Question

Kimpel Products makes pizza ovens for commercial use. James​ Kimpel, CEO, is contemplating producing smaller ovens for use in high school and college kitchens. The activities necessary to build an experimental model and related data are listed in the following​ table:

                                                                                                                                 

Activity	Normal Time​(weeks)	Crash Time​(weeks)	Normal Cost	Total Cost with Crashing	Immediate​ Predecessor(s)
A	3	2	$ 1,200	$1,800	—
B	2	1	$2,100	$2,900	__
C	1	1	$400	$400	—
D	7	3	$1,300	$1,620	A
E	6	3	$850	$1,075	B
F	2	1	$4,500	$5,500	C
G	4	2	$1,500	$2,000	​D, E

​a) Based on the given information regarding the activities for the​ project, the project completion date​ = ___ weeks

b) To reduce the duration of the project to 10 weeks at the least? cost, the following is the order of what needs to be followed while crashing the activities:

The first activity to crash is ___ by 2 weeks at a per week crashing cost of ____ dollars.

Next, one should crash activity (answer from above) and activity ____ by 2 weeks weeks for an additional total per week crashing cost of ____ dollars.

c) New requirements are to reduce the project to 7 weeks. Continuing to crash the activities from 10 weeks based on the above crashing already completed, the following is the order for crashing:

Next activity to crash in order is ___ by 2 weeks weeks at a per week crashing cost of ____ dollars.

Next, one should crash activity ___and activity E by 1 week at a total per week crashing cost of ___ dollars.

Total cost of crashing the project to 7 weeks = ___ dollars.

Answer 1

With the given data, we compute the below table :

a) The project completion date = 14 weeks

b) The critical path is the one having activities having zero slack and has the longest completion time.

There are 3 paths :

Path 1 = A-D-G = 3+7+4 = 14 weeks : CRITICAL PATH

Path 2 = B-E-G = 2+6+4 = 12 weeks

Path 3 = C-F = 1+2 = 3 weeks

A-D-G is the critical path and here Activity D has the lowest per week crashing cost.

So, we crash Activity D by 2 weeks at cost of $80 per week

Path 1 = A-D-G = 3+5+4 = 12 weeks : CRITICAL PATH

Path 2 = B-E-G = 2+6+4 = 12 weeks : CRITICAL PATH

Path 3 = C-F = 1+2 = 3 weeks

Now, B-E-G also becomes critical path along with A-D-G and we crash the activity G to reduce both these paths

So, we crash Activity G by 2 weeks at cost of $250 per week

c) Now the 3 paths are :

Path 1 = A-D-G = 3+5+2 = 10 weeks : CRITICAL PATH

Path 2 = B-E-G = 2+6+2 = 10 weeks : CRITICAL PATH

Path 3 = C-F = 1+2 = 3 weeks

To complete the project to 7 weeks, we need to crash both the critical paths by 3 weeks.

G had a maximum crashing of 2 weeks and already done above. So further crashing is not possible for G.

D can be crashed further by 2 weeks at cost of $80 per week

E can be crashed further by 2 weeks at cost $75 per week

A can be crashed by 1 week at cost of $600 per week

E can be crashed further by 1 week at cost $75 per week

The final paths are :

Path 1 = A-D-G = 2+3+2 = 7 weeks : CRITICAL PATH

Path 2 = B-E-G = 2+3+2 = 7 weeks : CRITICAL PATH

Path 3 = C-F = 1+2 = 3 weeks

Total Cost of crashing the project to 7 weeks = 2*80 + 2*250 + 2*80 + 2*75 + 1*600 + 1*75 = 160+500+160+150+600+75

Total Cost of crashing the project to 7 weeks = $1,645