Exam 1 Study Guide

PROBLEMS 1. Mabel's Ceramics spent $3000 on a new kiln last year, in the belief that it would cut energy usage 25% over the old kiln. This kiln is an oven that turns "greenware" into finished pottery. Mabel is concerned that the new kiln requires extra labor hours for its operation. Mabel wants to check the energy savings of the new oven, and also to look over other measures of their productivity to see if the change really was beneficial. Mabel has the following data to work with:

The year before Year just ended Production (finished units) 4000 4100 Labor (hrs) 350 375 Capital ($) 15000 18000 Energy (kWh) 3000 2600

Also, suppose that the average labor cost is $12 per hour, cost of capital is 20%, and cost of energy is $0.40 per kwh. a. Were the modifications beneficial? (Compute labor, energy, and capital productivity for the two years and compare.) b. Compute percentage change in multi-factor productivity of the year just ended from that of year before. c. If the multifactor productivity for next year must be restored to what it was the year before, assuming the same output next year as the year just ended, by how the input must be reduced from what it is this year?

2. An Appliance Service company made house calls and repaired 10 lawn-mowers, 2 refrigerators, and 3 washers in an 8-hour day with his standard crew of 3 workers. The retail price for each respective service is $50, $200, and $120. The average wage for the workers is $12 per hour. The materials cost for a day was $200 while the overhead cost was $50. a. What is the company’s labor productivity? b. What is the multifactor productivity? c. How much of a reduction in input is necessary for a 5% increase in multifactor productivity?

3. Consider the tasks, durations, and predecessor relationships in the following network. Draw the AON network and answer the questions that follow.

Activity Description Immediate Optimistic Most Likely Pessimistic Predecessor(s) (Weeks) (Weeks) (Weeks) A --- 4 7 10 B A 2 8 20 C A 8 12 16 D B 1 2 3 E D, C 6 8 22 F C 2 3 4 G F 2 2 2 H F 6 8 10 I E, G 4 8 12 J I 1 2 3

a. Schedule the activities of this project and determine (i) the expected project completion time, (ii) the earliest and latest start and finish times, and the slack for all the activities, and (iii) all the critical paths. b. What is the probability of completion of the project before week 42? c. What is the probability of completion of the project before week 35? d. With 99% confidence what is your estimate for the project completion time.

4. Consider the following project. All activity times are in weeks.

Immediate Normal Crash Normal Crash Activity Predecessor(s) Time time cost cost A - 7 4 20000 38000 B - 8 5 50000 74000 C A 9 7 80000 110000 D A, B 8 8 30000 30000 E B 9 8 10000 12000 F C 10 8 90000 124000 G D, E 5 5 25000 25000 H E 10 8 32000 40000 I F, G 5 4 28000 35000

a. Draw an AON network. b. Identify all the unique paths from start to finish and determine the critical path, normal project completion time, and normal project cost. c. Compute MAC, Cost of crashing/week. d. Which activity would you crash first and by how many weeks? e. Determine the project time and cost after crashing the activity selected in (d). 5. Consider the following CPM Solver model.

a) Determine the successor activities in cells I2 to I10. b) Determine the Excel formulas for the following cells: F2, G2, C15, C18, D18, D21, C25, G19, G16, G15, H15, B27, B28, and B29. c) What is the Solver Target cell for minimizing the project completion time? d) What is the Solver changing cell range? e) What are the Solver constraints?

6. What is the forecast for May based on a 3-period MA and a weighted 3-period moving average applied to the following past demand data? Let the weights be, 3, 3, and 4 (last weight is for most recent data). Compute MAD, MSE and MAPE for both cases and compare.

Nov. Dec. Jan. Feb. Mar. April 37 36 40 42 47 43

7. Sales of music stands at the local music store over the past ten days are shown in the table below. Forecast demand using exponential smoothing with an  of .6 (initial forecast = 16). a) Compute the forecast for period 6 and the MAD. b) Compute the tracking signal for periods 1 to 5. What do you recommend for this forecasting process?

t 1 2 3 4 5 Demand 13 21 28 37 25 8. Weekly sales of copy paper at Cubicle Suppliers are in the table below. Forecast week 8 with a trend projection model. Week Sales (cases) 1 17 2 22 3 27 4 32 5 35 6 37 7 41

9. The quarterly sales for specific educational software over the past three years are given in the following table. Compute the four seasonal indices and find forecast for Year 4 if the annual demand for year 4 is estimated to be 10% more than that of year 3.

