Chapter 3, Problem 20P

The estimated times and immediate predecessors for the activities in a project at George Kyparis’s retinal scanning company are given in the following table. Assume that the activity times are independent.

a) Calculate the expected time and variance for each activity.

b) What is the expected completion time of the critical path? What is the expected completion time of the other path in the network?

c) What is the variance of the critical path? What is the variance of the other path in the network?

d) If the time to complete path A–C is normally distributed, what is the probability that this path will be finished in 22 weeks or less?

e) If the time to complete path B–D is normally distributed, what is the probability that this path will be finished in 22 weeks or less?

f) Explain why the probability that the critical path will be finished in 22 weeks or less is not necessarily the probability that the project will be finished in 22 weeks or less.

Summary Introduction

To determine: The expected time and variance.

Introduction:

The activity times of various tasks in a PERT project network are assumed to follow a probability distribution. For calculating the expected times and standard deviation, the parameters required are the following:

• Optimistic time: This is the time denoted by “a”, which is the best possible or in other words, the quickest time in which an activity can be completed, assuming that everything goes as per plan.
• Pessimistic time: This is the worst case scenario, where all the conditions are adverse or unfavorable. The maximum time which may be taken in such a situation is the pessimistic time denoted by “b”.
• Most likely time: The most realistic estimate of the time, denoted by “m” in normal conditions, is another parameter which is important in the computation of expected times and variances.

Answer to Problem 20P

The expected times and varianceare shown in Table 1.

Explanation of Solution

Given information:

Activity	Immediate predecessor	Time in weeks
		a	m	b
A		9	10	11
B		4	10	16
C	A	9	10	11
D	B	5	8	11

Formula:

Calculate the expected time t and the variances for each activity by using the equations

$\text{t}=\frac{\left(\text{a}+\text{4m}+\text{b}\right)}{\text{6}}$ (1)

where ‘a’ is the optimistic time, ‘m’ is the most likely time and ‘b’ is the most pessimistic time.

$\text{Variance}={\left[\frac{\left(\text{b}-\text{a}\right)}{6}\right]}^2$ (2)

Calculation of expected time and variance:

The calculations are shown below.

Activity	Immediate predecessor	Time in weeks			t	Variance
		a	m	b
A		9	10	11	10	0.11
B		4	10	16	10	4
C	A	9	10	11	10	0.11
D	B	5	8	11	8	1

Table 1

Excel worksheet:

Hence, the expected times and variance are shown in Table 1.

Summary Introduction

To determine: The  expected completion of critical path and other path (non-ciritical paths).

Answer to Problem 20P

The critical path is AàC and its expected completion time is 20 weeks. The expected completion time of the other path BàD is 18 weeks.

Explanation of Solution

Given information:

Activity	Immediate predecessor	Time in weeks
		a	m	b
A		9	10	11
B		4	10	16
C	A	9	10	11
D	B	5	8	11

Calculation of expected completion of critical path and other path (non-ciritical paths):

AON diagram is constructed which is shown below.

The critical path is AàC and expected completion time is 20 weeks. The expected completion time of the other path BàD is 18 weeks.

Hence, the critical path is AàC and its expected completion time is 20 weeks. The expected completion time of the other path BàD is 18 weeks.

Summary Introduction

To determine: The  variance of critical path and other path (non-ciritical paths).

Answer to Problem 20P

The variance of the critical path AàC is 0.222 weeks and other path BàD is5 weeks.

Explanation of Solution

Given information:

Activity	Immediate predecessor	Time in weeks
		a	m	b
A		9	10	11
B		4	10	16
C	A	9	10	11
D	B	5	8	11

Calculation of variance of critical path and other path:

The variance of the critical path AàC is the sum of the variances of activities A and C.

$\begin{array}{l}=0.1111+0.1111\\ =0.2222\end{array}$

The variance of the critical path is the sum of 0.1111 and 0.1111 (refer table 1) which is 0.2222 weeks.

