C) Find the optimal way to crash the project by one day. What is the projected p
ID: 446936 • Letter: C
Question
C) Find the optimal way to crash the project by one day.
What is the projected project cost? What is the projected project duration? What is the critical path (or paths)? What task or tasks were crashed?
D) Calculate the shortest completion time for the project.
E) Explain the results of calculations above.
What is the projected project cost? What is the projected project duration? What is the critical path (or paths)? What task or tasks were crashed?
Activity Normal Time Normal Cost Crash Time Crash Cost Crash Cost per day (assuming crast costs are linear/has equal cost) a b c d e f g h i Activty PredecessoreCost Time Normal Crash Crash Cost $50 $40 $70 $20 $30 $80 $50 $60 $50 $450 $150 $200 $160 $50 $100 $290 $100 $180 $50 Start Critical Paths Time Cost selected A-B-C-D-G-I A-B-C-E-G-I A-B-F-H-I A-B-F-D-G- 25$280 26 $290 25 26 $290 Activity! Normal Time! C0s NormalCrashCrash Crash Time! Cost Cost DayExplanation / Answer
Identify the critical path of the network, the time, and cost of the normal level of activity for the project:
The critical path (or paths) is: a – b – f – d – g – i and a – b – c – e – g - i
The project duration: 4 + 4+ 7 + 3 +5 + 3 = 26 days
The project cost is: $50 + $40 + $70 +$20 + $30 + $80 +$50 +$60 + $50 = $450
Calculate the crash cost-per-day (all activities may be partially crashed). Enter your response to question 1(b) in the table provided below
Activity
Normal Time
Normal Cost
Crash Time
Crash Cost
Crash Cost/Day[1]
a
4
$ 50
2
$ 150
= (150 -50)/(4-2) = $50
b
4
$ 40
2
$ 200
= (200 -40)/(4-2) = $80
c
7
$ 70
4
$ 160
= (160 -70)/(7-4) = $30
d
2
$ 20
1
$ 50
= (50 -20)/(2-1) = $30
e
3
$ 30
2
$ 100
= (100 -30)/(3-2) = $70
f
8
$ 80
5
$ 290
= (290 -80)/(8-5) = $70
g
5
$ 50
3
$ 100
= (100 -50)/(5-3) = $25
h
6
$ 60
2
$ 180
= (180 -60)/(6-2) = $30
i
3
$ 50
3
$ 50
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[1] Assume that each task’s crash costs are linear, ie that each day a task is crashed has an equal cost.
(c). Find the optimal way to crash the project by one day.
Explain the results of your calculations.
For crashing the project by one day, activity with least crash cost/day is chosen. Activities a, b, g and i are common to both critical paths. Activity ‘g’ has the least crash cost at $25. For uncommon activities crashing needed to be done on both chains, which would be costlier.
Number of days of crashing left with activity ‘g’ = 2-1 = 1
(d). Find the optimal way to crash the project by two days.
Explain the results of your calculations.
For crashing the project by one day, activity with least crash cost/day is chosen. Activities a, b, g and i are common to both critical paths. Activity ‘g’ has the least crash cost at $25. For uncommon activities crashing needed to be done on both chains, which would be costlier.
Number of days of crashing left with activity ‘g’ = 2-1 = 1
(e). Calculate the shortest completion time for the project.
Explain the results of your calculations.
Shortest completion time is when all activities on critical path are crashed
Crashing activity ‘a’ by 2 days, additional cost = 2*50 = $100
Crashing activity ‘b’ by 2 days, additional cost = 2*80 = $160
Crashing activity ‘g’ by 2 days, additional cost = 2*20 = $50
Crashing activity ‘f’ and ‘c’ by 3 days, additional cost = 3*(70+30) = $300
Crashing activity ‘e’ and ‘d’ by 3 days, additional cost = 3*(70+30) = $300
So, total additional cost = 100 + 160 + 50 + 300 + 300 = $910
Days crashed = 2 +2+2+3+3 = 12
Activity
Normal Time
Normal Cost
Crash Time
Crash Cost
Crash Cost/Day[1]
a
4
$ 50
2
$ 150
= (150 -50)/(4-2) = $50
b
4
$ 40
2
$ 200
= (200 -40)/(4-2) = $80
c
7
$ 70
4
$ 160
= (160 -70)/(7-4) = $30
d
2
$ 20
1
$ 50
= (50 -20)/(2-1) = $30
e
3
$ 30
2
$ 100
= (100 -30)/(3-2) = $70
f
8
$ 80
5
$ 290
= (290 -80)/(8-5) = $70
g
5
$ 50
3
$ 100
= (100 -50)/(5-3) = $25
h
6
$ 60
2
$ 180
= (180 -60)/(6-2) = $30
i
3
$ 50
3
$ 50
-
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