Chapter 17, Problem 7P

Three recent college graduates have formed a partnership and have opened an advertising firm. Their first project consists of activities listed in the following table.

a. Draw the precedence diagram.

b. What is the probability that the project can be completed in 24 days or less? In 21 days or less?

c. Suppose it is now the end of the seventh day and that activities A and B have been completed while activity D is 50 percent completed. Time estimates for the completion of activity D are 5, 6, and 7. Activities C and H are ready to begin. Determine the probability of finishing the project by day 24 and the probability of finishing by day 21.

d. The partners have decided that shortening the project by two days would be beneficial, as long as it doesn’t cost more than about $20,000. They have estimated the daily crashing costs for each activity in thousands, as shown in the following table. Which activities should be crashed, and what further analysis would they probably want to do?

Summary Introduction

To draw: A precedence diagram.

Answer to Problem 7P

Precedence diagram:

Explanation of Solution

Given information:

Activity	Immediate Predecessor	Optimistic time	Most likely time	Pessimistic time
A		5	6	7
B		8	8	11
C	A	6	8	11
D		9	12	15
E	C	5	6	9
F	D	5	6	7
G	F	2	3	7
H	B	4	4	5
I	H	5	7	8
End	E, G, I

Activity	First crash	Second crash
C	$       8.00	$         10.00
D	$     10.00	$         11.00
E	$       9.00	$         10.00
F	$       7.00	$           9.00
G	$       8.00	$           9.00
H	$       7.00	$           8.00
I	$       6.00	$           8.00

Precedence diagram:

The precedence diagram is drawn from the first task till the last task. The activities are placed from left to right. The directions are represented with arrows to indicate the relationship between activities. The arrows are represented with the activity name.

Summary Introduction

To determine: The probability at which the projected can be completed in 24 days or less and 21 days or less.

Answer to Problem 7P

The probability at which the projected can be completed in:

24 days or less = 0.9686

21 days or less = 0.2350

Explanation of Solution

Given information:

Activity	Immediate Predecessor	Optimistic time	Most likely time	Pessimistic time
A		5	6	7
B		8	8	11
C	A	6	8	11
D		9	12	15
E	C	5	6	9
F	D	5	6	7
G	F	2	3	7
H	B	4	4	5
I	H	5	7	8
End	E, G, I

Activity	First crash	Second crash
C	$       8.00	$         10.00
D	$     10.00	$         11.00
E	$       9.00	$         10.00
F	$       7.00	$           9.00
G	$       8.00	$           9.00
H	$       7.00	$           8.00
I	$       6.00	$           8.00

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 expected time 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	5	6	7	6	0.333	0.111
B	8	8	11	8.5	0.500	0.250
C	6	8	11	8.17	0.833	0.694
D	9	12	15	12	1.000	1.000
E	5	6	9	6.33	0.667	0.444
F	5	6	7	6	0.333	0.111
G	2	3	7	3.5	0.833	0.694
H	4	4	5	4.17	0.167	0.028
I	5	7	8	6.83	0.500	0.250

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

A-C-E:

$\begin{array}{c}\text{Expected duration}=6+8.17+6.33\\ =20.50\end{array}$

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

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

D-F-G:

$\begin{array}{c}\text{Expected duration}=12+6+3.50\\ =21.50\end{array}$

$\begin{array}{c}\text{Variance}=1+0.111+0.694\\ =1.805\end{array}$

$\begin{array}{c}\text{Standard deviation}=\sqrt{\text{Variance}}\\ =\sqrt{1.805}\\ =1.3435\\ =1.344\end{array}$

B-H-I:

$\begin{array}{c}\text{Expected duration}=8.50+4.17+6.83\\ =19.50\end{array}$

$\begin{array}{c}\text{Variance}=0.250+0.028+0.250\\ =0.528\end{array}$

$\begin{array}{c}\text{Standard deviation}=\sqrt{\text{Variance}}\\ =\sqrt{0.528}\\ =0.726\end{array}$

Calculation of z value for all paths:

Formula:

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

24 days or less:

A-C-E:

$\begin{array}{c}\text{Z}=\frac{\text{24}-\text{20}\text{.50}}{\text{1}\text{.118}}\\ =3.13\end{array}$

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

D-F-G:

$\begin{array}{c}\text{Z}=\frac{\text{24}-\text{21}\text{.50}}{\text{1}\text{.344}}\\ =1.86\end{array}$

From the standard normal distribution table,

The probability value for (z = 1.86) is 0.9686.

