chapter 17: project management - … · chapter 17 - project management 17-3 blank forms to...

Chapter 17 - Project Management

17-1

CHAPTER 17: PROJECT MANAGEMENT

Solutions

1. a.

Path

Expected

Path Time

1–2–4–7–10–12 23

1–2–5–8–10–12 24

1–3–6–9–11–12 31*

b.

Path

Expected

Path Time

1–2–4–6–8–9 41

1–2–5–6–8–9 48

1–2–5–7–8–9 55*

1–3–7–8–9 40

c.

Path

Expected

Path Time

1–2–5–12–16 44*

1–3–6–13–16 39

1–3–7–14–16 41

1–4–8–9–10–11–15–16 35

38 1–4–8–10–11–15–16

2. a. Choose topic Shop

Library res. Select

Outline Install

Write Paper

Grammar ck

Submit paper

b. 1. AOA diagram

.5

Grammar

2

Library

Select

Write Submit

Install Shop

Choose Outline

3 .2

.8

1

2

.6 .4 Start End


17-2

Solutions (continued)

2. AON diagram

c. 7.5 Shop, Select, Install, Write, Grammar, Submit

d. Parallel paths.

Write Grammar Submit

Select Shop Install

Library Choose Outline

Start End


17-3

Blank forms to distribute to students if desired:

3a. Bank location.

Weeks after start

Activity 2 4 6 8 10 12 14 16 18 20

1-2

1-3

2-4

2-5

3-5

4-5

5-6

b. Solved problem #2.

Weeks after start

Activity 2 4 6 8 10 12 14 16 18

1-2

2-5

2-4

1-3

3-4

4-5

3. a. Bank location.

Weeks after start

Activity 2 4 6 8 10 12 14 16 18 20

1-2

1-3

2-4

2-5

3-5

4-5

5-6

b. Solved Problem #2.

Weeks after start

Activity 2 4 6 8 10 12 14 16 18

1-2

2-5

2-4

1-3

3-4

4-5


17-4


4. a.

Immediate

Activity Predecessor

A –

B –

C A

D A,B

E C

End D,E

4. b. Case 1: Activity-on-Node diagram

A

B

C

D

E

Dummy

a d

k

b e

f h

i g c

End

Start

Start


17-5

Case 2: Activity-on-Node diagram

m l

j

p n

k r

v

q

s

t w

End

Start


17-6

5. a.

Summary:

Activity ES EF LF LS Slack

1–2 0 4 11 7 7

2–4 4 13 21 12 8

4–7 13 18 26 21 8

7–10 18 20 28 26 8

10–12 21 24 31 28 7

2–5 4 12 19 11 7

5–8 12 19 26 19 7

8–10 19 21 28 26 7

1–3 0 10 10 0 0

3–6 10 16 16 10 0

6–9 16 20 20 16 0

9–11 20 25 25 20 0

11–12 25 31 31 25 0

12

4

28

21

19

12

26

18

1

2

5 8

9 6

3

12

11

26

19

28

21

19

12

11

4

0

0

10

10

16

16

20

20

25

25

31

31

25

25

20

20 16

16

10

10

26

18

28

20

31

24

21

13

7

0

21

13

26

19

10

4 7

9

5

2

3

6

5

4 6

10

4 8

7 2

11

4

LS

ES

LF

EF


17-7


5. b.

Summary:

