BADM3601 14 Final Exam Solutions copy PDF

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BADM 3601: FINAL EXAM SOLUTIONS

Operations Management

Professor Altug

December 13, 2012

Guidelines: 1. There are 7 questions and you have 120 minutes to finish the exam. 2. Show all your work and write your answers clearly. It is your responsibility to make sure that the grader is able to follow your work. 3. The mathematical formulas and tables required for some of the questions are at the end. 4. If the space provided after each question is not sufficient, you can use the back of each page to write your answers. 5. You are not allowed to ask any questions during the last 45 minutes. 6. It is closed-book, closed notes and you can use your calculator, if needed.

NAME:

.

1. [10 points] Twelve samples, each containing seven parts, were taken from a process that produces light bulbs. The lifetime of each bulb in the samples was determined. The results are tabulated and sample means and ranges are computed as follows: Sample # 1 2 3 4 5 6 7 8 9 10 11 12

Sample mean 28.2 27 26.6 26.1 26 28.5 27.6 25.9 26.3 27 25.6 28.2

Range 7 6 5 6 8 9 11 12 8 5 4 8

Determine the 3-sigma upper and lower control limits for both the mean and the range chart. Is the process in control? Explain your answer. X = 26.91 ; R = 7.416 Sample size n=7; we don’t know the population standard deviation, so we can use A2=0.419; D4 = 1.924 D3 = 0.076 X- Chart : UCL = 26.91 + 0.419(7.416) = 30.0173 LCL = 26.91 - 0.419(7.416) = 23.8027

R-Chart: UCL = (1.924)(7.416) = 14.268 LCL = (0.076)(7.416) = 0.5636 All the points are within the control limits in both charts, we can say that the process is in control.

2. [15 points] An organization is considering to generate forecast for January using exponential smoothing method with α =0.45 and simple 2-period moving average method. The actual sales are given in the table for October, November and December. The forecast for October using exponential smoothing method and the forecast for October and November using simple 2-period moving average method are given in the table as well.

Month

Actual sales

Forecast using Exponential Smoothing with α =0.45 100

October

120

November

85

90

December

70

?

January

Forecast using Simple 2-Period Moving Average 130

?

?

a) Find the forecast for January using exponential smoothing method with α =0.45 Using Ft = Ft-1 + α (At-1 – Ft-1) starting from November F(November) = 100 + 0.45(120-100) = 109 F(December) = 109 + 0.45(85 – 109) = 98.2 F(January) = 98.2 + 0.45 (70-98.2) = 85.51

b) Find the forecast for December and January using simple 2-period moving average method F(December) = (120+85)/2= 102.5 F(January) = (85+70)/2 =77.5

c) Based on the information given in the table and what you found in (a) and (b), which forecasting method would you choose according to MAD, if the actual sales of June is 55? MAD(ExpSm) = (|120-100| + |85-109|+|70-98.2|+|55-85.51|)/4= 25.67 MAD(MovAvg) = (|120-130| + |85-90|+|70-102.5|+|55-77.5|)/4= 17.5 We would choose 2-period moving average method since it has a lower MAD

3. [15 points] Consider a local bank. Currently, there is only one bank teller in a single-channel, single-phase system and assume time is 10:00AM. The bank has collected information on the inter-arrival time, and service time distributions from past transactions. They are represented in the tables below. Use the following two-digit random numbers given below to simulate 5 customers through the bank. What is the average time in line, and average time in system? (Set first arrival time to the inter-arrival time generated by first random number.)

