Qat1 Task 4

QAT 1 Task 4

(A1)

Using these formulas: Expected time to complete= (a+4m+b)/6 Variance= (b-a) ^2/36

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Expected time to complete the critical path is B+F+G+I+J=7+10+6.5+6.5+3.5=33.5 weeks

Expected variance of the critical path is B+F+G+I+J=1.78+0.44+1.36+2.25+0.69=6.52 weeks

(A2)

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Critical path is B-F-G-I-J

(A3 a)

Time path A-C-E-H-J = 3+4.5+5+11+3.5=27 weeks

Time path B-D-H-J = 7+11.5+11+3.5=33 weeks

Time path B-F-G-I-J = 7+10+6.5+6.5+3.5-33.5 weeks

Critical path is B-F-G-I-J = 7+10+6.5+6.5+3.5 making expected duration of entire project 33.5 weeks.

(A3 b,c)

Slack time for project task A= latest finish-earliest finish

9.5-3=6.5

Slack time for project task H=latest finish-earliest finish

30-29.5=0.5

(A3 d,e)

Project task F is schedule to start in the 8th week.

In the path B-F task B takes 7 weeks to complete. So task F will start in the 8th week.

Project task I is scheduled to finish in the 30th week.

Add the task in the path B-F-G-I to determine Project task I schedule finish.

7+10+6.5+6.5=30

(A4)

Take the sum of the variances of the critical path:

1.78+0.44+1.36+2.25+0.69 = 6.52 weeks

Standard deviation:

Sqrt(6.52) = 2.55 apprx

Z = 34-33.5 = 0.196

2.55

Probability 0.5777 or 57.77%

(B 1,2)

Maximum reduction in time: Normal duration-crash duration

(a) 3-2=1 (b) 7-5=2 (c) 4.5-3=1.5 (d) 11.5-7=4.5 (e) 5-3=2 (f) 10-6=4 (g) 6.5-5=1.5

(h) 11-8=3 (i) 6.5-4.5=2 (j) 3.5-3=0.5

Crash cost per week: Crash cost-normal cost/normal duration-crash duration.

(a) (11200-8400)/(3-2)=2800 (b) (40000-28000)/(7-5)=6000 (c) (24000-18000)/(4.5-3)=4000

(d) (44800-36800)/(11.5-7)=1778 (e) (16800-14000)/(5-3)=1400 (f) (24000-20000)/(10-6)=1000

(g) (28000-18200)/(6.5-5)=6533 (h) (64000-44000)/(11-8)=6667 (i) (36000-26000)/(6.5-4.5)=5000 (j) (36000-21000)/(3.5-3)=30000...