# Decision Analysis

Topics: Normal distribution, Probability density function, Probability theory Pages: 5 (807 words) Published: July 17, 2012
Case Problem RC Coleman

ActivityDescriptionImmediate Predecessor
BObtain vendor proposals-
CSelect vendorA, B
DOrder SystemC
EDesign new warehouse layoutC
FDesign warehouseE
GDesign computer interfaceC
HInterface computerD,F,G
IInstall systemD,F
JTrain system operatorsH
KTest systemI,J

(a)(m)(b)
Time
ActivityOptimisticMost ProbablePessimistic
A468
B6816
C246
D81024
E71013
F468
G4620
H468
I4614
J345
K246

Where mean (expected time), t = (a + 4m + b) / 6

tA = (4 + 4(6) + 8) / 6 = 6
tB = (6 + 4(8) + 16) / 6 = 9
tC = (2 + 4(4) + 6) / 6 = 4
tD = (8 + 4(10) + 24) / 6 = 12
tE = (7 + 4(10) + 13) / 6 = 10
tF = (4 + 4(6) + 8) / 6 = 6
tG = (4 + 4(6) + 20) / 6 = 8
tH = (4 + 4(6) + 8) / 6 = 6
tI = (4 + 4(6) + 14) / 6 = 7
tJ = (3 + 4(4) + 5) / 6 = 4
tK = (2 + 4(4) + 6) / 6 = 4

for Network Diagram, refer to attachment.

Critical Path for activities = B, C, E, F, H, J, K = 43

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

To calculate the probability to complete in 40 weeks:-

Z = (T –TE) / ((v)^2)^0.5

V = ((b-a)/6)^2

vB = ((16 – 6) / 6)^2
= 2.77

vC = ((6-2)/6)^2
= 0.44

vE = ((13-7)/6)^2
= 1

vF = ((8-4)/6)^2
= 0.44

vH = ((8-4)/6)^2
= 0.44

vJ = ((5-3)/6)^2
= 0.11

vK = ((6-2)/6)^2
= 0.44

Z = 3 / ((2.77)+(0.44)+(1)+(0.44)+(0.44)+(0.11)+(0.44))^0.5

= 3 / (5.64)^0.5

= 3 / 2.374

= 1.26

From the Normal Distribution Chart:-

Z (1.26) = 0.8962

= 89.62 %

Therefore; the probability to complete the project in 40 weeks is 89.62% If the 40 weeks completion time is required, R.C. Coleman can shorten the completion time from 43 weeks to 40 weeks by crashing the activity times. 2) Suppose the management request that activity times be shortened to provide an 80 percent chance of meeting the 40 weeks 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 percent chance of completion within 40 weeks?

Z = (T – TE) / ((v)^2)^0.5

= (40 – TE) / (5.64)^0.5, where 5.64 is the variance same as in part (1),

= (40 – TE) / 2.37

Z (x) = 0.8, from Normal Distribution Chart, we found that x = 0.84

Therefore,

0.84 = (40 – TE) / 2.37

1.994 = 40 - TE

TE = 38 weeks

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

ActivityCrashed Activity TimeCost (\$)
(weeks)NormalCrashed

A41,0001,900
B71,0001,800
C21,5002,700
D82,0003,200
E75,0008,000
F43,0004,100
G58,00010,250
H45,0006,400
I410,00012,400
J34,0004,400
K35,0005,500

Normal TimeCrash TimeMaximum Crash Cost per day
Crash Time(\$)

tA = 642900
tB = 972800
tC = 4221,200
tD = 12841,200
tE = 10733,000
tF = 6421,100
tG = 8532,250
tH = 6421,400
tI = 7432,400
tJ = 4311,400
tK = 4311,500

Where, Maximum crash time = normal time – crash time

The revised critical path activity = B, C, E, F, H, J, K
= 7+2+7+4+4+3+3
= 30 weeks

The project completion time can be shortened to 30 weeks with the crashing cost of:- = 800 + 1,200 + 3,000 + 1,100 + 1,400 +...