# The critical path Table 1.1 essay

Thecritical path

Table1.1

Thecritical path is

A&gtC&gtE&gtI

4. The expectedduration time for each of the nine project activities

I.e. the resultin the end box which is given by the value of preceding value of taskI. =47.73

Slack for projectactivity E

S of E= lateststart –earliest start

E=43.4-43.4

=0

C.slack forproject activity C

=16.9-16.9

=0

D.Earliest starttime for project F

Project F ispreceding from project A, therefore, the latest finish for project Fwill be the earliest start for project F

=3

Latest finish G

Project G is apreceding activity to H, therefore the relationship between G and His that project activity H latest start is the latest finishofproject G

=36.4

5. Probability ofcompleting project in time.

Using Z score,

Z=(X-µ)/SD

Z-score

X-days inquestion

µ-days in

SD-standarddeviation

µ=48

X=48.033

Sd =8.35

48.0333-48)/8.35

=0.004

Using Z squaretable, we have

0.5160,

Therefore, theprobability is

51.60%

PERT CHART

3 11.25 14.25

B

7.15 4.12 18.4

16.9 12 28.9

D

18.4 1.530.4

28.9 6 34.9

G

30.4 1.5 36.4

34.9 741.9

H

36.4 1.5 43.4

033

A

0 03

3 3.916.9

C

3 016.9

16.9 26.5 43.4

E

16.9 0 43.4

43.4 4.63 48.03

I

43.4 0 48

48.033

3 3336

F

10.4 7 43.4

The critical path isA C E I

Table 1.2

QuestionC

1.

Paths time

LOPSU 2+2+3+5.5+2=14.5

MPSU 10+3+5.5+2=20.5

MQSU 10+6+5.5+2=23.5

NRTU 1.5+3+7+2=13.5

The critical pathis M Q S U

=23.5

In our criticalpath, activity S has the lowest cost to crash. Therefore we take S .

Normal time foractivity S is 5.5, we reduce it to 4.5 and the cost is 988.

The total sum inour critical value reduces to 22.5 from 23.5.the time for crashingactivity S is one, therefore, since our goal is to complete the taskWITHIN the 22 weeks, hence achieved.

Achievement ofthelowest cost possible is

1*988

=\$988

QuestionC2

The number ofweeks each activity would be crashed is

For activity Scrash by one week

The rest ofactivities were not to be crashed since our goal was to complete thetask within 22 weeks.

But their crashtime was

L 1.5

M 4

N 0.5

O 0.5

P 1

Q 3

R 2

S 1

T 1

U 1.5

QUESTIONC3

The sum cost is

3000*1.5+1000*0.5+2291.667*4+1775*1+988*1+5350*3+23800*1.5+12500*0.5+6475*2+291*1

=\$ 88170.668

WorkCited

Ragsdale, C. T. (2001). Spreadsheetmodelling and decision analysis: A practical introduction tomanagement science (3rd ed.)South-Western College Publishing.