Question

Chapter 9 Reducing Project Durarion 327 7. Use the information contained below to compress one time unit per move using the least cost method. Assume the total indirect cost for the project is $2.000 and there is a savings of $100 per time unit reduced. Calculate the total direct, indirect, an graph. What is the optimum cost-time schedule for the project? Note: The correct normal project duration and total direct cost are provi ded. Normal Cost 200 1000 800 1000 800 1000 1000 200 Normal Time Act. Crash Cost (Slope)Maximum Crash Time 50 200 200 100 40 40 Project duration 20 HDirect cost 6,000 Indirect cost 2000 Total cost8000

Answer 1

Solution:

The below table shows about listing allowable Crash time and Crash cost per unit time:

Act.ID	Predecess	Normal time	Normal cost	Maximum crash time	Crash cost (slope)	ES	LS	EF	LF	Slack	Successor	Crash time remaining
A		2	$ 200	0	$ 0	0	0	2	2	0	B	0
B	A	4	$ 1,000	1	$ 50	2	2	6	6	0	C,D	0
C	B	5	$ 800	2	$ 200	6	6	11	11	0	E	0
D	B	5	$ 1,000	2	$ 200	6	6	11	11	0	E	0
E	C,D	3	$ 800	1	$ 100	11	11	14	14	0	F,G	0
F	E	5	$ 1,000	1	$ 40	14	14	19	19	0	H	1
G	E	4	$ 1,000	1	$ 40	14	15	18	19	1	H	1
H	F,G	1	$ 200	0	$ 0	19	19	20	20	0		0
			total=6,000

POSSIBLE PATHS	DUR​ATION
A-B-C-E-F-H	20	<-Critical path
A-B-C-E-G-H	19
A-B-D-E-F-H	20	<-Critical path
A-B-D-E-G-H	19

Normal Time = 20

Total Direct cost = $ 6,000

saving/Time unit reduced = $ 100

Project crashing using least cost method:

Iteration	Activity Crashed	Crash Time	Path-1 A-B-C-E-F-H	Path-2 A-B-C-E-G-H	Path-3 A-B-D-E-F-H	Path-4 A-B-D-E-G-H	Project Duration	Activity crash cost	Cumulative crash	Total Direct cost	Indirect cost	Total Cost
0			20	19	20	19	20			6,000	2,000	8,000
1	F	1	19	19	19	19	19	40	40	6,040	1,900	7,940
2	B	1	18	18	18	18	18	50	90	6,090	1,800	7,890
3	E	1	17	17	17	17	17	100	190	6,190	1,700	7,890
4	C	2	15	15	17	17	17	400	590	6,590	1,700	8,290
5	D	2	15	15	15	15	15	400	990	6,990	1,500	8,490

Optimum Cost - Time Schedule for the Project is:17 ,here both time and cost are lowest.

Time Period = 17

Total Cost = 7,890