...Running head: PROBLEM SET I Problem Set I Problem Set I 4-20 Juan’s Taco Company has restaurants in five college towns. Juan wants to expand into Austin and College Station and needs a bank loan to do this. Mr. Bryan, the banker, will finance construction if Juan can present an acceptable three-month financial plan for January through March. The following are actual and forecasted sales figures: Table 1 Actual November $120,000 December 140,000 Forecast January $190,000 February 210,000 March 230,000 April $230,000 Of Juan’s sales, 30 percent are for cash and the remaining 70 percent are on credit. Of credit sales, 40 percent are paid in the month after sale and 60 percent are paid in the second month after the sale. Materials cost 20 percent of sales and are paid for in cash. Labor expense is 50 percent of sales and it is paid in the month of sales. Selling and administrative expense is 5 percent of sales and is paid in the month of sale. Overhead expense is $12,000 in cash per month; depreciation expense is $25,000 per month. Taxes of $20,000 and dividends of $16,000 will be paid in March. Cash at the beginning of January is $70,000 and the minimum desired cash balance is $65,000.For January, February, and March, prepare a schedule of monthly cash receipts, monthly cash payments, and a complete monthly cash budget with borrowings and repayments. Table 2 Cash Receipts Schedule Sales Credit sales (70%) Cash sales (30%) Collections (month after credit...
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...Problem set 2 1) PA=FV(1+y1) 95.24=100(1+y1) Y1=5% PB=c(1+y1) +c+FV(1+y2)2 107.42=10(1+0.05) +10+100(1+y2)2 Y2=6% PC=c(1+y1) +c(1+y2)2+ c+FV(1+y3)3 140.51=20(1+0.05) +20(1+0.06)2+ 20+100(1+y3)3 Y3=5% PD=FV(1+y4)4 85.48=100(1+y4)2 Y4=4% 2) PE=c(1+y1) +c+FV(1+y2)2 FV=100 C=25 PE=25(1+0.05) +25+100(1+0.06)2=135.06 Build a replicating portfolio because arbitrage opportunity only available when selling bonds E and buying a replicating portfolio. So XA CF1A+XB CF1B=CF1c, XA*100+XB*10=25 XA CF2A+XB CF2B=CF2C, XA+0+XB*110=125 XA=0.1364 XB=1.1364 Arbitrage profit is 136-135.06=0.94 In conclusion, the arbitrage profit is $0.94 when building replicating portfolio whih consist of buying 0.1364 bond A and 1.1364 bond B. 3) P=c(1+y1) +c+FV(1+y2)2 105.6=10(1+y1) +10+100(1+y2)2 123.86=20(1+y1) +20+100(1+y2)2 (1+y2)2= 120123.86-20(1+y1) So, 105.6=10(1+y1) +10+100120123.86-20(1+y1) Y1=4.98% 123.86=20(1+0.0498) +20+100(1+y2)2 Y2=7% 4) FV=100 Coupon rate=10% A, if 1 year. CF1=C1+P1 CF1=C1+10+100(1+1Y2)2 Finally, we do not know 1Y2 so that we can’t make sure at what price need to sell the bind. This will lead the customer to liquidity risk. B, if 3 years. Cause is a 2 years bond. CF1=10*(1+1Y3)2+110*(1+2Y3) The investor will face reinvestment risk when they don’t not what rate their cahs...
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...sensitivity of a bond and its time to maturity are positive correlated. (c) To a zero-coupon bond Typically, the negative sign is ignored. This expression implicates that the elasticity of a zero-coupon bond is equal to its time to maturity (t). 2. Duration and elasticities (a) The higher the coupon payments, the ‘heavier’ the left hand side. To get the ‘system on balance’, we have to move the fulcrum to the left. This would lead to a shorter distance and hence a lower duration. And vice versa, the lower the coupon payments, the ‘heavier’ the right hand side. Now, to get the ‘system on balance’, we have to move the fulcrum to the right. This would lead to a longer distance from the origin and hence a higher duration. (b) Based on the preceding formula, the sigh of duration is positive. Modified duration is considered as a direct measure of price sensitivity to interest rate changes. Thus, the relationship of bond prices to interest rates is negative correlation. (C) -108*2.2*0.25%＄-0.594 3. Duration calculations (a) What is the price of the bond? According to the equation , we get the following: t | CF | PV | 0.5 | $25 | $23.95 | 1 | $25 | $22.94 | 1.5 | $25 | $21.97 | 2 | $25 | $21.04 | 2.5 | $25 | $20.15 | 3 | $25 | $19.30 | 3.5 | $25 | $18.49 | 4 | $1,025 | $726.14 | | Price | $873.98 | (b) Calculate and interpret the Macauley Duration for the bond. According to the Macauley Duration’s equation...
