Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences)
Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences)
12th Edition
ISBN: 9781259580093
Author: William J Stevenson
Publisher: McGraw-Hill Education
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Chapter 17, Problem 12P

A project manager has compiled a list of major activities that will be required to install a computer information system in her firm. The list includes estimated completion times for activities and precedence relationships.

Chapter 17, Problem 12P, A project manager has compiled a list of major activities that will be required to install a

a. Construct a network diagram. You can use either AOA or AON (see Example 5).

b. If the project is finished within 26 weeks of its start, the project manager will receive a bonus of $1,000; and if the project is finished within 27 weeks of its start, the bonus will be $500. Find the probability of each bonus.

a)

Expert Solution
Check Mark
Summary Introduction

To construct: A network diagram.

Answer to Problem 12P

Answer

Network diagram:

Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences), Chapter 17, Problem 12P , additional homework tip  1

Explanation of Solution

Given information:

Activity Immediate predecessor Optimistic time Most likely time Pessimistic time
A 2 4 6
D A 6 8 10
E D 7 9 12
H E 2 3 5
F A 3 4 8
G F 5 7 9
B 2 2 3
I B 2 3 6
J I 3 4 5
K J 4 5 8
C 5 8 12
M C 1 1 1
N M 6 7 11
O N 8 9 13
End H, G, K, O

Formula to calculate expected time and variance:

Expected time=Optimistic time+(4×Most likely time)+Pessimistic time6

Standard deviation=Pessimistic timeOptimistic time6

Variance=(Pessimistic timeOptimistic time)262

Calculation of mean and variance:

Activity Optimistic time Most likely time Pessimistic time Expected time Standard deviation Variance
A B C (A+(4*B)+C)/6 (C-A)/6 (C-A)^2/6^2
A 2 4 6 4 0.667 0.444
D 6 8 10 8 0.667 0.444
E 7 9 12 9.17 0.833 0.694
H 2 3 5 3.17 0.500 0.250
F 3 4 8 4.5 0.833 0.694
G 5 7 9 7 0.667 0.444
B 2 2 3 2.17 0.167 0.028
I 2 3 6 3.33 0.667 0.444
J 3 4 5 4 0.333 0.111
K 4 5 8 5.33 0.667 0.444
C 5 8 12 8.17 1.167 1.361
M 1 1 1 1 0.000 0.000
N 6 7 11 7.5 0.833 0.694
O 8 9 13 9.5 0.833 0.694
End

Network diagram:

Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences), Chapter 17, Problem 12P , additional homework tip  2

The activity on arrow diagram is constructed. The activities are mentioned on the arrows. The arrows in the diagram are mentioned in the precedence order. The arrows indicates the successor for each node.

b)

Expert Solution
Check Mark
Summary Introduction

To determine: The probability of each bonus.

Answer to Problem 12P

The probability of achieving a bonus of $1,000 is 0.4099. The probability of achieving a bonus of $500 is 0.6742.

Explanation of Solution

Given information:

  • If the project is finished in 26 weeks the bonus is $1,000.
  • If the project is finished in 27 weeks the bonus is $500.

Calculation of expected duration, variance and standard deviation for each path:

Path A-D-E-H:

Expected duration=4+8+9.17+3.17=24.34

Variance=0.444+0.444+0.694+0.250=1.832

Standard deviation=Variance=1.832=1.3535=1.354

Path A-F-G:

Expected duration=4+4.5+7=15.50

Variance=0.444+0.694+0.444=1.582

Standard deviation=Variance=1.582=1.2577=1.258

Path B-I-J-K:

Expected duration=2.17+3.33+4+5.33=14.83

Variance=0.028+0.444+0.111+0.444=1.027

Standard deviation=Variance=1.027=1.014

Path C-M-N-O:

Expected duration=8.17+1+7.50+9.50=26.17

Variance=1.361+0.0+0.694+0.694=2.749

Standard deviation=Variance=2.749=1.658

Bonus of $1,000 (less than or equal to 26 weeks):

Calculation of probability of completion in less than or equal to 26 weeks:

Calculation of z value for all paths:

Formula:

Z=Specified time-Path meanPath standard deviation

Path A-D-E-H:

Z=26-24.341.354=1.23

From the standard normal distribution table,

The probability value for (z = 1.23) is 0.8907.

Path A-F-G:

Z=26-15.501.258=8.35

Since z value is greater than +3.00, probability of completion is 1.0000.

Path B-I-J-K:

Z=26-14.831.014=11.02

Since z value is greater than +3.00, probability of completion is 1.0000.

Path C-M-N-O:

Z=26-26.171.658=0.10

From the standard normal distribution table,

The probability value for (z = -0.10) is 0.4602.

Probability of completion in less than or equal to 26 weeks:

Probability=0.8907×1.0000×1.0000×0.4602=0.4099

The probability of achieving a bonus of $1,000 is 0.4099.

Bonus of $500 (less than or equal to 27 weeks):

Calculation of probability of completion in less than or equal to 27 weeks:

Calculation of z value for all paths:

Formula:

Z=Specified timePath meanPath standard deviation

Path A-D-E-H:

Z=27-24.341.354=1.96

From the standard normal distribution table,

The probability value for (z = 1.96) is 0.9750.

Path A-F-G:

Z=27-15.501.258=9.14

Since z value is greater than +3.00, probability of completion is 1.0000.

Path B-I-J-K:

Z=27-14.831.014=12

Since z value is greater than +3.00, probability of completion is 1.0000.

Path C-M-N-O:

Z=27-26.171.658=0.50

From the standard normal distribution table,

The probability value for (z = 0.50) is 0.6915.

Probability of completion in less than or equal to 27 weeks:

Probability=0.9750×1.0000×1.0000×0.6915=0.6742

The probability of achieving a bonus of $500 is 0.6742.

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Chapter 17 Solutions

Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences)

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