Chapter 9

Project Management CPM and PERT

9.1 Introduction A. Each project is consist of numerous jobs or tasks performed by a variety of departments or individuals B. Project management — Identify project goals or objectives — Work breakdown structure — Establish activities relationship — Estimate activities times — Estimate project completion time — Post completion evaluation — Resource allocations C. Gantt chart

— Disadvantage of Gantt chart — Unable to reflect the interrelation among activities — No revelation of the delay of activity on other activities D. Disadvantage of Gantt chart a. PERT (Program Evaluation and Review Technique) : Probabilistic model (stochastical model) b. CPM (Critical Path Model) : deterministic model E. Nations in PERT a. Activity : effort need resource and time b. Event : milestones or checkpoints c. Project : a collection of activity d. Network : a logical and chronological set of activities and events e. Critical activity f. Path

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g. Critical path — Advantage of PERT/CPM — Commitments and communications — Detailed planning — Efficient monitoring and control — Identify potential problem areas — Proper use of resources — Rescheudling — Government requirement — Easily understood — Adaptive to computers — Decision making tools — Cost-time trade-off F. PERT/CPM process a. phase1 : formulation 1.analysis of the project 2.sequence the activities 3.estimate activity times and costs b. phase2 : solution 1.construct the network 2.event analysis 3.activity analysis c. Analysis and control 1.Monitoring and control 2.Resource utilization G. The network representation a. Activity (operation) b. Node (event) : the beginning and ending of each activity

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1 1

C

A

2

A

B

2 B

B

C C

1 A

A

C

B

A B 2

Ex : Activity A B C D E F G H I J

4

D

time 3 5 3 4 8 2 4 2 5 3

E

5

8

H

A

Prerequired B A,C D C F F B E,G,H

1

3

G

6

7

C

B

F

2

I

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J 9

9.2 CPM A. Find the “longest path” – “critical path” — forward pass ES i : the Earliest start time of activity i EFi : the Earliest finish time of activity i ES j = max[ES i + t i ] = max[EFi ] i

EF j = ES j + t j

— backward pass LS i : the Latest start time of activity i LFi : the Latest finish time of activity i

[

]

[ ]

LFi = min LFi − t j = min LS j j

LS i = LFi + t i — slack of free time (1)Total Float (TF) : slack for on event (2)Free Float (FF) : slack for a activity TFi = LS i − ES i = LFi − EFi FFi = ES j − ES i − t i = ES j − EFi Link (1,4) (1,2) (2,3) (4,5) (5,7) (3,6) (6,7) (8,6) (2,9) (7,9)

Activity A B C D E F G H I J

ES 0 0 5 8 12 8 10 10 5 20

EF 3 5 8 12 20 10 14 12 10 23

LS 5 0 5 8 12 14 16 18 18 20

LF 8 5 8 12 20 16 20 20 23 23

TF 5 0 0 0 0 6 6 8 13 0

B. Observations — TFi = 0 ⇒ FFi = 0 — FFi = 0 ≠ — if TFi ≠ 0

TFi = 0 noncritical activity

— if FFi ≤ TFi (noncritical event) i can be delayed as much as FFi

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FF 5 0 0 0 0 0 6 8 13 0

9.3 PERT (for uncertain activity time) A. Assumption — Time estimate for each activity is based on three difference aspects — Based on Beta distribution assumption (1) a: optimistic time (minimum time) (2) b: pessimistic time (maximum time) (3) m: most probable time (normal time) B. Average Time (expected time) — based on Beta probability distribution ( a + b) + 2m a + 4m + b 2 (1) E = = 3 6 (2) σ 2 = (

b−a 2 ) 6

— treat the expected time as a fixed time proceed with CPM — if there are more than one critical paths identify the one with the largest variance C. Compute the E[U i ] and Var[U i ] — U i = the earliest occurrence time of event — assume the activities are statistically independently

E[U i ] = ES i = ∑ E k

Var[U i ] = ∑ σ 2 k

— where k = the activities along the longest path leading to i — if we specify a certain project completion time ( STi ) , we can computes the probability of meeting this dead line ⎡U − E (U i ) STi − E (U i ) ⎤ * ≤ P(U i ≤ STi ) = P ⎢ i ⎥=PZ≤Z Var (U i ) ⎦⎥ ⎣⎢ Var (U i )

