Ingenieria Economica

Present Worth Analysis

Solutions to Problems

5.1 A service alternative is one that has only costs (no revenues).

5.2 (a) For independent projects, select all that have PW ≥ 0; (b) For mutually

exclusive projects, select the one that has the highest numerical value.

5.3 (a) Service; (b) Revenue; (c) Revenue; (d) Service; (e) Revenue; (f) Service

5.4 (a) Total possible = 25 = 32

(b) Because of restrictions, cannot have any combinations of 3,4, or 5. Only 12 are

acceptable: DN, 1, 2, 3, 4, 5, 1&3, 1&4, 1&5, 2&3, 2&4, and 2&5.

5.5 Equal service means that the alternatives end at the same time.

5.6 Equal service can be satisfied by using a specified planning period or by using the

least common multiple of the lives of the alternatives.

5.7 Capitalized cost represents the present worth of service for an infinite time. Real

world examples that might be analyzed using CC would be Yellowstone National

Park, Golden Gate Bridge, Hoover Dam, etc.

5.8 PWold = -1200(3.50)(P/A,15%,5)

= -4200(3.3522)

= $-14,079

PWnew = -14,000 – 1200(1.20)(P/A,15%,5)

= -14,000 – 1440(3.3522)

= $-18,827

Keep old brackets

5.9 PWA = -80,000 – 30,000(P/A,12%,3) + 15,000(P/F,12%,3)

= -80,000 – 30,000(2.4018) + 15,000(0.7118)

= $-141,377

PWB = -120,000 – 8,000(P/A,12%,3) + 40,000(P/F,12%,3)

= -120,000 – 8,000(2.4018) + 40,000(0.7118)

= $-110,742

Select Method B

5.10 Bottled water: Cost/mo = -(2)(0.40)(30) = $24.00

PW = -24.00(P/A,0.5%,12)

= -24.00(11.6189)

= $-278.85

Municipal water: Cost/mo = -5(30)(2.10)/1000 = $0.315

PW = -0.315(P/A,0.5%,12)

= -0.315(11.6189)

= $-3.66

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