Activity ES EF LS LF Slack

1 0 5 0 5 0

2 5 23 5 23 0

3 5 18 20 33 15

4 23 26 37 40 14

5 23 33 23 33 0

6 33 37 40 44 7

7 33 44 33 44 0

8 44 53 44 53 0

9 53 55 53 55 0

LS

ES

LF

EF

1

2

8 9

6

3

0

0

37

23

40

26

5

5

4

7

5

5

5

23

23

5

18

23

23

33

33

40

33

44

37

44

44

53

53 53

53

55

55

33

33

44

44

20

5

33

18

10

13

11

9 2

4

3


17-8


5. c. Activity ES EF LF LS Slack

1–2 0 15 16 1 1

2–4 15 27 28 16 1

4–7 27 33 34 28 1

7–9 33 36 37 34 1

4–9 27 32 37 32 5

1–3 0 8 8 0 0

3–6 8 16 30 22 14

6–9 16 23 37 30 14

3–5 8 17 17 8 0

5–8 17 31 31 17 0

8–9 31 37 37 31 0

32

27

15

a

34

33

37

36

37

32

30

16

16

15

1

0

8

8

17

17

22

8

28

27

34

33

31

31 17

17

8

8

37

37

28

27

16

15

0

0 30

16

1

2

8

9

6

3

4

7

5

31

31

37

23

f

8 8

g

h

9 j

14

i

7

k

6

12

b

6

c 3

e d

5

LS

ES

LF

EF


17-9


6. The network diagram is given in Problem 1, Part a.

For path 1–2–4–7–10–12, the 13 weeks required for 1–2–4 has been reduced to 12 weeks. On

path 1–2–5–8–10–12, the 12 weeks required for 1–2–5 has been increased to (about) 14 weeks

[12+(8–6)]. However, 1–2–5–8–10–12 is not along the critical path and has a slack of 7 weeks.

Along path 1–3–6–9–11–12, the 13 weeks required to complete 1–3 and half of 3–6 has been

reduced to 12 weeks; a reduction of one week along the critical path.

The project may be completed in (31–1) = 30 weeks, if all following activities are completed

according to the original estimate.

7. a.

b. Activity te 2 Path

Expected

Duration

Std.

Dev. 24 Prob.

a 6 4/36 a–c–e 20.5 1.118 3.13 0.9991

b 8.5 9/36 d–f–g 21.5 1.344 1.86 0.9686

c 8.17 25/36 b–h–i 19.5 0.726 6.19 1.0000

d 12 36/36

e 6.33 16/36 For 21 Probability

f 6 4/36 (21 – 20.5) = .447 0.45

0.6736

g 3.5 25/36 1.118

h 4.17 1/36

i 6.83 9/36 (21 – 21.5) = –0.3721 –0.37

0.3557

1.344

(21–19.5) = 2.065 2.07

0.9808

0.7265

0.6736 x 0.3557 x 0.9808 = 0.235

1

2

5

3 6 8

7 4

c

e

g

i

f d

a

b

h

Start

End


17-10


c. On the 8th day, the network could be reviewed as follows:

*Assume it is the beginning of the 8th day

Replace d by d and merge nodes (2) and (4) with (1).

d: te = 6: 2 = 4/36

7 days were used to complete activities a, b and one half of d. In the modified network:

Path

Expected Duration

from 8th day

Expected duration from

the start of the project Variance

1. c–e 14.50 14.50 + 7 = 21.5 41/36

2. d–f–g 15.5 15.5 + 7 = 22.5 33/36

3. h–i 11 11 + 7 = 18 10/36

Path Standard Dev. Z24 Probability (24)

1. c–e 1.0672 24 – 21.5 = 2.343

.9904

1.0672

2. d–f–g 0.9574 24 – 22.5 = 1.567

.9418

0.9574

3. h–i 0.5270 24 – 18 = 11.384

1.0000

0.527

.9904 x .9418 = .9328

Path Z21 Probability (21)

1. c–e 21 – 21.5 = –0.469

0.3192

1.0672

2. d–f–g 21 – 22.5 = –1.567

0.0582

0.9574

3. h–i 21 – 18 = 5.692

0.527

.3192 (.0582) = 0.0186

1

5

3

7

8 6

e (6.33)

f (6) d (6) g (3.5)

i (6.83) h (4.17)

c (8.17)

8th

day


17-11

d. Crash activities F, C, and G one day each for a total cost of $23,000. Decide if an additional

expenditure of $3,000 over budget would be worth the cost, or if it would be better to crash only

one day (Activity F for a cost of $7,000), or don’t crash at all.

