HW_3_sol - A-B-D-G, 25 weeks, 5+10+6+4. b. Activity Normal...

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Homework Solutions – Chapter 3 1. The following activities are part of a project to be scheduled using CPM: Activity Immediate Predecessor Time (wks) A - 6 B A 3 C A 7 D C 2 E B,D 4 F D 3 G E,F 7 a. Draw the network b. What is the critical path? c. How many weeks will it take to complete the project? d. How much slack does activity B have? Solution a. b. A-C-D-E-G, also shown in the network above as the bold path. c. 26 weeks, 6+7+2+4+7. d. 6 weeks, 15-9.
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2. For the project with the following information, a. Determine the critical path and the early completion time in weeks. b. Reduce the project completion time by three weeks. Assume a linear cost per week shortened, and show, step by step, how you arrived at your schedule. Activity Imm. Pred. Normal time Normal cost Crash time Crash cost A -- 5 7000 3 13000 B A 10 12000 7 18000 C A 8 5000 7 7000 D B 6 4000 5 5000 E C 7 3000 6 6000 F C 4 6000 3 7000 G D,E,F 4 7000 3 9000 Solution a.
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Unformatted text preview: A-B-D-G, 25 weeks, 5+10+6+4. b. Activity Normal Time (NT) Normal Cost (NC) Crash Time (CT) Crash Cost (CC) NT-CT Cost/week to expedite A 5 $7,000 3 $13,000 2 $3,000 B 10 12,000 7 18,000 3 2,000 C 8 5,000 7 7,000 1 2,000 D 6 4,000 5 5,000 1 1,000 E 7 3,000 6 6,000 1 3,000 F 4 6,000 3 7,000 1 1,000 G 4 7,000 3 9,000 1 2,000 First, reduce D (lowest cost activity on the critical path) by one week. This adds an additional critical path with activities C and E in it. Second, crash activity G by one week. Critical paths remain the same. Third, crash activity A by one week at a cost of $3,000, which is the least expensive. Summary of activities crashed: Step Activity Cost to crash Weeks reduced 1 D $1,000 1 2 G 2,000 1 3 A 3,000 1 Total cost $6,000...
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HW_3_sol - A-B-D-G, 25 weeks, 5+10+6+4. b. Activity Normal...

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