MGT 3110: Exam 1 Study Guide

PROBLEMS1. 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 endedProduction (finished units) 4000 4100Labor (hrs) 350 375Capital ($) 15000 18000Energy (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 ImmediatePredecessor(s)

Optimistic(Weeks)

Most Likely(Weeks)

Pessimistic(Weeks)

A --- 4 7 10B A 2 8 20C A 8 12 16D B 1 2 3E D, C 6 8 22F C 2 3 4G F 2 2 2H F 6 8 10I E, G 4 8 12J 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.

ActivityImmediate

Predecessor(s)Normal

TimeCrash

timeNormal

costCrash

costA - 7 4 20000 38000B - 8 5 50000 74000C A 9 7 80000 110000D A, B 8 8 30000 30000E B 9 8 10000 12000F C 10 8 90000 124000G D, E 5 5 25000 25000H E 10 8 32000 40000I 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. April37 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 5Demand 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)1 172 223 274 325 356 377 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 3Quarter 1 1690 1800 1850Quarter 2 940 900 1100Quarter 3 2625 2900 2930Quarter 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.

Store Sales Profits1 7 152 2 103 6 134 4 155 14 256 15 277 16 248 12 209 14 27

10 20 4411 15 3412 7 17

a. Calculate a regression line for the data. What is your forecast of profit for a store with sales of $24 million? $30 million?

b. Calculate the standard error of the estimate.

c. Determine the value of the coefficient of correlation between sales and profit and the value of the coefficient of determination.

11. 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 FrequencyFood taste 80Food temperature 9Order mistake 2Slow service 16Table/utensils dirty 47Too expensive 4

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

12. 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 accuracyAntiquated scales Equipment in disrepairLack of clear instructions Technician calculation offDamaged raw material Jars mislabeledOperator misreads display Temperature controls offInadequate cleanup Incorrect weightsIncorrect maintenance Priority miscommunicationInadequate flow controls Inadequate instructions

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

13. 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 Range1 33.8 1 6 34.3 0.42 34.4 0.3 7 33.9 0.53 34.5 0.5 8 34.0 0.84 34.1 0.7 9 33.8 0.35 34.2 0.2 10 34.0 0.3

14. 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 flaws1 6 6 22 1 7 1

3 3 8 04 2 9 05 1 10 2

15. 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 blemishes1 52 33 14 35 46 2

16. 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 basketb. If 3-sigma control limits are used, calculate the lower control limit, centerline, and upper control

limit.

17. 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.

18. 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.

19. 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 NowProduction 4000 4100

Labor 350 375Capital = 15000 18000Energy = 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 4500Capital $ = 15000 18000

Energy $ = $0.40 x Energy = 1200 1040Total input $ = 8400 9140

Multifactor productivity (Units/$) = 0.4762 0.4486 -0.0276 -5.8%Target productivity = 0.4762Target input = 4100/0.4762 = 8610Reduction in input needed = 9140 – 8610 = 530

#2 LM R WNumber serviced 10 2 3Dollar value/unit 50 200 120Production in $ 500 400 360 1260 <-- Total $

Labor hours = 3 workers x 8 hrs. = 24 per dayLabor productivity = 1260/24 = $ 52.50 per hour of labor

Multifactor productivityLabor cost = 3x8x$12 = 288

Material = 200Overhead = 50

Total input cost = 538 = 288 + 200 + 50Productivity = 1260/538 = $ 2.3420 per $ input

5% improvement in MF productivity = 0.1171Target productivity after 5% improvement = 2.4591

Input for improved productivity = $ 512.38 <-- Output/Productivity = 1260/2.4591Reduction in input needed = $ 25.62 <-- 538 – 512.38

3. (a)

Finish

J

I

H

G

F

E

DB

C

Start A

Start

B

A C

E

D

F

G

H

FinishI

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

Task t ES EF LS LF SlackStart 0 0A 7 0 7 0 7 0 CriticalB 9 7 16 8 17 1C 12 7 19 7 19 0 CriticalD 2 16 18 17 19 1E 10 19 29 19 29 0 CriticalF 3 19 22 24 27 5G 2 22 24 27 29 5H 8 22 30 31 39 9I 8 29 37 29 37 0 CriticalJ 2 37 39 37 39 0 CriticalFinish 39 39

Critical path = A-C-E-I-J, Project completion time TE = 39Variance for project completion time = 2

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 = 0.80785c. For P(T <=35), Z = (35 – 39)/3.4319 = -1.17, Table area = .879; Probability = 1 - .879 = 0.121d. Z for 99% confidence = 2.325, T = 39 + 2.325(3.4319) = 46.98

4.

Activity Normal Time

Crash time

Normal cost

Crash cost MCA Crashing

cost/week Predecessor(s)

A 7 4 20000 38000 3 6000 -B 8 5 50000 74000 3 8000 -C 9 7 80000 110000 2 15000 AD 8 8 30000 30000 0 A, BE 9 8 10000 12000 1 2000 B

F 10 8 90000 124000 2 17000 CG 5 5 25000 25000 0 D, EH 10 8 32000 40000 2 4000 EI 5 4 28000 35000 1 7000 F, G

Sum = 365,000

Paths Path timeA - C - F - I 31 Critical pathA - D - G - I 25

Normal project time = 31 weeksNormal project cost = 365,000

B - D - G - I 26B - E - G - I 27B - 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 weeksProject cost after crashing A = 365,000 + 3 x 6,000 = 383,000

5.(a)

Activity Successors(s)A C, DB D, EC FD GE G, HF IG IH FinishI Finish

(b)

F2 B2-C2G2 (E2-D2)/F2C15 B2-B15D18 Max(E15,E16)E18 D18+C18D21 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-E15B27 Sum(D2:D10)

B28Sumproduct(BG15:B23,G2:G10)

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:F10B15:B23 = Integer (Optional)

6.