8. Path Expected

Duration

Std. Dev. Z16 Z15 Z13

A 10 1.1 5.45 4.55 2.73

B 8 1.414 5.66 4.95 3.54

C 12 1. 4 3 1

D 15 1.7 0.59 0 –1.18

E 14 1.2 1.67 0.83 –0.83


17-12


(a) Prob. (T 16) = 1 x 1 x 1 x 0.7224 x 0.9525 = 0.6881

(b) Prob. (T 15) = 1 x 1 x 1 x 0.50 x 0.7977 = 0.3984

(c) Prob. (T 13) = .9968 x 1 x 0.8413 x .1197 x .2023 = .0203

9. Solution

Path

Expected

duration Std. dev. z Probability

1–2–3 4 + 5 = 9 1.64 0.61 .7291

1–3 8 1.60 1.25 .8944

P(duration 10 weeks): 1 – .7291 (.8944) = .3479

10. Solution

a. Path

Expected

duration Std. Dev. z for 11 wk. Probability

1–2–4 4 + 6 = 10 1.14 0.88 .8106

1–3–4 3 + 9 = 12 2.00 –0.50 .3085

P(duration 11 weeks): .8106(.3085) = .2501. Yes, the manager should be concerned

because the probability of finishing on schedule is only about .25.

b.

Path

Expected

duration Std. dev. z for 12 wk. Probability

1–2–4 4 + 6 = 10 1.14 1.75 .9599

1–3–4 3 + 9 = 12 2.00 0 .5000

P(duration 12 weeks) = 1 – .9599(.5000) = .52.

11. Project Management

a. Path Mean Var. Std. Dev. z49 Prob. z46 Prob.

1–2–3–8–11 37.34 1.056 1.027 11.35 1 8.43 1

1–2–4–6–11 43 5.11 2.26 2.65 .996 1.33 .9082

1–2–4–7–11 40 0.8056 0.898 10.02 1 3.34 1

1–2–5–10–9–11 46.83 1.25 1.118 1.94 .9738 –0.74 .2296

b. P(T 49 wk.) = 1 – [(.9738) (.996)] = .0301

c. P(T 46 wk.) = .9082 (.2296) = .2085


17-13


12.

Activity

Expected

Duration

Path

Mean Variance

Path

Variance

Path Std.

Dev.

a 4.

24.34

16/36

1.83

1.354 d 8. 16/36

e 9.17 25/36

h 3.17 9/36

f 4.5

15.5

25/36

1.58

1.258 g 7. 16/36

4.[a] 16/36

b 2.17

14.83

1/36

1.03

1.014 i 3.33 16/36

j 4. 4/36

k 5.33 16/36

c 8.17

26.17*

49/36

2.75

1.658 m 1. 0

n 7.5 25/36

o 9.5 25/36

*critical path duration

k

h

d

a

b

c

m

i j

g

f

e

End

n

Start

o


17-14


Path Mean Std. Dev. z27 Prob. z26 Prob.

a–d–e–h 24.34 1.354 1.96 .9750 1.23 .8907

a–f–g 15.5 1.258 9.14 8.35

b–i–j–k 14.83 1.014 12.00 11.00

c–m–n–o 26.17 1.658 .50 .6915 –0.10 .4602

Prob. ($500): .9750 (.6915) = .6742

Prob. ($1000): .8907 (.4602) = .4099

13.

Activity

Duration (wk.)

first crash

second crash

1-2 5 $8 #2 $10 2-4 6 7 9 4-7 3 14 #3 15 1-3 3 9 11 3-4 7 8 9 1-5 5 10 #1 15 5-6 5 11 #2 13 6-7 5 12 #3 14

2

1

3

1

5

6

4

7

7 5

6

3

3

5

5

5


17-15


Path Initial time After first crash After second crash After third crash

1-2-4-7

1-3-4-7

1-5-6-7

14 wk.

13 wk.

15 wk.

14 wk.

13 wk.

14 wk.

13 wk.

13 wk.

13 wk.

12 wk.

12 wk.

12 wk.

