## Review Problems

REVIEW PROBLEM 4.1

Tilson Dairies operates several cheese plants. The plants are all old and in need of renovation. Tilson's engineers have developed plans to renovate all the plants. Each project would have a positive present worth at the company's MARR. Tilson has S3.5 million available to invest in these projects. The following facts about the potential renovation projects are available:

 Project First Cost Present Worth A: Renovate plant 1 \$0.8 million \$1.1 million B: Renovate plant 2 SI.2 million \$1.7 million C: Renovate plant 3 \$1.4 million \$1.8 million D: Renovate plant 4 \$2.0 million \$2.7 million

Which projects should Tilson accept? ANSWER

Table 4.2 shows the possible mutually exclusive projects that Tilson can consider.

Tilson should accept projects A, B, and C. They have a combined present worth of \$4.6 million. Other feasible combinations that come close to using all available funds are B and D with a total present worth of \$4.4 million, and C and D with a total present worth of \$4.5 million.

Note that it is not necessary to consider explicitly the "leftovers" of the \$3.5 million budget when comparing the present worths. The assumption is that any leftover part of the budget will be invested and provide interest at the MARR, resulting in a zero present worth for that part. Therefore, it is best to choose the combination of projects that has the largest total present worth and stays within the budget constraints

 Total First Total Present Project Cost Worth Feasibility Do nothing SO.O million SO.O million Feasible A S0.8 million \$1.1 million Feasible B Si.2 million \$1.7 million Feasible C SI.4 million \$1.8 million Feasible D S2.0 million \$2.7 million Feasible A and B S2.0 million S2.8 million Feasible A and C \$2.2 million \$2.9 million Feasible A and D \$2.8 million S3.8 million Feasible B and C \$2.6 million \$3.5 million Feasible BandD \$3.2 million \$4.4 million Feasible CandD \$3.4 million S4.5 million Feasible A, B,and C S3.4 million S4.6 million Feasible A, B,and D \$4.0 million \$5.5 million Not feasible A, C, and D S4.2 million \$5.6 million Not feasible B, C,and D \$4.6 million \$6.2 million Not feasible A, B, C, and D \$5.4 million \$7.3 million Not feasible

REVIEW PROBLEM 4.2

City engineers are considering two plans for municipal aqueduct tunnels. They are to decide between the two, using an interest rate of 8%.

Plan A is a full-capacity tunnel that will meet the needs of the city forever. Its cost is \$3 000 000 now, and Si00 000 every 10 years for lining repairs.

Plan B involves building a half-capacity tunnel now and a second half-capacity tunnel in 20 years, when the extra capacity will be needed. Each of the half-capacity tunnels costs \$2 000 000. Maintenance costs for each tunnel are \$80 000 every 10 years. There is also an additional \$15 000 per tunnel per year required to pay for extra pumping costs caused by greater friction in the smaller tunnels.

(a) Which alternative is preferred? Use a present worth comparison.

(b) Which alternative is preferred? Use an annual worth comparison.

(a) Plan A: Full-Capacity Tunnel

First, the \$100 000 paid at the end of 10 years can be thought of as a future amount which has an equivalent annuity.

Thus, at 8% interest, Si00 000 every" 10 years is equivalent to S6903 every year.

Since the tunnel will have (approximately) an infinite life, the present cost of the lining repairs can be found using the capitalized cost formula, giving a total cost of

Plan B: Half-Capacity1 Tunnels

For the first tunnel, the equivalent annuity for the maintenance and pumping costs is

The present cost is then found with the capitalized cost formula, giving a total cost of

Now, for the second tunnel, basically the same calculation is used, except that the present worth calculated must be discounted by 20 years at 8%, since the second tunnel will be built 20 years in the future.

PW, = {2 000 000 + [15 000 + 80 000(0.06903)]/0.08}(P/F,8%,20) = 2 256 525(0.21455) = 484 137

Consequently, the two half-capacity7 aqueducts with a present worth of costs of S2 740 662 are economically preferable.

(b) Plan A: Full-Capacity Tunnel

First, the Si00 000 paid at the end of 10 years can be thought of as a future amount that has an equivalent annuity of

Thus, at 8% interest, S100 000 every 10 years is equivalent to S6903 every7 year.

Since the tunnel will have (approximately) an infinite life, an annuity equivalent to the initial cost can be found using the capitalized cost formula, giving a total annual cost of

AW(Plan A) = 3 000 000(0.08) + 6903 = 246 903 Plan B: Half-Capacity Tunnels

For the first tunnel, the equivalent annuity for the maintenance and pumping costs is

The annual equivalent of the initial cost is then found with the capitalized cost formula, giving a total cost of

Now, for the second tunnel, basically7 the same calculation is used, except that the annuity must be discounted by 20 years at 8%, since the second tunnel will be built 20 years in the future.

Consequently, the two half-capacity aqueducts with an annual worth of costs of \$219 253 are economically preferable. il

### REVIEW PROBLEM 4.3

Fernando Constantia, an engineer at Brandy River Vineyards, has a \$100 000 budget for winery improvements. He has identified four mutually exclusive investments, all of five years' duration, which have the cash flows shown in Table 4.3. For each alternative, he wants to determine the payback period and the present worth. For his recommendation report, he will order the alternatives from most preferred to least preferred in each case. Brandy River uses an 8% ALARR for such decisions.

 Alternative 0 Cash 1 Flow' at the End of Each Year 2 3 4 5 A -S100 000 S25 000 S25 000 S25 000 \$25 000 \$ 25 000 B -100 000 5000 10 000 20 000 40 000 80 000 C -100 000 50 000 50 000 10 000 0 0 D -100 000 0 0 0 0 The payback period can be found by decrementing yearly. The payback periods for the alternatives are then A: 4 years B: 4.3125 or 5 years C: 2 years D: 4.1 or 5 years The order of the alternatives from most preferred to least preferred using the payback period method with yearly decrementing is: C, A, D, B. The present worth computations for each alternative are: B: PW = -100 000 + 5000(P/F,8%,1) + 10 000(P/F,8%,2) + 20 000(P/F,8%,3) + 40 000(P/F,8%,4) + 80 000(P/F,8%,5) = -100 000 + 5000(0.92593) + 10 000(0.85734) + 20 000(0.79383) + 40 000(0.73503) + 80 000(0.68059) = 12 982 CHAPTER 4 Comparison Methods Part 1 C: PW = -100 000 + 50 000(P/F,8%,1) + 50 000(P/F,8%,2) + 10 000(P/F,8%,3) = -100 000 + 50 000(0.92593) + 50 000(0.85734) D: PW = -100 000 + 1 000 000(P/F,8%,5) = -100 000 + 1 000 000(0.68059) = 580 590 The order of the alternatives from most preferred to least preferred using the present worth method is: D, B, A, C.B 