly but a quarterly interest rate is given. The equivalent monthly rate must be found. (1 + j12/12)12 | = (1 + j4/4)4 | | = (1 + 0.06/4)4 | j12/12 | = (1.015)1/3 - 1 | | = 0.00498... |

Use the exact monthly interest rate. Store it in the memory of your calculator.
There are 2 years and 10 months between the date of the first payment and the date of the last payment. The number of monthly payments is (12×2 + 10 + 1) = 35.
The first payment is 3 years after the loan was granted i.e. it is at time 36. The 'a' function will find the value of the 35 payments at time 35 which is one time unit before the first payment. With focal date at time 0, the value at time 35 must be multiplied by (1.00498...)-35 to move it back to time 0. Loan | 12000 | | |____ | ____|__ | ... | __|____ | ____|____ | ____|__ | ... | __| | i = 0.00498... | Date | 3/2006 | 4/2006 | ... | 2/2009 | 3/2009 | 4/2009 | ... | 1/2012 | | Time (mths) | 0 | 1 | ... | 35 | 36 | 37 | ... | 70 | | Payments | | | | | R | R | ... | R | | | ↓ | ← | ↓ | | | | | |
Setting up an equation of value at 1 March 2006 to calculate the monthly instalment R 12000 | = Ra(35, 0.00498...)(1.00498...)-35 | 12000 | = R × 32.0492... × 0.84055... | R | = 445.45 |
Score: 0/10
12.

A term insurance policy pays a benefit if the insured life survives to the end of the term of the policy. | | | Student Response | Value | Correct Answer | Feedback | A. | True | | | | B. | False | 100% | | A term insurance policy only pays a benefit if the insured life dies before the end of the term of the policy. | | Score: | 1/1 | |
13.

Trauma insurance or crisis cover is classified as life insurance. | | | Student Response | Value | Correct Answer | Feedback | A. | True | 100% | | It is life insurance. | B. | False | | | | | Score: | 1/1 | |
14.

An accountant would take out a public liability policy to insure against losses sustained by a client due to negligence. | | | Student Response | Value | Correct Answer | Feedback | A. | True | 0% | | A professional indemnity policy would be taken out. | B. | False | | | | | Score: | 0/1 | |
15.

Whan a motor vehicle is registered in New South Wales it is compulsory to have a comprehensive motor vehicle insurance policy. | | | Student Response | Value | Correct Answer | Feedback | A. | True | | | | B. | False | 100% | | It is compulsory to have a Compulsory Third Party (CTP) policy or "Green Slip". This covers bodily injury to persons injured in accidents. A comprehensive motor vehicle policy covers damage to property. | | Score: | 1/1 | |
16.

An employee in Australia has an annual salary of \$37,000. Calculate the total minimum superannuation contribution p.a. that the employer is required to make on his/her behalf. | | | Student Response | Value | Correct Answer | Answer: | 3330 | 100% | 3,330.00 | | General Feedback: | |
From page 134 Viney 5th ed. or Table 3.4 Viney 4th ed., the contribution rate is 9% of salary. | | Score: | 1/1 | |
17.

A defined benefit scheme has a lump sum retirement benefit of
16% × years of service with the company × final salary at retirement. An employeee with a current salary of \$65,000 joined the company at age 27 and expects to retire at age 63. Calculate the lump sum retirement benefit based on the employee's current salary. | | | Student Response | Value | Correct Answer | Answer: | 374400 | 100% | 374,400.00 | | General Feedback: | |
Lump sum = 0.16 × 36 × 65,000 where years of service = (63 – 27) = 36 Score: 1/1

Title: | ACST101 Assignment 8 | Started: | 12 May 2009 12:32 PM | Submitted: | 12 May 2009 8:18 PM | Time spent: | 07:45:21 | Total score: | 30/100 = 30% Total score adjusted by 0.0 Maximum possible score: 100 | 1. | | | Catherine is buying a house and needs to borrow \$100,000. She applies to a bank which offers her a 20 year loan at 7.2% p.a. convertible quarterly, with repayment instalments of \$ 1800 per quarter for the first 5 years followed by \$R per quarter during the remaining 15 years. Calculate R. (Set up an equation of value at any convenient focal date equating the value of the sum borrowed to the value of the repayments.) | | | Student Response | Value | Correct Answer | Answer: | 1552.42 | 0% | 2,739.20 | | General Feedback: | | Loan | 100000 | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __| | i = 0.072/4 | Time (¼ yrs) | 0 | 1 | ... | 20 | 21 | ... | 80 | | Payments | | 1800 | ... | 1800 | R | ... | R | | | ↓ | | | | | | | |
The equation of value at time 0 is 100,000 | = 1800a(20, 0.018) + Ra(60, 0.018)(1.018)-20 | R | = 2739.2 | | Score: | 0/7 | |
2.

For the loan in Question 1, calculate the principal repaid in the first 5 years. You can use either the retrospective method or the prospective method when calculating the loan outstanding. | | | Student Response | Value | Correct Answer | Answer: | 0 | 100% | 0 | | General Feedback: | |
If i is the quarterly interest rate and P the instalment in the first 5 years from Q1
Loan outstanding after 5 years = 100,000(1 + i)20 - Ps(20, i) = 100,000
Principal repaid = 100,000 - 100,000 = 0.00
If you use the prospective method for the loan outstanding after 5 years the outstanding payments are R as calculated in Question 1 not P as used in the retrospective method. The prospective loan outstanding will also equal 100,000 allowing for a few cents rounding error which will be taken into account by the computer marking.
The amount of principal repaid in the first 5 years is zero because the quarterly instalment of P exactly pays the quarterly interest of i × 100,000. | | Score: | 7/7 | |
3.

The following timeline shows the payments already made for a \$29,000 loan at rate 0.005 per time unit. These payments are \$300 at time 6 followed by \$800 each time interval thereafter until time 18. Use the retrospective method to calculate the loan outstanding at time 18. Loan | 29000 | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __|____ | ____| | i = 0.005 | Time | 0 | 1 | ... | 6 | 7 | ... | 17 | 18 | | Payments | | | | 300 | 800 | ... | 800 | 800 | | | | | Student Response | Value | Correct Answer | Answer: | 19755.00 | 0% | 21,536.99 | | General Feedback: | |

Loan | 29000 | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __|____ | ____| | i = 0.005 | Time | 0 | 1 | ... | 6 | 7 | ... | 17 | 18 | | Payments | | | | 300 | 800 | ... | 800 | 800 | | | | | | | | | | ↓ | |
Stated in words the retrospective method says:
Find the future value of the loan and subtract the future value of the payments made. Loan o/s at time 18 | = 29,000(1.005)18 - 300(1.005)12 - 800s(12, 0.005) | | = 21536.99 |
Score: 0/7
4.

The following timeline shows the outstanding payments at time 4 for a loan at rate 0.007 per time unit. These payments are \$500 at time 11 followed by \$800 each time interval thereafter until time 16. Use the prospective method to calculate the loan outstanding at time 4. | |____ | ____|__ | ... | __|____ | ____|__ | ... | __|____ | ____| | i = 0.007 | Time | 4 | 5 | ... | 11 | 12 | ... | 15 | 16 | | Payments | | | | 500 | 800 | ... | 800 | 800 | | | | | Student Response | Value | Correct Answer | Answer: | not answered | 0% | 4,206.84 | | General Feedback: | |

| |____ | ____|__ | ... | __|____ | ____|__ | ... | __|____ | ____| | i = 0.007 | Time | 4 | 5 | ... | 11 | 12 | ... | 15 | 16 | | Payments | | | | 500 | 800 | ... | 800 | 800 | | | ↓ | | | | | | | | |
Stated in words the prospective method says:
Find the value at time 4 of the outstanding payments. Loan o/s at time 4 | = 500(1.007)-7 + 800a(5, 0.007)(1.007)-7 | | = 4206.84 |
Score: 0/7
5.

