Question

Question 16 4 points Save Answe The following provides information about a project, including a list of activities and precedence relationships, activity durations (normal and "crash" times), direct activity costs (normal and "crash"), indirect costs for the project, as well as calculated crash costs/day and the maximum amount that each activity can be shortened (crashed) by. There are no penalty costs for this project. Total project cost (using normal times), a network diagram, and the four paths and durations have also been provided. Your job is to "crash" the project to find the minimum cost schedule. Please provide: • The new project duration in days • The new project cost $ (Please enter only the amount, without commas or dollar sign.) A (7) B(6) E (3) Start 1 C(9) D (7) F (6) Finish (10) G H Indirect project costs $ 10.000 per day Immed Predis) Normal Time (days Acitvity 7 1.000 A o B C D E 9 7 7 с B,C Direct Costs - Crash Time Direct Costs - Crash Cost Max Crash Normal (day Crashed per Day (days S 60.000 5 $ 62,000 s 2 s 120.000 $ 120,000 NA 0 $ 16.000 8 $ 22,000S 6.000 1 $ 50.000 $ 58,000 $ 8.000 1 S 90.000 2 S 90,000 $ 0.000 $ 46,000 5 $ 50,000 $ $ 80.000 7 $ 92,000 $ 12.000 S 15.000 4 $ 20,000 $ 2.500 2 $ 100.000 9 $ 102,000S 2.000 1 $ 576,000 F 3 6 8 5,000 1 1 1 G с E.FH 10 Total Project Costs Indirect $ 330,000 $ Direct 578,000 S 906.000 "Crash" the project to find the MINIMUM COST schedule Activity Additional Cost Net to Crash Cost Savings Savings CHI 33 PATH Durations C-D-FI C-D-E-1 32 ABEI 20

Answer 1

As seen from the duration of the paths, the path C-G-H-I is the critical path that has the logest duration.

To reduce project duration, we have to crash activities on critical path first.

As seen from above table, we start with crashing activities with minimum crash cost per day to have minimum cost impact. SInce the savings in indirect cost is $10,000 per day, we will only crash activities with crash cost below $10,000 per day as this will result in net profit.

We find the answer using the format given in question as shown below:

Activity to Crash Additional Cost Cost Savings Net Savings C-G-H-1 C-D-F-I C-D-E-1 A-B-E-I 33 32 29 - 32 28 2000 2500 10000 10000 8000 7500 31 31 26 25 25 H 31 28

As seen from above, duration for C-G-H-I and C-D-F-I becomes same as the longest paths. Hence, path C-D-F-I also becomes critical. To crash the project further, we have to either crash 1 activity on each path or a common activity whichever alternative has the lowest total crash cost per week.

Activity to Crash Additional Cost Cost Savings Net Savings C-G-H-1 C-D-F-I C-D-E-I A-B-E-1 33 32 29 26 25 - 31 28 2000 2500 8000 7500 H 32 31 30 31 28 10000 10000 10000 10000 25 6000 4000 30 27 C F,H 25 25 7500 2500 29 29 27

We cannot crash further since the cost of next crash will be more than 10000 in any combination. Hence, this is the minimum duration.

Total Savings = 8000 + 7500 + 4000 + 2500 = $22000

New project cost = 906000 - 22000 = $884,000

New project duration = Duration of Critical paths = 29

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