CASE PROBLEMPERT/CPM

GROUP 3:SAIFUL BAHRI HJ AHMAD MOKHTAR

MOHD FAUZI MD FADZILWAN AISHAH WAN MOHD NOWALID

FARHANAH MOHD NOAH

CASE PROBLEM

List of the activities

The optimistic, most probable and pessimistic time (in week)

Question (a)1

• R.C Coleman’s top management established a required 40-week completion time for the project. Can this completion time be achieved?

I

FINISH

J

G

KEH

F

D

C

B

A

START

Project Network

6

4 bmat

Expected Time•Optimistic time (a) The minimum activity time if everything progresses ideally

•Most probable time (m)The most probable activity time under normal conditions

•Pessimistic time (b)The maximum activity time if significant delays are encountered

•Expected time (t)The average time for the activity

6

8)6(44 tA

6

16)8(46 tB

6

6)4(42 tC

6

24)10(48 tD

6

13)10(47 tE

6

8)6(44 tF

6

20)6(44 tG

6

8)6(44 tH

6

14)6(44 tI

6

5)4(43 tJ

6

6)4(42 tK

ACTIVITY EXPECTED TIME (WEEKS)

A 6

B 9

C 4

D 12

E 10

F 6

G 8

H 6

I 7

J 4

K 4

TOTAL 76

Expected Time In Weeks

I 29 367 32 39

FINISH

J 35 394 35 39

G 13 218 21 29

K 39 434 39 43

E 13 2310 13 23

H 29 356 29 35

F 23 296 23 29

D 13 2512 17 29

C 9 134 9 13

B 0 99 0 9

A 0 66 3 9

START

Project Network (Earliest Start, Finish Times, Latest Start and Finish Times)

ACTIVITY ES EF LS LF SLACK

A 0 6 3 9 3

B 0 9 0 9 0 (critical)

C 9 13 9 13 0 (critical)

D 13 25 17 29 4

E 13 23 13 23 0 (critical)

F 23 29 23 29 0 (critical)

G 13 21 21 29 8

H 29 35 29 35 0 (critical)

I 29 36 32 39 3

J 35 39 35 39 0 (critical)

K 39 43 39 43 0 (critical)

Activity Slack Time

Critical Path:B-C-E-F-H-J-K

Project Completion Time:43 weeks

R.C Coleman’s required 40-week completion time. This completion time can’t be achieved

Question (a)2

Include probability information in your discussion. What recommendations do you have if the 40-week completion time is required?

Variance (2)

• used to describe the dispersion or variation in the activity time values

2

6

2

ab

ACTIVITY VARIANCE

A 0.44

B 2.78

C 0.44

D 7.11

E 1.00

F 0.44

G 7.11

H 0.44

I 2.78

J 0.11

K 0.44

Variance of Each Activity

Variance in the project completion time (2) 2 = 2

B + 2C + 2

E + 2F + 2

H + 2J + 2

K = 2.78 + 0.44 + 1.00 + 0.44 + 0.44 + 0.11 + 0.44= 5.65

Standard deviation () = √5.65

= 2.38

Expected Completion Time

43

Time (weeks)

= 2.38

Normal Distribution of Project Completion Time

= 2.38

43

Time (weeks)

40

Probability the project will meet the 40-week completion time

At 40 ≤ T ≤ 43z = (40 – 43) / 2.38 = 1.26

using table,Pr (40 ≤ T ≤ 43) = 0.3962

Pr (T ≤ 40) = 0.5000 – 0.3962

= 0.103= 10.38%

We find that the probability of the project meeting the 40-weeks deadline is 10.38%. Thus, R.C Collman have a low chance if they want the project completed at 40-weeks

Question (b)

• Suppose that management request that activity times be shortened to provide an 80 percent chance of meeting the 40-week completion time. If the variance in the project completion time is the same as you found in (a), how much should the expected project completion time be shortened to achieved the goal of an 80 percent chance of completion within 40 weeks?

= 2.38

40X

Time (weeks)

Expected completion time to achieve 80% chance of completion within 40 weeks

Pr (T ≤ 40) = 0.8000Pr (x ≤ T ≤ 40) = 0.8000 – 0.5000 = 0.3000

using table,z (x ≤ T ≤ 40) = 0.84(40 – x) / 2.38 = 0.84

x = 38

It shows the project has to be shortened to 38 weeks to achieve the goal of an 80 % chance of completion within 40 weeks.

Question (c)

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

Crash Time and Cost’s Information

iiiM '

iM

iC

iC

iK

'

Maximum possible reduction in time, Mi

(3)

with i = expected time for activity I’i = time for activity i under maximum crashing

Crash cost per unit of time, Ki

with Ci = cost for activity I C’i = cost for activity i under maximum crashing

ACTIVITYTIME COST

Mj KjNORMAL CRASHED NORMAL CRASHED

A 6 4 1000 1900 2 450

B 9 7 1000 1800 2 400

C 4 2 1500 2700 2 600

D 12 8 2000 3200 4 300

E 10 7 5000 8000 3 1000

F 6 4 3000 4100 2 550

G 8 5 8000 10259 3 753

H 6 4 5000 6400 2 700

I 7 4 10000 12400 3 800

J 4 3 4000 4400 1 400

K 4 3 5000 5500 1 500

Linear Programming Model

EXCEL

ACTIVITY TIME (WEEKS) CRASHING (WEEKS)

A 6 0B 7 2C 4 0D 12 0E 10 0F 5 1G 8 0H 6 0I 7 0J 3 1K 3 1

Total Crashing Time 5 weeksTotal Crashing Cost $ 2,250.00

Activity Crashing Decision