Jane borrowed \$3000, to be repaid over 4 years by monthly instalments at 7% p.a. flat. Find : 1. the total loan interest; 2. the monthly repayment instalment; 3. the effective monthly rate of compound interest (as a %, 2 decimal places) using linear interpolation between 1% and 1.25%; 4. the effective annual rate of compound interest (as a %, 2 decimal places). | | | Student Response | Value | Correct Answer | 1. | 840 | 25% | Equals 840.00 (25%)
Equals 840 (25%) | 2. | 80.00 | 25% | Equals 80.00 (25%)
Equals 80 (25%) | 3. | 1.06 | 25% | Equals 1.06 (25%)
Equals 1.05 (25%)
Equals 1.07 (25%) | 4. | not answered | 0% | Equals 13.46 (25%)
Equals 13.44 (25%)
Equals 13.45 (25%)
Equals 13.47 (25%)
Equals 13.48 (25%) | | General Feedback: | |

1. | Total loan interest | = Prt since we have a 'flat' rate | | | = 3000 × 0.07 × 4 | | | = 840 |

2. | Monthly instalment | = (3000 + 840) / 48 | | | = 80 |
3. Draw a timeline showing the loan granted and the payments made. Loan | 3000 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i to be found | Time (mths) | 0 | 1 | 2 | ... | 47 | 48 | | Payment | | 80 | 80 | ... | 80 | 80 | | | ↓ | | | | | | |

3000 | = 80a(48, i) at i per month | a(48, i) | = 37.500000 | Now a(48, 0.01) | = 37.973959 | a(48, 0.0125) | = 35.931481 | i - 1%
1.25% - 1% | = 37.500000 - 37.973959 35.931481 - 37.973959 | i | = 1.0580% or 1.06% to 2 dec. places |

4. | Effective annual rate | = (1.01058)12 - 1 | | | = 13.46% converting j12/12 to j1 |

| | Score: | 12/16 | |
6.

Goods costing \$2,499 may be purchased on the following terms from a discount store : Deposit | : | \$99 | Repayments | : | \$65.90 per month | Term | : | 4 years |
What flat rate of interest p.a. is being charged for the loan (as a %, 2 decimal places)? | | | Student Response | Value | Correct Answer | Answer: | not answered | 0% | 7.95 | | General Feedback: | |
Using the simple interest formula I = Prt for the interest on the loan where r is the flat rate Instalment = | Loan + Interest on loan
Number of instalments |

65.90 = | 2,400 + 2,400 × r × 4
48 |

r = | 65.90 × 48 - 2,400
2,400 × 4 | = | 7.95% p.a. |
Score: 0/7
7.

A loan of \$1,200 at 4.9% p.a. flat over 2 years has monthly instalments of \$54.90. What effective rate of interest per annum is being charged (as a %, 2 decimal places)? When calculating the monthly rate, use linear interpolation with no more than a 0.25% gap between the higher and lower estimates. (The answer required is j1 and not the monthly rate and not j12.) | | | Student Response | Value | Correct Answer | Answer: | not answered | 0% | 9.54 | | General Feedback: | |
The equation of value at time 0 is 1,200 = 54.90a(24, i) at i per month
Rough approximation for i is 2 x 4.9% / 12 which indicates that i is just under 1%.
Calculate i = j12/12 using linear interpolation as in Question 5 part 3.
Possible starting points are i = 0.0075 and i = 0.01 giving a monthly rate of 0.762%
Then calculate j1 as in Question 5 part 4. giving an effective annual rate of 9.54%
Different starting points will give a slightly different (and possibly more accurate) answer.
Sufficient tolerance is allowed to accept any answer calculated by the instruction given in the question.
Make sure that your starting points are on either side of the true value. Score: 0/7
8.

Calculate the net present value (rounded to the nearest \$) for the following investment if the initial outlay is \$100,000 and the cost of capital is 7% per annum: Year | Cash flow at end of year (dollars) | | | 1 | 56,000 | 2 | 41,000 | 3 | 41,000 | 4 | -14,000 |
As an example, the answer should be shown as 8259 not 8258.63. | | | Student Response | Value | Correct Answer | Answer: | 10935.12 | 100% | 10,935 | | General Feedback: | |

| |____ | ____|____ | ____|____ | ____|____ | ____| | i = 0.07 | Time (yrs) | 0 | 1 | 2 | 3 | 4 | | Cash flows (\$,000) | -100 | 56 | 41 | 41 | -14 | | | ↓ | | | | | |

NPV | = -100000 + 56000(1.07)-1 + 41000(1.07)-2 + 41000(1.07)-3 - 14000(1.07)-4 | | = 10935 |
Score: 7/7
9.

A superannuation fund is able to earn 4.6% per half-year on its investment in government bonds. What is the net present value to the fund of an investment that costs \$84,000 initially and returns \$11,000 at the end of each half-year for the next 4.5 years? (Round your answer to the nearest \$. Your answer could be negative. If your answer is negative, type in the negative sign as part of your answer.
The interest rate given is already the half-yearly rate. It is not j2.) | | | Student Response | Value | Correct Answer | Answer: | 4402.26 | 0% | -4,402 | | General Feedback: | |

| |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.046 | Time (½ yrs) | 0 | 1 | 2 | ... | 8 | 9 | | Cash flows (\$000) | -84 | 11 | 11 | ... | 11 | 11 | | | ↓ | | | | | | |
Since the payments at times 1 to 9 are level their present value may be found at time 0 using the 'a' function rather than finding the present value of each payment separately using a (1 + i)-n factor. NPV | = -84,000 + 11,000a(9, 0.046) | | = -4402 |
Score: 0/7
10.

Calculate the internal rate of return on an investment where the initial outlay is \$100,000 and the future returns are expected to be: Year | Cash flow at end of year (dollars) | | | 1 | 20,000 | 2 | 40,000 | 3 | 55,000 | 4 | 80,000 |
(Express your answer as a rate % p.a., rounded to the nearest integer. As an example, the answer should be shown as 18 not 17.78.) | | | Student Response | Value | Correct Answer | Answer: | not answered | 0% | 26 | | General Feedback: | |
The IRR is the interest rate i which makes the NPV = 0. Solve the following for i
-100,000 + 20,000(1+i)-1 + 40,000(1+i)-2 + 55,000(1+i)-3 + 80,000(1+i)-4 = 0
A rough approximation for i is -100,000 + 20,000 + 40,000 + 55,000 + 80,000
4 × 100,000 | = 23.75% |
NPV at 24% = 4828
However this is only a starting point and some trial and error is required to find the integer value of i which makes the expression closest to 0. If the above NPV is positive we need to try a higher value of i to make the present value smaller and vice versa.
NPV at 26% = 303
NPV at 27% = -1849
The NPV is closer to 0 at 26%, hence the IRR is 26% p.a. as a % to the nearest integer. Score: 0/8
11.

Mary and Jim borrowed \$4600 at 5.5% p.a. flat repayable with monthly instalments over 5 years. 1. Find the level monthly instalment. The effective monthly rate of loan interest can be shown to be 0.8338%. 2. What is the effective annual rate of loan interest (as a %, 2 decimal places)? 3. Using the prospective method, find the amount of loan outstanding after two years. Mary and Jim repay the loan in full after two years with a lump sum payment based on 3. together with a \$50 penalty for early repayment. 4. What effective monthly rate of interest did they pay on the loan?
(as a %, 2 decimal places, using linear interpolation between 0.75% and 1.00%) 5. What effective annual rate of interest did they pay on the loan (as a %, 2 decimal places)? | | | Student Response | Value | Correct Answer | 1. | 97.75 | 20% | Equals 97.75 (20%) | 2. | 0.84 | 0% | Equals 10.48 (20%)
Equals 10.47 (20%)
Equals 10.49 (20%) | 3. | not answered | 0% | Equals 3029.15 (20%)
Equals 3029.14 (20%)
Equals 3029.16 (20%) | 4. | not answered | 0% | Equals 0.88 (20%)
Equals .88 (20%)
Equals 0.87 (20%)
Equals .87 (20%)
Equals 0.89 (20%)
Equals .89 (20%) | 5. | not answered | 0% | Equals 11.14 (20%)
Equals 11.12 (20%)
Equals 11.13 (20%)
Equals 11.15 (20%)
Equals 11.16 (20%) | | General Feedback: | |

