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Published byAustin Sutton Modified over 9 years ago
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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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Answers set 7: (1) Shortest Route L-P60=MIN L-S125
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L B V S M P 60 10 50 55 15 20 125
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Delete L-P, ADD PATHS FROM P L-S125 P-M (new) 60+10=70 =MIN P-V (new) 60+55=115
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Delete P-M, add path from m P-V115=MIN L-S125 M-B (new) 70+50=120
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Delete P-V, add path from V M-B120=MIN L-S125
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Delete both M-B,V-B L-S125 L-P 60 L-P-M 70 L-P-V115 L-P-M-B120
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Answer to (2) PERT
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1 4 3 A B CD E F 2
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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
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ANSWER TO (a) E(x)=12
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LF,LS ACTIVITYLFLS AE(x)=12 B C D E F
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LF,LS ACTIVITYLFLS A B C D EE(x)= 1212-3=9 FE(x)=1212-4=8
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LF,LS ACTIVITYLFLS A B C DLS(E)=9 EE(x)=1212-3=9 FE(x)=1212-4=8
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LF,LS ACTIVITYLFLS A B C DLS(E)=99-2=7 EE(x)=1212-3=9 FE(x)=1212-4=8
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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
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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
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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
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SLACK ACTIVITYLSESLS-ES A707 B101 C000 D770 E990 F871
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ANSWER TO (b) CRITICAL PATH: C-D-E
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NORMAL TABLE Kinderman Supplement, p 58 Row 1.0 Col.00
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Z = 1.00 Z.00 … 1.0.84134
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Answer to (c) P(finish project before deadline)=.84
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Excel Class demo NOT same as assignment should be in memo format Describe critical activities in sentence
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