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Slide 1 - PERT/CPM AGENDA MGT 606 1. Motivation PERT/CPM 2a. Digraphs (Project Diagrams) 2b. PERT 3a. CPM--Starts, Finishes, Slacks, 3b. Resource Allocation 4. CPM with Crashing 5. PERT Simulation
Slide 2 - 1. Motivation for PERT/CPM Use Definition of a Project Organizational Trajectories and Change The Environment Adaptation and Agility Reactive Strategy Proactive Strategy Ubiquity of Projects Dealing with Project Complexity 2
Slide 3 - 2. PERT Example A. AON Network Diagrams: General Foundry Diagramming a project’s network of activities Installation of air-pollution control equipment @ General Foundry, Milwaukee Immediate Activity Description Predecessor A build internal components ___ B modify roof and floor ___ C construct collection stack A D pour concrete and install frame B E build hi-temp burner C F install control system C G install air-pollution control device D,E H inspect and test F,G 3
Slide 4 - 2. PERT Example A. AON Network Diagrams (continued): General Foundry A C Node A represents Activity A Node C represents Activity C edge or arc The edge or arc represents the precedence relationship between the two activities 4
Slide 5 - 2. PERT Example A. AON Network Diagrams (continued): General Foundry continued A C B D A C B D F E A C B D 5
Slide 6 - 2. PERT Example A. AON Network Diagrams (continued): General Foundry concluded A B C D F E G H 6
Slide 7 - 2. PERT Example A. AON Network Diagrams (continued) The General Foundry project network began on more than one node and ended on a single node. Other variants are: A B C D E F Beginning with one node and ending with more than one node. A B C D E F Beginning with more than one node and ending with more than one node. 7
Slide 8 - 2. PERT Example A. AON Network Diagrams (concluded) Sometimes the following convention is used. Start Finish A B C D E F The “Start” and “Finish” boxes tie the network off at its ends and give one a sense that the network has defined points in time at which the project begins and ends . The use of such a convention is not necessary and will generally be avoided. 8
Slide 9 - 2. PERT Example A2. AOA Network Diagrams (continued): General Foundry 1 2 Node 1 represents the beginning of Activity A Node 2 represents the ending of Activity A edge or arc The edge or arc represents Activity A 8b Activity A
Slide 10 - 2. PERT Example A2. AOA Network Diagrams (continued): General Foundry continued 1 C 1 2 3 2 8c A B A 1 2 3 A B 4 5 C D E
Slide 11 - 2. PERT Example A2. AOA Network Diagrams (continued): General Foundry concluded 1 2 3 4 5 6 8d 7 A B C D E F G H
Slide 12 - 2. PERT Example A2. AOA Network Diagrams (concluded) Note that the General Foundry project network on slide #6 began on a single node and ended on a single node. When constructing Network diagrams using the AOA approach, this convention is followed--network diagrams begin on a single node and end on a single node. Note also a second convention in AOA Network diagram construction. Two nodes are connected by one and only one activity (edge or arc). Thus, Act. I.P. A __ B __ C A,B D B Finally, a third convention. An arc cannot emanate from or terminate at more than one node. Act. I.P. A __ B __ C A,B D A E B 8e A B C D No, and why not? B A D C d1 Rather, this is preferred and why? A B C D E A B C E d2 d1 D NO!