1. | Total loan interest | = Prt since we have a 'flat' rate | | | = 4600 × 0.055 × 5 | | | = 1265 | | Monthly instalment | = (4600 + 1265) / 60 | | | = 97.75 | 2. | Effective annual rate | = (1.008338)12 - 1 | | | = 10.48% converting j12/12 to j1 | 3. | Loan o/s after 2 years | = 97.75a(36, 0.008338) using the prospective method | | | = 3029.15 | 4. Draw a timeline showing the loan granted and the payments actually made in the 24 months. | Loan | 4600 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i to be found | Time (mths) | 0 | 1 | 2 | ... | 23 | 24 | | Payments | | 97.75 | 97.75 | ... | 97.75 | 97.75 | | | | | | | (3029.15+ 50) | | ↓ | | | | | | | | | 4600 | = 97.75a(24, i) + (3029.15 + 50)(1 + i)-24 at i per month | | RHS | = 4713.314 at i = 0.75% | | RHS | = 4501.575 at i = 1% | | i - 0.75%
1% - 0.75% | = 4600.000 - 4713.314 4501.575 - 4713.314 | | i | = 0.8838% using linear interpolation | | | | 5. | Effective annual rate | = (1.008838)12 - 1 | | | = 11.14% which is greater than the 10.48% of part 2. |

| | Score: | 4/20 |

Title: | ACST101 Assignment 9 | Started: | 19 May 2009 7:11 PM | Submitted: | 19 May 2009 11:47 PM | Time spent: | 04:36:18 | Total score: | 48/100 = 48% Total score adjusted by 0.0 Maximum possible score: 100 | 1. | | | A bond with face value \$500 redeemable at par in 13 years time, is paying interest at j2 = 8.1%. Find the purchase price of the bond to yield j2 = 9.9%. | | | Student Response | Value | Correct Answer | Answer: | 434.98 | 100% | 434.98 | | General Feedback: | | Price paid | P | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.099/2 | Time (½ yrs) | 0 | 1 | 2 | ... | 25 | 26 | | Interest | | 20.25 | 20.25 | ... | 20.25 | 20.25 | | Redemption | | | | | | 500 | | | ↓ | | | | | | | Half-yearly interest payment | = 500 × 0.081 / 2 | | = 20.25 | The redemption payment will be 500 and the required half-yearly yield is 0.099 / 2 | | | Price | = 20.25a(26, 0.0495) + 500(1.0495)-26 | | = 434.98 | | Score: | 7/7 | |
2.

A \$1,000 bond has coupons (interest payments) at j2 = 5.0% and is redeemable at 103 on 1 May 2017. Find the purchase price on 1 November 2007 if the desired yield is 8.1% per annum convertible half-yearly. (As there will not be an interest payment on 1 November 2007 the number of interest payments will equal the number of half-yearly time units between 1 November 2007 and 1 May 2017. The half-yearly yield is 8.1%/2.) | | | Student Response | Value | Correct Answer | Answer: | 770.25 | 0% | 811.40 | | General Feedback: | |

Price paid | P | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.081/2 | Date | 11/2007 | 5/2008 | 11/2008 | ... | 11/2016 | 5/2017 | | Time (½ yrs) | 0 | 1 | 2 | ... | 18 | 19 | | Interest | | 25 | 25 | ... | 25 | 25 | | Redemption | | | | | | 1030 | | | ↓ | | | | | | |

Half-yearly interest payment | = 1,000 × 0.05 / 2 = 25 | Redemption payment is 10 × 103 = 1030 since 103 per every 100 face value | Required half-yearly yield | = 0.081/2 | Number of interest payments | = 19 (these are paid half-yearly) | | | Price | = 25a(19, 0.0405) + 1030(1.0405)-19 | | = 811.4 |
Score: 0/7
3.

A \$1,000 bond, redeemable at par on 1 January 2020, is paying interest at j2 = 6.2%. Find the purchase price at 25 May 2010 to produce a yield of 7.8% per annum convertible half-yearly assuming compound interest for the fractional interest period. (First find the price on 1 January 2010 and then move the focal date forward a fraction of a time interval.) | | | Student Response | Value | Correct Answer | Answer: | 890.31 | 0% | 917.82 | | General Feedback: | |
Half-yearly interest payment = 1,000 × 0.062 / 2 = 31
The redemption payment will be 1,000
The required half-yearly yield is 0.078 / 2
To use the standard pricing formula find the price on the interest payment date before purchase i.e. on 1 January 2010. (Interest will be payable half-yearly on 1 January and 1 July each year because 1 January is the redemption date and six months from 1 January is 1 July).
There are 10 years between 1 January 2010 and redemption on 1 January 2020.
The number of half-yearly interest payments will be 2 × 10 = 20. Do not add 1 to find the number of payments as there is a payment on only one of the end dates. Price paid | | P | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.078/2 | Date | 1/1/10 | 1/7/10 | 1/1/11 | ... | 1/7/19 | 1/1/20 | | Time (½ yrs) | 0 | 1 | 2 | ... | 19 | 20 | | Interest | | 31 | 31 | ... | 31 | 31 | | Redemption | | | | | | 1000 | | | ↓ → ↓ | | | | | | | 25/5/2010 | | | | | |

Price on 1 January 2010 | = 31a(20, 0.039) + 1,000(1.039)-20 | | = 890.308 |
Move focal date forward a fraction of a time unit (half-year) from 1 January to 25 May.
Number of days between 1 January and 25 May = 144
Number of days between 1 January and 1 July = 181
Using S = P(1 + i)n Price on 25 May 2010 | = 890.308(1.039)144/181 | | = 917.82 |
Score: 0/10
4.

A bond with face value \$2,000, paying interest at j2 = 11%, is purchased 12 years before redemption for \$1,900. If redemption is at par, use the APPROXIMATE formula to calculate the yield, j2, to maturity (as a %, 2 decimal places). (The required answer is the j2 value and not the half-yearly yield.
When a question says to use the approximate formula, linear interpolation is not required.) | | | Student Response | Value | Correct Answer | Answer: | 0.18 | 0% | 11.71 | | General Feedback: | |

| Price paid | 1900 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i to be found | Time (½ yrs) | 0 | 1 | 2 | ... | 23 | 24 | | Interest | | 110 | 110 | ... | 110 | 110 | | Redemption | | | | | | 2000 | | | ↓ | | | | | | |

Half-yearly interest payment = | 2,000 × 0.11 / 2 | = | 110 | The redemption payment is 2,000 and the nominal annual yield is to be calculated | | | 1900 = | 110a(24, i) + 2,000(1 + i)-24 at i per half-year | Using formula (8) i = | 110 + (2,000 - 1900) / 24 (2,000 + 1900) / 2 | j2 = | 2i | = | 11.71% |
Score: 0/7
5.

A bond with face value \$2,000, paying interest at j2 = 11%, is purchased 12 years before redemption for \$1,977. If redemption is at par, use LINEAR INTERPOLATION to calculate the yield, j2, to maturity (as a %, 2 decimal places). (The required answer is the j2 value and not the half-yearly yield. For the half-yearly yield use starting points of 5.5% and 6%.) | | | Student Response | Value | Correct Answer | Answer: | 11.18 | 100% | 11.18 | | General Feedback: | |
The timeline and equation of value will be the same as in the previous question except that the price paid is 1977. 1977 | = 110a(24, i) + 2,000(1 + i)-24 at i per half-year | RHS | = 2000 at i = 5.5% | RHS | = 1874.496 at i = 6% | i - 5.5%
6% - 5.5% | = 1977.000 - 2000.000 1874.496 - 2000.000 | j2 | = 2i | | = 11.18% |
Score: 7/7
6.