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...•5. Suppose a Midwest Telephone and Telegraph (MTT) Company bond, maturing in 1 year, can be purchased today for $975. Assuming that the bond is held until maturity, the investor will receive $1,000 (principal) plus 6 percent interest (that is, 0.06 × $1000 = $60). Determine the percentage holding period return on this investment. INTERMEDIATE 6.•a. National Telephone and Telegraph (NTT) Company common stock currently sells for $60 per share. NTT is expected to pay a $4 dividend during the coming year, and the price of the stock is expected to increase to $65 a year from now. Determine the expected (ex ante) percentage holding period return on NTT common stock. b. Suppose that 1 year later, NTT’s common stock is selling for $75 per share. During the 1-year period, NTT paid a $4 common stock dividend. Determine the realized (ex post) percentage holding period return on NTT common stock. c. Repeat Part b given that NTT’s common stock is selling for $58 1 year later. d. Repeat Part b given that NTT’s common stock is selling for $50 1 year later. INTERMEDIATE 7. One year ago, you purchased a rare Indian-head penny for $14,000. Because of the recession and the need to generate current income, you plan to sell the coin and invest in Treasury bills. The Treasury bill yield now stands at 8 percent, although it was 7 percent one year ago. A coin dealer has offered to pay you $12,800 for the coin. Compute the holding period return on this...
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...Other information: Setup costs vary with the number of batches processed in each department. The budgeted number of batches for each product line in each department is as follows: Trophies Plaques Forming department 40 116 Assembly department 43 103 Supervision costs vary with direct labor costs in each department. 1. Calculate the budgeted cost of trophies and plaques based on a single plant-wide overhead rate, if total overhead is allocated based on total direct costs. 2. Calculate the budgeted cost of trophies and plaques based on department overhead rates, where forming department overhead costs are allocated based on direct labor costs of the forming department , and assembly department overhead costs are allocated based on total direct costs of the assembly department. 3. Calculate the budgeted cost of trophies and plaques if Tarquin allocated overhead costs in each department using activity-based costing. 4. Explain how the disaggregation of information could improve or reduce decision quality. Overhead rate = Total overhead 56,346 0.712113744 Total Direct costs $79,125 Tarquin's Trophies Budgeted Information For the Year Ended November 30, 2011 Forming Department Trophies Plaques Total Direct materials $13,000 $11,250 $24,250 Direct labor 15,600 9,000 24,600 Overhead Costs Setup 12,000 Supervision ...
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...Problem Set II P2-6A A. Earnings per share. | | Earnings per share = Net income/Average No. of outstanding shares | | 2011-$2.00 = 60,000/30,0002012- $2.12=70,000/33,000 | | B. Working capital. | | Working capital = current assets - current liabilities | | 2011- 85,000 = 155,000-70,0002012- 113,000 = 188,000- 75,000 | | C. Current ratio. | | Current ratio = Current assets/Current liability | | 2011 :2.2 =155,000/70,0002012: 2.5 = 188,000/75,000 | | D. Debt to total assets ratio. | | | | Debt to total asset ratio = Total debt/Total assets | | 2011- 23.4% = 160,000/685,0002012- 20.4% = 155,000/760,000 | | E. Free cash flow. | | Free cash flow = Cash from operating activities-Cash used for cap. Exp.-Dividends | 2011- 3,0002012-17,000 | | F.) Based on the ratios calculated, discuss briefly the improvement or lack thereof in | Financial position and operating results from 2011 to 2012 of Sievert Corporation. | | | There was an increase of net income and earnings per share in which showed the growth of the company. The liquidity showed in the working capital and ratio. The solvency improved the debt of total assets ratio and the free cash flow | | | | | P13-2A A. Earnings per share: Earnings per share = Net income/Average No. of outstanding shares $218,000/59,000 = $3.69 No. of outstanding share for 2012 = 290,000/5: 58000 No. of outstanding share for 2011 = 300,000/5: 60000 Average no. of outstanding...
