# D 3 h 1 a 3 start b 2 e 2 f 4 g 2 i 2 j 4 end c 4 j

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Unformatted text preview: (2) F (4) G (2) I (2) J (4) END C (4) J. G. Carlsson, U of MN ISyE Lecture 7: PERT & CPM October 3/17, 2013 16 / 38 Time to completion If everything is completed on time, when will the project be completed? D (3) H (1) A (3) START B (2) E (2) F (4) G (2) I (2) J (4) END C (4) To solve this, look at every possible path from START to END: START→D→H→I→J→END = 10 START→A→E→F→G→I→J→END = 17 START→B→E→F→G→I→J→END = 16 START→C→J→END = 8 J. G. Carlsson, U of MN ISyE Lecture 7: PERT & CPM October 3/17, 2013 16 / 38 Critical path The (estimated) project duration equals the length of the longest path from START to END The longest path(s) is(are) called the critical path(s) D (3) H (1) A (3) START B (2) E (2) F (4) G (2) I (2) J (4) END C (4) J. G. Carlsson, U of MN ISyE Lecture 7: PERT & CPM October 3/17, 2013 17 / 38 Earliest start/earliest finish: “forward pass” D (3) H (1) A (3) START B (2) E (2) F (4) G (2) I (2) J (4) END C (4) How do we compute the earliest start time (ES) of an activity? Simple, just look at its predecessors: ES of activity = largest EF of all predecessors OK, how do we compute the earliest finish time (EF) of an activity? Simple, it’s just EF of activity = ES of activity + activity duration J. G. Carlsson, U of MN ISyE Lecture 7: PERT & CPM October 3/17, 2013 18 / 38 Earliest start/earliest finish: “forward pass” D (3) H (1) A (3) START B (2) E (2) F (4) G (2) I (2) J (4) END C (4) ES(A) = 0; EF(A) = 3 ES(B) = 0; EF(B) = 2 ES(C) = 0; EF(C) = 4 ES(D) = 0; EF(D) = 3 ES(E) = 3; EF(E) = 5 ES(F) = 5; EF(F) = 9 ES(G) = 9; EF(G) = 11 ES(H) = 3; EF(H) = 4 ES(I) = 11; J. G. Carlsson, U of MN ISyE EF(I) = 13 ES(J) = 13; EF(J) = 17 Lecture 7: PERT & CPM October 3/17, 2013 19 / 38 Latest start/latest finish: “backward pass” D (3) H (1) A (3) START B (2) E (2) F (4) G (2) I (2) J (4) END C (4) By examining the critical path, we can determine how long a project should take If we’re given the maximum time limit, how do we compute the latest finish time (LF) of an activity? In other words, “if we have to finish the project by a certain time, when is the latest we can finish this activity?” Simple, just look at its successors: LF of activity = smallest LS of all successors Lecture 7: start time October OK, how do we compute the latest PERT & CPM (LS) of an activity? 3/17, 2013 J. G. Carlsson, U of MN ISyE 20 / 38 Latest start/latest finish: “backward pass” D (3) H (1) A (3) START B (2) E (2) F (4) G (2) I (2) J (4) END C (4) Assume that we need to complete by time 17 (the length of the critical path): LF(J) = 17; LS(J) = 13 LF(I) = 13; LS(I) = 11 LF(H) = 11; LS(H) = 10 LF(G) = 11; LS(G) = 9 LF(F) = 9; LS(F) = 5 LF(E) = 5; LS(E) = 3 LF(D) = 10; LS(D) = 7 LF(C) = 13; LS(C) = 9 LF(B) = 3; J. G. Carlsson, U of MN ISyE LS(B)...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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