HW+_7+Solutions

HW+_7+Solutions - 4.17 days 4 days 14.33 days = 22.5 days C...

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ECE 4001 Homework 7 Solutions Problem 1 A. AEM 3+3+11w+4 = 21 BCGM 4+3+1w+2+7w+4 = 21 AFGM 3+5+2+72+4 = 21 BCHNM 4+3+1w6+3+4 = 21 AFHNM 3+5+6+3+4=2 1 BCHP 4+3+1w+6+3 = 17 AFHP 3+5+6+3 = 17 BCHLR 4+3+1w+6+3+1w+ 6 = 24 AFHLR 3+5+6+3+1w+ 6 = 24 BCJKLR 4+3+1w+5+5+1w+ 6 = 25 AFJKR 3+5+5+5+6 = 24 BDKR 4+3+5w+5+1w+6 = 24 AFJKR is the critical path and has a duration of 24 days. B. TASK ES LS Float TASK ES LS Float A 0 0 0 L 14 15 1 B 0 2 2 M 17 20 3 C 4 5 1 N 14 17 3 D 4 10 6 P 14 21 7 E 3 17 14 R 18 18 0
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F 3 3 0 G 8 18 10 H 8 9 1 J 8 8 0 K 12 12 0 Problem 2
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Problem 3
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A. TASK Expected Time ES LS Float A 7 0 6.5 6.5 B 5.33 0 2.84 2.84 C 4.17 0 0 0 D 4 4.17 4.17 0 E 9 7 13.5 6.5 F 14.33 8.17 8.17 0 G 8.83 4.17 5.17 1 H 8.5 13 14 1 The critical path is the set of task items that have 0 float time. CDF is the critical path. B. The project duration can be determined from the length of time for the critical path.
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Unformatted text preview: 4.17 days + 4 days + 14.33 days = 22.5 days. C. Pr(T > 25 days) σ C 2 = ((6 – 3)/6) 2 = 0.25 σ D 2 = ((7-1)/6) 2 = 1 σ F 2 = ((25-5)/6) 2 = 11.11 σ T = [σ C 2 + σ D 2 + σ F 2 ] 1/2 = [0.25+1+11.11] 1/2 = 3.52 days z s = (25 – 22.5)/3.52 = 0.71 Pr(z s > 0.71) = 0.283352 The project has a 28% probability of taking longer than 25 days. Problem 4 (Hyman Chapter 7 Problem 1) Problem 5 (Hyman Chapter 7 Problem 8) Problem 6 (Hyman Chapter 7 Problem 12) Problem 7 (Hyman Chapter 7 Problem 18)...
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This note was uploaded on 02/08/2011 for the course ECE 4001 taught by Professor Frazier during the Fall '09 term at Georgia Tech.

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HW+_7+Solutions - 4.17 days 4 days 14.33 days = 22.5 days C...

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