Question

A manufacturer of personal computers has recently identified time estimates, activities and predecessors to design, manufacture and market a new laptop computer. Their estimates are listed below Immediate Activity Description Estimated completion time (days) 90 15 Predecessors Prototype design Purchase of materials Manufacture of prototype Revision of design Initial production run Staff training Staff input on prototype Sales training Preproduction advertising campaign Post production advertising campaign 20 21 25 14 28 30 C,F D,I Management has also provided you with cost estimates for each activity showing normal costs and crash costs as well as normal time to complete each activity and crash times Activity time Normal Cost Crash (days) Crash Cost Normal Time (days) ($,000) ($,000) 90 $2,250 85 $2,520 15 $180 13 $210 $90 4 $120 20 16 $300 $340 21 17 $231 $315 25 $375 $250 14 $70 $112 28 $392 25 $448 30 28 $570 $510 45 $1,350 41 $1,530

Answer 1

(b)

The critical path of a project is composed off activities with zero total slack. The critical path(s) is/ are the longest path(s) in a project network. The slacks for each activity can be computed by carrying out forward pass and backward pass in the project network.

Forward Pass:

ES of the starting activities = 0

ES of all other activities = Max. (EF of their immediate predecessors)

EF of an activity = Its ES + Its duration

Backward Pass:

LF of ending activities = Max. (All the EFs)

LF of all other activities = Min. (LS of their immediate successors)

LS of an activity = Its LF - Its duration

Total slack (or, float) = LF - EF for each activity.

Activity	Duration	ES	EF	LS	LF	Slack
A	90	0	90	0	90	0
B	15	90	105	95	110	5
C	5	105	110	110	115	5
D	20	129	149	129	149	0
E	21	149	170	173	194	24
F	25	90	115	90	115	0
G	14	115	129	115	129	0
H	28	149	177	166	194	17
I	30	90	120	119	149	29
J	45	149	194	149	194	0

The critical path is constituted by activities with slack=0. So, the critical path is A-F-G-D-J with project completion time=194 days.

(c)

First, compute the cost slope using the following formula:

Slope = (Crash cost - Normal cost) / (Normal time - Crash time)

Activity	Normal time	Normal cost	Crash time	Crash cost	Slope	Max crash limits
A	90	2250	85	2520	54.00	5
B	15	180	13	210	15.00	2
C	5	90	4	120	30.00	1
D	20	300	16	340	10.00	4
E	21	231	17	315	21.00	4
F	25	250	22	375	41.67	3
G	14	70	12	112	21.00	2
H	28	392	25	448	18.67	3
I	30	510	28	570	30.00	2
J	45	1350	41	1530	45.00	4

Always select activities on the critical paths for crashing which has the minimum slope.

Initial state 21 20 14 15 Start End 90 28 25 45 30 Project duration 194 Duration Project paths 165 A-B-C-G-D-E 172 A-B-C-G-D-H 189 A-B-C-G-D-J 170 A-F-G-D-E A-F-G-D-H 194 A-F-G-D-J 165

Step-1: Crash 'D' by 4 days 15 14 21 Start 90 End 25 28 30 45 Duration Project duration Project paths 161 190 A-B-C-G-D-E 168 A-B-C-G-D-H 185 5623 A-B-C-G-D-J Normal Cost 166 A-F-G-D-E Crash cost 40 173 A-F-G-D-H Total cost 5663 190 A-F-G-D-J 165

Step-2: Crash 'G' by 2 days 21 x 16 x 12 15 Start End 90 45 Project duration 188 Duration Project paths 159 A-B-C-G-D-E 166 A-B-C-G-D-H 5623 Normal Cost 183 A-B-C-G-D-J Crash cost 164 A-F-G-D-E 5705 Total cost A-F-G-D-H 188 A-F-G-D-J 165

Step-3: Crash 'F' by 2 days x 16 15 x 12 21 Start 90 End 23 28 45 30 Project duration 186 Project paths Duration 159 A-B-C-G-D-E 166 A-B-C-G-D-H 183 A-B-C-G-D-J Normal Cost 5623 A-F-G-D-E 162 Crash cost 165.333 169 A-F-G-D-H Total cost 5788.3 186 A-F-G-D-J 165

The step-3 is the last step because the project duration has been reduced to 186 days. The activities crashed were D (4 days), G (2 days), and F (2 days). The total minimum cost is 5,788.33 ($,000).