Assignment 1

3.2

When PW is present value worth, F is future value, I is interest per annum, n is number of years then.

PW = F * (1+i)^-n

F = $20,000, n= 20 years,

a. When i = 8% per annum

PW = 20000*(1+.08)^-20 = $4290.96.

b. When I = 2% per quarter, interest rate per annum will be 8% (4 quarters per year, 2* 4 = 8%)

PW = 20000*(1+.08)^-20 = $4290.96

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3.30

Formula for PA(n) = A1(P/F,i,1) + A2(P/F,i,2) ….+ An(P/F,i,n)

NAC(n) = [P + PA(n)](A/P,I,n)-S(n)(A/F,I,n).

Thus for this problem

NAC 1= 50,000(A/P, 0.1, 1) + 10,000 = 50,000×1.1 + 10000 = 65,000

NAC 2 = 50,000(A/P, 0.1, 2) + 10,000 + 30,000(A/G, 0.1, 2)

= 50,000×0.5762 + 10,000 + 30,000×0.476 = 28810 + 10,000 + 14,280

= 53,090

NAC 3 = 50,000(A/P, 0.1, 3) + 10,000 + 30,000(A/G, 0.1, 3)

= 50,000×0.4021 + 10,000 + 30,000×0.937= 20,105 + 10,000 + 28,110

= 58,215

Minimum of NAC 1, NAC 2, and NAC 3 is NAC 2 so economic life time is 2years.

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4.9

Given Average monthly cost of model A = $500,

Total cost of Model b for 4 years = Purchase price – resale value + yearly maintenance*4 + Monthly operational cost * 12 * 4 + Yearly car insurance * 4 + Repair lost per failure *(4*12/14)

= 23000 - 13000 + 1100*4 + 90 * 12 *4 + 1400 *4 + 650 * 48/14

=10000 + 4400 + 4320 + 5600 + 2228.57 = 26548.57

Average monthly cost of model B = 26548.57 / 48 = 553.095 (48 = 12*4 years)

Minimum is for model A.SO Natasha should choose Model A.