[

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]

Example: Activity A B C D E F G H I J E=

a 1 3 2 2 4 1 2.5 1 4 1.5

m 3 4.5 3 4 7 1.5 3.5 2 5 3

b 5 9 4 6 16 5 7.3 3 6 4.5

E 3 5 3 4 8 2 4 2 5 3

S2 4/9 1 1/9 4/9 4 4/9 25/36 1/9 1/9 1/4

S 2/3 1 1/3 2/3 2 2/3 5/6 1/3 1/3 1/2

a + 4m + b 1 + 4 * 3 + 5 = =3 6 6 2

2

⎛2⎞ ⎛b−a⎞ ⎟ =⎜ ⎟ ⎝3⎠ ⎝ 6 ⎠ E (U i ) = 5 + 3 + 4 + 8 + 3 = 23

σ2 =⎜

σ 2 (U i ) = 1 + 1 / 9 + 4 / 9 + 4 + 1 / 4 = 5.836 S (U i ) = 2.41 STi = 23 STi = 25

⎡ ST − E (U i ) ⎤ P ⎢Z ≤ i ⎥ = P( Z ≤ 0) = 0 .5 S U ( ) i ⎣ ⎦ 25 − 23 ⎤ ⎡ P ⎢Z ≤ = P ( Z ≤ 0.83) = 0.7963 2.41 ⎥⎦ ⎣

9.4 Considering Time-Cost Trade-Offs A. The duration of activity can be shorten by adding resource to that activity B. The need to identify the least cost to crash activity time to meet the desired date C. Notation — T j = normal time for activity j — T j = time for activity j under maximum crashing '

— C j = cost for activity j under maximum crashing — K j = crash cost for activity j Kj =

C 'j − C j T j − T j'

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Cost (4,900)

(8,500)

Time Kj =

C 'j − C j Tj − T

' j

=

900 − 500 = 100 8−4

D. Observation — other path may be critical — check the revised network E. LP model for crashing consideration — Variables — X i = the occurrence of event i — Yi =

amount of crash time used for activity j

— objective function — min ∑ K j Y j j

— constraints (1)describe the network (2)limiting the activity crash time (3)meeting the project complete time

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4 4 3

［0,3］ ［5,8］ 1

5

8

5

［12,20］ ［12,20］

7 4

［10,14］ 1 ［8,10］ ［16,20］ 3 6 ［14,16］2 ［10,12］ 3 ［18,20］

［5,8］ ［5,8］

［0,5］ ［0,5］

［8,12］ ［8,12］

2

8

3

［20,23］ ［20,23］

［5,10］ ［18,23］

9 5 min K AYA + K B YB + K C YC + K DYD + K E YE + K F YF + K G YG + K H YH + K I YI + K j Y j X 2 ≥ TB − YB + X 1 st

X 3 ≥ TC − YC + X 2 X 4 ≥ TA − YA + X 1 X 6 ≥ TF − YF + X 3 X 7 ≥ T E − YE + X 5 X 7 ≥ TG − YG + X 6 X 7 ≥ TH − YH + X 8 X 9 ≥ TJ − YJ + X 7 X9 X4 X6 X5

≥ TI − Y I + X 2 ≥ X3 ≥ X8 ≥ T0 − Y0 + X 4 Activity A B C D E F G H I J Yi ≤ Ti

∀i

YA ≤ 2

Y0 ≤ 1

Prerequired B A,C D C F F B E,G,H

X 9 ≤ 20

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time 3 5 3 4 8 2 4 2 5 3

9.5 PERT/COST A. Concentrate on time aspect of a project B. Budget cost — identify cost — project when cost occurs C. Assumption — cost occurs at a constant rate D. Procedure — forecast the project cost based on the earliest and latest start time — divide the cost uniformly into unit-time cost the shadow area between two cost curves shows the feasible budget

Ex.

for earliest start Budget cost for an earliest starting pate schedule Month Activity 1 2 3 4 5 A 5 5 B 10 10 10 C 3 D 2 2 E 10 10 F G Monthly cost 15 15 13 12 12 Total cost 15 30 43 55 67

