Question

21) Determine the information missing from Table 7.7; then answer the following questions. How many week(s) of crashing are available for activity D in Table 2.9?

22) How many week(s) of crashing are available for activity B in Table 7.7?
23) How many weeks of crashing are available for activity C in Table 7.7?

24) What is the crashing cost per week for activity A in Table 7.7?

25) What is the crashing cost per week for activity E in Table 7.7?

26) What is the crashing cost per week for activity F in Table 7.7?

27) If a decision is made to crash activity D in Table 7.7 by one week, what is the cost for this one week of crashing?

28) Which activity should be crashed first for the project shown in Table 7.7?

Answer 1

[Note: All the cost has been expressed as ($000 s) such as cost value 54 is considered for cost value 54000]

Activity	Normal Time	Crash Time	Normal cost($000s)	Crash cost ($000s)	Slope
A	4	2	8	14	3
B	3	2	9	11	2
C	4	4	10	10	0
D	5	3	10	15	2.5
E	4	1	11	14	1
F	1	1	6	6	0
			54

Total Normal cost = $54

Crashing cost per week = Slope of an activity = (Crash cost – normal cost)/(Normal Time – crash time)

So, Slope of A = (14-8)/(4-2) = 3

Likewise, slope for other activities have been calculated.

Let’s see how many paths are available to complete the activities and duration of each paths to find the critical path

Path	Duration
A-B-D-F	4+3+5+1 = 13
A-C-D-F	4+4+5+1 = 14
A-C-E-F	4+4+4+1 = 13

Critical path is the path having maximum duration here critical path is A-C-D-F

Normal project completion time = 14 weeks

1^st Iteration

Crash limit and Slope

Critical Path	Critical Activities	Crash Limit(Normal time - crash time)	Cost Slope
A-C-D-F	A	2	3
	C	0	0
	D	2	2.5
	F	0	0

Least cost slope is of activity D .So, first crash activity D by 1 week as difference between critical path and next path is 1 day .

Critical activities,

Path	Duration
A-B-D-F	4+3+4+1 = 12
A-C-D-F	4+4+4+1 = 13
A-C-E-F	4+4+4+1 = 13

Now project completion time is 13 weeks and we have two critical paths A-C-D-F and A-C-E-F

2^nd Iteration

Crash limit and Slope

Critical Path	Critical Activities	Crash Limit(Normal time - crash time)	Cost Slope
A-C-D-F	A	2	3
	C	0	0
	D	1	2.5
	F	0	0

Critical Path	Critical Activities	Crash Limit(Normal time - crash time)	Cost Slope
A-C-E-F	A	2	3
	C	0	0
	E	3	1
	F	0	0

Crash combinations between critical paths as we have to crash in such a way that both the critical paths are reduced

Crash combinations	Cost slope
A	3
D,E	3.5

Crash A by 1 day as difference between critical path and next path is 1 day so crash cost = 3 and project duration is now 12 weeks

Critical activities,

Path	Duration
A-B-D-F	3+3+4+1 = 11
A-C-D-F	3+4+4+1 = 12
A-C-E-F	3+4+4+1 = 12

3^rd iteration

Crash limit and Slope

Critical Path	Critical Activities	Crash Limit(Normal time - crash time)	Cost Slope
A-C-D-F	A	1	3
	C	0	0
	D	1	2.5
	F	0	0

Critical Path	Critical Activities	Crash Limit(Normal time - crash time)	Cost Slope
A-C-E-F	A	1	3
	C	0	0
	E	3	1
	F	0	0

Crash combinations between critical paths

Crash combinations	Cost slope
A	3
D,E	3.5

Crash A by 1 day so crash cost = 3 and project duration is now 11 weeks

Critical activities,

Path	Duration
A-B-D-F	2+3+4+1 = 10
A-C-D-F	2+4+4+1 = 11
A-C-E-F	2+4+4+1 = 11

4^th iteration

Crash limit and Slope

Critical Path	Critical Activities	Crash Limit(Normal time - crash time)	Cost Slope
A-C-D-F	A	0	3
	C	0	0
	D	1	2.5
	F	0	0

Critical Path	Critical Activities	Crash Limit(Normal time - crash time)	Cost Slope
A-C-E-F	A	0	3
	C	0	0
	E	3	1
	F	0	0

Crash combinations between critical paths

Crash combinations	Cost slope
D,E	3.5

Now we have to crash D and E simultaneously as D has 1 crash limit so let’s Crash D by 1 day and E by 1 day and project duration is now 10 weeks

Critical activities,

Path	Duration
A-B-D-F	2+3+3+1 = 9
A-C-D-F	2+4+3+1 = 10
A-C-E-F	2+4+3+1 = 10

Now we can not crash more as crash limit of D = 0

Based on the above table;

Activity	Normal Time	Crash Time	Normal cost($000s)	Crash cost ($000s)	Available weeks of crashing	Crashing cost/week
A	4	2	8	14	2	3
B	3	2	9	11	0	2
C	4	4	10	10	0	0
D	5	3	10	15	2	2.5
E	4	1	11	14	1	1
F	1	1	6	6	0	0
			54

20.

Critical path is the path having maximum duration here critical path is A-C-D-F

21. Available week for crashing for D = 2

22. Available week for crashing for B = 0

23. Available week for crashing for C= 0

24. Crashing cost per week for activity A = 3

25. Crashing cost per week for activity E = 1       

26. Crashing cost per week for activity F = 0

27. If a decision is made to crash activity D in Table 7.7 by one week, what is the cost for this one week of crashing? Ans is 2.5   

28. Which activity should be crashed first for the project shown in Table 7.7? Ans is Activity D