Solution-Review Problems of Chapter 17

Review Problems of Chapter 17 1.

For the following network diagram, determine both the critical path and the expected project duration. The numbers on the arrows represent expected activity times.

Solution:

Path

2.

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

Chris received new word processing software for her birthday. She also received a check, with which she intends to purchase a new computer. Chris’s college instructor assigned a paper due next week. Chris decided that she will prepare the paper on the new computer. She made a list of the activities she will need to do and their estimated times. a. b. c. d.

Arrange the activities into two logical sequences. Construct an AON diagram. Determine the critical path and the expected duration time. What are some possible reasons for the project to take longer than the expected duration?

Solution:

2.

a.

Choose topic Library res. Outline

Shop Select Install

Write Paper Grammar ck Submit paper b. AON diagram Choose

Library

Outline

Start

Write Shop

c. 7.5

Select

Grammar

Submit

Install

Shop, Select, Install, Write, Grammar, Submit

d. Parallel paths, probabilistic activity durations.

3.

For problem 1, determine the following quantities for each activity: the earliest start time, latest start time, earliest finish time, latest finish time, and slack time. List the critical activities, and determine the expected duration of the project.

Solution:

End

Summary: Activity 1

ES 0

EF 5

LS 0

LF 5

Slack 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

37 40 23 26 5 23 5 23

2 0 0

5 5

40 44 33 37

4 3

6

23 33 23 33

18

4 44 53 44 53

5 10 5

1

20 33 5 18

LF EF

33 44 33 44

89

53 55 53 55

92

13

3

11

7

4.

The project manager of a task force planning the construction of a domed stadium had hoped to be able to complete construction prior to the start of the next college football season. After reviewing construction time estimates, it now appears that a certain amount of crashing will be needed to ensure project completion before the season opener. Given the following time and cost estimates, determine a minimum-cost crashing schedule that will shave five weeks off the project length. Note: No activity can be crashed more than two weeks.

Solution:

B 14

A 12

Start

K 9 End M

C 10

G D 17

3 N

15

11

E 18 F 12

H 8 J 12

I 7

Path A–B–K

5.

N:

P 8

Length after crashing N weeks 0 1 2 35

3

4

5

42

40

C–E–H–P

44

43

42

42

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

A construction project has indirect costs totaling $40,000 per week. Major activities in the project and their expected times are shown in this precedence diagram:

Crashing costs for each activity are

a. b.

Determine the optimum time-cost crashing plan. Plot the total-cost curve that describes the least expensive crashing schedule that will reduce the project length by six weeks.

Solution:

Activity 1–2

1st week $18 [2]

2nd week 3rd week $22 [6] –

2–5

24

25

25

5–7

30

30

35

7–11

15 [1]

20 [3]

–

11–13

30 [4]

33 [5]

36

1–3

12 [6]

24

26

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

4–6

8 [4]

13

–

6–10

5 [3]

12 [6]

–

15

–

10–12

14

Summary: Project Cum. wk. Length Shortened 35 0

1

2 $18

15

Weeks Crashed 3 4 5

6 $22

20 30

33 12

10 8 5 $15

$18

$25

Path 1–2–5–7–11–13

0 35 wk.

1 34

1–3–8–11–13

32

1–3–9–12–13

20

1–4–6–10–12–13

33

33

Cum. Crash Costs ($0000) 0

Indirect Costs ($000) 35(40) = 1,400

Total Cost ($000) 1,400

12 $38

$43

$46

Weeks Crashed 2 3 4 33 32 31

5 30

6 29

32

31

30

29

32

31

30

29

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

1,400 Total Cost ($000) 1,300

0

30

31

32

33

34

Product Length (wk)

35

6.

Chuck’s Custom Boats (CCB) builds luxury yachts to customer order. CCB has landed a contract with a mysterious New York lawyer (Mr. T). Relevant data are shown below. The complication is that Mr. T wants delivery in 32 weeks or he will impose a penalty of $375 for each week his yacht is late. Note: No activity can be crashed more than two weeks.

Develop a crashing schedule.

Solution:

Start

7 L 125, -

9 K 410, 415

5 N 45, 45

4 M 300, 350 6 J 50,-

8 P -, -

5 Q 200, 225

7 Y 85, 90

6 Z 90, -

Project duration = 39 wk Project Shorten Crash length activity cost 39 wk – 0 38

Z

90

37

N, L

36

Q

200

35

Q

225

34

M, N

345

170 = (125 + 45)

Stop here; additional crashing will cost more than the $375 weekly penalty. 7.

Given the accompanying network diagram, with times shown in days, a. Determine the expected duration of the project. b. Compute the probability that the project will take at least 18 days.

Solution: a. 18.5 (See table in part b.) b. Path

Expected Duration

1-2-4-6

5+8.17 +5.33 = 18.5

1-3-5-6

8.33 + 3 + 3.83 = 15.16

Standard Deviation 1.17

Z17

Probability

–.43

.3336

1.12

2.54

.9945

En d

1 – .3336 = .67.