Question

The following table provides the crash data for the project network . the normal activity times are considered to be deterministic and not probabilistic .

	Activity time(week)		Activity Cost($)
Activity	Normal	Crash	Normal	Crash
a	9	7	$4,800	$6,300
b	11	9	$9,100	$15,500
c	7	5	$3,000	$4,000
d	10	8	$3,600	$5,000
e	1	1	$0	$0
f	5	3	$1,500	$2,000
g	6	5	$1,800	$2,000
h	3	3	$0	$0
i	1	1	$0	$0
j	2	2	$0	$0
k	8	6	$5,000	$7,000

Crash the network the maximum amount , indicate how much it would cost the bank and identify the critical path(s). “ don’t use any software, solve the problem manually with explaining the steps “.

Above details are based on :

A bank is planning to install a new computerized accounts system. Bank management has

determined the activities required to complete the project, the precedence relationships of

the activities, and activity time estimates as follows:

			Time Estimates
Activity	Description	Predecessor	a	m	b
a	Position recruting	-	5	8	17
b	system development	-	3	12	15
c	system training	a	4	7	10
d	equpiment training	a	5	8	23
e	manual system test	b,c	1	1	1
f	preliminary system change over	b,c	1	4	13
g	computre-personnel interface	d,e	3	6	9
h	equpiment modification	d,e	1	2.5	7
i	equipment testing	h	1	1	1
j	system debugging nd installation	f,g	2	2	2
k	equipment changeover	g,i	5	8	11

Answer 1

We will first calculate expected time for each activity by using formula

Te= (a+4*m+b)/ 6

			Time Estimates
Activity	Description	Predecessor	a	m	b	Expected time
a	Position recruting	-	5	8	17	(5+4*8+17)/6=9
b	system development	-	3	12	15	11
c	system training	a	4	7	10	7
d	equpiment training	a	5	8	23	10
e	manual system test	b,c	1	1	1	1
f	preliminary system change over	b,c	1	4	13	5
g	computre-personnel interface	d,e	3	6	9	6
h	equpiment modification	d,e	1	2.5	7	3
i	equipment testing	h	1	1	1	1
j	system debugging nd installation	f,g	2	2	2	2
k	equipment changeover	g,i	5	8	11	8

We will draw network diagram and then find critical path. Critical path is the path with longest completion time among alll possible paths.

Possible paths:

• a-d-h-i-k: 9+10+3+1+8=31
• a-d-g-k:33
• a-d-g-j:27
• a-c-e-h-i-k:29
• a-c-e-g-k:31
• a-c-e-g-j:25
• a-c-f-j:23
• b-e-h-i-k:24
• b-e-g-k:26
• b-e-g-j:20
• b-f-j:18

Hence, critical path is a-d-g-k with total project duration of 33 weeks (same as critical path)

For crashing the project, we will first calculate crash slope for all activities by using formula

Crash slope = (Crash cost - Normal cost)/ (Normal time - crash time)

	Activity time(week)		Activity Cost($)
Activity	Normal	Crash	Normal	Crash	Crash slope
a	9	7	$4,800	$6,300	(6300-4800)/(9-7)=750
b	11	9	$9,100	$15,500	3200
c	7	5	$3,000	$4,000	500
d	10	8	$3,600	$5,000	700
e	1	1	$0	$0
f	5	3	$1,500	$2,000	250
g	6	5	$1,800	$2,000	200
h	3	3	$0	$0
i	1	1	$0	$0
j	2	2	$0	$0
k	8	6	$5,000	$7,000	1000

Now we will crash activities on critical path to reduce the project duration. We will first crash activities with lowest crash slope.

- On critical path a-d-g-k, the activity g has lowest crash slope ($200) and will be crashed first by 1 week. Project duration is now 32 weeks. Total crash cost is $200.

- On critical path a-d-g-k, next we will crash activity d ($700) by 1 week as it has lowest crash slope since activity g cannot be crashed further. Project duration is now 31 weeks. Total crash cost is $200+$700=$900. Critical path is still same a-d-g-k

- On critical path a-d-g-k, next we will crash activity d ($700) by second week as it has lowest crash slope. Project duration is now 30 weeks. Total crash cost is 900+700=$1600. Critical paths are a-d-g-k and a-c-e-g-k

- Next we will crash activity a ($750 each) by two week as it is common activity and lowest cost slope. Project duration is now 28 weeks. Total crash cost is $1600+750*2=3100. Critical paths are a-d-g-k and a-c-e-g-k

- Next we will crash activity k (1000 each) by two week as it is common activity and lowest cost slope. Project duration is now 26 weeks. Total crash cost is $3100+1000*2=5100. Critical paths are a-d-g-k and a-c-e-g-k

This is maximum crashing which is possible since have crashed all possible activity on critical path (a-d-g-k ) to maximum crashing limit.

Hence, the final crashed project duration is 26 weeks. Total minimum crashing cost is $5,100. Two critical paths are a-d-g-k and a-c-e-g-k

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