Submitted by: Submitted by cathrynanne
Views: 1263
Words: 592
Pages: 3
Category: Business and Industry
Date Submitted: 04/25/2013 01:23 PM
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)
[pic]
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...