YEAR 1 YEAR 2 YEAR 3 Quarter 1 1690 1800 1850 Quarter 2 940 900 1100 Quarter 3 2625 2900 2930 Quarter 4 2500 2360 2615

10. Arnold Tofu owns and operates a chain of 12 vegetable protein "hamburger" restaurants in northern Louisiana. Sales figures and profits for the stores are in the table below. Sales are given in millions of dollars; profits are in hundred thousand dollars. 11. Store Sales Profits 1 7 15 2 2 10 3 6 13 4 4 15 5 14 25 6 15 27 7 16 24 8 12 20 9 14 27 10 20 44 11 15 34 12 7 17

a.a. Calculate a regression line for the data. What is your forecast of profit for a store with sales of $24 million? $30 million? a.b. Calculate the standard error of the estimate. a.c. Determine the value of the coefficient of correlation between sales and profit and the value of the coefficient of determination.

12. A restaurant manager tracks complaints from the diner satisfaction cards that are turned in at each table. The data collected from the past week’s diners appear in the following table.

Complaint Frequency Food taste 80 Food temperature 9 Order mistake 2 Slow service 16 Table/utensils dirty 47 Too expensive 4

Prepare a Pareto chart. To cover 80% of problems which complaints must be address first?

13. A list of issues that led to incorrect formulations in Tuncey Bayrak’s jam manufacturing unit in New England is provided below: Incorrect measurement Variability in scale accuracy Antiquated scales Equipment in disrepair Lack of clear instructions Technician calculation off Damaged raw material Jars mislabeled Operator misreads display Temperature controls off Inadequate cleanup Incorrect weights Incorrect maintenance Priority miscommunication Inadequate flow controls Inadequate instructions

Create a fish-bone diagram and categorize these issues using the “four Ms” method.

14. Cartons of Plaster of Paris are supposed to weigh exactly 32 oz. Inspectors want to develop process control charts. They take ten samples of six boxes and weigh them. Based on the following data, compute the lower and upper control limits and determine whether the process is in control.

Sample Mean Range Sample Mean Range 1 33.8 1 6 34.3 0.4 2 34.4 0.3 7 33.9 0.5 3 34.5 0.5 8 34.0 0.8 4 34.1 0.7 9 33.8 0.3 5 34.2 0.2 10 34.0 0.3

15. McDaniel Shipyards wants to develop control charts to assess the quality of its steel plate. They take ten sheets of 1" steel plate and compute the number of cosmetic flaws on each roll. Each sheet is 20' by 100'. Based on the following data, develop limits for the control chart and determine whether the process is in control. Sheet Number of flaws Sheet Number of flaws 1 6 6 2 2 1 7 1 3 3 8 0 4 2 9 0 5 1 10 2

16. Rancho No Tengo Orchards wants to establish control limits for its mangos before they are sent to the retailers. They randomly take six containers (assume it is enough) of one hundred mangos in an attribute testing plan and find some mangos with blemishes. What should be the limits on the control chart? Is the process in control? Container Number of mangos with blemishes 1 5 2 3 3 1 4 3 5 4 6 2

17. A woodworker is concerned about the quality of the finished appearance of her work. In sampling units of a split-willow hand-woven basket, she has found the following number of finish defects in ten units sampled: 4, 0, 3, 0, 1, 0, 1, 1, 0, 2. a. Calculate the average number of defects per basket b. If 3-sigma control limits are used, calculate the lower control limit, centerline, and upper control limit.

18. The specifications for a plastic liner for concrete highway projects call for a thickness of 6.0 mm ± 0.1 mm. The standard deviation of the process is estimated to be 0.02 mm. What are the upper and lower specification limits for this product? The process is known to operate at a mean thickness of 6.04 mm.

Determine the values of Cpk and Cp for this process. Is the process capable? Explain.

