answer the survey Quations .short 3 answers

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Selby’s Construction Data

ACTIVITY IMMED. PRED. TIME ESTIMATES EXPECTED VARIA- a m b TIME NCE

A ------- 1 2 9 3 1.78 B ------- 2.5 4.5 9.5 5 1.36 C ------- 2 5 14 6 4 D C 4 6.5 18 8 5.44 E A 2 4 18 6 7.11 F B, D, E 22 30 50 32 21.78 G F 15 20 37 22 13.44 H F 4.5 10 21.5 11 8.03 I G, H 12 15 24 16 4 J G 14 14.5 48 20 32.11 K I, J 5 5 5 5 0

Expected Project duration (m) = 93 weeks; Variance of the project (s2) = vC + vD + vF + vG + vJ + vK = 76.78; Standard Deviation (s ) = (76.78)1/2 = 8.76

# 2 (ii) What is the probability that the project will be completed in 100 days?

(100 -93)/(8.76) = 0.8

Z-Value = 0.2881

Thus, the probability

= 0.5 + 0.2881 = 0.7881

Probability of Completion

# 2 (ii) What is the probability that the project will be completed in 84 days?

(84-93)/(8.76) = -1.027

= -1.03

Z-Value = 0.3485

Thus, the probability =

0.5 - 0.3485 = 0.5-0.3485

= 0.1515 (Use the

Symmetric property of

The normal curve)

Probability of Completion

Selby Builders, Inc.; Linear Programming Formulation (CPM)

9 B = 4.5 F = 30 K = 5

Objective Function: Minimize Z = (Duration) = x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9

4 5

Selby Builders, Inc.; Linear Programming Formulation (CPM)

9 B = 4.5 F = 30 K = 5

Objective Function: Minimize Z = (Duration) = x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9

4 5

Subject to x2 – x1 > t A = 2 x4 – x1 > t B = 4.5

x3 – x1 > tC = 5 x4 – x2 > tE = 4 x4 – x3 > tD = 6.5

x5 – x4 > tF = 30 x7 – x5 > tG = 20 x6 – x5 > tH = 10

x6 – x7 > tDUM = 0 x8 – x6 > tI = 15 x8– x7 > tJ= 14.5 x9– x8 > tK= 5

Time and Cost Trade-Off

Crashing a Project Time (Wks) Cost ($)

Crash Cost Critical Activity Normal Crash Normal Crash Per Wk ($) Path?

A 7 5 50,000 62,000 6,000 No B 9 6 80,000 110,000 10,000 No C 10 9 40,000 45,000 5,000 Yes D 8 6 30,000 42,000 6,000 Yes

1 4

10 wk.

9 wk.

8 wk.

7 wk. B

CD: Critical path is longest path: 18 Wks.

u Nj = Normal time for activity j

u Nj / = Time for activity j under maximum crashing

u Mj = Maximum possible reduction in time for activity j due to crashing, i.e. , Nj - Nj

u Cj = Normal cost for activity j

u Cj / = Crash cost if the activity j is crashed to Nj

u Kj = Crash cost for activity j per unit time

u Linear Programming Formulation for an Optimal Crash schedule:

Min Kj yj subject to constraints of the project where yj = the crash applied to activity j to

minimize total crash cost

TIME and COST TRADE-OFF

Selby’s Time & Cost Schedule for Crashing (Use CPM Data of Prob. 1)

ACTIVITY NORMAL TIME UNDER NORMAL CRASH MAX CRASH TIME MAX. CRASH COST COST CRASH COST/ DAY

(Nj ) (Nj/ ) (Cj ) (Cj/ ) DAY (Mj ) (Kj )

A 2 1 500 1200 1 700 B 4.5 2.5 1200 2300 2 550 C 5 4 1500 2000 1 500 D 6.5 5 1500 3000 1.5 1000 E 4 2 1100 2900 2 900 F 30 20 15000 22000 10 700 G 20 17 12000 15000 3 1000 H 10 7 8000 9800 3 600 I 15 8 9000 13900 7 700 J 14.5 12 10000 12000 2.5 800 K 5 4 6000 6650 1 650

Selby Builders, Inc.; Designation of the Critical Path

B = 4.5 F = 30 K = 5

Enumeration of Possible Paths

A-E-F-H-I-K = 66 A-E-F-G-DUM-I-K = 76 A-E-F-G-J-K = 75.5

B-F-H-I-K = 64.5 B-F-G-DUM-I-K = 74.5 B-F-G-J-K = 74

C-D-F-H-I-K = 71.5 C-D-F-G-DUM-I-K = 81.5 C-D-F-G-J-K = 75.5

4 5

8 9

Problem #3 Problem #2 Problem #1

Linear Programming Formulation for an Optimal Crash schedule:

Min Kj yj subject to constraints of the project

(1) Maximum crash applied to an activity j will not exceed the allowable level yj.

(2) Time to get to a node has to be less than the time at the previous node plus the activity time – crash applied to the activity.

(3) The objective of achieving a target project completion date due to crashing must be achieved.

Linear Programming Formulation for Crashing Schedule

Min Kj yj where yj = the crash applied to activity j to minimize total crash cost

Minimize 700 yA + 550yB + 500yC + 1000yD + 900yE + 700yF + 1000yG + 600yH + 700yI + 800yJ + 650yK

subject to constraints of the maximum crashing time allowed for each activity.

(1) Maximum crash applied to an activity will not exceed the allowable level.

yA < 1; yB < 2 ; yC < 1; yD < 1.5; yE < 2; yF < 10; yG < 3; yH < 3; yI < 7; yK < 1; yDUM = 0

(1) Time to get to a node has to be less than the time at the previous node plus the activity time – crash applied to the activity.

(1) x2 – x1 > 2 – yA or yA + x2 – x1 > 2 (Node #2)

Linear Programming Formulation for Crashing Schedule (#3)

(2) yB + x4 – x1 > 4.5 (Node #4); (3) yE + x4 – x2 > 4 (Node # 4) (4) yB + x4 – x1 > 4.5 (Node #4); (5) yc + x3 – x1 > 5 (Node #3) (6) yF + x5 – x4 > 30 (Node #5); (7) yH+ x6 – x5 > 10 (Node #6); yDUM + x6 – x7 > 0 or x6 > x7 (Node #6); (8) yG + x7 – x5 > 20 (Node #7); yI + x8 – x6 > 15 (Node #8); yJ + x8 – x7 > 14.5 (Node #8); yK+ x9 – x8 > 5 (Node 9); x9 < 75 (Completing the project in 75 days or less)

Non-negativity constraints yA >0; yB > 0 ; yC > 0; yD > 0; yE > 0; yF > 0; yG > 0; yH > 0;

yI > 0; yK > 0; yDUM = 0; x1 > 0; x2 > 0; x3 > 0; x4 > 0; x5 > 0;

x6 > 0; x7 > 0 ; x8 > 0; x9 > 0;

Linear Programming Formulation for Crashing Schedule