. Determine the logical steps to reduce SEVEN (7) days of the following project $100/day. with the least amount of cost increasing. The indirect (overhead) costs are D Lag: FS + 2 B G A E F H Lag: FS + 3 C Activity ES Duration EF LS TF FF LF 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 $100/day. Activit Time Time Cost Cost Time Cost Slope yID Normal Crash Normal Crash Difference | Difference ($/Day) A - 4 4 $100 $100 B | A 7 | 5 $200 | $110 Cc A 10 10 $300 $300 D B 7 2 $500 | $300 | E B (FS +3) 4 2 $100 $50 F E 3 | 1 $150 | $50 | G | D(FS+2),F 7 | 4 $280 $100 | H | C,G 7 5 $500 | $300 . Cycle #0. Activity to Can Be Days Cost per Cost for Direct Project Indirect Total Cost Cycle # Shorten Shortened Shortened Day Cycle Cost Duration Cost O W N 5 6Activity Time Time Slope . Cycle #1. Reduce Activity ( days) ID Normal Crash ($/Day) A 4 4 B 7 5 C 10 10 D 7 2 E 4 2 F 3 D Lag: FS + 2 G A 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) Activity to Can Be Days Cost per Cost for Direct Project Indirect Cycle # Total Cost Shorten Shortened Shortened Day Cycle Cost Duration Cost YOUAWN-OActivity Time Time Slope . Cycle #2. Reduce Activity ( days) ID Normal Crash ($/Day) A 4 4 B 7 5 C 10 10 D 7 2 E 4 2 F 3 D Lag: FS + 2 G 4 H 7 5 B G A E F H Lag: FS + 3 C Activity ES Duration EF LS TF FF LF. Cycle #2. Reduce Activity ( days) Activity to Can Be Days Cost per Cost for Direct Project Indirect Cycle # Total Cost Shorten Shortened Shortened Day Cycle Cost Duration Cost O YOUAWN Cycle #3. Reduce Activity ( #_H, Lag: FS+3 C iFF days) Activity Time Time Slope ID Normal Crash ($/Day) A | 4 4 B ' 7 5 c | 10 10 D | 7 2 E | 4 2 F | 3 1 e | 7 4 Ho| 7 5 G H Activity Duration Cycle #3. Reduce Activity ( days) Activity to| Can Be Days Cost per| Cost for Direct Project Indirect Cycle# | chorten | Shortened | Shortened Day Cycle Cost Duration Cost Total Cost N ON - Activity Time Time Slope . Cycle #4. Reduce Activity ( days) ID Normal Crash ($/Day) A 4 4 B 7 5 C 10 10 D 7 2 E 4 2 F 3 D Lag: FS + 2 G 7 4 H 7 5 B G A E F H Lag: FS + 3 C Activity ES Duration EF LS TF FF LF Cycle #4. Reduce Activity ___ ( days) Activity to Shorten Cycle # Can Be Shortened Days Shortened | Cost per Day Cost for Cycle Direct Cost Project Indirect Duration = Cost | Total Cost Nooa b~ N = . Cycle #5. Reduce Activity Activity Time Time Slope days) ID Normal Crash ($/Day) A 4 4 B 7 5 C 10 10 D 7 2 E 4 2 F D 3 Lag: FS + 2 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 # Shorten | Shortened | Shortened Day Cycle Cost Duration = Cost Total Cost 0 N A WON = Activity Time Time Slope . Cycle #6. Reduce Activity days) ID Normal Crash ($/Day) A 4 4 B 7 5 C 10 10 D 7 2 E 4 2 F 3 D Lag: FS + 2 G 7 4 H 7 5 B G A E F H Lag: FS + 3 C Activity ES Duration EF LS TF FF LF Cycle #6. Reduce Activity ___ ( Activity to | Shorten | Shortened | Shortened days) Cost per Day Cost for Cycle Direct Cost Project Duration Indirect Cost Total Cost N O A WON - Activity Time Time Slope . Cycle #7. Reduce Activity ( days) ID Normal Crash ($/Day) A 4 4 B 7 5 C 10 10 D 7 2 E 4 2 F 3 D Lag: FS + 2 G 4 H 5 B G A E F H Lag: FS + 3 C Activity ES Duration 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 O 2 3 YOU A