Mr Morgan buys a \$1,000 bond that pays bond interest at j2 = 6% and is redeemable at par in 15 years. The price he pays will give him a yield of j2 = 7% if he holds the bond to maturity. After 6 years, Mr Morgan sells this bond to Ms Grenfell who wants a yield to maturity of j2 = 5% on her investment. 1. What price did Mr Morgan pay? 2. What price did Ms Grenfell pay? (She is assumed to hold the bond for (15 - 6) = 9 years to maturity.) 3. Use the APPROXIMATE formula to calculate the yield, j2, earned by Mr Morgan on his investment (as a %, 2 decimal places). (The C term in the formula will be the sale price in 2. and the n term will be the time between purchase by Mr Morgan and sale by Mr Morgan.) | | | Student Response | Value | Correct Answer | 1. | 908.04 | 30% | Equals 908.04 (30%)
Equals 908.03 (30%)
Equals 908.05 (30%) | 2. | 1071.77 | 30% | Equals 1071.77 (30%)
Equals 1071.76 (30%)
Equals 1071.78 (30%) | 3. | 8.82 | 40% | Equals 8.82 (40%)
Equals 8.81 (40%)
Equals 8.83 (40%) | | General Feedback: | |
1. Half-yearly interest payment = 1,000 × 0.06 / 2 = 30
The redemption payment will be 1,000.
The time to maturity is 15 years or 30 half-years.
The required half-yearly yield for Mr Morgan is 0.07 / 2 = 0.035. Price paid by Mr Morgan | = 30a(30 ,0.035) + 1,000(1.035)-30 | | = 908.04 |
2. The interest payment is fixed at \$30 per half-year for the term of the bond.
Ms Grenfell is buying a bond with a term to maturity of (15 - 6) = 9 years or 18 half-years. Her required half-yearly yield is 0.05 / 2 = 0.025. Price paid by Ms Grenfell | = 30a(18, 0.025) + 1,000(1.025)-18 | | = 1071.77 |
3. Draw a timeline showing all the payments made and received by Mr Morgan over the 6 years that he held the bond. Then set up an equation of value and solve it using the approximate formula which is formula (8) on the formula sheet. Price paid | 908.04 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i to be found | Time (½ yrs) | 0 | 1 | 2 | ... | 11 | 12 | | Interest | | 30 | 30 | ... | 30 | 30 | | Sale proceeds | | | | | | 1071.77 | | | ↓ | | | | | | |

908.04 = | 30a(12, i) + 1071.77(1 + i)-12 at i per half-year | i = | 30 + (1071.77 - 908.04) / 12 (1071.77 + 908.04) / 2 | = | 4.409% | j2 = | 8.82% |
Score: 20/20
7.

A \$10,000 bond is issued by ACME Pty Ltd. Each bond has yearly coupon payments of \$960 and is redeemable at par in 19 years. Joh buys a \$10,000 bond at a price to give him a yield to maturity of j4 = 10.8%. What price did Joh pay for the bond? (As the interest payments are annual you will need to use the equivalent j1 value for the yield when calculating the price. Use the unrounded yield (store it in the memory of your calculator) or your answer will probably be marked as incorrect.) | | | Student Response | Value | Correct Answer | Answer: | 7760.84 | 0% | 8,730.04 | | General Feedback: | |
The interest payments are yearly, but the yield given is a nominal rate convertible quarterly. Since the annuity payments (i.e. the interest payments) are annual, they should be valued using the equivalent annual effective yield. Effective annual yield j1 | = (1 + 0.108 / 4)4 - 1 = 0.11245... | Price paid by Joh | = 960a(19, 0.11245...) + 10,000(1.11245...)-19 | | = 960 × 7.71857 + 10,000 × 0.132022 | | = 8730.04 |
Score: 0/9
8.

A \$10,000 bond was issued to be redeemable at par in 18 years. The bond has yearly coupon payments of \$870. Immediately after the 5th coupon payment, the bond is sold to Russ at a price which gives Russ a yield to maturity of j1 = 10.3%. What price did Russ pay for the bond? | | | Student Response | Value | Correct Answer | Answer: | 8800.91 | 0% | 8,880.91 | | General Feedback: | |
Russ is assumed to hold the bond for (18 - 5) = 13 years to maturity.
Both the interest payments and the required yield are annual. Price paid by Russ | = 870a(13, 0.103) + 10,000(1.103)-13 | | = 8880.91 |
Score: 0/7
9.

A \$1,000 bond issued by company A has half-yearly coupons at j2 = 6.3%. It is redeemable at par in 15 years. Find the price of a \$1,000 bond from company A to yield j1 = 7%. (Use the unrounded equivalent half-yearly yield.) | | | Student Response | Value | Correct Answer | Answer: | 946.11 | 100% | 946.12 | | General Feedback: | |

Half-yearly interest payment | = 1,000 × 0.063 / 2 = 31.5 | Number of interest payments | = 2 × 15 = 30 | Half-yearly yield j2/2 | = 1.071/2 - 1 = 3.4408...% | | | Price paid | = 31.5a(30, 0.034408...) + 1,000(1.034408...)-30 | | = 31.5 × 18.52921 + 1,000 × 0.362446 | | = 946.12 |
Score: 9/9
10.

Company B also issues \$1,000 bonds, paying coupons of \$X annually. They are redeemable at 105 in 9 years. Find the value of \$X that will also yield j1 = 7% on a \$1,000 bond from company B, if such a bond costs \$954. | | | Student Response | Value | Correct Answer | Answer: | 113.03 | 0% | 58.77 | | General Feedback: | |

Price paid | 954 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.07 | Time (yrs) | 0 | 1 | 2 | ... | 8 | 9 | | Interest | | X | X | ... | X | X | | Redemption | | | | | | 1050 | | | ↓ | | | | | | |
The redemption payment is 10 × 105 = 1,050.
The interest payments and the required yield are both annual
Setting up the equation of value at time 0 954 | = Xa(9, 0.07) + 1,050(1.07)-9 | X | = 58.77 |
Score: 0/10
11.

In a government bond tender the coupon rate was j2 = 6.5% and the weighted average yield was j2 = 6%. At this yield the bonds were issued | | | Student Response | Value | Correct Answer | Feedback | A. | at a premium to the face value. | 100% | | | B. | at a discount to the face value. | | | | | General Feedback: | |
Since the yield is less than the coupon interest rate, the price will be above the face value if the bonds are issued on an interest payment date. If the time to the first interest payment is less than the normal time of half a year, this will also increase the price. | | Score: | 1/1 | |
12.

A Treasury Bond is | | | Student Response | Value | Correct Answer | Feedback | A. | Short-term equity | | | | B. | Short-term debt | | | | C. | Long-term equity | | | | D. | Long-term debt | 100% | | | | Score: | 1/1 | |
13.

Look up the financial pages of a newspaper. | | | Student Response | Value | Correct Answer | Feedback | A. | The yield on Australian government Treasury Bonds is less than the yield on NSW Treasury Bonds. | 100% | | | B. | The yield on Australian government Treasury Bonds is greater than the yield on NSW Treasury Bonds. | | | | | General Feedback: | |
Australian government Treasury Bonds have a higher credit rating and higher liquidity and hence trade at lower yields. | | Score: | 1/1 | |
14.

A capital indexed bond with face value of \$1,000 pays interest each quarter at 1% of the indexed capital value. The index increases by 0.41% compound at the end of each of the first 6 quarters and by 0.73% compound at the end of each of the next 4 quarters. Calculate the interest payment at the end of the 10th quarter. (First calculate the indexed capital value at the time the coupon is to be paid.
This is a question from Week 3 of Techniques of accumulating a single payment using 2 rates of compound interest.
Then multiply by the quarterly interest rate.) (Table 12.4 (5th ed) or Table 12.6 (4th ed) of Viney sets out the calculations.
For example at the end of the 4th quarter in Table 12.4 (5th ed) Indexed capital value | = 1,000(1.01)4 | | = 1,040.60 | Interest payment | = 0.015 × 1,040.60 | | = 15.61 |
For example at the end of the 4th quarter in Table 12.6 (4th ed) Indexed capital value | = 100(1.02)4 | | = 108.24 | Interest payment | = 0.03 × 108.24 | | = 3.25) | | | | Student Response | Value | Correct Answer | Answer: | 513.76 | 0% | 10.55 | | General Feedback: | |

Indexed capital value | = 1,000(1.0041)6 × (1.0073)4 | | = 1055.11 | Interest payment | = 0.01 × 1055.11 | | = 10.55 |
Score: 0/2
15.

