Project Management with PERT/CPM

PERT/CPM PERT : program evaluation and review technique CPM : critical path method Use a project network, Activity-on-Node (AON): Nodes: activities, or tasks, to be performed Arcs: show immediate predecessors to an activity Times: duration times of activities are written next to the node

Reliable Construction Company Example Activity list for the Reliable Construction Co. project Activity Activity Description Immediate Predecessors Estimated Duration A Excavate - 2 weeks B Lay the foundation 4 weeks C Put up the rough wall 10 weeks D Put up the roof 6 weeks E Install the exterior plumbing F Install the interior plumbing 5 weeks G Put up the exterior siding 7 weeks H Do the exterior painting E,G 9 weeks I Do the electrical work J Put up the wallboard F,I 8 weeks K Install the flooring L Do the interior painting M Install the exterior fixtures N Install the interior fixtures K,L

The project network for the Reliable Construction Co. project START Activity Code A. Excavate B. Foundation C. Rough wall D. Roof E. Exterior plumbing F. Interior plumbing G. Exterior siding H. Exterior painting I. Electrical work J. Wallboard K. Flooring L. Interior painting M. Exterior fixtures N. Interior fixtures A 2 B 4 C 10 I D 6 E 4 7 G F 5 7 J 8 H 9 K 4 L 5 M 2 N 6 FINISH

Bake a Cake Example Task Immediate predecessors Task Time A: Buy frosting ingredients — ½ hr B: Clean up kitchen 1 hr C: Buy cake ingredients D: Prepare frosting A,B ¼ hr E: Prepare batter & bake B,C 2 hrs F: Frost cake D,E 1/2 A B D E F 1 1/4 2 Start C Finish

Critical Path A path through a project network is a route from START to FINISH. The length of path is the sum of the task times (durations) of the nodes (activities) on the path. The critical path is the longest path. The project duration is the length of the longest path.

The paths and lengths through Reliable’s project network START →A →B →C →D →G →H →M→FINISH START →A →B →C →E →H →M →FINISH START →A →B →C →E →F →J →K →N →FINISH START →A →B →C →E →F →J →L →N →FINISH START →A →B →C →I →J →K →N →FINISH START →A →B →C →I →J →L →N →FINISH 2 + 4 + 10 + 6 + 7 + 9 + 2 =40 weeks 2 + 4 + 10 + 4 + 9 + 2 =31 weeks 2 + 4 + 10 + 4 + 5 + 8 + 4 + 6 =43 weeks 2 + 4 + 10 + 4 + 5 + 8 + 5 + 6 =44 weeks 2 + 4 + 10 + 7 + 8 + 4 + 6 =41 weeks 2 + 4 + 10 + 7 + 8 + 5 + 6 =42 weeks

To Find the Critical Path and Slacks Work from top to bottom in the network, calculating ES = earliest start time for an activity EF = earliest finish time for an activity ES for activity i = largest EF of the immediate predecessors ES = 0 if no immediate predecessors EF = ES + activity duration time Work from bottom to top in the network, calculating LS = latest start time for an activity LF = latest finish time for an activity LS = LF – activity duration time LF for activity i = smallest LS of the immediate successors LF at Finish = EF at Finish if no immediate successors Slack = LF - EF = LS - ES If slack is zero, the activity is on the critical path.

The complete project network showing ES, LS, EF and LF for each activity of the baking example S = (ES, LS) F = (EF, LF) Slack = LS – ES = LF - EF S=( ) F=( ) Slack= Start S=( ) F=( ) Slack= ½ S=( ) F=( ) Slack= B 1 A S=( ) F=( ) Slack= ½ C S=( ) F=( ) Slack= ¼ D E 2 S=( ) F=( ) Slack= F ½ S=( ) F=( ) Slack= Finish S=( ) F=( ) Slack= Work down the network calculating ES and EF (ES at Start = 0, EF at Start = 0) Work backward up the network calculating LS and LF (LF and LS at Finish is EF and ES at Finish)

