Comparison Exercise Answers

1. Compare the three options with a monthly cost comparison.

The cash flow of a is an initial investment of 30000 with a salvage of 5000 after six years.

Aa = (30000 - 5000)(A/P, 0.0175, 72) + 5000(0.0175) = 25000*0.0245 + 5000*0.0175 = 612.50 + 87.50 = 700.00 per month.

The cash flow of b is an initial investment of 3000 monthly payments of 400 for 36 months.

Ab = 3000(A/P, 0.0175, 36) + 400 = 3000*0.0377+ 400 =113.1000+ 400 = 513.10 per month.

The cash flow of c is the same as for b plus a payment of 20,000 at year 3 and a salvage of 5000 at 72 months. To obtain a monthly cost this must be spread over 72 months. We take the monthly cost of b and bring it and the 20,000 to the present. Then we spread it over the 72 month life.

Pc = 513.10(P/A, 0.0175, 36) + 20000(P/F, 0.0175, 36) - 5000(P/F, 0.0175, 72)

Pc = 513.10*26.543+ 20000*0.5355 - 5000*0.2868 = 13619.21+ 10710.0 - 1434.= 22895.21

Ac = 22895.21(A/P, 0.0175, 72) = 22895.21*0.0245 = 560.9326

The straight lease option has the lowest monthly cost. It also has the advantage of a newer car for the second three years.

2. Compare the three options with an annual cost comparison. Cost below are in millions. i is 8%.

The cash flow of a is an initial investment of 3 with an additional cost of 1 after six years.

Aa = (3)(A/P, i, 6) +1(A/F, i 6) = 3*0.2163 + 1*0.1363 = 0.6489 + 0.1363 = 0.7852 per year.

The cash flow of b is an initial investment of 1 and annual payments of 0.5 for 3 years.

Ab = 1(A/P, i, 3) + 0.5 = 1*0.3880 + 0.5 = 0.3880 + 0.5 = 0.8880 per year.

The cash flow of c is the same as for b plus a payment of 2 at year 3. To obtain an annual cost this must be spread over 6 years. We take the annual cost of b and bring it and the 2 million to the present. Then we spread it over the 6 year life.

Pc = Ab(P/A, i, 3) + 2(P/F, i, 3)= 0.8880*2.577 + 2*0.7938 = 2.2884 +1.5876 .= 3.8760

Ac = Pc(A/P, i, 6) = 3.8760*0.2163 = 0.8384 per year.

The cash purchase plan is the best option with the lowest cost per year.

A Present worth comparison requires a 6 year study period.

3.

a. Compare on an annual cost basis.

NAC_A = 20000(A/P, .2, 6) = 20000*0.3007 = 6014

NAC_B = 10000(A/P, .2, 4) + 1000 + 500(A/G, .2, 4) - 2000(A/F, .2, 4) = 5127

Choose B

b. NAC_A = 6014 per year

NAC_C = 8000(A/P, .2, 2) + 2000 = 6236

Choose A

4. a

 Investment Evaluated  B-C  A-C
 NAW Computed  -30*.35 + 10 = -0.50/yr  -50*.35 + 20 = 2.50/yr
 Conclusion  Reject B - C  Accept A - C

Select A as the best.

b.

 Investment Evaluated  B-C  A - B
 NPW Computed

-20 + 10(P/A,.25,5) = 6.890

-30 + 10(P/A,.25,5) = -3.110

 Conclusion  Accept B - C  Reject A - B


Choose B