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...1. Refer to Problem 1.8 on page 59 of the Hubbard/O’Brien textbook. To answer questions a, b, and c, carefully study the graph and identify the points that are unattainable, efficient, and inefficient. Be sure to explain how you identified the points in each case. To answer question d, note that the graph shows the trade-off between producing capital goods (machines, tools, and factories) and producing consumption goods (food, clothing, and movies); which of these will add to an economy’s ability to produce in the future? a. Point E is currently unattainable because that point is beyond the line meaning that there are not enough resources to reach that point. b. Points B, C, and D are efficient because the points are on the line in the graph. c. Point A is inefficient because that point is below the line meaning all resources are not being used. d. Point D is likely to have the highest growth rate because there are more consumption goods being used at that point. 2. Refer to Problem 3.7 on page 62 of the Hubbard/O’Brien textbook. George represents households while BMW and McDonald’s represent firms. For each transaction, identify whether the transaction takes place in the factor market or the product market. • If the transaction takes place in the factor market, indicate whether households are supplying or demanding factors of production and whether firms are supplying or demanding factors of production. • If...
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...Text Problem Sets - Week Two Vaughan Thompson FIN/571: Corporate Finance April 22, 2013 Text Problem Sets: FIN/571 - Week Two Chapter Five Question # 4 Define the following terms: bond indenture, par value, principal, maturity, call provision, and sinking fund. Bond indenture. Bond indenture is a legal contract for a publicly traded bond. The structure of this contract outline incentives explicitly by detailing responsibilities, constraints, penalties, and oversight required. For example, contracts may specify interest and principal payment timing and amounts. Par value. Par value denotes face value or designated value of a bond or stock. Par value of a bond is typically $1,000 and the sum investors pay upon issue. It is also the sum received when they redeem the bond matures. Conversely, stock par value is frequently set at $1. In this case, par value is an accounting tool that shows no connection to the stocks’ market value. Principal. The term “principal” refers to a sum of money one borrows or invests. The face amount of a bond - the value printed on a stock or bond, or a debt balance. Principal does not encompass finance charges. Principal also describes an investor represented by a broker who executes trades on that investor’s behalf or an investor who trades for his or her own benefit. Principal also refers to a party affected by an agent’s decisions in a principal-agent relationship. Maturity. Maturity is the end of a bond’s life. In finance, maturity (or...
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...FIN 571 Week 2 Individual Assignment Text Problem Sets Get Tutorial by Clicking on the link below or Copy Paste Link in Your Browser https://hwguiders.com/downloads/fin-571-week-2-individual-assignment-text-problem-sets/ For More Courses and Exams use this form ( http://hwguiders.com/contact-us/ ) Feel Free to Search your Class through Our Product Categories or From Our Search Bar (http://hwguiders.com/ ) FIN 571 W2 Ch 5 Problem A1 (Bond Valuation) FIN 571 W2 Ch 5 Problem A10 (Dividend discount model) FIN 571 W2 Ch 5 Problem A12 (Required return for a preferred stock) FIN 571 W2 Ch 5 Problem A14 (Stock Valuation) FIN 571 W2 Ch 5 Problem B16 (Interest-rate risk) FIN 571 W2 Ch 5 Problem B18 (Default risk) FIN 571 W2 Ch 5 Problem B20 (Constant growth model) FIN 571 W2 Ch 7 Problem C1 (Beta and required return) FIN 571 Week 2 Individual Assignment Text Problem Sets Get Tutorial by Clicking on the link below or Copy Paste Link in Your Browser https://hwguiders.com/downloads/fin-571-week-2-individual-assignment-text-problem-sets/ For More Courses and Exams use this form ( http://hwguiders.com/contact-us/ ) Feel Free to Search your Class through Our Product Categories or From Our Search Bar (http://hwguiders.com/ ) FIN 571 W2 Ch 5 Problem A1 (Bond Valuation) FIN 571 W2 Ch 5 Problem A10 (Dividend discount model) FIN 571 W2 Ch 5 Problem A12 (Required return for a preferred stock) FIN 571 W2 Ch 5 Problem A14 (Stock Valuation) FIN 571 W2...