The complete project network showing ES, LS, EF and LF for each activity of the baking example S=(0, 0) F=(0, 0) Slack=0 S = (ES, LS) F = (EF, LF) Slack = LS – ES = LF - EF Start S=(0, 0) F=(1,1) Slack=0 ½ S=(0, 2 ¼ ) F=(½, 2 ¾ ) Slack=2 ¼ B 1 A S=(0, ½) F=(½, 1) Slack= ½ ½ C S=(1, 2 ¾) F=(1 ¼, 3) Slack=1 ¾ ¼ D E 2 S=(1, 1) F=(3, 3) Slack=0 F ½ S=(3, 3) F=(3 ½, 3 ½) Slack=0 Finish S=(3 ½, 3 ½) F=(3 ½, 3 ½) Slack=0 Critical Path is Start →B→E →F →Finish Activity D has slack of 1¾ hours (Start of D could be delayed without affecting total project duration) Also, activities A and C have slack of 2 ¼ and ½ respectively.

The complete project network showing ES, LS, EF and LF for each activity of the Reliable Construction Co. project S= (0, 0) F= (0, 0) START S = (ES, LS) F = (EF, LF) A S= (0, 0) F= (2, 2) 2 B 4 S= (2, 2) F= (6, 6) S= (6, 6) F= (16, 16) C 10 S= (16,20) F= (22,26) D S= (16, 16) F= (20, 20) 6 E 4 I S= (16,18) F= (23,25) 7 S= (22,26) F= (29,33) S= (20,20) F= (25,25) G 5 7 F S= (25,25) F= (33,33) J 8 S= (29,33) F= (38,42) H 9 S= (33,34) F= (37,38) K 4 L S= (33,33) F= (38,38) 5 S= (38,42) F= (40,44) M 2 N S= (38,38) F= (44,44) 6 S= (44,44) F= (44,44) FINISH

Slack for Reliable’s activities Activity Slack (LF - EF) On Critical Path? A Yes B C D 4 No E F G H I 2 J K 1 L M N

The spreadsheet used by MS project for entering the activity list for the Reliable Construction Co. project

Incorporate Uncertain Activity Duration Times (Probabilistic) PERT Three-Estimate Approach m = most likely estimate of activity duration time o = optimistic estimate of activity duration time p = pessimistic estimate of activity duration time Assume Beta distribution of activity time Approximately:

Expected value and variance of the duration of each activity for Reliable’s project Optimistic Estimate o Most Likely Estimate m Pessimistic Estimate p A 1 2 3 1/9 B 3 ½ 8 4 C 6 9 18 10 D 5 ½ E 4 ½ 5 4/9 F G 6 ½ 11 7 H 17 I 7 ½ J K L M N

The paths and path lengths through Reliable’s project network when the duration of each activity equals its pessimistic estimate Path Length START →A →B →C →D →G →H →M→FINISH START →A →B →C →E →H →M →FINISH START →A →B →C →E →F →J →K →N →FINISH START →A →B →C →E →F →J →L →N →FINISH START →A →B →C →I →J →K →N →FINISH START →A →B →C →I →J →L →N →FINISH 3 + 8 + 18 + 10 + 11 + 17 + 3 =70 weeks 3 + 8 + 18 + 5 + 17 + 3 =54 weeks 3 + 8 + 18 + 5 + 10 + 9 + 4 + 9 =66 weeks 3 + 8 + 18 + 5 + 10 + 9 + 7 + 9 =69 weeks 3 + 8 + 18 + 9 + 9 + 4 + 9 =60 weeks 3 + 8 + 18 + 9 + 9 + 7 + 9 =63 weeks

mean length of path µp = sum of mean activity times on the path For a path (typically the critical path), find the mean length (time) µp and the variance σp2 mean length of path µp = sum of mean activity times on the path (because E[X+Y] = E[X] + E[Y]) Variance of length of path σp2 = sum of variances of activity times on path (because we assume independence: Var [X+Y] = Var [X] + Var [Y] if X,Y independent) Example: critical path Start →A →B →C →E →F →J →L →N →Finish µp = 2 + 4 + 10 + 4 + 5 + 8 + 5 + 6 = 44 σp2 = 1/9 + 1 + 4 + 4/9 + 1 + 1 + 1 + 4/9 =9