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6

7

8

5

5

5 82

5 87

2

8 10 77

Budget cost for an latest starting pate schedule Month Activity 1 2 3 4 5 A 5 5 B 10 10 10 C D 2 2 E F G Monthly cost 10 10 10 7 7 Total cost 10 20 30 37 44

6

3 2 10

15 49

Budgeted cost for an earliest starting date (Schedule)

Cost

Budgeted cost for an latest starting date (Schedule)

Possible Budgets for total project cost

Months

9.6 Project Control A. Purpose a. Identify the actual of activities b. Periodically information collections — Activity cost update — Percent completion to date B. Objective — Identify cost overrun and underruns take necessary correction actions — Notations — Vi = value of work completed for activity i — Pi = percentage completed for activity i — Bi = budget cost activity i

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7

10 5 15 64

8

5 8 23 87

— Vi =

( Pi / 100) Bi

— Cost overruns — ACi = actual cost to date of activity i — Di = difference between actual cost and value of work completed for activity i — Di = AC i − Vi Vi = completion%* Bi Activity

AC i 12000 3000 1000 2000 10000 0 0 55000

A B C D E F G

6500/48500=13.4%

completion% 100 100 50 33 25 0 0

SERIOUS

Bi 10000 30000 3000 6000 20000 5000 4000

Vi 10000 30000 1500 2000 5000 0 0 485000

PROBLEM

— Identify cost over runs A,E — A is completed and E is only 25%then adjust activity E — Drawback — need tremendous clerical efforts — cost allocation problem — need different cost accounting system

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Di 2000 0 -500 0 5000 0 0 6500

例. 競標 工程代號 A B C D E F G H I J K L M N

工作情況 事先準備 通知財產所有人 選擇有關競標人 與競標人聯絡 發出競標通知 取得競標者保薦書 取得財產所有權狀 等競標人準備競標 取得競標 評估競標 準備工作分配 與自來水公司取得聯絡 通知已入選的競標者 商恰簽約核定事宜 完成合約 E

2

3

2

4

H 10

5

2

1

9 K

1

2

8

4

L 4

G

B

14

2

7

ES i = max{ES k + Dk ,i } k

EFij = ES j + Di , j

[

6

5

D

LFi = min LFi − Di , j

C A, D E B F H I G, J G, J K L, M N

J

5

C

前置作業

I

4

A

0

3

F

工作天數 5 2 4 4 3 2 14 10 2 5 2 4 1 2 0

]

LS i = LFi − Di ,i

TFi = LS i − ES i = LFi − EFi FFi = ES j − ES i − t i = ES j − EFi

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M

10

N 2

11

作業代號 A B C D E F G H I J K L M N

0-2 0-7 0-1 1-2 2-3 3-4 7-8 4-5 5-6 6-8 8-9 8-10 9-10 10-11

工作天數 ES i

EFij

LS ij

LF j

TF

FF

5 2 4 4 3 2 14 10 2 5 2 4 1 2

5 2 4 8 11 13 16 23 25 30 32 34 33 36

3 14 0 4 8 11 16 13 23 25 31 30 33 34

8 16 4 8 11 13 30 23 25 30 33 34 34 36

3 14 0 0 0 0 14 0 0 0 1 0 1 0

3 0 0 0 0 0 14 0 0 0 0 0 1 0

0 0 0 4 8 11 2 13 23 25 30 30 32 34 E

2

3

F

4

H

I 5

6 9 J

A

0

C

1

M

K

D 8

L

10

G

B 7

要徑：C-D-E-F-H-I-J-L-N （0-1-2-3-4-5-6-8-10-11） 9.7 Resource Utilization A. purpose ─resource smoothing ─allocate resource for smooth usage ─resource leveling ─reduce peak resource requirements ─resource allocation ─for a desired resource level, find the shortest possible project schedule B. Trial and error heuristic approach ─Step1：resource accumulation by the earliest start schedule ─Step2：draw the Gantt chart ─Step3：start from the last nonciritical activity not considered move it in total float period ─Strp4：if no more noncritical activity to be considered stop, otherwise go to step3

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N

11