Slide 13 - 2. PERT Example B. Project Completion Times and Probabilities: General Foundry of Milwaukee The duration time for each of a Project’s activities in a PERT environment are estimated on the basis of most likely, pessimistic, and optimistic completion times. These times can be arrived at in various ways. A number of these ways contain substantial subjective components because it is often the case that little historical information is available to guide those estimates. The expected (value) duration time of an individual activity and its variation follow what is called a beta distribution and are calculated as follows: E(ti) = (a + 4m + b) / 6 and var(ti) = (b - a)2 / 36 where “a” is the activity’s optimistic completion time, “m” is the activity’s most likely completion time, and “b” is the activity’s most pessimistic completion time. 9
Slide 14 - 2. PERT Example B. Project Completion Times and Probabilities: General Foundry of Milwaukee (continued) Consider the following table of activities; immediate predecessor(s) (I.P.); optimistic, most likely, and pessimistic completion times for General Foundry; and the E(ti) and var(ti) for each activity. ACT I.P. Optimistic (a) Most Likely (m) Pessimistic(b) E(ti) var(ti) A __ 1 week 2 weeks 3 weeks 2 wks. 4/36 B __ 2 3 4 3 4/36 C A 1 2 3 2 4/36 D B 2 4 6 4 16/36 E C 1 4 7 4 36/36 F C 1 2 9 3 64/36 G D,E 3 4 11 5 64/36 H F,G 1 2 3 2 4/36 10
Slide 15 - 2. PERT Example B. Project Completion Times and Probabilities: General Foundry of Milwaukee (continued) A 2 B 3 C 2 D 4 F 3 E 4 G 5 H 2 11 From the reduced version of General Foundy’s PERT table., the E(ti) for each activity can be entered into the project’s network diagram. ACT I.P. (a) (m) (b) E(ti) var(ti) A __ 1 2 3 2 4/36 B __ 2 3 4 3 4/36 C A 1 2 3 2 4/36 D B 2 4 6 4 16/36 E C 1 4 7 4 36/36 F C 1 2 9 3 64/36 G D,E 3 4 11 5 64/36 H F,G 1 2 3 2 4/36 Inspection of the network discloses three paths thru the project: A-C-F-H; A-C-E-G-H; and B-D-G-H. Summing the E(ti) on each path yield time thru each path of 9, 15, and 14 weeks, respectively. With an E(t) = 15 for A-C-E-G-H, this path is defined as the critical path (CP) being the path that govens the completion time of the project . Despite the beta distribution of each activity, the assumption is made that the number of activities on the CP is sufficient for it to be normally distributed with a variance equal to the sum of the variances of its activities only, var(t) = 112/36 = 3.11.
Slide 16 - From the reduced version of General Foundy’s PERT table., the E(ti) for each activity can be entered into the project’s network diagram. ACT I.P. (a) (m) (b) E(ti) var(ti) A __ 1 2 3 2 4/36 B __ 2 3 4 3 4/36 C A 1 2 3 2 4/36 D B 2 4 6 4 16/36 E C 1 4 7 4 36/36 F C 1 2 9 3 64/36 G D,E 3 4 11 5 64/36 H F,G 1 2 3 2 4/36 Inspection of the network discloses three paths thru the project: A-C-F-H; A-C-E-G-H; and B-D-G-H. Summing the E(ti) on each path yield time thru each path of 9, 15, and 14 weeks, respectively. With an E(t) = 15 for A-C-E-G-H, this path is defined as the critical path (CP) being the path that governs the completion time of the project . Despite the beta distribution of each activity, the assumption is made that the number of activities on the CP is sufficient for it to be normally distributed with a variance equal to the sum of the variances of its activities only, var(t) = 112/36 = 3.11. 2. PERT Example B2. Project Completion Times and Probabilities: General Foundry of Milwaukee (continued) 11b 5 6 7 4 2 1 3 A2 B3 C2 D4 E4 F3 G5 H2 critical path