B-H-I:

$\begin{array}{c}\text{Z}=\frac{\text{24}-\text{19}\text{.50}}{\text{0}\text{.726}}\\ =6.20\end{array}$

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

Probability of completion in 24 days or less:

$\begin{array}{c}\text{Probability}=1.00\times 0.9686\times 1.00\\ =0.9686\end{array}$

The probability at which the project can be completed in 24 days or less is 0.9686.

21 days or less:

A-C-E:

$\begin{array}{c}\text{Z}=\frac{\text{21}-\text{20}\text{.50}}{\text{1}\text{.118}}\\ =0.45\end{array}$

From the standard normal distribution table,

The probability value for (z = 0.45) is 0.6736.

D-F-G:

$\begin{array}{c}\text{Z}=\frac{\text{21}-\text{21}\text{.50}}{\text{1}\text{.344}}\\ =-0.37\end{array}$

From the standard normal distribution table,

The probability value for (z = -0.37) is 0.3557.

B-H-I:

$\begin{array}{c}\text{Z}=\frac{\text{21}-\text{19}\text{.50}}{\text{0}\text{.726}}\\ =2.07\end{array}$

From the standard normal distribution table,

The probability value for (z = 2.07) is 0.9808.

Probability of completion in 21 days or less:

$\begin{array}{c}\text{Probability}=0.6736\times 0.3557\times 0.9808\\ =0.2350\end{array}$

The probability at which the project can be completed in 21 days or less is 0.2350.

Summary Introduction

To determine: The probability of completing the project by day 24 and day 21.

Answer to Problem 7P

The probability at which the projected can be completed in:

Day 24 = 0.9328

Day 21 = 0.0186

Explanation of Solution

Given information:

• At the end of 7^th day activities A and B are completed and D is 50% completed.
• Time estimates of activity D completion are 5, 6 and 7.
• Activities C and H are ready to begin.

Activity	Immediate Predecessor	Optimistic time	Most likely time	Pessimistic time
A		5	6	7
B		8	8	11
C	A	6	8	11
D		5	6	7
E	C	5	6	9
F	D	5	6	7
G	F	2	3	7
H	B	4	4	5
I	H	5	7	8
End	E, G, I

Activity	First crash	Second crash
C	$       8.00	$         10.00
D	$     10.00	$         11.00
E	$       9.00	$         10.00
F	$       7.00	$           9.00
G	$       8.00	$           9.00
H	$       7.00	$           8.00
I	$       6.00	$           8.00

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 expected time 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	5	6	7	6	0.333	0.111
B	8	8	11	8.5	0.500	0.250
C	6	8	11	8.17	0.833	0.694
D	5	6	7	6	0.333	0.111
E	5	6	9	6.33	0.667	0.444
F	5	6	7	6	0.333	0.111
G	2	3	7	3.5	0.833	0.694
H	4	4	5	4.17	0.167	0.028
I	5	7	8	6.83	0.500	0.250

Revised project diagram:

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

C-E:

$\begin{array}{c}\text{Expected duration}=7+\left(8.17+6.33\right)\\ =21.50\end{array}$

$\begin{array}{c}\text{Variance}=0.694+0.444\\ =1.138\end{array}$

$\begin{array}{c}\text{Standard deviation}=\sqrt{\text{Variance}}\\ =\sqrt{1.138}\\ =1.0667\\ =1.067\end{array}$

D-F-G:

$\begin{array}{c}\text{Expected duration}=7+\left(6+6+3.50\right)\\ =22.50\end{array}$

$\begin{array}{c}\text{Variance}=0.111+0.111+0.694\\ =0.916\end{array}$

$\begin{array}{c}\text{Standard deviation}=\sqrt{\text{Variance}}\\ =\sqrt{0.916}\\ =0.9570\\ =0.957\end{array}$

H-I:

$\begin{array}{c}\text{Expected duration}=7+\left(4.17+6.83\right)\\ =18\end{array}$

$\begin{array}{c}\text{Variance}=0.028+0.250\\ =0.278\end{array}$

$\begin{array}{c}\text{Standard deviation}=\sqrt{\text{Variance}}\\ =\sqrt{0.278}\\ =0.527\end{array}$

Calculation of z value for all paths:

Formula:

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

24 days or less:

C-E:

$\begin{array}{c}\text{Z}=\frac{\text{24}-\text{21}\text{.50}}{\text{1}\text{.067}}\\ =2.34\end{array}$

From the standard normal distribution table,

The probability value for (z = 2.34) is 0.9904.

