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# Quiz6AKey - A(6 pts Assuming each activity takes its normal...

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15 10 5 D 11 6 10 6 4 C 10 6 0 0 0 ST 0 0 10 4 6 E 10 4 17 15 2 F 14 12 4 0 4 B 4 0 6 0 6 A 6 0 17 10 7 G 17 10 17 17 0 FIN 17 17 OSCM 230 Quiz 6A (Key)  Name: Score:           /10 pts  Activity Normal  Time  (Ti) (weeks) Budgeted  Cost  (Ci)  (\$000) Crash Time =(T’ i ) (weeks) Crash  Cost (C’ i ) (\$000) Cost = C’ - Ci  (\$000) Time = Ti -T’ (week) Cost/ Time (\$000/week) A 6 10 3 11.5 1.5 3 0.5 B 4 5 3 6 1 1 1.00 C 4 4 3 4.8 0.8 1 0.80 D 5 2 3 3.2 1.2 2 0.60 E 6 7 5 7.7 0.7 1 0.7 F 2 4 1 5 1 1 1.00 G 7 3 4 9 6 3 2.00 The project network diagram is provided below. This is how to read the table:  If a total of  \$10,000 is spent on activity A, then it normally takes 6 weeks to finish activity A. If a total of  \$11,500 is spent on activity A, then activity A can be finished within 3 weeks.  We assume a  linear cost structure, i.e., it costs (11,500-10,000)/(6-3) =  Cost/ Time =\$ 500 to crash one week  out of activity A.

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Unformatted text preview: A. (6 pts) Assuming each activity takes its normal time. Please complete the forward and backward passes on the following project diagram . The project can be completed within ______ 17_[4pts, including numbers on graph]_____ weeks. The critical path(s) of the project is(are) __ B –E –G and A – C –G _ [2pts]. B. (4 pts) If we want to complete the project one week ahead of the current plan, what is the most cost effective crashing plan? That is, which activity (activities) would you pick to crash for one week__ A and E __[ 3pts ]_____? How much additional cost does crashing incur?______ \$1200 ___[ 1pt]...
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