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Each answer must provide a managerial explanation of the outcomes. 1. Office Automation Inc., has developed a proposal for introducing a new computerized office system that will improve word processing and interoffice communications for a particular company. Contained in the proposal is a list of activities that must be accomplished in order to complete the new office system project. Information about the activities is shown in the table below. Times are in weeks. Data for Office Automation, Inc. Activity Description Immediate Normal Crash Normal Crash Predecessors Time Time Cost (\$) Cost (\$) A Plan needs ----------10 8 30 70 B Order equipment A 8 6 120 150 C Install equipment B 10 7 100 160 D Set up training lab A 7 6 40 50 E Training course D 10 8 50 75 F Testing system C, E 3 3 60 ------a). Show the network for the project. b). Develop an activity schedule for the project using normal times. c). What are the critical path activities and what is the expected project completion time? 2. Use the Trial and Error Method to solve this problem. What is the Total Cost of Crashing including the Normal Cost? 3. Suppose that 12 components are sampled (n = 12) from a very large lot of components that have just been received. Assume that management decides to accept the lot only if no more than two nonconforming items are found in the sample; that is, c = 2. Although we do not have the actual percentage of nonconforming components in the lot, let us assume that this value is 5%. What is the probability that the lot will be accepted using this particular sampling plan? What happens if c = 1? b). Use the information above and develop the OC curve. If AQL is 3% and LTPD is 15%, what will be the producer and consumer risks for the sampling plan (n=12, c=2) to be appropriate? 4. The Goodman Tire and Rubber Company periodically tests its tires for tread wear under simulated road conditions. To study and control its manufacturing process, 20 samples, each containing 3 radial tires, were chosen from different shifts over several days of operation. The results are presented in the table below. Assuming that these data were collected when the manufacturing process was believed to be operating in control, develop the appropriate control charts. Discuss and make recommendations based on your charts. Tread Wear is measured in the hundredths of an inch. Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Tread Wear 31 42 26 18 25 30 17 25 38 29 41 42 21 17 32 26 41 34 29 17 26 31 23 19 17 24 43 35 18 25 30 42 28 36 40 29 18 29 22 34 28 35 34 21 35 36 29 28 33 30 40 25 32 17 29 31 32 31 28 26 5. A product is ordered once each year, and the reorder point without safety stock (dL) is 100 units. Inventory carrying cost is \$10 per unit per year, and the cost of a stockout is \$50 per year. Given the following demand probabilities during the reorder period, how much safety stock should be carried? DEMAND DURING REORDER PERIOD 0 50 ROP 100 150 200 PROBABILITY .1 .2 .4 .2 .1 ...
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MoTutor
School: UT Austin

Hello, the final answer is uploaded. Good Luck to you. :)

Solution 1
a) The network of the project is as shown below

b) Activity schedule using normal times is

c) There are 2 paths in the project
Path A-B-C-F with time = 10+8+10+3 = 31 weeks
Path A-D-E-F with time = 10+7+10+3 = 30 weeks

As the time taken is higher for A-B-C-F, the critical path is A-B-C-F and the project takes 31
weeks to complete

Solution 2
The crash cost per week and crash time is given below

1

No. of
Cash
Normal Crash
Normal Crash
weeks to
cost per
Time
Time
Cost
cost
be crashed week
10
8
30
70
2
40
8
6
120
150
2
30
10
7
100
160
3
60
7
6
40
50
1
10
10
8
50
75
2
25
3
3
60
0
0
0

Activity
A
B
C
D
E
F

As, Activity A is common to both paths, crashing A by 2 weeks
Additional cost = 40*2 = \$80

As, B-C-F is the critical path, crashing B by 1 week as this has a minimum cost
Additional cost = \$30

Now, both the paths have the same duration and hence activities on both paths are crashed.
Maximum crash possible in A-B-C-F = 1+2 = 3 week, as B has already been crashed by 1 week
Maximum crash possible in A-D-E-F = 1 + 2 = 3 week
Additional cost = \$30 + \$60*2+\$10+\$25*2 = \$210

So, total cost of crashing = \$80 + \$30 + \$210 = \$320
Total cost of project = Total normal cost + \$210 = 30+120+100+40+50+60+210 = \$610

Solution 2
a) n = 12
c=2
p = 0.05
Probability of accepting the lot = (1-0.05)12 + 12*0.05*(1-0.05)11 + 12*11/2*0.052*(1-0.05)10 =
0.44

When c=1,

2

3

Probability of accepting the lot = (1-0.05)12 + 12*0.05*(1-0.05)11= 0.34

Solution 4
The table with mean and range of data is given below
Sample

Obs 1

Obs 2

Obs 3

Average (xbar)

Range
(R)

1

31

42

28 33.67

14

2

26

18

35 26.33

17

3

25

30

34 29.67

9

4

17

25

21 21.00

8

5

38

29

35 34.00

9

6

41

42

36 39.67

6

7

21

17

29 22.33

12

8

32

26

28 28.67

6

9

41

34

33 36.00

8

10

29

17

30 25.33

13

11

26

31

40 32.33

14

12

23

19

25 22.33

6

13

17

24

32 24.33

15

14

43

35

17 31.67

26

15

18

25

29 24.00

11

16
17

30
28

42
36

31 34.33
32

12
8

4
32.00

18

40

29

31 33.33

11

19

18

29

28 25.00

11

20

22

34

26 27.33

12

R chart
Average range:
R-bar = sum of all range (R)/20 = 11.4

Using Table of factors for...

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