Slide 17 - 2. PERT Example B. Project Completion Times and Probabilites: General Foundry (concluded) Given General Foundry’s E(t) = 15 weeks and var(t) = 3.11, what is the probability of the project requiring in excess of 16 weeks to complete? P( X > 16) = 1 - P ( X < 16) = 1 - P ( Z < (16 - 15) / 1.76 = .57) = 1 - 0.716 where the value 1.76 is the square root of 3.11, the standard deviation of the expected completion time of General Foundry’s project. The assumption being made is that summing up a sufficient number of activities following a beta distribution yield a result which approximates or a approaches a variable which is normally distributed--~N(,2) 12
Slide 18 - 3. CPM Example A. Starts, Finishes, Slacks E(arly)S(tart) -- earliest possible commencement time for a project activity. ES calculation -- beginning at the initial node(s) of a project’s network diagram, conduct a forward pass where ESj = ESi + ti A B C D 3 4 2 5 0 3 7 9 0 = 0 + 0 3 = 0 + 3 7 = 3 + 4 9 = 7 + 2 B C D 4 2 5 3 7 ? What if? ESD = MAX ESB + tB = 7 ESC + tC = 9 13
Slide 19 - 3. CPM Example A. Starts, Finishes, Slacks E(arly)F(inish) -- earliest possible time a project’s activity can be completed. EF calculation -- beginning at the initial node(s) of a project’s network diagram , conduct a forward pass where EFi = ESi + ti A B C D 3 4 2 5 0 3 7 9 3 = 0 + 3 7 = 3 + 4 9 = 7 + 2 14 = 9 + 5 3 7 9 14 14
Slide 20 - 3. CPM Example A. Starts, Finishes, Slacks E(arly) S(tarts), E(arly F(inishes) for Milwaukee Foundry . A 2 B 3 C 2 D 4 F 3 E 4 G 5 H 2 0 2 0 3 4 8 4 13 2 3 4 7 8 13 15 7 EF ES 15
Slide 21 - A 3 B 4 C 2 D 5 0 3 7 9 14 14 9 7 3 9 = 14 - 5 7 = 9 - 2 3 = 7 - 4 C 2 D 5 9 14 B 4 A 3 LFC -tC = 7 LFD -tD =9 MIN LFB = ? 16 3. CPM Example A. Starts, Finishes, Slacks L(ate) F(inish) -- latest possible time an activity can be completed without delaying the completion time of the project. LF calculation -- beginning at the final node, conduct a backward pass through the network’s paths where LFi = LFj - tj
Slide 22 - 3. CPM Example A. Starts, Finishes, Slacks LSi = LFi - ti 14 9 7 3 3 4 2 5 A B C D 9 = 14 -5 7 = 9 -2 3 = 7 -4 0 = 3 - 3 0 3 7 9 17 LS calculation -- beginning with the final node(s) of the network make a backward pass through the network where L(ate) S(tart) -- the latest possible time and activity can begin without delaying the completion time of the project.
Slide 23 - E(arly) S(tarts)/ F(inishes) and L(ate) S(tarts)/F(inishes) for Milwaukee Foundry . A B C D E F G H 2 3 2 4 3 4 5 2 0 0 2 3 4 4 8 13 15 13 7 8 4 7 2 3 EF ES 15 13 8 4 13 8 2 4 0 1 LF LS 13 8 10 4 2 4 critical path 18 3. CPM Example A. Starts, Finishes, Slacks
Slide 24 - 3. CPM Example A2. Starts, Finishes, Slacks E(arly)S(tart) -- earliest possible commencement time for a project activity. ES calculation -- beginning at the initial node(s) of a project’s network diagram, conduct a forward pass where: ESj = ESi + ti A3 C2 D5 1 2 3 0 = 0 + 0 3 = 0 + 3 7 = 3 + 4 9 = 7 + 2 B4 C2 D5 3 7 4 What if? ESD = MAX ESB + tB = 7 ESC + tC = 9 18b B4 0 3 7 4 9 2 3 9 node # ESj
