Chapter 3, Problem 17P

Ross Hopkins, president of Hopkins Hospitality, has developed the tasks, durations, and predecessor relationships in the following table for building new motels. Draw the AON network and answer the questions that follow.

a) What is the expected (estimated) time for activity C?

b) What is the variance for activity C?

c) Based on the calculation of estimated times, what is the critical path?

d) What is the estimated time of the critical path?

e) What is the activity variance along the critical path?

f) What is the probability of completion of the project before week 36?

Summary Introduction

To determine: The expected time for activity C.

Introduction: In the activity on nodes (AON) project network diagram, the nodes denote activities and the arrows show only the precedence and succession sequence.

Answer to Problem 17P

The expected time for activity C is 12 weeks.

Explanation of Solution

Given information:

Activity	Immediate predecessor	Time estimates
		Optimistic	Most likely	Pessimistic
A		4	8	10
B	A	2	8	24
C	A	8	12	16
D	A	4	6	10
E	B	1	2	3
F	E,C	6	8	20
G	E,C	2	3	4
H	F	2	2	2
I	F	6	6	6
J	D,G,H	4	6	12
K	I,J	2	2	3

Calculation of expected time for activity C:

AON project network diagram shown below, given the precedence relationships of various activities,

Figure 1

With the given predecessor AON diagram is constructed. To Calculate the expected time for activity C given the data that the optimistic time estimate “a” of activity C of 8 weeks, the most likely time estimate “m” of 12 weeks and the pessimistic time estimate “b” of 16 weeks using the formula.

$\text{Expected}\text{time}=\frac{\text{a}+\text{4m}+\text{b}}{\text{6}}$ (1)

Where ‘a’ is the optimistic time estimate, ‘m’ is the most likely time estimate and ‘b’ is the pessimistic time estimate

Substitute into the equation (1), the values of $\text{a}=8$ weeks, $\text{m}=12$ weeks and $\text{b}=16$ weeks to obtain the expected time for activity C.

$\begin{array}{c}\text{t}=\frac{8+\left(4\times 12\right)+16}{6}\\ =\frac{72}{6}\\ =12\text{weeks}\end{array}$

Therefore the expected time for activity C is calcualted by adding 8, 48 and16 and dividing the sum with 6 which gives $12\text{weeks}$

Hence, the expected time for activity C is 12 weeks.

Summary Introduction

To calculate: The variance for activity C.

Answer to Problem 17P

The variance for activity C is 1.78 weeks.

Explanation of Solution

Given information:

Activity	Immediate predecessor	Time estimates
		Optimistic	Most likely	Pessimistic
A		4	8	10
B	A	2	8	24
C	A	8	12	16
D	A	4	6	10
E	B	1	2	3
F	E,C	6	8	20
G	E,C	2	3	4
H	F	2	2	2
I	F	6	6	6
J	D,G,H	4	6	12
K	I,J	2	2	3

Calculation of variance for activity C:

Calculate the variance for activity C using the formula

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

Substitute in equation (2) the values of $\text{b}=16$ weeks and $\text{a}=8$ weeks and obtain

$\begin{array}{c}{\text{σ}}^{\text{2}}={\left(\frac{16-8}{6}\right)}^2\\ =\frac{16}{9}\\ =1.78\end{array}$

The variance of activity C is calculated by squaring the value obtained by dividing the difference of 16 and 8 with 6 which gives the resultant as 1.28 weeks.

The variance in activity C is 1.78 weeks.

Summary Introduction

To determine: The critical path.

Answer to Problem 17P

The critical path AàCàFàHàJàK.

Explanation of Solution

Given information:

Activity	Immediate predecessor	Time estimates
		Optimistic	Most likely	Pessimistic
A		4	8	10
B	A	2	8	24
C	A	8	12	16
D	A	4	6	10
E	B	1	2	3
F	E,C	6	8	20
G	E,C	2	3	4
H	F	2	2	2
I	F	6	6	6
J	D,G,H	4	6	12
K	I,J	2	2	3

Calculation of critical path:

Using equation (1) and (2)calculate the expected times and variances for all the activitiesand obtain the values shown in table.

