Question

Chuck’s Custom Boats (CCB) builds customer luxury yachts. CCB has landed a contract with a mysterious New York real estate developer, Mr. T. Relevant data on the project are shown below. The complication is that Mr. T wants delivery in 30 weeks or he will impose a penalty of $250 for each week his yacht is late. Determine the lowest cost plan for crashing this project.

Note: All numbers are in thousands, but keep them that way (i.e., don't enter the last three zeros).

1: What is the expected project duration before crashing any activities?

2: What is the cost of the second crash?

3: What is the cost of the seventh crash?

4: What is the total cost of all crashes that should be done to minimize costs?

5: How much should be spent on penalties?

Activity	Immediate Predecessor	Normal Time (weeks)	# Possible Crashes	Crash Cost Per Week
J	-	10	1	175
K	J	8	2	45
L	J	5	2	60
M	K	4	2	20
N	L	6	2	55
P	M, N	5	2	265
Q	P	7	1	70
Y	P	8	2	35
Z	Q	6	1	95

Note: If you want a thumbs up, please answer all five questions above. Thanks

Answer 1

1.

Initial State 一、 | Critical? 60 | Project paths | |-K-M-P-Y | D-K-M-P-Q-Z | |-L-N-P-Q-Z | J-L-N-P-Ý | Duration 35 40 39 34 . Crash cost Penalty Total $2,500 $2,500 Yes | | Project length 40

So, the expected project duration before crashing any activities = 40 weeks

2.

Step-1: Crash 'M' by 1 week Duration Critical? $20 Project paths J-K-M-P-Y J-K-M-P-Q-Z J-L-N-P-Q-Z J-L-N-P-Y Crash cost Penalty Total $2,250 $2,270 39 Yes Yes 34 Project length 39

Step-2: Crash 'Q' by 1 week Critical? Project paths J-K-M-P-Y J-K-M-P-Q-Z J-L-N-P-Q-Z J-L-N-P-Y Duration 34 38 38 34 Crash cost Penalty Total $90 $2,000 $2,090 Yes Yes Project length 38

So, the cost of the second crash = Cost of crashing 'Q' by one week = $70

3.

Step-3: Crash 'M' and 'N' both by 1 week Duration | Critical? Project paths J-K-M-P-Y J-K-M-P-Q-Z J-L-N-P-Q-Z J-L-N-P-Y Crash cost Penalty Total $165 $1,750 $1,915 Yes 37 37 33 Project length 37

Step-4: Crash 'Z' by 1 week Duration Critical? Project paths J-K-M-P-Y J-K-M-P-Q- Z J-L-N-P-Q-Z J-L-N-P-Y Crash cost Penalty Total $260 $1,500 $1,760 3 6 Yes Yes 33 Project length 36

Step-5: Crash 'K' and 'N' both by 1 week Duration | Critical? Project paths J-K-M-P-Y J-K-M-P-Q-Z J-L-N-P-Q-Z J-L-N-P-Y Crash cost Penalty Total $360 $1,250 $1,610 Yes Project length 35

Step-6: Crash 'K' and 'L' both by 1 week Duration Critical? Project paths J-K-M-P-Y J-K-M-P-Q-Z J-L-N-P-Q-Z J-L-N-P-Y Crash cost Penalty Total $465 $1,000 $1,465 34 Yes 34 Yes Project length

Step-7: Crash'' by 1 week Duration Critical? Project paths J-K-M-P-Y J-K-M-P-Q-Z J-L-N-P-Q-Z J-L-N-P-Y Crash cost Penalty Total $640 $750 $1,390 33 Yes Yes 30 Project length 33

So, the cost of the seventh crash = cost of crashing 'J' by 1 week = $175

4.

Step-8: Crash 'P' by 1 week 6 x Duration Critical? Project paths J-K-M-P-Y J-K-M-P-Q-Z J-L-N-P-Q-Z J-L-N-P-Y Crash cost Penalty Total $905 $500 $1,405 32 32 29 Yes Yes Project length 32

Note that in step-8, the total cost increases from what it was in step-7. So, step-7 is the optimality point.

So, the total cost of all crashes at this point = $640.

5.

At this point (i.e. at step-7), the penalty paid was = $750