D-F-G:

$\begin{array}{c}\text{Z}=\frac{\text{24}-\text{22}\text{.50}}{\text{0}\text{.957}}\\ =1.57\end{array}$

From the standard normal distribution table,

The probability value for (z = 1.57) is 0.9418.

H-I:

$\begin{array}{c}\text{Z}=\frac{\text{24}-\text{18}}{\text{0}\text{.527}}\\ =11.39\end{array}$

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

Probability of completion in 24 days or less:

$\begin{array}{c}\text{Probability}=0.9904\times 0.9418\times 1.00\\ =0.9328\end{array}$

The probability at which the project can be completed in 24 days is 0.9328.

21 days or less:

C-E:

$\begin{array}{c}\text{Z}=\frac{\text{21}-\text{21}\text{.50}}{\text{1}\text{.067}}\\ =-0.47\end{array}$

From the standard normal distribution table,

The probability value for (z = -0.47) is 0.3192.

D-F-G:

$\begin{array}{c}\text{Z}=\frac{\text{21}-\text{22}\text{.50}}{\text{0}\text{.957}}\\ =-1.57\end{array}$

From the standard normal distribution table,

The probability value for (z = -1.57) is 0.0582.

H-I:

$\begin{array}{c}\text{Z}=\frac{\text{21}-\text{18}}{\text{0}\text{.527}}\\ =5.69\end{array}$

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

Probability of completion in 21 days or less:

$\begin{array}{c}\text{Probability}=0.3192\times 0.0582\times 1.000\\ =0.0186\end{array}$

The probability at which the project can be completed in 21 days is 0.0186.

Summary Introduction

To determine: The activities that should be crashed and further analysis.

Explanation of Solution

Given information:

• The partners want to shorten the project by 2 days as long as the cost is not more than $20,000.

Activity	Immediate Predecessor	Optimistic time	Most likely time	Pessimistic time
A		5	6	7
B		8	8	11
C	A	6	8	11
D		5	6	7
E	C	5	6	9
F	D	5	6	7
G	F	2	3	7
H	B	4	4	5
I	H	5	7	8
End	E, G, I

Activity	First crash	Second crash
C	$       8.00	$         10.00
D	$     10.00	$         11.00
E	$       9.00	$         10.00
F	$       7.00	$           9.00
G	$       8.00	$           9.00
H	$       7.00	$           8.00
I	$       6.00	$           8.00

Paths and expected duration:

Paths	Expected Duration
C-E	21.50
D-F-G	22.50
H-I	18.00

The critical path is D – F – G.

The activities are crashed based on the cost of crash given.

Activity	Cost
F	$7
G	$8
D	$10

Step 1:

Activity F has the lowest crashing cost ($7,000) and will be crashed first for 1 day. The expected duration of D-F-G will be 21.50 days.

Step 2:

Path	Expected Duration
C-E	21.50
D-F-G	21.50
H-I	18.00

Now there are two critical paths C-E and D-F-G.

The critical activities are arranged in the order of low crash costs.

Path	Activity	Cost
C-E	C	$8
	F	$9

Path	Activity	Cost
D-F-G	D	$8
	F	$9
	G	$10

One activity in each path is chosen to crash.

Activity C is crashed for 1 day since it has the lowest crashing cost ($8,000) on path C-E. The expected duration of path C-E is now 20.50 days.

Activity G is crashed for 1 day since it has the lowest crashing cost ($8,000) on path D-F-G. The expected duration of path D-F-G is now 20.50 days.

Calculation of total crashing cost:

$\begin{array}{c}\text{Total cost}=\$7,000+\$8,000+\$8,000\\ =\$23,000\end{array}$

The total cost of crashing is over the budget of $20,000 ($23,000 > $20,000). Hence, the partners will have to determine if crashing the project by 1 day or 2 days is really beneficial or not.