19. A medical supplies company buys its supplies in bulk and redistributes them to doctor’s offices and clinics. The receive thermometers in lots of 500 from the vendor. They are considering a sampling plan of n = 50 and c = 1. a. Develop a OC curve for this sampling plan. (Use Poisson Tables) b. Determine the producer’s risk if the AQL is 2%. c. Determine the consumer’s risk if the LTPD is 14%. d. Develop a curve for AOQ and determine the value of AOQL.

20. An acceptance sampling plan has lots of 5000 units, a sample size of 200 and c is 5. Suppose that the incoming lots have percentage defective of 3%. What is AOQ?

Answers:

1. The energy modifications did not generate the expected savings; labor and capital productivity decreased.

Given data Last year Now Production 4000 4100 Labor 350 375 Capital = 15000 18000 Energy = 3000 2600 Change Change % Labor productivity (Units/hr) = 11.4286 10.9333 -0.4952 -4.33%

Capital productivity (units/$) = 0.2667 0.2278 -0.0389 -14.58% Cost of capital = 3000 3600 Energy productivity (Units/KWH) = 1.3333 1.5769 0.2436 18.27%

Labor cost = Hours x $12 = 4200 4500 Capital $ = 15000 18000 Energy $ = $0.40 x Energy = 1200 1040 Total input $ = 8400 9140 Multifactor productivity (Units/$) = 0.4762 0.4486 -0.0276 -5.8% Target productivity = 0.4762 Target input = 4100/0.4762 = 8610 Reduction in input needed = 9140 – 8610 = 530

#2 LM R W Number serviced 10 2 3 Dollar value/unit 50 200 120 Production in $ 500 400 360 1260 <-- Total $ Labor hours = 3 workers x 8 hrs. = 24 per day Labor productivity = 1260/24 = $ 52.50 per hour of labor

Multifactor productivity Labor cost = 3x8x$12 = 288 Material = 200 Overhead = 50 Total input cost = 538 = 288 + 200 + 50 Productivity = 1260/538 = $ 2.3420 per $ input 5% improvement in MF productivity = 0.1171 Target productivity after 5% improvement = 2.4591 Input for improved productivity = $ 512.38 <-- Output/Productivity = 1260/2.4591 Reduction in input needed = $ 25.62 <-- 538 – 512.38

3. (a)

Task a m B t Variance A 4 7 10 7 1.0000 B 2 8 20 9 C 8 12 16 12 64/36 D 1 2 3 2 E 6 8 22 10 256/36 F 2 3 4 3 G 2 2 2 2 H 6 8 10 8 I 4 8 12 8 64/36 J 1 2 3 2 4/36

Task t ES EF LS LF Slack Start 0 0 A 7 0 7 0 7 0 Critical B 9 7 16 8 17 1 C 12 7 19 7 19 0 Critical D 2 16 18 17 19 1 E 10 19 29 19 29 0 Critical F 3 19 22 24 27 5 G 2 22 24 27 29 5 H 8 22 30 31 39 9 I 8 29 37 29 37 0 Critical J 2 37 39 37 39 0 Critical Finish 39 39

Critical path = A-C-E-I-J, Project completion time TE = 39 2 Variance for project completion time =  p = 1 + 388/36 = 11.7778; p =3.4319 b. For P(T <=42), Z = (42 – 39)/3.4319 = 0.87, Table area = 0.80785, Probability = 0.80785 c. For P(T <=35), Z = (35 – 39)/3.4319 = -1.17, Table area = .879; Probability = 1 - .879 = 0.121 d. Z for 99% confidence = 2.325, T = 39 + 2.325(3.4319) = 46.98

4.