The successful bidders for Treasury bonds were:
Bidder A for \$110 million face value at a yield of 6.61% p.a.
Bidder B for \$70 million face value at a yield of 6.66% p.a.
Bidder C for \$ 220 million face value at a yield of 6.77% p.a. Calculate the weighted average yield (as a %, 2 decimal places) at which the Reserve Bank would be allocated bonds. (This applies principles from first year statistics. Weighted average yield = | sum of each yield times the face value of bonds issued at that yield total face value of bonds issued | ie Weighted average yield = | Σxifi / Σfi |

The answer will be similar to but not the same as adding up the three yields and dividing by 3.
You might try to verify the weighted average yield of 6.42% in Table 12.1 (5th ed) or Table 12.2 (4th ed).
The Grade Point Average (GPA) calculation is another example of finding a weighted average.) | | | Student Response | Value | Correct Answer | Answer: | 6.71 | 100% | 6.71 | | General Feedback: | |

Weighted average yield = | 110×6.61 + 70×6.66 + 220×6.77
110 + 70 + 220 | % |
Score: 2/2

Title: | ACST101 Assignment 10 | Started: | 26 May 2009 5:54 PM | Submitted: | 26 May 2009 11:26 PM | Time spent: | 05:31:13 | Total score: | 14/100 = 14% Total score adjusted by 0.0 Maximum possible score: 100 | 1. | | | A \$500 bond is redeemable at par in 8 years. The coupon (interest) rate is 9.3% p.a. payable half-yearly. What price should be paid for the bond by an investor (who pays tax at 30.0% on interest) to produce a net yield to maturity of 5.5% p.a. convertible quarterly? (As the interest payments are half-yearly, the yield will need to be converted to a half-yearly rate to match the frequency of the payments. Use the unrounded yield (store it in the memory of your calculator) or your answer will probably be marked as incorrect. This also applies in Question 4.) | | | Student Response | Value | Correct Answer | Answer: | 645.12 | 0% | 531.08 | | General Feedback: | | Gross half-yearly interest | = 500 × 0.093 / 2 | | = 23.25 | Net half-yearly interest | = Gross interest - Tax | | = 23.25 - 0.3 × 23.25 | | = 16.275 | Redemption payment | = 500 | Equivalent half-yearly yield | = (1 + 0.055/4)2 - 1 | | = 0.02769... (use the exact value) | | | Price paid | P | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.02769... | Time (½ yrs) | 0 | 1 | 2 | ... | 15 | 16 | | Net Interest | | 16.275 | 16.275 | ... | 16.275 | 16.275 | | Redemption | | | | | | 500 | | | ↓ | | | | | | | | | | Price | = 16.275a(16, 0.02769...) + 500(1.02769...)-16 | | = 16.275 × 12.78592 + 500 × 0.64597 | | = 531.08 | | Score: | 0/10 | |
2.

A debenture pays interest at j4 = 14% and will mature 7 years from today at par. Ignoring the interest payment due today, what net yield, j4 (as a %, 2 decimal places), would an investor taxed at 30% on interest obtain if the debenture was purchased for \$83.00 per \$100 face value? Use the APPROXIMATE yield formula. (The j4 value is required and not the quarterly rate.
When a question says to use the approximate formula, linear interpolation is not required.) | | | Student Response | Value | Correct Answer | Answer: | 13.36 | 100% | 13.36 | | General Feedback: | |
Let the face value and redemption payment be \$100 Gross quarterly interest = | 100 × 0.14 / 4 | = | 3.50 | Tax on interest = | 0.3 × 3.50 | = | 1.05 | Net quarterly interest = | 3.50 - 1.05 | = | 2.45 | | | Price paid | 83 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i to be found | Time (¼ yrs) | 0 | 1 | 2 | ... | 27 | 28 | | Net Interest | | 2.45 | 2.45 | ... | 2.45 | 2.45 | | Redemption | | | | | | 100 | | | ↓ | | | | | | | | 83 = | 2.45a(28, i) + 100(1 + i)-28 at i per quarter | Using formula (8) i = | 2.45 + (100 - 83) / 28 (100 + 83) / 2 | j4 = | 4i | = | 13.36% |
Score: 9/9
3.

A debenture pays interest at j4 = 14% and will mature 7 years from today at par. Ignoring the interest payment due today, what net yield, j4 (as a %, 2 decimal places) would an investor taxed at 30% on interest obtain if the debenture was purchased for \$85.80 per \$100 face value? Use LINEAR INTERPOLATION between 3% and 3.5% for the quarterly yield. (The j4 value is required and not the quarterly rate.) | | | Student Response | Value | Correct Answer | Answer: | 14.37 | 0% | 12.94 | | General Feedback: | |
The timeline and equation of value will be the same as in Q2 except that the price paid is 85.80. 85.8 | = 2.45a(28, i) + 100(1 + i)-28 at i per quarter | RHS | = 89.680 at i = 3% | RHS | = 81.450 at i = 3.5% | i - 3.0%
3.5% - 3.0% | = 85.800 - 89.680 81.450 - 89.680 | j4 | = 4i | | = 12.94% |
Score: 0/9
4.

A \$100 bond pays interest at j2 = 8.5% and is redeemable at par on 1 June 2028. What price should be paid for the bond on 10 August 2010 by an investor liable to tax on interest at 25 cents in the dollar who wants to earn a net yield of 6.5% p.a. effective? Use compound interest for the fractional interest period. | | | Student Response | Value | Correct Answer | Answer: | 108.64 | 0% | 100.97 | | General Feedback: | |

Interest date before purchase | = 1 June 2010 | Number of interest payments | = 36 | Gross half-yearly interest | = 100 × 8.5 / 200 = 4.25 | Tax on Interest | = 0.25 × 4.25 = 1.0625 | Net half-yearly interest | = 4.25 - 1.0625 = 3.1875 | Redemption payment | = 100 | Equivalent half-yearly yield | = (1.065)1/2 - 1 | | = 0.03199... (use the exact value) | | | Price paid | | P | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.03199... | Date | 1/6/10 | 1/12/10 | 1/6/11 | ... | 1/12/27 | 1/6/28 | | Time (½ yrs) | 0 | 1 | 2 | ... | 35 | 36 | | Net Interest | | 3.1875 | 3.1875 | ... | 3.1875 | 3.1875 | | Redemption | | | | | | 100 | | | ↓ → ↓ | | | | | | | 10/8/02010 | | | | | | | | | To use the standard pricing formula find the price on 1 June 2010. | Price on 1 June 2010 | = 3.1875a(36, 0.03199...) + 100(1.03199...)-36 | | = 99.76 | | | Move focal date forward a fraction of a time unit from 1 June to 10 August.
Number of days between 1 June and 10 August = 29 + 31 + 10 = 70
Number of days between 1 June and 1 December = 183
Using S = P(1 + i)n | Price on 10 August 2010 | = 99.76(1.03199...)70/183 | | = 100.97 |
Score: 0/10
5.

A \$100 bond pays interest at j2 = 7.4% and is redeemable at par in 4.5 years' time. Calculate the price which should be paid for the bond if a net yield to maturity of 6.5% p.a. convertible half-yearly is required and tax on both interest and capital gains is at the rate of 25%. (The capital gain and hence the net redemption payment is based on the unknown purchase price P. A term involving P will appear on both sides of the equation of value.) | | | Student Response | Value | Correct Answer | Answer: | 96.21 | 0% | 95.50 | | General Feedback: | |

Gross half-yearly interest payment | = 100 × 0.074 / 2 | | = 3.7 | Net half-yearly interest payment | = Gross interest - Tax | | = 3.7 - 0.25 × 3.7 | | = 2.775 | | | Gross redemption payment | = 100 | Price paid | = P | Capital gain | = 100 - P | Tax on capital gain | = 0.25 × (100 - P) | Net redemption payment | = 100 - 0.25 × (100 - P) | | = 75 + 0.25P | | | Price paid | P | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.065/2 | Time (½ yrs) | 0 | 1 | 2 | ... | 8 | 9 | | Net Interest | | 2.775 | 2.775 | ... | 2.775 | 2.775 | | Net Redemption | | | | | (75 + 0.25P) | | ↓ | | | | | | | | | | P = | 2.775a(9, 0.0325) + (75 + 0.25P)(1.0325)-9 | P(1 - 0.25(1.0325)-9) = | 2.775a(9, 0.0325) + 75(1.0325)-9 | 0.81253P = | 77.597 | P = | 95.5 |
Score: 0/10
6.

A \$100 bond pays interest at j2 = 7% and is redeemable at par in 10 years' time. The bond is purchased for \$94.40. Calculate the net redemption payment allowing for capital gains tax at the rate of 25% if the bond is held to maturity. (Net redemption payment = Gross redemption payment - Capital gains tax.) | | | Student Response | Value | Correct Answer | Answer: | 6.11 | 0% | 98.60 | | General Feedback: | |

Gross redemption payment | = 100 | Capital gain | = 100 - 94.4 | Tax on capital gain | = 0.25 × (100 - 94.4) | Net redemption payment | = 100 - 0.25 × (100 - 94.4) | | = 98.6 |
Score: 0/4
7.