Slide 25 - 3. CPM Example A2. Starts, Finishes, Slacks E(arly)F(inish) -- earliest possible time a project’s activity can be completed. EF calculation -- beginning at the initial node(s) of a project’s network diagram , conduct a forward pass where: EFi = ESi + ti 3 = 0 + 3 7 = 3 + 4 9 = 7 + 2 18c A3 C2 1 2 3 B4 0 3 7 4 9 D5 0 3 7 9 E2 4 3 D4 3 4 5 7 3 4 7/6 F5 What if? 6 12 12 G3 critical path node # ESj EFi
Slide 26 - 3. CPM Example A2. Starts, Finishes, Slacks E(arly) S(tarts), E(arly F(inishes) for Milwaukee Foundry . 18d 1 2 3 4 5 6 7 A2 B3 C2 D4 E4 F3 G5 H2 0 2 4 3 8 13 15 critical path 3 8/7 2 4 7/13 15 node # ESj EFi
Slide 27 - 3. CPM Example A2. Starts, Finishes, Slacks L(ate) F(inish) -- latest possible time an activity can be completed without delaying the completion time of the project. LF calculation -- beginning at the final node, conduct a backward pass through the network’s paths where: A3 1 B4 2 C2 3 D5 4 3 7 9 = 14 - 5 7 = 9 - 2 3 = 7 - 4 C2 4 D5 9 B4 3 2 LFC -tC = 7 LFD -tD =9 MIN LFB = ? 18e 0 0 LFi = LFj - tj 5 14 9 14 0 = 3 - 3 14 5 14 = 14 - 0 What if? EFi LFi ESi node #
Slide 28 - 3. CPM Example A2. Starts, Finishes, Slacks L(ate) S(tart) -- the latest possible time and activity can begin without delaying the completion time of the following activity. LS calculation -- beginning with the final node(s) of the network make a backward pass through the network where: LSi = LFi - ti 0 1 2 3 4 A3 B4 C2 D5 9 = 14 -5 7 = 9 -2 3 = 7 - 4 0 = 3 - 3 0 3 7 9 18f 9 5 14 14 7 3 C2 4 D5 9 B4 3 2 LFC -tC = 7 LFD -tD = 9 LSC,D = 7/9 14 5 What if? EFi LFi ESi LSj node #
Slide 29 - 3. CPM Example A2. Starts, Finishes, Slacks E(arly) S(tarts)/ F(inishes) and L(ate) S(tarts)/F(inishes) for Milwaukee Foundry . 18g 1 2 3 4 5 6 7 A2 B3 C2 D4 E4 F3 G5 H2 0 2 4 3 8 13 15 critical path 3 8/7 2 4 7/13 15 0 0 0 15 13 8 4 2 4 13 10/4 15 2 8 4
Slide 30 - T(otal) S(lack) -- the amount of time an activity’s completion can be delayed without delaying the project’s completion where, TSi = LFi - ESi - ti and where i = the ith activity. F(ree) S(lack) -- the amount of time an activity’s completion can be delayed without delaying the commencement of the next activity where, FSi = ESj - ESi - ti and where i = the ith activity and j = the jth activity. 3. CPM Example A. Starts, Finishes, Slacks 19
Slide 31 - 3. CPM Example A. Starts, Finishes., Slacks T(otal) S(lacks) and F(ree) S(lacks) for Milwaukee Foundry . A B C D E F G H 2 3 2 4 3 4 5 2 0 0 2 3 4 4 8 13 ES 15 13 8 4 13 8 2 4 LF critical path TSH=0=15-13-2 FSH=0=15-13-2 TSE=8-4-4=0 FSE=8-4-4=0 TSD=1=8-4-3 FSD=1=8-4-3 TSB=1=4-3-0 FSB=0=3-3-0 20
Slide 32 - Shared Slack -- the slack in a project along a “non-critical” segment of a path which all activities on that non-critical segment share. Consider the lower path for Milwaukee Foundry. 3. CPM Example A. Starts, Finishes., Slacks B 3 D 4 G 5 0 4 3 8 8 13 TSB = 1 FSB = 0 TSD = 1 FSD = 1 TSG = 0 FSG = 0 The slack of one time period along this non-critical path segment is shared between activities B and D, i.e., 7 time periods of activities are scheduled over an 8 time period segment. 8 time periods (TPs)* A -- 3 TPs B -- 4 TPs A -- 3TPs B -- 4TPs B -- 4TPs A -- 3TPs *Other variations are possible if the one TP of slock is divided up. 21
