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Unformatted text preview: Chapter 13 Solutions 13.1 Drill Problems P;AJS%._12
PW(18%)G = —$2.5M + $700,000{ 4.7932 ) = 38551240 PfAJéS%.,lU PfF:18“/L_.10
PI—$’(18%]M = —$2.5M + $5001000{ 4.4941 )+ $100,000( 0.1911 ) = $233,840 P/A._18%:8 PfFJS‘ﬁﬂ
PW(18%JP = —$2.5M + $400., 000( 4.0716 }+ $300,000( 0.2660 ) = —$780, 100
The expected present worth is:
E(PH’) = U.25($855._ 2410] + DEM—$233,840) + 0.15(—$?89, 160} = —$44._ 868 The investment shouldn’t be made since the PW is negative. 4. The delay option should only be considered if the prices will change with time. 5. After 1 year delay:
E(PH’) = (U.3[$855.240} — 0.6($233, 8840) — 0.1(8789. 160})/1.18 = $20, 750
After two year delay:
E(PlV) = (0.35($855, 240} — l].6($233. 884} — 0.U5($T89,160})/1.182 = $36,904
It is best to delay the project for two periods. 5. The only option that is worthy of investment is the good scenario. At time zero. with perfect information:
(0.25)[$855. 240} — —$=14._868 = $258. 678
At time 1:
(0.3)($855, 240) — $20, 750 = $235, 822
At time 2: (0.35)($855, 240) — $36. 904 = $202, 430 13.2 Application Problems £1. The amount of coal deﬁnes the investment rewards:
{a} P/A.25%.,2000 PxF.25%..2000
PW{25%)H=—18M+({250.OIJIJ}(5I])—5M)( 4.0000 1—200“ 0.0000 )=YUAN12M. P;A,25%._1200 P/F,25%._1200
PW(25%)M=—18M+((25EI,OOO)(50}—5M}( 4.0000 )—20M( 0.0000 )=YUAN12M. P/A.25%.400 PxF.25%..4ee
PW(25%)L=—18M+((250.IJIJO}(5IJ}—5M}( 4.0000 )—20M( 0.0000 )=YUAN12M. As all scenarios result in the same present worth, the expected value is 12 million yuan. {h} The value of perfect information is zero because there is no uncertainty in the outcome. (c) The joint and marginal probabilities for testing as follows: Test Results Low Medium High Pr(Test Results}
Favorable 0.01 0.125 0.255 0.39 Ineonclusive 0.03 0.3 0.042 0.372 Unfavorable 0.10 0.0?5 0. 003 0. 238 The conditional probabilities obtained from joint and marginal probabilities are: Favorable Inconelusive Unfavorable Low 0.025 0.08 0672
Medium 0.32 0.8 0315
High 0.655 0.12 0.013 E(PH"Fav) = Pr{SFav)PW(H} + Pr{MFav)PH—’(M} + Pr[DFav)PIV(L) This can be applied for other test results. too. Emu1mg) n1ax{0, 0.655(1263M} + 032(1080M) — 0.025(11MJ) = YUANllTQM
E(P1’VIn) = max(0,0.12(1263M) + 0.8(1080M) — I].08(11M}} = YUANIOlSM
E(PW on) mas[(0, 0.013(1263M} + 0.315(108IJM} — 0.672(11MJ) = YUAN349M The expected value is then:
E(PW) = Pr(Fa.v}Pl’l’(Fav) + PrUan‘WUn} + Pr(Un}PW{Un} E(PVV} = 030(1171M) + 0.372(1013M) + 0.238(349M) = YUANQITM
The testing results in the same value for the decision. It is not worth performing the test.
(f) See above. (g) You would not pay anything for the test. which is less than the value of perfect informa— tion. {11) W'e would invest in the facility, regardless of the test results. Chapter 14 Solutions 14.1 Drill Problems 1. The feasible portfolios and their returns are given in the spreadsheet in Figure 14.1. A rill Problem 14.1 or DID rDJECtS nuestment COTE
1 I S 2 1 529? 0.519?68?0?
511 5 v. 5 . . . . 0426935844
552,453 O.469?133?3
599,9?0 0580624358 Figure 14.1: Feasible investment. portfolios with analysis. (a) Portfolio 6 is the best under certainty.
(b) The variance is 3399.970. (c) Portfolio 4 (project 3] is dominated according to the efﬁciency frontier. It is drawn in
Figure 14.2 (d) Only portfolios 1. 2. and 4 should be considered due to the payback resitriction. (0) Scores of 3;”6. 2;”6. and MG were used for measures. Porfolio 6 has the best score. (a) The spreadsheet in Figure 14.8 shows that the difference in the AEC values (between the two assets] is more sensitive to the initial 0&11 costs when compared to the investment and increase in 0353.1 costs. (b) Using Goal Seek with the spreadsheet in Figure 14.9. the break—even number of batches was 91. (c) Using Goal Seek with the spreadsheet in Figure 14.9, the break—even rate for Asset 3‘s increasing costs is 10.31% per year.
A 8 C D E F G H I J L00 brill Problem 145 Input
L01 Asset 1 Error Asset 2 Error
L02 Period Asset 1 Asset 2 Purchase $350,000.00 30.00% Purchase $245,000.00 30.00%
L03. 0 $350,000.00 $245,000.00 Malntenance $400,000.00 0.00% Maintenance $450,000.00 0.00%
L04 1 54W 54W Sale W 000%.,0 (Maim 1.00211: 0100911.
L05 2 $400,000.00 $463,500.00 Life 8 Salvage Value $0.00 0.00%
L06 3 $400,000.00 $477,405.00 MARR 900% Life 6
g 4 $400,000.00 $491,727.15
L08 5 $400,000.00 $506,478.96 Output
L09 6 $400,000.00 $521,673.33 AEC (1] $458,702.31
L10 7 $400,000.00 AEC (2] $536,164.93
l11 s $350,000.00 AEC (12] (517 452.51] 501%
L1; AEC (12] (5.73 mam) Base
L13
L14 Sensitivity of AEC
L15 Error Investment Initial can 0&M Increase
L16 30% 5.01% 33.16% 13.17%
L17 15% 2.50% 16.58% 6.62%
L18 0% 0% 0% 0%
L19 15% 2.50% 16.58% 6.70% 30% 5.01% 33.16% 13.47% Figure 14.8: Comparisons of Assets A and B WlTh sensltwity results.
A 8 C D E F G H 122 brill Problem 14.50c Input 123 Asset 1 Asset 2 124 Period Asset 1 Asset 1 Purchase $500,000.00 Purchase $350,000.00 125 0 $500,000.00 $350,000.00 Malntenance $300,000.00 Maintenance $250,000.00 126 1 $5,300,000.00 $4,250,000.00 Batch $5,000.00 Batch $4,000.00 127 2 5530000000 $4,662,261.14 SalvageValue $50,000.00 g (batch‘l 10.31% 128 3 $5,300,000.00 $5,117,012.09 Life 8 Salvage Value $0.00 129 4 $5,300,000.00 $5,618,632.08 MARR 9.00% LlFe 6 130 5 55,300,00000 $6,171,951.68 Batches 1000 131 6 $5,m,000.00 $6,782,299.32 Output 132 7 $5,300,000.00 AEC(1] $5,385,803.47 133 8 $5,250,000.00 AEC (2] $5,385,803.47 AEC 12 50.00 Figure 14.9: Comparisons of Assets A and B with breakeven analysis. ...
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This note was uploaded on 01/24/2011 for the course IE 226 taught by Professor Tonkay during the Spring '09 term at Lehigh University .
 Spring '09
 TONKAY

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