Ch 10 practice problems - 2

# D the new critical paths are still a c f g i and a d

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Unformatted text preview: F-G-I; and A-D-F-G-I. Project completion time is reduced from 18 to 16 weeks. 1 HW4 Solution OPER3100 Spring 2013 2. For the following diagram, (a) indicate the critical path when normal activity times are used; (b) How many days will the project take if we willing to pay crashing cost as long as it is less than \$1,000 (daily indirect cost)? a. Activity Normal Time A B C D E F 6 10 5 4 9 2 Cost per Day to Crash \$500 \$300 \$700 \$600 \$800 Crash Time 6 8 4 1 7 1 A(6) Available Days 2 1 3 2 1 B(10) Start F(2) C(5) D(4) E(9) Length After Crashing n Days Path n=0 1 2 3 4 A -B -F 18 18 18 17 16 C -D -E -F 20 19 18 17 16 Activity Crashed C E F B&E Direct Costs to Crash \$300<\$,1000 \$600<\$1,000 \$800<\$1,000 \$1,100>\$1,000 We can only crash 3 days since the direct cost to crash the 4th day is \$500+\$600=\$1,100, which is greater than the indirect cost benefit \$1,000. Therefore, the optimal project duration of 18 days. 2...
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## This note was uploaded on 02/26/2013 for the course OPER 3100 taught by Professor Staff during the Spring '08 term at UNC Charlotte.

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