Normal Crash Normal Crash Crashing Activity MCA Predecessor(s) Time time cost cost cost/week A 7 4 20000 38000 3 6000 - B 8 5 50000 74000 3 8000 - C 9 7 80000 110000 2 15000 A D 8 8 30000 30000 0 A, B E 9 8 10000 12000 1 2000 B F 10 8 90000 124000 2 17000 C G 5 5 25000 25000 0 D, E H 10 8 32000 40000 2 4000 E I 5 4 28000 35000 1 7000 F, G Sum = 365,000

Paths Path time A - C - F - I 31 Critical path A - D - G - I 25 B - D - G - I 26 Normal project time = 31 weeks B - E - G - I 27 Normal project cost = 365,000 B - E - H 27

Activity to crash = A – among the critical activities (A, C, F, I) the crashing cost/week for A is the smallest. Weeks to crash = Minimum{MTR of A, Project time – time of second longest path} i.e. = Minimum{3, 31-27} = 3

Project time after crashing A 3 weeks = 31 – 3 = 28 weeks Project cost after crashing A = 365,000 + 3 x 6,000 = 383,000

5. (a)

Activity Successors(s) A C, D B D, E C F D G E G, H F I G I H Finish I Finish

(b)

F2 B2-C2 G2 (E2-D2)/F2 C15 B2-B15 D18 Max(E15,E16) E18 D18+C18 D21 Max(E18,E19) C25 Max(E22,E23) G19 Min(F21,F22) F19 G19-C19 G16 Min(F18,F19) G15 Min(F17,F18) H15 F15-D15 or G15-E15 B27 Sum(D2:D10) Sumproduct(BG15:B23,G2:G B28 10) B29 B27+B28

(c) Solver Target cell for minimizing the project completion time = C25 (d) Changing cell range = B15:B23 (e) What are the Solver constraints? B15:B23 <= F2:F10 B15:B23 = Integer (Optional)

6.

Demand 3-MA Weighte 2 Month (At) Forecast |Et| |Et|/At Et Weight d 3-MA |Et| |Et|/At Nov. 37 3 Dec. 36 3 Jan. 40 4 Feb. 42 37.67 4.33 0.1031 18.7489 37.90 4.1 0.0976 Mar. 47 39.33 7.67 0.1632 58.8289 39.60 7.4 0.1574 April 43 43.00 0 0.0000 0 43.40 0.4 0.0093 MAD = MAPE = MSE Forecast MAD MAPE = Forecast = 44.00 4 8.88% =25.8593 = 43.90 = 3.97 8.81%

7.

Period Demand Ft Et |Et| 1 13 16.00 -3.00 3.00 2 21 14.20 6.80 6.80 3 28 18.28 9.72 9.72 4 37 24.11 12.89 12.89 5 25 31.84 -6.84 6.84

F6 = 27.74 MAD = 7.85

Tracking signal Period Demand Ft Et |Et| CFEt CAEt MADt TS 1 13 16.00 -3.00 3.00 -3.00 3.00 3 -1 2 21 14.20 6.80 6.80 3.80 9.80 4.9 0.78 3 28 18.28 9.72 9.72 13.52 19.52 6.51 2.08 4 37 24.11 12.89 12.89 26.41 32.41 8.1 3.26 5 25 31.84 -6.84 6.84 19.57 39.25 7.85 2.49

8.

Week Sales XY X2 1 17 17 1 n = 7 X2 = 140 2 22 44 4 X = 28 XY = 954 3 27 81 9 Y = 211 b = 3.9286 4 32 128 16  = 4.0000 a = 14.4286 5 35 175 25  = 30.14 6 37 222 36 7 41 287 49 28 211 954 140

Regression equation: Ŷ = 14.4286 + 3.9286t

F8 = 14.4286 + 3.9286(8) = 45.8571

9. Demand Quarter Year 1 Year 2 Year 3 Average Index 1 1690 1800 1850 1780.00 0.8823 2 940 900 1100 980.00 0.4857 3 2625 2900 2930 2818.33 1.3969 4 2500 2360 2615 2491.67 1.2350 Overall average = 2017.50

Year 3 sum = 8495 Annual demand for year 4 = 1.1 x 8495 = 9345 Demand/season = 2336

Forecast for year 4

Quarter Average demand Seasonal Index Forecast 1 2336 0.8823 2061 2 2336 0.4857 1135 3 2336 1.3969 3263 4 2336 1.2350 2885

10. Store Sales (X) Profits (Y) XY X2 Y2 1 7 15 105 49 225 2 2 10 20 4 100 3 6 13 78 36 169 4 4 15 60 16 225 5 14 25 350 196 625 6 15 27 405 225 729 n = 12 X2 = 1796 7 16 24 384 256 576 X = 132 XY = 3529 8 12 20 240 144 400 Y = 271 Y2 = 7159 9 14 27 378 196 729  = 11 b = 1.5930 10 20 44 880 400 1936  = 22.5833 a = 5.0601