First Crash Second Crash Third Crash

Activity Cost

1 - 5 $10

Activity Cost

2 - 4 $7

5 - 6 11

$18

Activity Cost

4 - 7 $14

6 - 7 12

$26

14.

Length after crashing N weeks

Path N: 0 1 2 3 4 5

A–B–K 35

C–E–H–P 44 43 42 42 42 40

C–D–G–M 45 44 43 43

C–E–H–N 47 46 45 45 45 44

C–F–I–J–P 49 48 47 46 45 44

Activity

Crashed

– c c f f e,p

Cost – $5,000 5,000 12,000 15,000 36,000

Cum. Cost – $5,000 10,000 22,000 37,000 73,000

D

17 P

B

14

G

15

H

8

I

7

A

12

K

9

M

3 N

11

J

12

F

12

E

18

C 10

8

Start End


17-16


15.

Weeks Crashed

Activity 1st week 2nd week 3rd week 1 2 3 4 5 6

1–2 $18 [2] $22 [6] – $18 $22

2–5 24 25 25

5–7 30 30 35

7–11 15 [1] 20 [3] – 15 20

11–13 30 [4] 33 [5] 36 30 33

1–3 12 [6] 24 26 12

3–8 – – –

8–11 40 40 40

3–9 3 10 12

9–12 2 7 10

12–13 26 – –

1–4 10 [5] 15 25 10

4–6 8 [4] 13 – 8

6–10 5 [3] 12 [6] – 5 12

10–12 14 15 –

$15 $18 $25 $38 $43 $46

Weeks Crashed

Path 0 1 2 3 4 5 6

1–2–5–7–11–13 35 wk. 34 33 32 31 30 29

1–3–8–11–13 32 32 31 30 29

1–3–9–12–13 20

1–4–6–10–12–13 33 33 32 31 30 29


17-17


Summary:

Project

Length

Cum. wk.

Shortened

Cum. Crash

Costs ($0000)

Indirect Costs

($000)

Total Cost

($000)

35 0 0 35(40) = 1,400 1,400

34 1 15 34(40) = 1,360 1,375

33 2 33 33(40) = 1,320 1,353

32 3 58 32(40) = 1,280 1,338

31 4 96 31(40) = 1,240 1,336

30 5 139 30(40) = 1,200 1,339

29 6 185 29(40) = 1,160 1,345

1,300

Total

Cost

($000)

1,400

0 30 31 32 33 34 35

Product Length

(wk)


17-18


16.

Project duration = 39 wk

Project

length

Shorten

activity

Crash

cost

39 wk – 0

38 Z 90

37 N, L 170 = (125 + 45)

36 Q 200

35 Q 225

34 M, N 345

Stop here; additional crashing will cost more than the $375 weekly penalty.

17. a. 18.5 (See table in part b.)

b.

Path

Expected Duration

Standard Deviation

Z17

Probability

1-2-4-6

1-3-5-6

5+8.17 +5.33 = 18.5

8.33 + 3 + 3.83 = 15.16

1.17

1.12

–.43

2.54

.3336

.9945

1 – .3336(.9945) = .6682 or approximately .67.

9

K

410, 415

5

Q

200, 225

7

Y

85, 90

7

L

125, -

4

M

300, 350

8

P

-, -

6

Z

90, -

5

N

45, 45

6

J

50,-

En

d

Start


17-19

18.

Event Probability Cost ($000) Exp. Cost

1 .25 15 3.75

2 .35 25 8.75

3 .20 55 11.00

4 .80 10 8.00

5 .10 77 7.70

6 .40 55 22.00

7 .60 50 30.00

80

60

5

3 6

7

40

Cost 20

2

1

4

0 .25 .50 .75 1.00

Probability of Occurring

The manager should probably be most concerned about event #7, which has a greater than 50%

chance of occurring and a relatively high cost. (Note its expected cost in table above is the highest.)

19.

200

160

120

3

2

Cost 80

40

1

6

5

4

0 .25 .50 .75 1.00

Probability of Occurring

The weather problems (4) and funding delays (5) are placed conservatively (i.e., using the worst likely

probabilities).

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