Slide 33 - Nested Slacks -- when one segment of a non-critical path is imbedded in another segment of a non-critical path, the “free slack” of the terminal activity in the imbedded non-critical segment will not necessarily be 0. 3. CPM Example A. Starts, Finishes., Slacks C A B D E 8 10 7 5 15 Activity I.P. A B C D E __ A __ C D 25 25 10 10 0 0 13 20 8 15 TS,FS = 5 TS=5, FS=0 TS=5, FS=0 TS,FS=0 TS,FS=0 NOTE: Path C-D-E is the non-critical path and is nested in A-B (the critical path). Hence, the ES for activity A is 0 and the LF for activity E is 25, the ES and LF for activities A and B , respectively. 22
Slide 34 - Nested Slacks -- when one segment of a non-critical path is imbedded in another segment of a non-critical path, the “free slack” of the terminal activity in the imbedded non-critical segment will not necessarily be 0. 3. CPM Example A. Starts, Finishes., Slacks C A B D E 8 10 7 5 15 Activity I.P. A B C D E __ A __ C D,F 25 25 10 10 0 0 13 20 8 15 TS,FS = 3 TS=5, FS=2 TS=5, FS=0 TS,FS=0 TS,FS=0 F 17 0 20 F __ TS=3, FS=0 NOTE: Path A-B is the one critical path in this poject while paths F-E and C-D-E are non-critcal with C-D-E nested in F-E and shorter in path length by two time periods. As a consequence, in figuring the total and free slacks on C-D-E, ref;ect the two additional time periods of slack shared by them beginning with activity D where TSD = FSD or FSD = 0. 23
Slide 35 - 3. CPM Example A2. Starts, Finishes., Slacks T(otal) S(lacks) and F(ree) S(lacks) for Milwaukee Foundry . 23b 1 2 3 4 5 6 7 A2 B3 C2 D4 E4 F3 G5 H2 0 2 4 3 8 13 15 3 7/8 2 4 7/13 15 0 0 0 15 13 8 4 2 4 13 10/8 2 8 4 early start late finish TSC = 4 - 2 - 2 = 0 FSC = 4 - 2 - 2 = 0 TSD = 8 - 3 - 4 = 1 FSD = 8 - 3 - 4 = 1 TSB = 4 - 0 - 3 = 1 FSB = 3 - 0 - 3 = 0
Slide 36 - Shared Slack -- the slack in a project along a “non-critical” segment of a path which all activities on that non-critical segment share. Consider the lower path for Milwaukee Foundry. 3. CPM Example A2. Starts, Finishes., Slacks B3 2 D4 3 5 4 1 5 4 8 8 TSB = 1 FSB = 0 TSD = 1 FSD = 1 The slack of one time period along this non-critical path segment is shared between activities B and D, i.e., 7 time periods of activities are scheduled over an 8 time period segment. 8 time periods (TPs)* B -- 3 TPs D -- 4 TPs B -- 3TPs D -- 4TPs D -- 4TPs B -- 3TPs *Other variations are possible if the one TP of slock is divided up. 23c 4 0 0 C8 TSC = 0 FSC = 0
Slide 37 - Nested Slacks -- when one segment of a non-critical path is imbedded in another segment of a non-critical path, the “free slack” of the terminal activity in the imbedded non-critical segment will not necessarily be 0. 3. CPM Example A2. Starts, Finishes., Slacks C8 A10 B15 D7 E5 1 2 4 3 Activity I.P. A B C D E __ A __ C D 10 20 0 10 0 13 8 15 TS,FS = 5 TS=5, FS=0 TS=5, FS=0 TS,FS=0 TS,FS=0 NOTE: Path C-D-E is the non-critical path and is nested in A-B (the critical path). Hence, the ES for activity A is 0 and the LF for activity E is 25, the ES and LF for activities A and B , respectively. 23d 5 25 25
Slide 38 - Nested Slacks -- when one segment of a non-critical path is imbedded in another segment of a non-critical path, the “free slack” of the terminal activity in the imbedded non-critical segment will not necessarily be 0. 3. CPM Example A2. Starts, Finishes., Slacks Activity I.P. A B C D E __ A __ C D,F 20 F17 F __ TS=3, FS=0 NOTE: Path A-B is the one critical path in this project while paths F-E and C-D-E are non-critical with C-D-E nested in F-E and shorter in path length by two time periods. As a consequence, in figuring the total and free slacks on C-D-E, they reflect the two additional time periods of slack shared by them beginning with activity D. 23e C8 A10 B15 D7 E5 1 2 4 3 10 20 0 10 0 13 8 17 TS,FS = 3 TS=5, FS=2 TS=5, FS=0 TS,FS=0 TS,FS=0 5 25 25