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...000/ $5= $58,000 ($60,000 + $58,000) / 2=$59,000 $218,000/ $59,000= $3.69 $3.69 (B) Return on Common Stockholders' Equity 40.80% Return Common Stockholders' Equity Ratio = Net Income - Preferred Stock Dividends Average Common Stockholders' Equity $218,000/ [($465,400+$603,400)]/ 2= $218,000/ $534,400= 40.80% (C) Return on Assets Net Imcome x Net Sales Net Sales Average Total Assets $218,000/ [($1,026,900+ $852,800)]/ 2= 23.2 23.20% Net Income Average Total Assets (D) Current Ratio Current Ratio = Current Assets ÷ Current Liabilities $377,900/ $203,500= 1.86 1.86 Page 1 Chapter_13_P13-2A (E) Receivables Turnover 17.1 Receivables Turnover Ratio = Net Credit Sales ÷ Average Net Receiable $1,890,540/[($102,800+$117,800)/2]= $1,890,540/$110,300= 17 (F) Average Collection Period Average Collection Period = 365 ÷ Receivables Turnover Ratio 365 days/ 17.1= 21.3 days 21.3 (G) Inventory Turnover 8.8 Inventory Turnover Ratio = Cost of Goods Sold ÷ Average Inventory $1,058,540/[($115,500+$126,000)]/2= $1,058,540/$120,750= 8.8 times (H) Days in Inventory Days in Inventory = 365 ÷ Inventory Turnover Ratio 365 days/ 8.8= 41.5 days 41.5 (I) Times Interest Earned Net Income + Interest Expense + Tax Expense $332,000/ $22,000= 15.1 times 15.1 ÷ Interest Expense Page 2 Chapter_13_P13-2A (J) Asset Turnover Ratio 2.01 Asset Turnover Ratio = Net Sales ÷ Average Total Assets $1,890,540/[($1,026,900+$852,800)]/2= $1,890,540/ $939,850 =2.01 (K) Debt...
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...Edition Solutions to Text Problems: Chapter 1 Chapter 1: Problem 1 A. Opportunity Set With one dollar, you can buy 500 red hots and no rock candies (point A), or 100 rock candies and no red hots (point B), or any combination of red hots and rock candies (any point along the opportunity set line AB). then: Algebraically, if X = quantity of red hots and Y = quantity of rock candies, 0.2 X + 1Y = 100 That is, the money spent on candies, where red hots sell for 0.2 cents a piece and rock candy sells for 1 cent a piece, cannot exceed 100 cents ($1.00). Solving the above equation for X gives: X = 500 − 5Y which is the equation of a straight line, with an intercept of 500 and a slope of −5. Elton, Gruber, Brown, and Goetzmann Modern Portfolio Theory and Investment Analysis, 7th Edition Solutions to Text Problems: Chapter 1 1-1 B. Indifference Map Below is one indifference map. The indifference curves up and to the right indicate greater happiness, since these curves indicate more consumption from both candies. Each curve is negatively sloped, indicating a preference of more to less, and each curve is convex, indicating that the rate of exchange of red hots for rock candies decreases as more and more rock candies are consumed. Note that the exact slopes of the indifference curves in the indifference map will depend an the individual’s utility function and may differ among students. Chapter 1: Problem 2 A. Opportunity Set The individual can consume...
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... Preface Although this may seem a paradox, all exact science is dominated by the idea of approximation. Bertrand Russell (1872–1970) Most natural optimization problems, including those arising in important application areas, are NP-hard. Therefore, under the widely believed conjecture that P = NP, their exact solution is prohibitively time consuming. Charting the landscape of approximability of these problems, via polynomial time algorithms, therefore becomes a compelling subject of scientific inquiry in computer science and mathematics. This book presents the theory of approximation algorithms as it stands today. It is reasonable to expect the picture to change with time. The book is divided into three parts. In Part I we cover a combinatorial algorithms for a number of important problems, using a wide variety of algorithm design techniques. The latter may give Part I a non-cohesive appearance. However, this is to be expected – nature is very rich, and we cannot expect a few tricks to help solve the diverse collection of NP-hard problems. Indeed, in this part, we have purposely refrained from tightly categorizing algorithmic techniques so as not to trivialize matters. Instead, we have attempted to capture, as accurately as possible, the individual character of each problem, and point out connections between problems and algorithms for solving them. In Part II, we present linear programming based algorithms. These are categorized under two fundamental techniques: rounding and...
