Question

R. C. Coleman

R. C. Coleman distributes a variety of food products that are sold through grocery store and supermarket outlets. The company receives orders directly from the individual outlets, with a typical order requesting the delivery of several cases of anywhere from 20 to 50 different products. Under the company's current warehouse operation, warehouse clerks dispatch order-picking personnel to fill each order and have the goods moved to the warehouse shipping area. Because of the high labor costs and relatively low productivity of hand order picking, management has decided to automate the warehouse operation by installing a computer-controlled order picking system, along with a conveyor system for moving goods from storage to the warehouse shipping area.

R. C. Coleman's director of material management has been named the project manager in charge of the automated warehouse system. After consulting with members of the engineering staff and warehouse management personnel, the director compiled a list of activities associated with the project. The optimistic, most likely, and pessimistic times (in weeks) have also been provided for each activity.

Activity	Description	Predecessor	Time (weeks)
			Optimistic a	Most Likely m	Pessimistic b
a b c d e f g h i j k	Determine Equipment needs Obtain vendor proposals Select vendor Order system Design new warehouse layout Design warehouse Design computer interface Interface computer Install system Train system operators Test system	─ ─ A, B C C E C D, F, G D, F H I, J	4 6 2 8 7 4 4 4 4 3 2	6 8 4 10 10 6 6 6 6 4 4	8 16 6 24 13 8         20 8 14 5 6

Develop a report that presents the activity schedule and expected project completion time for the warehouse expansion project. Include a project network in the report. In addition, take into consideration the following issues:

1. R. C. Coleman's top management established a required 40-week completion time for the project. Can this completion time be achieved? Include probability information in your discussion. What recommendations do you have if the 40-week completion time is required?

2. Suppose that management requests that activity times be shortened to provide an 80% chance of meeting the 40-week completion time. If the variance in the project completion time is the same as you found in part (1), how much should the expected project completion time be shortened to achieve the goal of an 80% chance of completion within 40 weeks?

3. Using the expected activity times as the normal times and the following crashing information, determine the activity crashing decisions and revised activity schedule for the warehouse expansion project:

Activity	Crashed Activity Time (weeks)	Normal Cost ($)	Crashed Cost ($)
b c d e f g h i j k	4 7 2 8 7 4 5 4 4 3 3	1,000 1,000 1,500 2,000 5,000 3,000 8,000 5,000 10,000 4,000 5,000		1,900 1,800 2,700 3,200 8,000 4,100 10,250 6,400 12,400 4,400 5,500

                       

Answer 1

The network diagram:

Activity	Optimistic	most probable	pessimist
A	4	6	8
B	6	8	16
C	2	4	6
D	8	10	24
E	7	10	13
F	4	6	8
G	4	6	20
H	4	6	8
I	4	6	14
J	3	4	5
K	2	4	6

Expected time = (a+4m+b)/6

Variance = [(b-a)/6]2

Let's derive expected time and variance for each activity as follows:

Expected time of activity A = (4+4*6 + 8)/6 = 36/6 = 6

Variance of activity A = ](8-4)/6]2 = (4/6)2 = 0.44

Activity	expected time (weeks) = (a + 4m+ b)/6	variance = [(b-a)/6]2
A	6	0.44
B	9	2.78
C	4	0.44
D	12	7.11
E	10	1
F	6	0.44
G	8	7.11
H	6	0.44
I	7	2.78
J	4	0.11
K	4	0.44

ES,LS,EF,LF,Slack :

activity	ES	LS	EF	LF	Slack
A	0	3	6	9	3
B	0	0	9	9	0
C	9	9	13	13	0
D	13	17	25	29	4
E	13	13	23	23	0
F	23	23	29	29	0
G	13	21	21	29	8
H	29	29	35	35	0
I	29	32	36	39	3
J	35	35	39	39	0
K	39	39	43	43	0

Slack = difference between LS & ES or LF & EF

?Activities with zero slack are the activities on critical path

Critical path = B + C+ E+ F+ H+J+K = 43 WEEKS

Variance of above critical path = 5.65

z-score = 40 - 43 / (5.65)1/2 = -1.26

Z score of -1.26 = 0.1038

We can conclude that RC Coleman has 10.4% of chance of completion in 40 weeks

2.

When probability is 80%, Z= 0.84

0.84 = 40 - X / 5.651/2

X = 38 weeks

if we want 80% of the project completion by 40 weeks, then the expected project time should be reduced to 38 weeks

3.

	Weeks			Cost $
Activity	Normal time	Crash time	Normal cost	Crash cost	Time to be crashed	Cost of crash	Cost of crash per period
A	6	4	1,000.00	1,900.00	2	900.00	450.00
B	9	7	1,000.00	1,800.00	2	930.00	400.00
C	4	2	1,500.00	2,700.00	2	1,200.00	600.00
D	12	8	2,000.00	3,200.00	4	1,200.00	300.00
E	10	7	5,000.00	8,000.00	3	3,000.00	1,000.00
F	6	4	3,000.00	4,100.00	2	1,100.00	550.00
G	8	5	8,000.00	10,250.00	3	2,250.00	750.00
H	6	4	5,000.00	6,400.00	2	1,400.00	700.00
I	7	4	10,000.00	12,400.00	3	2,400.00	800.00
1	4	3	4,000.00	4,400.00	1	400.00	400.00
k	4	3	5,000.00	5,500.00	1	500.00	500.00
			45,500.00	60,650.00

The total expected time to complete the project is 43 week and the chance of 80% completion is 38 weeks. Thus, the difference is 5 weeks and the project cannot be crashed more than 5 weeks, if it is crashed more than 5 weeks then it cannot be achieved.

Potential Path	Length	Crash J - 1 week	Crash B-2 weeks	Crash K- 1 week	Crash F-1 week
A-C-D-I-K	33	33	33	32	32
A-C-D-H-J-K	36	35	35	34	34
A-C-E-F-I-K	37	37	37	36	35
A-C-E-F-H-J-K	40	39	39	38	37
A-C-G-H-J-K	32	31	31	30	30
B-C-D-I-K	36	36	34	33	33
B-C-D-H-J-K	39	38	36	35	35
B-C-E-F-I-K	40	40	38	37	36
B-C-E-F-H-J-K	43	42	40	39	38
B-C-G-H-J-K	35	34	32	31	31
	Cost	400	800	500	550
Total crashing cost	2250
Total crashing week	5 week

Revised schedule:

Potential Path	Length
A-C-D-I-K	32
A-C-D-H-J-K	34
A-C-E-F-I-K	35
A-C-E-F-H-J-K	37
A-C-G-H-J-K	30
B-C-D-I-K	33
B-C-D-H-J-K	35
B-C-E-F-I-K	36
B-C-E-F-H-J-K	38
B-C-G-H-J-K	31