A bond was bought for \$81.60 and sold for \$99.00. The capital gains tax which will be payable is \$4.30. Calculate the rate of capital gains tax (as a %, 2 decimal places). | | | Student Response | Value | Correct Answer | Answer: | 4.71 | 0% | 24.71 | | General Feedback: | |

Capital gain | = Selling price - Buying price | | = 99 - 81.6 | Tax on capital gain | = Tax rate × Capital gain | 4.3 | = Tax rate × (99 - 81.6) | Tax rate | = 24.71% |
Score: 0/4
8.

A \$100 bond pays interest at j2 = 7.4% and is redeemable at par 6 years from now. Three years ago, Joanna purchased the bond for \$94. Joanna pays tax on interest at the rate of 30% and tax on capital gains at the rate of 25%. If she holds the bond to maturity, use the APPROXIMATE formula (suitably adjusted to take account of taxation) to calculate the net yield, j2, earned by Joanna over the 9 years of her investment (as a %, 2 decimal places). | | | Student Response | Value | Correct Answer | Answer: | 68.87 | 0% | 5.90 | | General Feedback: | |

Gross half-yearly interest = | 100 × 0.074 / 2 = 3.70 | Tax on interest = | 0.3 × 3.70 = 1.11 | Net half-yearly interest = | 3.70 - 1.11 = 2.59 | | | Gross redemption = | 100 | Capital gain = | 100 - 94 = 6 | Tax on capital gain = | 0.25 × 6 = 1.5 | Net redemption = | 100 - 1.5 = 98.5 | | | Price paid | 94 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i to be found | Time (½ yrs) | 0 | 1 | 2 | ... | 17 | 18 | | Net Interest | | 2.59 | 2.59 | ... | 2.59 | 2.59 | | Net Redemption | | | | | | 98.5 | | | ↓ | | | | | | | | 94 = | 2.59a(18, i) + 98.5(1 + i)-18 at i per half-year | Using formula (8) i = | 2.59 + (98.5 - 94) / 18 (98.5 + 94) / 2 | j2 = | 2i | = | 5.9% |
Score: 0/9
9.

Continuation of previous question. If Joanna sells the bond today for \$90.30, use the APPROXIMATE formula (suitably adjusted to take account of taxation) to calculate the net yield, j2, earned by Joanna over the 3 years of her investment (as a %, 2 decimal places). Ignore taxation on the capital loss. (The C value will be the sale payment.) | | | Student Response | Value | Correct Answer | Answer: | 70.10 | 0% | 4.28 | | General Feedback: | |

Ignore tax impications of capital loss of (90.3 - 94) | Net half-yearly interest payment = | 2.59 from previous question. | Net sale payment = | 90.3 | | | Price paid | 94 | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i to be found | Time (½ yrs) | 0 | 1 | 2 | ... | 5 | 6 | | Net Interest | | 2.59 | 2.59 | ... | 2.59 | 2.59 | | Sale payment | | | | | | 90.3 | | | ↓ | | | | | | | | 94 = | 2.59a(6, i) + 90.3(1 + i)-6 at i per half-year | Using formula (8) i = | 2.59 + (90.3 - 94) / 6 (90.3 + 94) / 2 | j2 = | 2i | = | 4.28% |
Score: 0/9
10.

A \$100 bond has interest payments of \$3.5 on 1 December 2010 and \$4 each half-year thereafter. The bond matures at par on 1 December 2015. The bond is bought on 14 October 2010 at a price to give a gross yield to maturity of j2 = 7.2%. Calculate the price paid. Use the Reserve Bank method.
(First calculate the price on 1 December 2010, ignoring the interest payment on that date. Then add in the interest payment and move the resulting price back to 14 October 2010.) | | | Student Response | Value | Correct Answer | Answer: | 112.1 | 0% | 105.82 | | General Feedback: | |

Price paid | P | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.072/2 | Date | 1/6/10 | 1/12/10 | 1/6/11 | ... | 1/6/15 | 1/12/15 | | Time (½ yrs) | 0 | 1 | 2 | ... | 10 | 11 | | Interest | | 3.5 | 4 | ... | 4 | 4 | | Redemption | | | | | | 100 | | | ↓ | ← ↓ | | | | | | | 14/10/10 | | | | | |
Number of \$4 interest payments = 10
Price on 1 December 2010 ignoring \$3.5 interest payment on that date
= 4a(10, 0.036) + 100(1.036)-10 = 103.31
Price on 1 December 2010 including \$3.5 interest payment on that date
= 103.31 + 3.5 = 106.81
Number of days between 14 October and 1 December = 48
Number of days between 1 June and 1 December = 183
Price on 14 October 2010 = 106.81(1.036)-48/183 = 105.82 Score: 0/10
11.

Twenty years ago Jim Bow bought a \$1,000 bond paying interest annually at 10.1%. The bond is redeemable at par in 12 years from now. An interest payment has just been made. For the past twenty years, tax on interest only has been at a rate of 30%. Legislation about to be introduced, however, means that tax on all future interest will be at the rate of 35%. Assuming this new rate will apply until the bond matures, and that there will be no capital gains tax, then Jim Bow will achieve a net yield to maturity of 9% p.a. effective. Find the amount he originally paid for the bond. (Write down an equation of value equating the price P paid at the beginning to the value of the net interest payments, which will have to be valued in two parts, plus the value of the redemption payment.) | | | Student Response | Value | Correct Answer | Answer: | 312 | 0% | 792.71 | | General Feedback: | |

Gross annual interest | = 1,000 × 0.101 = 101 | Net annual interest for past 20 years | = 101 - 0.3 × 101 = 70.7 | Net annual interest for next 12 years | = 101 - 0.35 × 101 = 65.65 |

Price | P | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __| | i = 0.09 | Time (yrs) | -20 | -19 | ... | 0 | 1 | ... | 12 | | Net Interest | | 70.7 | ... | 70.7 | 65.65 | ... | 65.65 | | Redemption | | | | | | | 1000 | | | ↓ | | | | | | | |

P | = 70.7a(20, 0.09) + 65.65a(12, 0.09)(1.09)-20 + 1,000(1.09)-32 | | = 792.71 |
Score: 0/10
12.

A redeemable preference share may be redeemed for | | | Student Response | Value | Correct Answer | Feedback | A. | cash | | | | B. | ordinary shares | 0% | | | | General Feedback: | |
A convertible preference may be converted into ordinary shares.
A redeemable preference share may be redeemed for cash. | | Score: | 0/1 | |
13.

A rights issue is the issue of additional ordinary shares on a pro-rata basis to selected institutional investors. | | | Student Response | Value | Correct Answer | Feedback | A. | True | | | | B. | False | 100% | | | | General Feedback: | |
In a rights issue shares are offered to all shareholders. | | Score: | 1/1 | |
14.

Commercial bills are sometimes called commercial paper. | | | Student Response | Value | Correct Answer | Feedback | A. | True | | | | B. | False | 100% | | | | General Feedback: | |
Commercial paper is another name for promissory notes. | | Score: | 1/1 | |
15.

Look up the financial pages of a newspaper. | | | Student Response | Value | Correct Answer | Feedback | A. | The yield on State Government securities is less than the yield on similar Australian Government securities. | | | | B. | The yield on State Government securities is greater than the yield on similar Australian Government securities. | 100% | | State Government securities have a slightly lower credit rating and trade at higher yields. | | Score: | 1/1 | |
16.

The price of fixed interest securities is not affected by changes in market interest rates. | | | Student Response | Value | Correct Answer | Feedback | A. | True | | | | B. | False | 100% | | | | General Feedback: | |
The market rates are the rate of return or yield required by investors. Fixed interest securities have payments of a fixed size but the price investors will pay for these payments is determined by the return that they require. There is an inverse relationship between price and yield (Techniques Week 9). | | Score: | 1/1 | |
17.