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...Pharmacy Technicians: A Worktext Showall your calculations in a Microsoft® Word document Completethe following conversions: 1. Test Your Knowledge (metric system), p. 41: Problems 2, 4, 6, 8, 12 &14 2. Test Your Knowledge (apothecary system), p. 42: Problems 21, 29, 34, 37 & 40 3. Test Your Knowledge (household system), p. 42: Problems 42, 44, 49, 52, 54 & 56 Postyour work and answers to all three sets of problems along with a signed copy of the Certificate of Originality as an attachment under the Assignment link. HCP 220 Week 4 Assignment- Measurements Using Metric, Apothecary, and Household Systems https://hwguiders.com/downloads/hcp-220-week-4-assignment-measurements-using-metric-apothecary-household-systems/ HCP 220 Week 4 Assignment- Measurements Using Metric, Apothecary, and Household Systems Resource: Ch. 5 of Pharmaceutical Calculations for Pharmacy Technicians: A Worktext Showall your calculations in a Microsoft® Word document Completethe following conversions: 1. Test Your Knowledge (metric system), p. 41: Problems 2, 4, 6, 8, 12 &14 2. Test Your Knowledge (apothecary system), p. 42: Problems 21, 29, 34, 37 & 40 3. Test Your Knowledge (household system), p. 42: Problems 42, 44, 49, 52, 54 & 56 Postyour work and answers to all three sets of problems along with a signed copy of the Certificate of Originality as an attachment under the Assignment link. HCP 220 Week 4 Assignment- Measurements Using Metric, Apothecary, and...
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...Face Any Problem E- Mail Us At JohnMate1122@gmail.com WEEK 1 Quiz 1 Set 1(20 Questions and Answers) Quiz 1 Set 2 (20 Questions and Answers) Quiz 1 Set 3 (20 Questions and Answers) WEEK 2 Discussion Question 1: Decision Tree Discussion Question 2: The Science of Probability Quiz 2 Set 1 (20 Questions and Answers) Quiz 2 Set 2 (20 Questions and Answers) WEEK 3 Assignment 1, Jet Copier Case Problem Discussion Question 1: Pseudorandom Discussion Question 2: Statistical Analysis WEEK 4 Assignment 2, Internet Field Trip Presentation Discussion Question 1: Forecasting Discussion Question 2: Forecasting Models WEEK 5 Midterm Exam 1 (40 Questions and Answers) Midterm Exam 2 (40 Questions and Answers) Midterm Exam 3 (40 Questions and Answers) Midterm Exam 4 (40 Questions and Answers) WEEK 6 Discussion Question 1: The Linear programming Model Discussion Question 2: The LP Model Discussion Question 3: Graphical Method Discussion Question 4: Minimization model Quiz 3 Set 1 (20 Questions and Answers) Quiz 3 Set 2 (20 Questions and Answers) Quiz 3 Set 3 (20 Questions and Answers) WEEK 7 Assignment 3, Case Problem, Julia's Food Booth Discussion Question 1: Shadow Price Discussion Question 2: Sensitivity Analysis Discussion Question 3: LP model and optimal solution. WEEK 8 Discussion Question 1: Objective Function Discussion Question 2: Applications of Linear Programming Quiz 4 Set 1 (20 Questions and Answers) Quiz 4 Set 2 (20 Questions...
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...Exercise 4.1, problem 5a for i := 1 to 123 do for j := 1 to i do print i * j a) How many times is the print statement of the third line executed? Since we have to count iterations starting from one until 123, the first count would be 1 then 3 then 6 and so forth. The segment can be translated to (n)(n+1)/2 where 123 would be (n). (123)(123 + 1)/2 The statement is executed 7626 times. Exercise 4.2, problem 18a a) How many permutations of 1, 2, 3 have k ascents, for k = 0, 1, 2? Ascent can be determined by simply looking at its numbers. In the case of 123, 3 > 2 and 2 >1 so there are 2 ascents. 123 = 2 321 = 0 231 = 1 213 = 1 132 = 1 312 = 1 k = 0 1 k = 1 4 k = 2 1 Exercise 4.3, problem 22a Solve each problem (if possible), and then convert the results to base 10 to check your answers. Watch for any overflow errors. 8 4 2 1 0101 5 + 0001 1 0110 6 Exercise 4.4, problem 1a 1. For each of the following pairs a, b ∈ Z+, determine gcd (a, b) and express it as a linear combination of a, b. a) 231, 1820 1820 = 7 (231) + 203 0 < 203 < 231 231 = 1 (203) + 28 0 < 203 < 28 203 = 7 (28) + 7 0 < 28 < 7 28 = 4 (7) + 0 7 gcd(1820, 231) = 7 7 = 203 – 7 (28) 203 – 7 (231 – 203) 8 (203) – 7 (231) 8 (1820 – 7 (231)) – 7(231) 8 (1820) – 63 (231) Exercise 5.1, problem 4 For which sets A, B is it true that A X B = B...
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