Question

Bowman Builders manufactures steel storage sheds for commercial use. Joe Bowman, president of Bowman Builders, is contemplating producing sheds for home use. The activities necessary to build an experimental model and related data are given in the accompanying table.

Activity	Normal Time	Crash Time	Normal Cost ($)	Crash Cost ($)	Immediate predecessors
A	3	2	1000	1600	-
B	2	1	2000	2700	-
C	1	1	300	300	-
D	7	3	1300	1600	A
E	6	3	850	1000	B
F	2	1	4000	5000	C
G	4	2	1500	2000	D, E

a: What is expected project complete time (before crashing)?

b: Which activities should be crashed and by how much?

c: What is the additional project cost to crash to 10 weeks?

Answer 1

Before proceeding to mail solution we have to understand few terms in brief :

1. Normal time : This is minimum time required to complete an activity at normal cost.

2. Normal Cost : This is the lowest possible direct cost required to complete an activity.

3. Crash Time : This is the minimum time required to complete an activity.

4. Crash Cost : This is the direct cost that is anticipated in completing an activity within the crash time.

Solution:

a.)

Possible paths to complete the activity

S.no. Paths Normal Duration (in weeks) 1. A-D-G 3+7+4=14 2. B-E-G 2+6+4=12 3. C-F-G 1+2+4=7

The normal duration for the activity is the time taken for completing critical path ( Longest path) i.e., 14 Weeks in A-D-G. Therefore the Normal duration to complete the project is 14 weeks.

b.) Crashing activity: Which activity should be crashed and by how much:

Calculation of Cost slope and possible reduction :

Activity	Normal Time	Crash Time	Normal Cost ($)	Crash Cost ($)	Cost Slope	Possible Reduction (in weeks)
A	3	2	1000	1600	600	1
B	2	1	2000	2700	700	1
C	1	1	300	300	-	-
D	7	3	1300	1600	75	4
E	6	3	850	1000	50	3
F	2	1	4000	5000	1000	1
G	4	2	1500	2000	250	2
		Total Cost	10950

Cost slope= (Crash Cost - Normal Cost) / (Normal time- Crash time)

The activities having least cost slope should be crashed first so that the total cost should be less. The first crashing should be done in critical path. The above table shows the cost slope of each activity and the possible reduction in weeks. Here we have no saving of Indirect cost occurs per day therefore the crashing will surely increase our cost and that is not advisable until and unless we have time constraint .

The possible crashing to adopt the time constraint to 10 Weeks is given below in part (c)

c.) Crashing Process:

Crashing No.1 : First crash critical path A-D-G , Crash D Having least slope of 75 by 2 weeks which will increase the cost by 2*75= 150 $ and remaining possible reduction in D activity will be 2 Weeks.

Crashing No.2 : Now we have Two critical path having duration of 12 Weeks , so we need to reduce the common element of both paths so as to reduce the duration to 10 weeks and not creating other critical path .

Therefore now crash activity G having cost slope of 250$ by 2 weeks , this will increase the cost by 250*2 =500$ and will make the activity time to 10 Weeks

S.no.	Paths	Normal Duration (in weeks)	Duration after First crashing (in weeks)	Duration after Second crashing (in weeks)
1.	A-D-G	3+7+4=14	3+5+4=12	3+5+2=10
2.	B-E-G	2+6+4=12	2+6+4=12	2+6+2=10
3.	C-F-G	1+2+4=7	1+2+4=7	1+2+2=5
	Total Cost	10950	11100	11600

Therefore addition cost incurred in crashing to 10 weeks will be 11600- 10950 = 650$