Find the probability the project is completed in 47 weeks using an assumption of a Normal distribution

Time-Cost Trade-offs If one can expedite the project, use money/resources to reduce task times, what is the best way to allocate money? For an activity, could pay extra to reduce time (crash) How much would it cost to reduce total project duration from 44 weeks to 40 weeks? Which activities should be “crashed”? Could calculate crash cost and crash time for all possible paths – but can also apply LP! Crash cost Normal cost Crash time Normal time Activity duration time $ assume linear

Crash Cost per Week Saved Time-cost trade-off data for the activities of Reliable’s project Activity Time Cost Maximum Reduction in Time Crash Cost per Week Saved Normal Crash A 2 weeks 1 week $180,000 $280,000 $100,000 B 4 weeks $320,000 $420,000 $50,000 C 10 weeks 7 weeks $620,000 $860,000 3 weeks $80,000 D 6 weeks $260,000 $340,000 $40,000 E $410,000 $570,000 $160,000 F 5 weeks G $900,000 $1,020,000 H 9 weeks $200,000 $380,000 $60,000 I $210,000 $270,000 $30,000 J 8 weeks $430,000 $490,000 K L $250,000 $350,000 M N $330,000 $510,000 If do all tasks normal, 44 weeks, cost is 4.55 million. If do all tasks crash, 28 weeks, cost is 6.15 million.

Marginal Cost Analysis For small networks, may reduce the project 1 week at a time, and observe the changes. Activity to Crash Crash Cost Length of Path ABCDGHM ABCEHM ABCEFJKN ABCEFJLN ABCIJKN ABCIJLN 40 31 43 44 41 42 J $ 30,000 39 F $ 40,000

Linear Programming to Make Crashing Decisions Let Z = total cost of crashing on any activity xj = reduction in the duration of activity j due to crashing j = A,B,C,…,N xj ≤ maximum reduction time = normal time – crash time yFINISH= project duration, time at which FINISH node is reached yj = start time of activity j yj ≥ yi + normal timei – xi i is an immediate predecessor of j F I J yJ yJ ≥ yF + 5 - xF yJ ≥ yI + 7 - xI yF 5 yI 7

Linear Programming Model

The schedule of cumulative project costs when all activities begin at their earliest start times or at their latest start times

Time-cost trade-off data for the activities of the baking project Activity Normal Time Crash Time Normal Cost Crash Cost A: Buy frosting 0.5 0.25 5 B: Clean kitchen 1.0 10 C: Buy cake D: Prepare frosting E: Prepare batter bake 2.0 1.5 F: Frost cake

The project network showing Normal Time and Crash Time for each activity of the baking project Start N: Normal time C: Crash time If all are normal: ADF 1 ¼ BDF 1 ¾ BEF 3 ½ CEF 3 Total time is 3 ½. If all are crashed: ADF ¾ BDF ¾ BEF 2 CEF 2 Total time is 2. N C 1 , 1/4 N C 1/2 , 1/4 A C N C 1/2 , 1/4 B D E N C 1/4 , 1/4 N C 2 , 1 1/2 F N C 1/2, 1/4 Finish

Could solve the crash LP for finish times between 3 Could solve the crash LP for finish times between 3.5 and 2 to evaluate alternatives 25 16.67 Cost 11.67 8.33 5 2.5 2 2.25 2.5 2.75 3 3.25 3.5 T= y Finish

