# Decision Analysis

**Topics:**Normal distribution, Probability density function, Probability theory

**Pages:**5 (807 words)

**Published:**July 17, 2012

ActivityDescriptionImmediate Predecessor

ADetermine equipment needs-

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 +...

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