Set 7 due today Set 8 due April

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

Answers set 7: (1) Shortest Route L-P 60=MIN L-S 125

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

Delete L-P, ADD PATHS FROM P L-S 125 P-M (new) 60+10=70 =MIN 60+55=115 P-V (new)

Delete P-M, add path from m P-V 115=MIN L-S 125 M-B (new) 70+50=120

Delete P-V, add path from V M-B 120=MIN L-S 125

Delete both M-B, V-B • • • L-S L-P-M L-P-V L-P-M-B 125 60 70 115 120

Answer to (2) PERT

A 4 1 E B 3 C D F 2

ES AND EF ACTIVITY ES EF A 0 5 B 0 8 C 0 7 D EF (C)=7 7+2=9 E MAX[(EF(B), EF(D)]=9 7 9+3=12=MAX= E(x) 7+4=11 F

ANSWER TO (a) E(x)=12

LF, LS ACTIVITY LF A E(x)=12 B C D E E(x)=12 F E(x)=12 LS

LF, LS ACTIVITY LF LS A B C D E E(x)= 12 12 -3=9 F E(x)=12 12 -4=8

LF, LS ACTIVITY LF LS A B C D LS(E)=9 E E(x)=12 12 -3=9 F E(x)=12 12 -4=8

LF, LS ACTIVITY LF LS A B C D LS(E)=9 9 -2=7 E E(x)=12 12 -3=9 F E(x)=12 12 -4=8

LF. LS ACTIVITY LF A LS E(x)=12 B C D MIN[LS(D) LS(F)]= MIN(7, 8)=7 LS(E)=9 9 -2=7 E E(x)=12 12 -3=9 F E(x)=12 12 -4=8

LF, LS ACTIVITY LF LS A E(x)=12 12 -5=7 B LS(E)=9 9 -8=1 C D MIN[LS(D) 7 -7=0 LS(F)]= MIN(7, 8)=7 LS(E)=9 9 -2=7 E E(x)=12 12 -3=9 F E(x)=12 12 -4=8

LF, LS ACTIVITY LF LS A E(x)=12 12 -5=7 B LS(E)=9 9 -8=1 C D MIN[LS(D) 7 -7=0 LS(F)]= MIN(7, 8)=7 LS(E)=9 9 -2=7 E E(x)=12 12 -3=9 F E(x)=12 12 -4=8

SLACK ACTIVITY LS ES LS-ES A 7 0 7 B 1 0 1 C 0 0 0 D 7 7 0 E 9 9 0 F 8 7 1

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

NORMAL TABLE • Kinderman Supplement, p 58 • Row 1. 0 • Col. 00

Z = 1. 00 Z . 00 … 1. 0 . 84134

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

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