Lasagna Dinner Example Task Tasks that must precede Time A: Buy the mozzarella cheese 30 mins B: Slice the mozzarella A 5 mins C: Beat 2 eggs 2 mins D: Mix eggs and ricotta cheese C 3 mins E: Cut up onions and mushrooms 7 mins F: Cook the tomato sauce E 25 mins G: Boil large quantity of water 15 mins H: Boil the lasagna noodles G 10 mins I: Drain the lasagna noodles H J: Assemble all the ingredients I, F, D, B K: Preheat the oven L: Bake the lasagna J, K

Construct project network Start E 7 G 15 A C 2 K 30 15 H 10 F 25 5 B D 3 I 2 A→B →J →L 75 * C→D →J →L 45 E→F →J →L 73 G→H →I →J →L 45 J 10 L 30 Finish

Start E 7 G 15 30 A C 2 K 15 H 10 D 3 F 25 5 B I 2 J 10 L 30 Finish EF = ES + activity time (or duration) if no predecessors, ES = 0; otherwise ES = max (EF) (immediate predecessors) work forward through network Start ES=0EF=30 ES=0EF=2 ES=0EF=7 ES=0EF=15 E 7 G 15 30 A C ES=0EF=15 2 K 15 H 10 ES=15EF=25 ES=7EF=32 ES=30EF=35 D 3 ES=2EF=5 F 25 5 B ES=25EF=27 I 2 ES=35EF=45 J 10 ES=45EF=75 L 30 Finish

The complete project network showing ES, LS, EF, LF and Slack for each activity of the Lasagna dinner example S = (ES, LS) F = (EF, LF) Slack = LF-EF=LS-ES Start S=(0,0) F=(30,30) Slack=0 30 A C S=(0,30)F=(2,32) slack=30 S=(0,3) F=(7,10) Slack=3 S=(0,10) F=(15,25) Slack=10 S=(0,30) F=(15,45) Slack=30 2 E 7 G 15 K 15 S=(30,30) F=(35,35) Slack=0 S=(15,25) F=(25,35) Slack=10 5 B S=(2,32) F=(5,35) slack=30 S=(7,10) F=(32,35) Slack=3 D F 25 H 10 3 S=(25,33) F=(27,35) Slack=8 I 2 S=(35,35) F=(45,45) Slack=0 J 10 S=(45,45) F=(75,75) Slack=0 L 30 Critical path has zero slack: A→B →J →L Finish

Because of a phone call, you will delayed by 6 minutes to cut onions and mushrooms (Task E) . By how much will dinner be delayed? slack is 3, delay of 6 minutes will delay dinner by 6-3=3 If you use your food processor instead to reduce cutting time from 7 minutes to 2 minutes, will dinner still be delayed? ES=0 LS=8 LS=2 LF=10 E 2 slack = 8, so phone of 6 won’t delay dinner

All of the critical path, ES, LS, EF, LF, are based on estimates of the activity times. How can you incorporate uncertainty into planning? PERT 3 – estimate approach Most Likely Estimate (m) most probable event Optimistic Estimate (σ) if everything goes perfectly Pessimistic Estimate (p) if everything goes wrong

What is the probability of meeting your deadline? Assume the distribution for activity time is a Beta distribution density f(t) t: time µ - 3σ µ + 3σ µ ± 3σ interval captures 99.73% of distribution

Consider Critical Path μ= 30 Buy mozzarella cheese σ2 = (6 2/3)2 o=10 m=30 p=50 B μ= 5 Slice cheese σ2 = (11/3)2 o=3 m=4 p=11 J μ= 10 Assemble σ2 = (2)2 = 4 L μ= 30 Bake σ2 = (3 2/3)2 o=20 m=30 p=40

Could calculate pessimistic length: 50 + 11 + 14 + 40 = 115 Longest pessimistic path may not be the critical mean path The mean length is sum of means: 30 + 5 + 10 + 30 = 75 = µp Assume all task times are independent, variance for path is sum of variances:

Assume distribution of path time is normal (central limit theorem if lots of tasks on a path) 67 75 83