Slide 39 - 3. CPM Example B. Resource Allocation Scheduling (ES) Activity *A (2) B (3) *C (2) D (4) *E (4) F (3) *G (5) *H (2) TP1 TP2 TP3 TP4 TP5 TP6 TP7 TP8 TP9 TP10 TP11 TP12 TP13 TP14 TP15 Activity Cost A -- 22K ($22,000) B -- 30K C -- 26K D -- 48K E -- 56K F -- 30K G -- 80K H -- 16K 11K 11K 13K 13K 14K 14K 14K 14K 16K 16K 16K 16K * -- critical path 16K 8K 8K 10K 10K 10K 12K 12K 12K 12k 10K 10K 10K 21 21 21 42 23 65 25 36 36 36 14 16 16 16 16 16 8 8 90 126 162 198 212 228 244 260 276 292 300 308 24
Slide 40 - 3. CPM Example B. Resource Allocation Scheduling (LS) Activity *A (2) B (3) *C (2) D (4) *E (4) F (3) *G (5) *H (2) TP1 TP2 TP3 TP4 TP5 TP6 TP7 TP8 TP9 TP10 TP11 TP12 TP13 TP14 TP15 Activity Cost A -- 22K ($22,000) B -- 30K C -- 26K D -- 48K E -- 56K F -- 30K G -- 80K H -- 16K 11K 11K 13K 13K 14K 14K 14K 14K 16K 16K 16K 16K * -- critical path 16K 8K 8K 10K 10K 12K 12K 12k 11 11 21 32 23 55 23 26 26 26 26 16 16 26 26 26 8 8 78 104 130 156 182 198 214 240 266 292 300 308 10K 12K 10K 10K 10K 25
Slide 41 - 3. CPM Example B. Resource Allocation Scheduling Early/Late Start Resource Allocation Schedules Project Time Periods C u m. P r o j. C o s t s 26
Slide 42 - 4. CPM Example CPM with Crashing --a It is sometime necessary to accelerate the completion time of a project. This usually leads to greater cost in completing the project than what might have otherwise been the case because of opportunity costs incurred as the result of diverting resources away from other pursuits. As a result of this increase in cost, it is incumbent upon project managers to reduce the completion time of the project in the most cost effective way possible. The following guidelines are designed to achieve that end. 1) Reduce duration times of critical activities only. 2) Do not reduce the duration time of critical activity such that its path length (in time) falls below the lengths (in time) of other paths in the network. 3) Reduce critical activity duration times on the basis of the least costly first and in case of a tie, the least costly activity closest to the completion node(s) of the project. 4) When two or more critical paths exist, reduce the length (in time) of all critical paths simultaneously. 5) Given two or more critical paths and a cost tie between a joint activity and a subset of disjoint activities on the same critical path, generally reduce the duration time of the one joint to the most paths. 6) Note, if reducing a joint activity means that more critical paths emerge that what would otherwise be the case, reduce disjoint activities. 27
Slide 43 - 4. CPM Example CPM with Crashing --b Consider the crashing of the Milwaukee Foundry Project Act. N.T. C.T. N.C. C.C. U.C.C. * A 2 1 22K 23K 1K B 3 1 30K 34K 2K * C 2 1 26K 27K 1K D 4 3 48K 49K 1K * E 4 1 56K 59K 1K F 3 2 30K 30.5K 0.5K * G 5 2 80K 86K 2K * H 2 1 16K 19K 3K * - Critical Path An inspection of Milwaukee Foundry’s project network identifies the following three paths and of duration, A - C - F - H : 9 *A - C - E - G -H : 15 B - D - G - H : 14 28
