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. Determine the logical steps to reduce SEVEN (7) days of the following project with the least amount of cost increasing. The indirect (overhead) costs
. Determine the logical steps to reduce SEVEN (7) days of the following project with the least amount of cost increasing. The indirect (overhead) costs are $100/day. Activity ID IPA Time Time Cost Cost Time Normal Cost Crash Slope Normal Crash Difference Difference ($/Day) A 4 4 $100 $100 B A 7 5 $200 $110 C A 10 10 $300 $300 D B 7 2 $500 $300 E B (FS + 3) 4 2 $100 $50 F E 3 $150 $50 G D (FS + 2), F 4 $280 $100 H C, G 7 5 $500 $300* Cycle #0. Cycle # Activity to Shorten Can Be Shortened Days Shortened Cost per Day Cost for Cycle Direct Cost Project Duration Indirect Cost Total Cost N O h WON - Activity Time Time Slope . Cycle #1. Reduce Activity days) ID Normal Crash ($/Day) A 4 4 B 7 5 C 10 10 D 7 2 E 4 2 F 3 D Lag: FS + 2 G 4 H 7 5 B G A E F H Lag: FS + 3 C Activity ES Duration EF LS TF FF LF. Cycle #1. Reduce Activity days) Can Be Days Cost per Cost for Direct Project Indirect Cycle # Activity to Total Cost Shorten Shortened Shortened Day Cycle Cost Duration Cost YOUAWN- OActivity Time Time Slope . Cycle #2. Reduce Activity ( days) ID Normal Crash ($/Day) A 4 4 B 7 5 C 10 10 D 7 2 E 4 2 F D Lag: FS + 2 G A H 7 5 B G A E F H Lag: FS + 3 C Activity ES Duration EF LS TF FF LF* Cycle #2. Reduce Activity ( days) Activity to Can Be Days Cost per Cost for Direct Project Indirect Cycle Shorten | Shortened | Shortened Day Cycle Cost Duration Cost JotaliCost 0 N o g A ON - . Cycle #3. Reduce Activity Activity Time Time days) Slope ID Normal Crash ($/Day) A 4 4 B 7 5 C 10 10 D 7 2 E 4 2 F D 3 Lag: FS + 2 G 7 4 H 7 5 B G A E F H Lag: FS + 3 C Activity ES Duration EF LS TF FF LF. Cycle #3. Reduce Activity ( days) Activity to Can Be Days Cost per Cost for Direct Project Indirect Cycle # Total Cost Shorten Shortened Shortened Day Cycle Cost Duration Cost O DUI AWN. Determine the logical steps to reduce SEVEN (7) days of the following project with the least amount of cost increasing. The indirect (overhead) costs are $100/day. D Lag: FS + 2 B G A E F H Lag: FS + 3 C Activity ES Duration EF LS TF FF LFActivity Time Time Slope . Cycle #4. Reduce Activity ( days) ID Normal Crash ($/Day) A 4 4 B 7 5 C 10 10 D 7 2 E 4 2 F D Lag: FS + 2 G -A H 7 5 B G A E F H Lag: FS + 3 C Activity ES Duration EF LS TF FF LF. Cycle #4. Reduce Activity ( days) Activity to Can Be Days Cost per Cost for Direct Project Indirect Cycle # Total Cost Shorten Shortened Shortened Day Cycle Cost Duration Cost JOUAWNActivity Time Time Slope . Cycle #5. Reduce Activity _( days) ID Normal Crash ($/Day) A 4 4 B 7 5 C 10 10 D 7 2 E 4 2 F D Lag: FS + 2 G 7 -A H 7 5 B G A E F H Lag: FS + 3 C Activity ES Duration EF LS TF FF LF. Cycle #5. Reduce Activity ( days) Activity to Can Be Days Cost per Cost for Direct Project Indirect Cycle # Total Cost Shorten Shortened Shortened Day Cycle Cost Duration Cost 2 OUA. Cycle #6. Reduce Activity Activity Time Time days) Slope ID Normal Crash ($/Day) A 4 4 B 7 5 C 10 10 D 7 2 E 4 2 F D 3 Lag: FS + 2 G 7 4 H 7 5 B G A E F H Lag: FS + 3 C Activity ES Duration EF LS TF FF LF. Cycle #6. Reduce Activity - ( days) Activity to Can Be Days Cost per Cost for Direct Project Indirect Cycle # Total Cost Shorten Shortened Shortened Day Cycle Cost Duration Cost 0 JOUAWNActivity Time Time Slope . Cycle #7. Reduce Activity days) ID Normal Crash ($/Day) A 4 4 B 7 5 C 10 10 D 7 2 E 4 2 F 3 D Lag: FS + 2 G A H 5 B G A E F H Lag: FS + 3 C Activity ES Duration I EF LS TF FF LF. Cycle #7. Reduce Activity _( days) Activity to Can Be Days Cost per Cost for Direct Project Indirect Cycle # Total Cost Shorten Shortened Shortened Day Cycle Cost Duration Cost YOUAWN-O
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