The variance of the other path BàD is the sum of the variances of activities B and D.

$\begin{array}{l}=4+1\\ =5\end{array}$

The variance of the other path is the sum of 4 and 1 (refer table 1) which is 5 weeksThe variance of the other path BàD is 5 weeks.

Hence, the variance of the critical path AàC is 0.222 weeks and other path BàD is 5 weeks.

Summary Introduction

To determine: The probability of finishing the project in 22 weeks, when A-C is normally distributed.

Answer to Problem 20P

Probability of finishing the project in 22 weeks is almost 1.

Explanation of Solution

Given information:

Activity	Immediate predecessor	Time in weeks
		a	m	b
A		9	10	11
B		4	10	16
C	A	9	10	11
D	B	5	8	11

Probability of finishing the project in 22 weeks, when A-C is normally distributed:

The expected completion time of critical path AàC is 20 weeks with a variance of 0.22 weeks.

The standard deviation:

$\begin{array}{c}\text{σ}=\sqrt{0.22}\\ =0.469\end{array}$

The standard deviation is calculated by taking square root of the variance which is 0.22 which yields 0.469.

$\begin{array}{c}\text{z}=\frac{22\text{weeks}-20\text{weeks}}{0.469}\\ =4.264\end{array}$

Z value is calculated by dividing the difference of 22 and 20 with 0.469 which yields 4.264.

Reading from normal distribution tables given the mean is 20 weeks, the variance $\sigma =0.469$ and $Z=4.264$, the area under the probability distribution curve is almost 1. Therefore the probability that the project along the critical path AàC would be completed in 22 weeks is almost equal to 1.

Hence, probability of finishing the project in 22 weeks is almost 1.

Summary Introduction

To determine: The probability of finishing the project in 22 weeks, when B-D is normally distributed.

Answer to Problem 20P

Probability of finishing the project in 22 weeks is 0.963.

Explanation of Solution

Given information:

Activity	Immediate predecessor	Time in weeks
		a	m	b
A		9	10	11
B		4	10	16
C	A	9	10	11
D	B	5	8	11

Probability of finishing the project in 22 weeks, when B-D is normally distributed:

The expected completion time of path BàD is 18 weeks with a variance of 5 weeks.

The standard deviation:

$\begin{array}{c}\sigma =\sqrt{5}\\ =2.236\end{array}$

The standard deviation is calculated by taking square root of the variance which is 5 which yields 2.236.

$\begin{array}{c}\text{z}=\frac{22\text{weeks}-18\text{weeks}}{2.236}\\ =1.79\end{array}$

Z value is calculated by dividing the difference of 22 and 18 with 2.236 which yields 1.79.

Reading from normal distribution tables given the mean is 18 weeks, the variance $\sigma =2.236$ and $z=1.79$, the area under the probability distribution curve is 0.963.

Hence, the probability of finishing the project in 22 weeks is 0.963.

Summary Introduction

To explain: The reason for the probability that the critical path will be finished in 22 weeks or less is not necessarily the probability that the project will be finished in 22 weeks or less.

Explanation of Solution

Given information:

Activity	Immediate predecessor	Time in weeks
		a	m	b
A		9	10	11
B		4	10	16
C	A	9	10	11
D	B	5	8	11

Explanation for the reason for the probability that the critical path will be finished in 22 weeks or less is not necessarily the probability that the project will be finished in 22 weeks or less:

The critical path AàC has a very small variance of only 0.22 weeks compared to the variance along path BàD which is 5 weeks. That is why, despite having lower expected time of completion of 18 weeks, the probability that the path BàD would be completed in 22 weeks is0.963.and is lower than the probability $\simeq 1$ of project completion in 22 weeks along path AàC.

The project is completed only when all the activities are completed.

Despite the probability of the critical path AàC, being completed in 22 weeks being almost 1, the probability that the project is completed in 22 weeks is0.963.