Slide 44 - 4. CPM Example CPM with Crashing --c ACT N.T. C.T. N.C. C.C. U.C.C. *A 2 1 22K 23K 1K B 3 2 1 1 30K 34K 2K 5 3K *C 2 1 1 26K 27K 1K 6 3K D 4 3 3 48K 49K 1K 3 2K *E 4 3 2 1 1 56K 59K 1K 1 1K F 3 2 30K 30.5K 0.5K *G 5 2 2 80K 86K 2K 2 6K *H 2 1 1 16K 19K 3K 4 3K 308 K 18K  - - SECOND CRITICAL PATH A - C - F - H : 9 8 *A - C - E - G -H : 15 14 B - D - G - H : 14 14 A - C - F - H : 9 9 9 9 8 7 7 *A - C - E - G -H : 15 14 11 10 9 8 7 B - D - G - H : 14 14 11 10 9 8 7 1 1 2 3 4 6 5 Act. E has the lowest U.C.C. and closest to the final node of the network-- crash one time period and two CPs emerge. One can now reduce Acts. E & D or G. Reduce G, three time periods. It is A joint activity. Now E & D can be reduced one time period for the same U.C.C. as G. Now reduce H by one time unit. Now reduce C & B by one time unit. Finally, reduce E & B by one time unit. 1 2 3 4 5 6 29
Slide 45 - 4. CPM Example CPM with Crashing --d ( Summary) Why would reduce Activity E in crashing step one ( 1 ) and by only one time unit? Why do you reduce Activity G and not Activities E & D in 2 and by three time units? Why do you now reduce Activities E & D in 3 and by only one time unit? Why do you reduce Activity H in 4 ? Why do you now reduce E & B in 5 but by only one time unit? Why do you now reduce C & B in 6 and how do you know now that you have crashed the project down to the minimum possible completion time? Why were Activities A & F never reduced? Why should we concern ourselves with crashing a project by always reducing the least costly activities first? 30
Slide 46 - 5. PERT Simulation --agenda-- Motivation Illustrative Examples Performing Simulations with WinQSB 31
Slide 47 - 5. PERT Simulation Motivation ▪ non-stochasticity/stochasticity & non-critical paths ▪ independence of/interdependence between paths ▪ strength of an assumption 32
Slide 48 - 5. PERT Simulation—motivation --non-stochasticity/stochasticity & non-critical paths-- Reconsider General Foundry of Milwaukee The E(ti) for each activity in this reduced version of General Foundy’s PERT table has been entered into the project’s network diagram. ACT I.P. (a) (m) (b) E(ti) var(ti) A __ 1 2 3 2 4/36 B __ 2 3 4 3 4/36 C A 1 2 3 2 4/36 D B 2 4 6 4 16/36 E C 1 4 7 4 36/36 F C 1 2 9 3 64/36 G D,E 3 4 11 5 64/36 H F,G 1 2 3 2 4/36 Inspection of the network discloses three paths thru the project: A-C-F-H; A-C-E-G-H; and B-D-G-H. Summing the E(ti) on each path yields the time thru each path to be 9, 15, and 14 weeks, respectively. With an E(t) = 15 for A-C-E-G-H, this path is defined as the critical path (CP) having a path variance of 3.11 (112.36). The other paths however, also have variances of 2.11 for A-C-F-H and 2.44 for B-D-G-H. We assume all three paths to be normally distributed. • Implications of assuming non-stochasticity on the (two) non-critical paths-- how serious? • How strong is the assumption of non-stochasticity in this case? 33 A 2 B 3 C 2 D 4 F 3 E 4 G 5 H 2
Slide 49 - 5. PERT Simulation--motivation --independence of/interdependence between paths-- Reconsider General Foundry of Milwaukee 34 A 2 B 3 C 2 D 4 F 3 E 4 G 5 H 2 A 2 B 3 C 2 D 4 E 4 G 5 H 2 F 3 Digraph with independent paths Digraph with interdependent paths Which paths are interdependent and why? Why might this path interdependence complicate estimating the expected completion time of this project? How strong might the assumption of path independence be in this case?