A lease in which the ownership of the asset passes to the lessee at the end of the lease after the payment of a specified residual amount is called | | | Student Response | Value | Correct Answer | Feedback | A. | a finance lease | 100% | | | B. | an operating lease | | | | | General Feedback: | |
Page 410 of text in 5th edition or page 246 of text in 4th edition. | | Score: | 1/1 | |

View Attempt 1 of 1 Title: | ACST101 Assignment 11 | Started: | 1 June 2009 8:51 PM | Submitted: | 2 June 2009 9:07 PM | Time spent: | 24:15:59 | Total score: | 78/100 = 78% Total score adjusted by 0.0 Maximum possible score: 100 | 1. | | | The following payments were made into an account for 10 years: \$400 at the end of year 1 increasing by \$160 p.a. Calculate the future value at the end of the 10 years if interest is at j1 = 6.7%. | | | Student Response | Value | Correct Answer | Answer: | 14916.28 | 0% | 14,098.95 | | General Feedback: | |
Since the payments are increasing by a constant 160, the given annuity must be expressed as a standard arithmetic progression with first payment 160 and payments increasing by a constant 160 together with a level annuity with payments of (400 - 160) = 240. | |____ | ____|____ | ____|____ | ____|__ | ... | __| | i = 0.067 | Time (yrs) | 0 | 1 | 2 | 3 | ... | 10 | | Payments | | 400 | 560 | 720 | ... | 1840 | | | | | | | | | | | = | 160 | 320 | 480 | ... | 1600 | | | + | 240 | 240 | 240 | ... | 240 | | | | | | | | ↓ | |
Using formula (10) for the future value of the standard arithmetic progression Future value = | 160 | 1.067s(10, 0.067) - 10
0.067 | + 240s(10, 0.067) | = | 160 × 67.68509 + 240 × 13.62221 | = | 14098.95 | | Score: | 0/7 | |
2.

An annuity has payments increasing by 3.7% each year. The payments are: \$19,000(1+3.7/100) at the end of year 1, \$19,000(1+3.7/100)2 at the end of year 2, \$19,000(1+3.7/100)3 at the end of year 3, and so on for 15 years. Find the cost (present value) of this 15 year annuity if interest is at 6.8% p.a. effective. (The first payment R is \$19,000(1+3.7/100) and not \$19,000.
Use the following to check that you are using the exact value of the special interest rate j. If the first payment is 10,000(1.03) and payments increase by 3% p.a. each year, the present value at 5% p.a. effective of the 15 year annuity is 129054.38.) | | | Student Response | Value | Correct Answer | Answer: | 226994.89 | 100% | 226,994.89 | | General Feedback: | |
In the timeline P = 19,000 | |____ | _____|_____ | ____|__ | ... | __|____ | _ | ____| | i = 0.068 | Time (yrs) | 0 | 1 | 2 | ... | 14 | | 15 | | Payments | | 1.037P | 1.0372P | ... | 1.03714P | | 1.03715P | | | ↓ | | | | | | | |
Use formula (11) for the present value of payments increasing in geometric progression with first payment R = 19,000(1.037) Cost | = [19,000(1.037)](1.037)-1a(15, j) | | = 226994.89 |

where j = | 0.068 - 0.037
1 + 0.037 | = 0.029894... |

The numerical value of a(15, j) is 11.9471 |
Score: 7/7
3.

A sum of \$370,000 is invested into a fund earning interest at j1 = 8.5%. A payment of \$37,000 will be made from the fund at the end of year 1, with subsequent payments from the fund increasing by 3.9% p.a. compound at the end of each later year. How many full payments can be made? (Your answer should be an integer.) (Set up an equation of value equating the value of the payment into the fund with the value of the n say payments from the fund.) | | | Student Response | Value | Correct Answer | Answer: | 17.92 | 0% | 14 | | General Feedback: | |
In the timeline R = 37,000 Payment in | 370000 | | |____ | _____|_____ | ____|__ | ... | __|____ | _ | ____| | i = 0.085 | Time (yrs) | 0 | 1 | 2 | ... | n-1 | | n | | Payments out | | R | 1.039R | ... | 1.039n-2R | | 1.039n-1R | | | ↓ | | | | | | | |
Solve for n from the equation of value at time 0 370,000 | = 37,000(1.039)-1a(n, j) where j = | 0.085 - 0.039
1 + 0.039 | = 0.044273... | a(n, j) | = 10.39 | | | 1 - (1+j)-n j | = 10.39 | | | (1 + j)-n | = 1 – j × 10.39 | | | (1 + j)-n | = 0.54 | | | n | = 14.22 taking logs | | |
The number of full payments will be n rounded down to the next integer. Score: 0/7
4.

A sum of \$360,000 is invested into a fund earning interest at j1 = 8%. A payment of \$36,000 will be made from the fund at the end of year 1, with subsequent payments from the fund increasing by 4.1% p.a. compound at the end of each later year. The number of full payments which can be made is 13. Calculate the size of the smaller final payment, one year after the last full payment. (Set up an equation of value equating the value of the payment into the fund with the value of the payments from the fund.
"Smaller" is relative to the size of the full payment of 36,000(1+4.1/100)13 that would have been made on that date.) | | | Student Response | Value | Correct Answer | Answer: | 26953.76 | 100% | 26,953.76 | | General Feedback: | |
In the timeline R = 36,000 Payment in | 360000 | | |____ | _____|_____ | ____|__ | ... | __|____ | _ | ____| | i = 0.08 | Time (yrs) | 0 | 1 | 2 | ... | 13 | | 14 | | Payments out | | R | 1.041R | ... | 1.04112R | | Y | | | ↓ | | | | | | | |
Solve for the final payment Y from the equation of value at time 0 360,000 | = 36,000(1.041)-1a(13, j) + Y(1.08)-14 | Y | = 26953.76 |

where j = | 0.08 - 0.041
1 + 0.041 | = 0.037464... and a(13, j) = 10.14464 |
Score: 7/7
5.

A sum of \$380,000 is invested into a fund earning interest at j1 = 7.6%. A payment of \$36,000 will be made from the fund at the end of year 1, with subsequent payments from the fund increasing by 3.1% p.a. compound at the end of each later year. How much is in the fund just after the fifth payment from the fund? (At time 5, find the future value of the investment of 380,000 and subtract the future value of the payments already made from the fund.) | | | Student Response | Value | Correct Answer | Answer: | 326156.02 | 100% | 326,156.02 | | General Feedback: | |
In the timeline R = 36,000 Payment in | 380000 | | |____ | ____|____ | ____|____ | ____|____ | ____|____ | ____| | i = 0.076 | Time (yrs) | 0 | 1 | 2 | 3 | 4 | 5 | | Payments out | | R | 1.031R | 1.0312R | 1.0313R | 1.0314R | | | | | | | | ↓ | |
At time 5, find the future value of the investment of 380,000 and subtract the future value of the payments made using formula (12) Amount | = 380,000(1.076)5 - 36,000(1.031)4s(5, j) | | = 326156.02 |

where j = | 0.076 - 0.031
1 + 0.031 | = 0.043647... and s(5, j) = 5.45594 |
Score: 8/8
6.

A loan for \$61,000 is obtained at an interest rate of j1 = 9.2%. The loan will be repaid by 25 annual repayments which will increase by 2.4% p.a. compound, so that if the payment at the end of the first year is \$R, succeeding annual payments will be \$(1+2.4/100)R, \$(1+2.4/100)2R, ..., \$(1+2.4/100)24R. Calculate R. | | | Student Response | Value | Correct Answer | Answer: | 5187.70 | 100% | 5,187.70 | | General Feedback: | |

Loan | 61000 | | |____ | _____|_____ | ____|__ | ... | __|____ | _ | ____| | i = 0.092 | Time (yrs) | 0 | 1 | 2 | ... | 24 | | 25 | | Payments | | R | 1.024R | ... | 1.02423R | | 1.02424R | | | ↓ | | | | | | | |
Calculate the instalment R in the first year from the equation of value at time 0 61,000 | = R(1.024)-1a(25, j) | R | = 5187.7 |

where j = | 0.092 - 0.024
1 + 0.024 | = 0.066406... and a(25, j) = 12.04079 |
Score: 8/8
7.

