Chapter 17, Problem 11P

The following precedence diagram reflects three time estimates in weeks for each activity. Determine:

a. The expected completion time for each path and its variance.

b. The probability that the project will require more than 49 weeks.

c. The probability that the project can be completed in 46 weeks or less.

Summary Introduction

To determine: The expected completion time for each path and its variance.

Explanation of Solution

Given information:

Activity	Optimistic time	Most likely time	Pessimistic time
1 to 2	8	8	8
2 to 3	11	12	13
2 to 4	5	6	7
2 to 5	10	12	14
3 to 8	9	10	12
4 to 6	14	18	26
4 to 7	13	13	13
5 to 9	7	10	12
6 to 11	8	10	14
7 to 11	10.5	13	15.5
8 to 11	5	7	10
9 to 10	10	11	12
10 to 11	6	6	6

Formula to calculate expected time and variance:

$\text{Expected time}=\frac{\text{Optimistic time}+\left(4\times \text{Most likely time}\right)+\text{Pessimistic time}}{6}$

$\text{Standard deviation}=\frac{\text{Pessimistic time}-\text{Optimistic time}}{6}$

$\text{Variance}=\frac{{\left(\text{Pessimistic time}-\text{Optimistic time}\right)}^2}{6^2}$

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
1 to 2	8	8	8	8	0.000	0.000
2 to 3	11	12	13	12	0.333	0.111
2 to 4	5	6	7	6	0.333	0.111
2 to 5	10	12	14	12	0.667	0.444
3 to 8	9	10	12	10.17	0.500	0.250
4 to 6	14	18	26	18.67	2.000	4.000
4 to 7	13	13	13	13	0.000	0.000
5 to 9	7	10	12	9.83	0.833	0.694
6 to 11	8	10	14	10.33	1.000	1.000
7 to 11	10.5	13	15.5	13	0.833	0.694
8 to 11	5	7	10	7.17	0.833	0.694
9 to 10	10	11	12	11	0.333	0.111
10 to 11	6	6	6	6	0.000	0.000

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

Path 1-2-3-8-11:

$\begin{array}{c}\text{Expected duration}=8+12+10.17+7.17\\ =37.34\end{array}$

$\begin{array}{c}\text{Variance}=0+0.111+0.250+0.694\\ =1.055\end{array}$

$\begin{array}{c}\text{Standard deviation}=\sqrt{\text{Variance}}\\ =\sqrt{1.055}\\ =1.027\end{array}$

Path 1-2-4-6-11:

$\begin{array}{c}\text{Expected duration}=8+6+18.67+10.33\\ =43\end{array}$

$\begin{array}{c}\text{Variance}=0+0.111+4+1\\ =5.111\end{array}$

$\begin{array}{c}\text{Standard deviation}=\sqrt{\text{Variance}}\\ =\sqrt{5.111}\\ =2.2607\\ =2.261\end{array}$

Path 1-2-4-7-11:

$\begin{array}{c}\text{Expected duration}=8+6+13+13\\ =40\end{array}$

$\begin{array}{c}\text{Variance}=0+0.111+0+0.694\\ =0.805\end{array}$

$\begin{array}{c}\text{Standard deviation}=\sqrt{\text{Variance}}\\ =\sqrt{0.805}\\ =0.898\end{array}$

Path 1-2-5-9-11:

$\begin{array}{c}\text{Expected duration}=8+12+9.83+11+6\\ =46.83\end{array}$

$\begin{array}{c}\text{Variance}=0+0.444+0.694+0.111\\ =1.249\end{array}$

$\begin{array}{c}\text{Standard deviation}=\sqrt{\text{Variance}}\\ =\sqrt{1.249}\\ =1.1175\\ =1.118\end{array}$

Summary Introduction

To determine: The probability that the project will require more than 49 weeks.

Answer to Problem 11P

The probability that the project will require more than 49 weeks is 0.0301.

Explanation of Solution

Calculation of probability of completion in more than 49 weeks:

Calculation of z value for all paths:

Formula:

$\text{Z}=\frac{\text{Specified time}-\text{Path mean}}{\text{Path standard deviation}}$

Greater than 49 weeks:

Path 1-2-3-8-11:

$\begin{array}{c}\text{Z}=\frac{\text{49}-\text{37}\text{.34}}{\text{1}\text{.027}}\\ =11.35\end{array}$

Since z value is greater than +3.00, probability of completion is 1.00.

Path 1-2-4-6-11:

$\begin{array}{c}\text{Z}=\frac{\text{49}-\text{43}}{\text{2}\text{.261}}\\ =2.65\end{array}$

From the standard normal distribution table,

The probability value for (z = 2.261) is 0.9960.

Path 1-2-4-7-11:

$\begin{array}{c}\text{Z}=\frac{\text{49}-\text{40}}{\text{0}\text{.898}}\\ =10.02\end{array}$

Since z value is greater than +3.00, probability of completion is 1.00.

Path 1-2-5-9-10-11:

$\begin{array}{c}\text{Z}=\frac{\text{49}-\text{46}\text{.83}}{\text{1}\text{.118}}\\ =1.94\end{array}$

From the standard normal distribution table,

The probability value for (z = 1.94) is 0.9738.

Probability of completion in less than or equal to 49 weeks:

$\begin{array}{c}\text{Probability}=1.0000\times 0.9960\times 1.00\times 0.9738\\ =0.9699\end{array}$

Calculation of probability of completion in greater than 49 weeks:

$\begin{array}{c}\text{Probability}=1.0000-0.9699\\ =0.0301\end{array}$

The probability that the project will require more than 49 weeks is 0.0301.

Summary Introduction

To determine: The probability that the project will be completed in 46 weeks or less.

Answer to Problem 11P

The probability that the project will be completed in 46 weeks or less is 0.2085.

Explanation of Solution

Calculation of probability of completion in 46 weeks or less:

Calculation of z value for all paths:

Formula:

$\text{Z}=\frac{\text{Specified time}-\text{Path mean}}{\text{Path standard deviation}}$

Path 1-2-3-8-11:

$\begin{array}{c}\text{Z}=\frac{\text{46}-\text{37}\text{.34}}{\text{1}\text{.027}}\\ =8.43\end{array}$

Since z value is greater than +3.00, probability of completion is 1.00.

Path 1-2-4-6-11:

$\begin{array}{c}\text{Z}=\frac{\text{46}-\text{43}}{\text{2}\text{.261}}\\ =1.33\end{array}$

From the standard normal distribution table,

The probability value for (z = 1.33) is 0.9082.

Path 1-2-4-7-11:

$\begin{array}{c}\text{Z}=\frac{\text{46}-\text{40}}{\text{0}\text{.898}}\\ =6.68\end{array}$

Since z value is greater than +3.00, probability of completion is 1.00.

Path 1-2-5-9-10-11:

$\begin{array}{c}\text{Z}=\frac{\text{46}-\text{46}\text{.83}}{\text{1}\text{.118}}\\ =-0.74\end{array}$

From the standard normal distribution table,

The probability value for (z = -0.74) is 0.2296.

Probability of completion in less than or equal to 46 weeks:

$\begin{array}{c}\text{Probability}=1.0000\times 0.9082\times 1.0000\times 0.2296\\ =0.2085\end{array}$

The probability that the project will be completed in 46 weeks or less is 0.2085.