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Set 7 due today Set 8 due April 18 C-3 due April 18 Exam May 7.

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Presentation on theme: "Set 7 due today Set 8 due April 18 C-3 due April 18 Exam May 7."— Presentation transcript:

1 Set 7 due today Set 8 due April 18 C-3 due April 18 Exam May 7

2 Answers set 7: (1) Shortest Route L-P60=MIN L-S125

3 L B V S M P 60 10 50 55 15 20 125

4 Delete L-P, ADD PATHS FROM P L-S125 P-M (new) 60+10=70 =MIN P-V (new) 60+55=115

5 Delete P-M, add path from m P-V115=MIN L-S125 M-B (new) 70+50=120

6 Delete P-V, add path from V M-B120=MIN L-S125

7 Delete both M-B,V-B L-S125 L-P 60 L-P-M 70 L-P-V115 L-P-M-B120

8 Answer to (2) PERT

9 1 4 3 A B CD E F 2

10 ES AND EF ACTIVITYESEF A05 B08 C07 DEF (C)=77+2=9 EMAX[(EF(B), EF(D)]=9 9+3=12=MAX= E(x) F77+4=11

11 ANSWER TO (a) E(x)=12

12 LF,LS ACTIVITYLFLS AE(x)=12 B C D E F

13 LF,LS ACTIVITYLFLS A B C D EE(x)= 1212-3=9 FE(x)=1212-4=8

14 LF,LS ACTIVITYLFLS A B C DLS(E)=9 EE(x)=1212-3=9 FE(x)=1212-4=8

15 LF,LS ACTIVITYLFLS A B C DLS(E)=99-2=7 EE(x)=1212-3=9 FE(x)=1212-4=8

16 LF.LS ACTIVITYLFLS AE(x)=12 B CMIN[LS(D) LS(F)]= MIN(7,8)=7 DLS(E)=99-2=7 EE(x)=1212-3=9 FE(x)=1212-4=8

17 LF,LS ACTIVITYLFLS AE(x)=1212-5=7 BLS(E)=99-8=1 CMIN[LS(D) LS(F)]= MIN(7,8)=7 7-7=0 DLS(E)=99-2=7 EE(x)=1212-3=9 FE(x)=1212-4=8

18 LF,LS ACTIVITYLFLS AE(x)=1212-5=7 BLS(E)=99-8=1 CMIN[LS(D) LS(F)]= MIN(7,8)=7 7-7=0 DLS(E)=99-2=7 EE(x)=1212-3=9 FE(x)=1212-4=8

19 SLACK ACTIVITYLSESLS-ES A707 B101 C000 D770 E990 F871

20 ANSWER TO (b) CRITICAL PATH: C-D-E

21

22

23

24 NORMAL TABLE Kinderman Supplement, p 58 Row 1.0 Col.00

25 Z = 1.00 Z.00 … 1.0.84134

26 Answer to (c) P(finish project before deadline)=.84

27 Excel Class demo NOT same as assignment should be in memo format Describe critical activities in sentence


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