A loan for \$100,000 is obtained at an interest rate of j1 = 9%. The loan will be repaid by 25 annual repayments which will increase by 3.5% p.a. compound, so that if the payment at the end of the first year is \$R, succeeding annual payments will be \$1.035R, \$1.0352R, ..., \$1.03524R. If R = 7,576.39, calculate the loan outstanding at the end of year 5 using the retrospective method. | | | Student Response | Value | Correct Answer | Answer: | 105519.84 | 100% | 105,519.84 | | General Feedback: | |
In the timeline R = 7,576.39 Loan | 100000 | | |____ | _____|_____ | ____|__ | ... | __|____ | _ | ____| | i = 0.09 | Time (yrs) | 0 | 1 | 2 | ... | 4 | | 5 | | Payments | | R | 1.035R | ... | 1.0353R | | 1.0354R | | | | | | | | | ↓ | |
At time 5, find the future value of the loan of 100,000 and subtract the future value of the payments made using formula (12) Retrospective loan outstanding | = 100,000(1.09)5 - 7,576.39(1.035)4s(5, j) | | = 105519.84 |

where j = | 0.09 - 0.035
1 + 0.035 | = 0.05314... and s(5, j) = 5.5604 |
Score: 8/8
8.

A loan for \$100,000 is obtained at an interest rate of j1 = 8%. The loan will be repaid by 25 annual repayments which will increase by 3.5% p.a. compound, so that if the payment at the end of the first year is \$R, succeeding annual payments will be \$1.035R, \$1.0352R, ..., \$1.03524R. If R = 6,871.03, calculate the loan outstanding at the end of year 11 using the prospective method. (The first outstanding payment is \$1.03511R and not \$R.) | | | Student Response | Value | Correct Answer | Answer: | 181580.05 | 0% | 100,069.20 | | General Feedback: | |
In the timeline R = 6,871.03 | |____ | _____|_____ | ____|__ | ... | __|____ | _ | ____| | i = 0.08 | Time (yrs) | 11 | 12 | 13 | ... | 24 | | 25 | | Payments | | 1.03511R | 1.03512R | ... | 1.03523R | | 1.03524R | | | ↓ | | | | | | | |
The unpaid payments are valued at time 11 using formula (11) Prospective loan outstanding | = [6,871.03(1.035)11](1.035)-1a(14, j) | | = 100069.2 |

where j = | 0.08 - 0.035
1 + 0.035 | = 0.043478... and a(14, j) = 10.32464 |
The loan outstanding will be greater than \$100,000 in the early years.
The instalment of 6,871.03 does not cover the interest of 0.08 × 100,000 = 8,000 in the first year. Score: 0/8
9.

On a debt of \$9,000, interest is paid monthly at j12 = 11.0% and monthly deposits are made into a sinking fund to repay the debt at the end of 7 years. If the sinking fund earns interest at j4 = 8.8%, what is the monthly expense of the debt? (Monthly expense = monthly interest payment + monthly sinking fund payment.
When finding the monthly sinking fund payment first find the equivalent monthly rate earned by the sinking fund.) | | | Student Response | Value | Correct Answer | Answer: | 160.57 | 100% | 160.58 | | General Feedback: | |
For the sinking fund (1 + j12/12)12 | = (1 + 0.088/4)4 | Monthly rate j12/12 | = 1.0221/3 - 1 = 0.00728... |

Deposits | | R | R | ... | R | R | | |____ | ____|____ | ____|__ | ... | __|____ | ____| | i = 0.00728... | Time (mths) | 0 | 1 | 2 | ... | 83 | 84 | | Required fund | | | | | | 9000 | | | | | | | | ↓ | |

Rs(84, 0.00728...) | = 9,000 | 115.26925R | = 9,000 | Monthly sinking fund payment R | = 78.08 | to which monthly interest of 9,000 × 0.11 / 12 = 82.5 is added Score: 7/7
10.

For the sinking fund of the previous question, calculate the amount in the sinking fund at the end of 21 months. (Make sure you accumulate the payment being made into the sinking fund and not the total expense.) | | | Student Response | Value | Correct Answer | Answer: | 1764.70 | 100% | 1,764.74 | | General Feedback: | |
For the sinking fund from the previous question
Monthly sinking fund payment R = 78.08
Monthly interest rate earned by sinking fund = 0.00728... Sinking fund at the end of 21 months | = 78.08s(21, 0.00728...) | | = 1764.74 |
Score: 7/7
11.

A Company wishes to establish one sinking fund to repay \$36,000 due in 7 years time, and \$76,000 due in 10 years time. What is the size of the level annual deposit for 10 years if the first deposit is made in one years time and the fund earns 10.2% p.a. effective? | | | Student Response | Value | Correct Answer | Answer: | 7717.19 | 100% | 7,717.19 | | General Feedback: | |

Payments in | | R | ... | R | R | ... | R | | |____ | ____|__ | ... | __|____ | ____|__ | ... | __| | i = 0.102 | Time (yrs) | 0 | 1 | ... | 6 | 7 | ... | 10 | | Payments out | | | | | 36000 | | 76000 | | | | | | | | | ↓ | |
With focal date at time 10 in years and R the annual deposit Rs(10, 0.102) | = 36,000(1.102)(10 - 7) + 76,000 | R | = 7717.19 |
Score: 7/7
12.

A certain machine costs \$10,000 and has an expected lifetime of 9 years and a scrap value of \$700. The annual maintenance cost is \$1100. Find the capitalised cost of the machine if the cost of funds is 9% p.a. | | | Student Response | Value | Correct Answer | Answer: | 30158.10 | 100% | 30,158.10 | | General Feedback: | |

Capitalised cost | = 10000 + | 10000 - 700
1.099 - 1 | + | 1100
0.09 | | = 30158.1 |
Score: 7/7
13.

A certain machine costs \$13,000 and has an expected lifetime of 7 years and a scrap value of \$1200. The annual maintenance cost is \$1200. Find the annual cost of the machine if the cost of funds is 11% p.a. | | | Student Response | Value | Correct Answer | Answer: | 3836.14 | 100% | 3,836.14 | | General Feedback: | |

Annual cost | = 13000 × 0.11 + | 13000 - 1200 s(7, 0.11) | + 1200 | | = 3836.14 |
Score: 7/7
14.

90-day bank-accepted bill futures contracts are quoted at 95.30. The settlement price of a contract of \$1,000,000 face value is | | | Student Response | Value | Correct Answer | Feedback | A. | \$953,000.00 | | | | B. | \$988,410.96 | | | | C. | \$988,543.73 | 100% | | | D. | \$1,000,000.00 | | | | | General Feedback: | |
The price is calculated using a simple interest rate of (100 - 95.3) = 4.7%.
Price = 1,000,000(1 + 0.047 × 90/365)-1 | | Score: | 1/1 | |
15.

An investor anticipates having surplus funds in 3 months time and intends to buy T-notes. Interest rates are expected to fall by the time the funds are available for investment. To hedge the value of this investment the investor should | | | Student Response | Value | Correct Answer | Feedback | A. | buy 90-day bank-accepted bill futures contracts | 100% | | | B. | sell 90-day bank accepted bill futures contracts | | | | | General Feedback: | |
As the investor is going to buy short-term T-notes in the future, he/she should buy a futures contract now. | | Score: | 1/1 | |
16.

It is March and an investor has a long position in the June Share Price Index (SPI) futures contract. To close out this contract the investor should | | | Student Response | Value | Correct Answer | Feedback | A. | buy a March SPI contract | | | | B. | sell a March SPI contract | | | | C. | buy a June SPI contract | | | | D. | sell a June SPI contract | 100% | | | | General Feedback: | |
A long contract is a buy contract. The closing contract must be the same as the original except opposite in direction. | | Score: | 1/1 | |
17.

The difference between the price in the physical market and the price of the related futures contract at the commencement of the hedging strategy is called | | | Student Response | Value | Correct Answer | Feedback | A. | initial basis risk | 100% | | | B. | initial margin | | | | | Score: | 1/1 | |
18.

A forward rate agreement would be traded on | | | Student Response | Value | Correct Answer | Feedback | A. | the Sydney Futures Exchange | | | | B. | the London Futures Exchange | | | | C. | both of a. and b. | | | | D. | Neither of a. and b. | 100% | | | | General Feedback: | |
A FRA is negotiated directly between the parties. Only standardised policies issued by an exchange can be traded on that exchange. | | Score: | 1/1 |

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