# Lecture2_AdditionalProblem3_Solution - Step 2: Step 3:...

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Lecture 2 – Solution to Additional Problem 3 1 Addl. Problem # 3 Project Activity and Cost Data (note: this is table 3.7 from the Problems) Activity Normal Time (days) Normal Cost (\$) Crash Time (days) Crash Cost (\$) Immediate Predecessor(s) A 5 1,000 4 1,200 -- B 5 800 3 2,000 -- C 2 600 1 900 A, B D 3 1,500 2 2,000 B E 5 900 3 1,200 C, D F 2 1,300 1 1,400 E G 3 900 3 900 E H 5 500 3 900 G Indirect project costs= \$250 per day and penalty cost = \$100 per day for each day the project lasts beyond day 14. 13 19 F 2 15 21 13 13 G 3 16 16 5 5 D 3 8 8 8 8 E 5 13 13 5 6 C 2 7 8 0 0 B 5 5 5 0 1 A 5 5 6 Start ES LS ID DUR EF LF 16 16 H 5 21 21 Finish Direct cost and time data for the activities: Maximum Crash Activity Crash Cost/Day Time (days) A 200 1 B 600 2 C 300 1 D 500 1 E 150 2 F 100 1 G 0 0 H 200 2 Solution: Original costs: Normal Total Costs = Total Indirect Costs = Penalty Cost =

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Lecture 2 – Solution to Additional Problem 3 2 Total Project Costs = Step 1: The critical path is and the project duration is
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Unformatted text preview: Step 2: Step 3: Costs Last Trial = Crash Cost Added = Total Indirect Costs = Penalty Cost = Total Project Costs = Step 4: repeat until direct costs greater than savings (step 2) (step 3) Costs Last Trial = Crash Cost Added = Total Indirect Costs = Penalty Cost = Total Project Costs = (step 4):repeat (step 2) (step 3) Costs Last Trial = Crash Cost Added = Total Indirect Costs = s Penalty Cost = Total Project Costs = A summary of the cost analysis follows. The recommended completion date is day ____ by crashing Resulting Reduc- Project Costs Crash Total Total Total Crash Critical Tion Duration Last Cost Indirect Penalty Project Trial Activity Paths (days) (days) Trial Added Costs Costs Costs 0 1 2 Further reductions will cost more than the savings in indirect costs and penalties. The critical path is BDEGH....
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## Lecture2_AdditionalProblem3_Solution - Step 2: Step 3:...

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