Figure 1, shows the AON diagram which gives the critical path. The critical path is AàCàFàHàJàK.

Hence, the critical path AàCàFàHàJàK.

Summary Introduction

To Compute: The estimated time of the critical path.

Answer to Problem 17P

The critical path AàCàFàHàJàK which is 40.18 weeks.

Explanation of Solution

Given information:

Activity	Immediate predecessor	Time estimates
		Optimistic	Most likely	Pessimistic
A		4	8	10
B	A	2	8	24
C	A	8	12	16
D	A	4	6	10
E	B	1	2	3
F	E,C	6	8	20
G	E,C	2	3	4
H	F	2	2	2
I	F	6	6	6
J	D,G,H	4	6	12
K	I,J	2	2	3

Calculation of estimated time of the critical path:

Figure 1, shows the AON diagram which gives the critical path. The critical path is AàCàFàHàJàK.

The various paths and the expected completion times are listed in the table

Various paths	Completion times
AàBàEàFàIàK	37.18 weeks
AàBàEàGàJàK	31.18 weeks
AàCàFàHàJàK	40.18 weeks
AàCàGàJàK	31.51 weeks
AàDàJàK	22.84 weeks

Hence, the estimated time of the critical path AàCàFàHàJàK is 40.18 weeks.

Summary Introduction

To determine: The variance of the critical path.

Answer to Problem 17P

The variance of the critical path is 10.03 weeks.

Explanation of Solution

Given information:

Activity	Immediate predecessor	Time estimates
		Optimistic	Most likely	Pessimistic
A		4	8	10
B	A	2	8	24
C	A	8	12	16
D	A	4	6	10
E	B	1	2	3
F	E,C	6	8	20
G	E,C	2	3	4
H	F	2	2	2
I	F	6	6	6
J	D,G,H	4	6	12
K	I,J	2	2	3

Calculation of variance of the critical path:

The activity variance σ²along the critical path AàCàFàHàJàK is the sum of the variances of individual activities A, C, F, H, J and K.

Therefore,

$\begin{array}{c}{\sigma }^2=1+1.78+5.44+0+1.78+0.03\\ =10.03\end{array}$

The activity variance along the critical path is the sum of 1, 1.78, 5.44, 0, 1.78, 0.03 is 10.03 weeks.

Hence, the variance of the critical path is 10.03 weeks

Summary Introduction

To compute: The probability of completion of the project before week 36 weeks.

Answer to Problem 17P

The probability of completion of the project before week 36 weeks is 9.34%.

Explanation of Solution

Given information:

Activity	Immediate predecessor	Time estimates
		Optimistic	Most likely	Pessimistic
A		4	8	10
B	A	2	8	24
C	A	8	12	16
D	A	4	6	10
E	B	1	2	3
F	E,C	6	8	20
G	E,C	2	3	4
H	F	2	2	2
I	F	6	6	6
J	D,G,H	4	6	12
K	I,J	2	2	3

Calculation of probability of completion of the project before week 36 weeks:

Use the normal distribution tables given that the estimated mean time is 40.18 weeks and the standard deviation is square root of 10.03 weeks.

First compute the standard deviation σ by computing the square root of 10.03 weeks.

$\begin{array}{c}\text{σ}=\sqrt{\text{10}\text{.03}}\\ =\text{3}\text{.17}\text{weeks}\end{array}$

Calculate the z value as shown below

$\begin{array}{c}\text{z}=\frac{\text{36}-\text{40}\text{.18}}{\text{3}\text{.17}}\\ =-\text{1}\text{.32}\end{array}$

For a z value of -1.32, the probability is 9.34%.

Therefore, there is a probability of 9.34% that the project may be completed on or before 36 weeks.

Hence, the probability of completion of the project before week 36 weeks is 9.34%.