Analisis

Chapter 5





Present Worth Analysis



Solutions to Problems


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.

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

6. Equal service can be satisfied by using a specified planning period or by using the
least commonmultiple of the lives of the alternatives.

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.

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

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

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 = $0.315PW = -0.315(P/A,0.5%,12)
= -0.315(11.6189)
= $-3.66

11. PWsingle = -4000 - 4000(P/A,12%,4)
= -4000 - 4000(3.0373)
= $-16,149

PWsite = $-15,000

Buy the sitelicense

12. PWvariable = -250,000 – 231,000(P/A,15%,6) – 140,000(P/F,15%,4)
+ 50,000(P/F,15%,6)
= -250,000 – 231,000(3.7845) – 140,000(0.5718) + 50,000(0.4323)
= $-1,182,656

PWdual = -224,000 –235,000(P/A,15%,6) –26,000(P/F,15%,3)
+10,000(P/F,15%,6)
= -224,000 –235,000(3.7845) –26,000(0.6575) + 10,000(0.4323)
= $-1,126,130

Select dual speed machine

13. PWJX = -205,000 – 29,000(P/A,10%,4) – 203,000(P/F,10%,2)
+ 2000(P/F,10%,4)
= -205,000 – 29,000(3.1699) – 203,000(0.8264) + 2000(0.6830)= $-463,320

PWKZ = -235,000 – 27,000(P/A,10%,4) + 20,000(P/F,10%,4)
= -235,000 – 27,000(3.1699) + 20,000(0.6830)
= $-306,927

Select material KZ
14. PWK = -160,000 – 7000(P/A,2%,16) –120,000(P/F,2%,8) + 40,000(P/F,2%,16)
= -160,000 – 7000(13.5777) –120,000(0.8535) + 40,000(0.7284)= $-328,328

PWL = -210,000 – 5000(P/A,2%,16) + 26,000(P/F,2%,16)
= -210,000 – 5000(13.5777) + 26,000(0.7284)
= $-258,950
Select process L

15. PWplastic = -75,000 - 27,000(P/A,10%,6) - 75,000(P/F,10%,2)
- 75,000(P/F,10%,4)
= -75,000 - 27,000(4.3553) - 75,000(0.8264) -...