Assignment 6 Solutions

Assignment 6 Solutions - Chapter 10 Solutions 10.1 Drill...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 10 Solutions 10.1 Drill Problems 1. $12M 5 Z $2.3Ix.-1(1.08)i-1 6:1 defines the payback period 11* = 5. ‘Nith interest: “‘ $2.31x.-1(1+n.08)i-1 $12313 2; (1+ 0.14)f defines n" = G. We require some measure of worth in which to make a decision. l'jl PXA,18%,? P/G,18%,? PI-5-'(18%} = —3400, 000 + {3100, 000 — 350, 000}( 3 . 3115 J + 350, 000( 3 .9570 } = 3238, 025 Investment % error PVV(18%} % error 3 —250000 37.5% 3 383,924 62.8 970- 3 —300000 25% 3 338,924 41.9 970 3 —350000 —l2.5 97:3 3 288,924 20.9 97:: 3 400000 0 90 3 238,924 970 0 3 -450000 12.5% 3 188,924 —20.9 % 3 —500000 25% 3 138,924 41.9% 3 —550000 37.5% 3 88,924 —52.8 % Gradient 9% error P1111890) % error 3 35,000 -30 95 3 104,410 -503 95 3 40,000 -20 95 3 140,254 -335 5 3 45,000 —10 95 3 104,080 48.895 3 50,000 0 95 3 238,024 0 95 3 55,000 10 95 3 283,759 18.8% 3 00,000 20 95 3 328,503 37.595 3 35,000 30 95 3 373,428 56.3 95 Since the slope of the investment is greater than the gradient‘s value, PW is more sensitive to investment. , 25 $10K] +15($25K + 1 35001: _, 130(595} = —l - MW; 1,, ( ) =3.45 $2.5M 0.1295 ) For the value of a. life: r 25 $10K + 15(325K + I Bcl‘fié'} = l l A/P,595,10l 2 1 $2.5M 0.1295 } 1911111.r lion—negative value results in a positive BC ratio. Thus, it is not a critical estimate. 20. The deterministic present worth is $1.244 million. From 100 Simulation runs, 98 of them produced positive present worths. The mean and standard deviation were $1.262 million and $700,000, respectively. 10.2 Application Problems 5. {a} PXA,12%,10 PW(12%) = —$76101+(125,000($1900 — $1750} — $500,000)( 5.5502 J = 827.1 million. The payback period is 6 years. {b} For all scenarios, the facility will produce at the capacity. For the pessimistic scenario, revenues are $1870/ton and the profit margin is 5%. The present worth is: P/A,12%._10 P1711295) = —$76M + {125, 000(0.05}(81870) — $500,000)( 5.5502 )= —$12.8 million. For the average scenario, revenues are $2000/ton and the profit margin is 10%: P,»'.4,12%.,10 PW(12%} = —$76M + (125, 000(O.10)($2000) — $500, 000} ( 5.6502 } = $62.41 million. For the optimistic scenario. revenues are $2200/ton and the profit margin is 15%: P7A,12%,10 P1111295) = —$’7GM + (125,000(U.15)($2200} — 8500, 000)( 5.5502 ) = $154 million. Difference Error % P‘W {12%) Error% 15 97; 155.54 % $ 122,453,275 351.52 % {c} 12.5% 112.95% $ 88,015,212 227.01 % 10 (70 70.36 970 $ 55,367,150 104.19% 7.5% 27.77 % $ 21,810,088 -1053 95 5 975 —14.82 970 3 -11,728,075 —143.25 % The hreakeven difference is 5.87% 10. One run of the simulation is given in Figure 10.12. __-_- 17 Eplication Problem 10.10 18 19 Year Total Inv Investment Sales Price Costs Fixed Cost Salva e Value Cash Flow 20 0 356.2632604 118.7544201 ($ 18.751 21 1 1187544201 (5}; 18.751 22 2 1187544201 ($ 18.751 23 3 0.65 352.54 293.83 0.00 . 28.15 24 4 0.69 590.94 458.58 0.30 _ 80.42 25 5 0.64 312.39 234.31 0.48 _ 39.26 26 6 0.70 584.99 439.49 0.71 $91.14 27 7 0.69 505.71 406.33 0.86 . 57.38 28 8 0.66 263.04 209.04 1.08 . 24.30 29 9 0.67 609.13 502.07 1.30 _ 60.15 30 10 0.70 400.04 305.56 1.57 . 54.57 31 11 0.70 600.20 473.64 1.82 _ 76.76 32 12 0.70 589.28 463.40 2.09 $76.02 33 13 0.70 359.87 296.83 2.36 $31.77 34 14 0.70 325.37 266.07 2.54 $28.97 35 15 0.70 348.86 274.58 2.81 _ 39.18 36 16 0.70 603.95 481.56 1.75 . 73.92 37 17 0.70 321.18 269.87 2.13 _ 23.78 38 18 0.70 587.97 460.31 2.36 $77.00 39 19 0.70 528.32 430.26 2.60 $56.05 40 20 0.70 514.44 427.27 2.74 $48.28 11 21 0.70 251.93 202.42 3.05 . 21.60 42 22 0.70 333.17 261.61 3.29 21.04 . 57.85 43 449.17 357.85 20.33% PW 10% . 52.72 Figure 10.12: Aluminum investment simulation. (a) The average present worth is $30.00 million (standard deviation $36.58 million). The hiin standard deviation suggests that it is a risky project. (b) The probability of success is 8007:. which points to it being not too risky. (C) It is clear from Figure 10.13 that the present worth increases with a high margin. (cl) It is clear from Figure 10.14 that the present worth increases with a higher per ton priee. 5200.00 Tr!" c o E .E ... 59 o F‘ ... 3 n. t I I. I I I 18.50931: 19.00%: I 19.50911: I 20.00031: .2005091'0 21.00931: 21.5094: (550.00] Figure 10.13: Present worth as a function of margin on ton price. Pw(10% } (in millions} 5200.00 a: $150.00 I 3 5100.00 3 g ' I I i x x I I I I “x 3" a x i I 1 550.00 x“ xx: lxgx K”! I K W x a" I. H“ “‘u " I: x "x I" § 8 I: a ’i In: x 50.00 3‘ h" . I: g . . K . . 5390.00 3 541050 ’é430.00 ” 5450.00 S4?0.00 5490.00 5510.00 $530.00 I: a I (550.00] Ton Price Figure 10.14: Present. worth as a function of ton price. ...
View Full Document

This note was uploaded on 04/26/2009 for the course IE 226 taught by Professor Tonkay during the Spring '09 term at Lehigh University .

Page1 / 5

Assignment 6 Solutions - Chapter 10 Solutions 10.1 Drill...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online