Inter-arrival time 1 2 3 4 5

Probability 0.1 0.2 0.3 0.3 0.1

Service time 1 2 3 4

Probability 0.3 0.2 0.3 0.2

Random numbers for interarrival times: 27, 67, 57, 13, 4

Random numbers of service times: 88, 62, 38, 83, 48 Inter-arrival time 1 2 3 4 5

Probability distribution 0.1 0.2 0.3 0.3 0.1

Cumulative Prob. Dist 0.1 0.3 0.6 0.9 1.0

Random Number Int 01-10 (4) 11-30 (27,13) 31-60 (57) 61-90 (67) 91-100

Service time 1 2 3 4 Simulation

Probability distribution 0.3 0.2 0.3 0.2

Cumulative Prob. Dist 0.3 0.5 0.8 1.0

Random Number Int 01-30 31-50 (38,48) 51-80 (62) 81-100 (88,83)

Customer # 1 2 3 4 5

RN

Arrives

RN

Leaves

27 67 57 13 4

10:02 10:06 10:09 10:11 10:12

88 62 38 83 48

10:06 10:09 10:11 10:15 10:17

Avg time in line = (0+0+0+0+3)/5 = 0.6 min. Avg time in the system (4+3+2+4+5)/5 = 3.6 min

Waits (time in line) 0 0 0 0 3

Time in system 4 3 2 4 5

4. [22 points] A manager must decide on the mix of products to produce for the coming month. Product A requires three minutes per unit for assembly, two minutes per unit for painting, and one minute for packing. Product B requires two minutes per unit for assembly, four minutes for painting, and three minutes per unit for packing. There will be 900 minutes available for assembly, 1000 minutes for painting, and 600 minutes for packing. The amount of B produced has to be greater than or equal to 100 units. Both products have profit margins of $20 per unit. How many product A and product B should be produced to maximize the profit? a) Formulate a Linear Programming model to help the manager with his decision (First define the decision variables and then write the objective function and constraints). (6 points) b) Plot all the constraints and highlight the feasible region (You can use the graph below to help you with the scale while plotting the constraints) (8 points) c) Using the corner-point method, find the optimum solution. (6 points) d) Would your optimum solution change if you removed the “painting” constraint (i.e. if you had ample painting time)? Just provide an argument by observing the feasible region. (You are not expected to solve the problem again) (2 points)

X1: Units to produce of Product A; X2 : Units to produce of Product B

Max{20 X1 + 20 X2 } s.t. 3X1 + 2X2 ≤ 900 2X1 + 4X2 ≤ 1000 X1 + 3X2 ≤ 600 X2 ≥ 100 ; X1 ≥ 0 ;

X2 ≥ 0

X2=100

X1 + 3X2 = 600 3X1 + 2X2 = 900

2X1 + 4X2 = 1000

1: (0,200): $4000 2.(0,100): $2000 3.(214.66,128.57): $6864 (OPTIMUM) 4.(233.33, 100) = $6666

d)No the optimal solution would not change as the painting constraint is a redundant constraint that does not determine the feasible region.

5. [15 points] ABC Inc. has the following aggregate demand requirements and other data for the upcoming four quarters. Quarter

Demand

Previous quarter's output

600 units

1

500

Beginning inventory

300 units

2

1000

Stockout cost

$100 per unit

3

1300

Inventory holding cost

$10 per unit at end of quarter

4

700

Hiring workers

$20 per unit

Firing workers

$40 per unit

Subcontracting cost

$120 per unit

Unit cost

$100 per unit

Overtime

$50 extra per unit

Which of the following production plans is better: Plan A—chase demand by hiring and firing; Plan B—pure level strategy, or Plan C—200 level with the remainder by subcontracting? (NOTE: There is an extra blank page provided after this question in case it is needed)

Plan A: Quarter Demand Production Increase (units) 1 500 200 200 2 1000 800 1000 3 1300 1300 300 4 700 700 Regular Production Cost: 3200*$100 = $320,000 Cost of Hiring: 1100*$20 = $22,000 Cost of Firing: 1000*$40=$40,000 TOTAL COST (Plan A)= $382,000

Decrease (in units) 400

600

Plan B: Quarter 1 2 3 4

Demand

Production

500 200 1000 1300 700

800 800 800 800

Inventory (units) 600 400

Stock-out (units)