11 15 34 510 225 1156 Ft = 5.0601 + 1.593 X 12 7 17 119 49 289 Sum = 132 271 3529 1796 7159

Estimated X Y profit 24 43.2923 $ 4,329,230 30 52.8503 $ 5,285,030

(b) = 4.0738

(c) = =0.9167

r2 = 0.8403

11. Complaint Frequency % Cum % Food taste 80 50.6% 50.6% Table/utensils dirty 47 29.7% 80.4% Slow service 16 10.1% 90.5% Food temperature 9 5.7% 96.2% Too expensive 4 2.5% 98.7% Order mistake 2 1.3% 100.0% 158 100.0%

To cover 80% of complaints, Food Taste and dirty utensils must be addressed first.

12.

13. Sample  R 1 33.8 1.0 n = 6 2 34.4 0.3 0.483 A2 = 3 34.5 0.5 A2  = 0.2415 4 34.1 0.7 LCL = - A2 = 33.86 5 34.2 0.2 UCL = + A2 = 34.34 6 34.3 0.4 7 33.9 0.5 D = 0 8 34.0 0.8 2 D = 2.004 9 33.8 0.3 3 10 34.0 0.3 LCLR = 0 UCLR = 1.002

= 34.1  = 0.5

The process is not in control, since the values for samples 1, 2, 3, and 9 fall outside the control limits. Although all the sample ranges fall within 0 and 1.002, the assignable causes should be investigated and eliminated.

14. Use c-chart = total defects/number of sheets = 1.8 UCLc = 1.8 + 3 = 1.8 + 4.02 = 5.825

LCLc = 1.8 - 3 = 1.8-4.02 = converts to zero Sheet number 1 has too many flaws; investigate the cause.

15. LCLp = 0 = 0.03 - (3 * 0.017) = -0.02; can’t be zero, so, round to 0

LCLp = 0 = 0.03 + (3 * 0.017) = 0.081

Limits are LCL = 0 and UCL = 0.081. All six points are in control; there is no pattern or trend in the data.

16. (a)  = 1.2; (b) LCLc = 1.2 – 3 = -2.0862, or zero

UCLc = 1.2 + 3 = 4.49.

17. LSL = 5.9 mm, USL = 6.1 mm.

Cpk = min{(6.1-6.04)/(3*0.02), (6.04 - 5.9)/(3*0.02) = min{1.00, 2.33} = 1.

Cp = (6.1 – 5.9)/(6*.02) = 1.67

Since Cpk is < 1.333 the process is not capable. Since Cp = 1.67, the process variability is small enough to be within the desired specification range. Therefore, the process needs to be centered to achieve a Cpk of at least 1.33.

18.

Pd nPd Pa

0.00 0.00 1.000 For AQL = 2%, Pa= 0.736 i.e., Producer’s risk = 1 – 0.736 = 0.264 0.01 0.50 0.910 0.02 1.00 0.736 For LTPD = 14%, nPd = 50 x 0.14 = 7.0, 0.03 1.50 0.558 Pa from Poisson table = 0.007 0.04 2.00 0.406 0.05 2.50 0.288 i.e. Consumer’s risk = 0.007 0.06 3.00 0.199 0.07 3.50 0.137 0.08 4.00 0.092 0.09 4.50 0.061 0.10 5.00 0.040

(d)

Pd Pa AOQ 0.00 1.000 0.00000 0.01 0.910 0.00819 0.02 0.736 0.01325 0.03 0.558 0.01507 0.04 0.406 0.01462 0.05 0.288 0.01294 0.06 0.199 0.01075 0.07 0.137 0.00860 0.08 0.092 0.00662 0.09 0.061 0.00494 0.10 0.040 0.00360 AOQL = 0.01507

19.

N = 5000 nPd = 6

n = 200 Pa = 0.446 <-- from Poisson table c = 5 AOQ = .03(.446)(5000-200)/5000

Pd = 3% AOQ = 0.0128448 i.e. = 1.28%