Assignment 6 Solutions

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

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

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

View Full Document

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

View Full Document
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
Ask a homework question - tutors are online