Slide 50 - 5. PERT Simulation—illustrative examples Reconsider General Foundry of Milwaukee CASE 1 ACT I.P. (a) (m) (b) E(ti) var(ti) A __ 1 2 3 2 4/36 B __ 2 3 4 3 4/36 C A 1 2 3 2 4/36 D B 2 4 6 4 16/36 E C 1 4 7 4 36/36 F C 1 2 9 3 64/36 G D,E 3 4 11 5 64/36 H F,G 1 2 3 2 4/36 CASE 2 ACT I.P. (a) (m) (b) E(ti) var(ti) A __ 1 2 3 2 4/36 B __ .1 3 5.9 3 33.64/36 C A 1 2 3 2 4/36 D B .1 4 7.9 4 60.84/36 E C 1 4 7 4 36/36 F C 1 2 9 3 64/36 G D,E 3 4 11 5 64/36 H F,G 1 2 3 2 4/36 The PERT table above merely replicates the results of a previous page–paths A-C-F-H; A-C-E-G-H; and B-D-G-H have E(t)s of 9, 15, and 14 weeks with variances of 2.11, 3.11, and 2.44, respectively. The PERT table above in contrast to the one at its left while replicating the same path expected duration times path as before B-D-G-H now has a variance of 4.513. Relaxing the two assumptions that 1) non-critical paths are non-stochastic and 2) paths are independent of each other, how might estimation results of completion times with respect to Case 1 and Case 2 differ one from the other? 35
Slide 51 - 5. PERT Simulation—running WinQSB Why simulation? Consider Case 2 Running WinQSB ▪ Open PERT/CPM > Select PERT > enter problem > Solve & Analyze > perform simulation ▪ The simulation input menu will drop defaulting to “random seed” with the estimated completion time of the critical path based on the standard method presented in 2b. ▪ Enter 10,000 for the # of simulated observations to be made. ▪ Enter the desired completion time … ▪ Click on the simulation button. ▪ View the results 36
Slide 52 - 5. PERT Simulation—running WinQSB Results: Foundry Project • independent paths/non-stochastic non-critical paths assumptions in play E(t) = 15; CP ≡ A – C – E – G - H; prob(X < 17) = 87.16% A-C-F-H: 9, σ2 = 2.11; A-C-E-G-H: 15, σ2 = 3.11; and B-D-G-H: 14, σ2 = 2.44 • independent paths/non-stochastic non-critical paths assumptions NOT in play Simulation E(t) = 15.19; prob(X < 17) = 86.10% A-C-F-H: 9, σ2 = 2.11; A-C-E-G-H: 15, σ2 = 3.11; and B-D-G-H: 14, σ2 = 2.44 (strength of the above assumptions if in play with Case 1 ????) • independent paths/non-stochastic non-critical paths assumptions NOT in play E(t) = 15.39; prob(X < 17) = 82.51% A-C-F-H: 9, σ2 = 2.11; A-C-E-G-H: 15, σ2 = 3.11; and B-D-G-H: 14, σ2 = 4.513 (strength of the above assumptions if in play with Case 2 ????) 37
Slide 53 - 5. PERT Simulation—running WinQSB Generalizations??? 38