200 100

0

Regular Production Cost: 3200*$100 = $320,000 Cost of Inventory: 1000*$10 = $10,000 Cost of stock-out: 100*$100=$10,000 Initial Cost of Hiring: 200* $20=$4000 TOTAL COST (Plan B)= $344,000 (BEST STRATEGY)

PlanC: Quarter 1 2 3 4

Demand 500 200 1000 1300 700

Production 200 200 200 200

Subcontract 0 800 1100 500

Regular Production Cost: 800*$100 = $80,000 Cost of Subcontracting: 2400*$120 = $288,000 Initial Cost of Firing: 400*$40=$16,000 TOTAL COST (Plan B)= $384,000

Increase(units)

Decrease(units) 400

6. [15 points] A manager is evaluating the four employees in the department: A, B, C, and D. The following table provides data on the time of each project when performed by a specific employee. Employee Project

A

B

C

D

1

27

29

28

30

2

30

32

31

34

3

33

25

29

26

4

29

31

24

28

Using the assignment method, determine the set of assignments that minimizes the total time. (Note: Make sure that you write all the steps clearly in order to get full credit and if there is more than one set of assignment that minimizes the total time, you should write them all). Row reduction: Subtract the minimum of the rows:

Employee Job

A

B

C

D

1

0

2

1

3

2

0

2

1

4

3

8

0

4

1

4

5

7

0

4

Column reduction: Subtract minimum of columns Employee Job

A

B

C

D

1

0

2

1

2

2

0

2

1

3

3

8

0

4

0

4

5

7

0

3

Only 3 lines are sufficient, so continue (subtract lowest among uncrossed ones and add that to the crossed cells): STEP2: Employee Job

A

B

C

D

1

0

1

0

1

2

0

1

0

2

3

9

0

4

0

4

6

7

0

3

Employee Job

A

B

C

D

1

0

0

0

0

2

0

0

0

1

3

10

0

5

0

4

6

6

0

2

There are three sets of optimal assignment leading to total time of : 109 4->C 2->B 3->D 1->A 4->C 2->A 3->D 1->B 4->C 2->A 3->B 1->D

7. [10 points] A work center has five jobs to be scheduled, shown in the order in which they arrived: Job A B C D E

Processing Time (hours) 5 10 3 6 8

Due (hours) 10 20 15 17 21

a) Apply the Shortest Processing Time (SPT) rule to schedule the above jobs and fill in the table below: (4 points): Job C A D E B

Processing time 3 5 6 8 10

Due Date

Flow Time

Late

15 10 17 21 20

3 8 14 22 32

0 0 0 1 12

b) Apply the Earliest Due Date (EDD) rule to schedule the above jobs and fill in the table below (4 points): Job A C D B E

Processing time 5 3 6 10 8

Due Date

Flow Time

Late

10 15 17 20 21

5 8 14 24 32

0 0 0 4 11

c) Find the average lateness and average flow time for both rules and explain which rule performs better (2 point): According to SPT: Average lateness: 2.6; Average flow time: 15.8 According to EDD: Average lateness: 3; Average flow time: 16.6 EDD performs better according to lateness and SPT performs better according to flow time

FORMULAS/TABLE: I. When we don’t know the population standard deviation, control limits for x-chart: UCLx = x + A2 R LCLx = x - A2 R where x is the mean of the sample means and R is the mean of sample ranges Control limits for the R-chart: UCLR = D4R LCLR = D3R where the coefficients A2, D4 and D3 can be found in the table below for a particular sample size “n”:

II. Forecast for period t using Exponential Smoothing = αAt-1 + (1-α)Ft-1 where Ft-1 is the forecast for the earlier period and At-1 is the actual demand for the earlier period and α is the smoothing constant. Forecast for period t using simple 2-period moving average method = ( At-1 + At-2)/2 where At-1 is the most recent actual demand and At-2 is the second most recent actual demand

MAD =

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