MonthDemand

(At)3-MA

Forecast |Et| |Et|/At Et2 Weight

Weighted 3-MA |Et| |Et|/At

Nov. 37 3Dec. 36 3Jan. 40 4Feb. 42 37.67 4.33 0.1031 18.7489 37.90 4.1 0.0976Mar. 47 39.33 7.67 0.1632 58.8289 39.60 7.4 0.1574April 43 43.00 0 0.0000 0 43.40 0.4 0.0093

Forecast = 44.00MAD =

4MAPE = 8.88%

MSE =25.8593

Forecast = 43.90

MAD = 3.97

MAPE = 8.81%

7.

Period Demand Ft Et |Et|1 13 16.00 -3.00 3.002 21 14.20 6.80 6.803 28 18.28 9.72 9.724 37 24.11 12.89 12.895 25 31.84 -6.84 6.84

F6 = 27.74 MAD = 7.85

Tracking signalPeriod Demand Ft Et |Et| CFEt CAEt MADt TS

1 13 16.00 -3.00 3.00 -3.00 3.00 3 -12 21 14.20 6.80 6.80 3.80 9.80 4.9 0.783 28 18.28 9.72 9.72 13.52 19.52 6.51 2.084 37 24.11 12.89 12.89 26.41 32.41 8.1 3.265 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 = 1402 22 44 4 X = 28 XY = 9543 27 81 9 Y = 211 b = 3.92864 32 128 16 = 4.0000 a = 14.42865 35 175 25 = 30.146 37 222 367 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 Index1 1690 1800 1850 1780.00 0.88232 940 900 1100 980.00 0.48573 2625 2900 2930 2818.33 1.39694 2500 2360 2615 2491.67 1.2350

Overall average = 2017.50

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

Forecast for year 4

Quarter Average demand Seasonal Index Forecast1 2336 0.8823 20612 2336 0.4857 11353 2336 1.3969 32634 2336 1.2350 2885

10.Store Sales (X) Profits (Y) XY X2 Y2

1 7 15 105 49 2252 2 10 20 4 1003 6 13 78 36 1694 4 15 60 16 2255 14 25 350 196 625

a=Y−b X

b=954−7( 4 )(30 .14 )140−7 (4 )2 =3.9286b=∑ XY−n X Y

∑ X2−n X2

a=30 .14−3 . 9286( 4 )=14 . 4286

6 15 27 405 225 729 n = 12 X2 = 17967 16 24 384 256 576 X = 132 XY = 35298 12 20 240 144 400 Y = 271 Y2 = 71599 14 27 378 196 729 = 11 b = 1.5930

10 20 44 880 400 1936 = 22.5833 a = 5.060111 15 34 510 225 1156 Ft = 5.0601 + 1.593 X12 7 17 119 49 289

Sum = 132 271 3529 1796 7159

X YEstimated

profit24 43.2923 $ 4,329,23030 52.8503 $ 5,285,030

(b)√ 7159−5.0601 (271 )−1.593(3529)12−2

= 4.0738

(c) 12(3529)−(132)(271)

√ [12 (1796 )−1322 ] ¿¿¿ = 6576

7173.8258 =0.9167

r2 = 0.8403

11.Complaint Frequenc

y % 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%

Food t

aste

Table/

utens

ils di

rty

Slow se

rvice

Food t

empe

rature

Too ex

pens

ive

Order m

istak

e0

102030405060708090

0.0%

20.0%

40.0%

60.0%

80.0%

100.0%

Pareto Chart: ComplaintsFr

eque

ncy

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

12.

13.Sample R

1 33.8 1.0 n = 6A2 = 0.483

A2 = 0.2415LCL = X́ - A2 = 33.86UCL = X́ + A2 = 34.34

D2 = 0D3 = 2.004LCLR = 0

UCLR = 1.002

2 34.4 0.33 34.5 0.54 34.1 0.75 34.2 0.26 34.3 0.47 33.9 0.58 34.0 0.89 33.8 0.310 34.0 0.3

X́ = 34.1 = 0.5

The process is not in control, since the x 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 c = total defects/number of sheets = 1.8

UCLc = 1.8 + 3√1. 8 = 1.8 + 4.02 = 5.825LCLc = 1.8 - 3√1.8 = 1.8-4.02 = converts to zeroSheet number 1 has too many flaws; investigate the cause.

15. LCLp = 0 .03−3√ 0.03(1−0.03)100

= 0.03 - (3 * 0.017) = -0.02; can’t be zero, so, round

to 0

LCLp = 0.03+3√ 0.03 (1−0.03)100

= 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 √1.2 = -2.0862, or zeroUCLc = 1.2 + 3 √1.2 = 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 Pa0.00 0.00 1.000 For AQL = 2%, Pa= 0.736

i.e., Producer’s risk = 1 – 0.736 = 0.264

For LTPD = 14%, nPd = 50 x 0.14 = 7.0,Pa from Poisson table = 0.007i.e. Consumer’s risk = 0.007

0.01 0.50 0.9100.02 1.00 0.7360.03 1.50 0.5580.04 2.00 0.4060.05 2.50 0.288

0.06 3.00 0.1990.07 3.50 0.1370.08 4.00 0.0920.09 4.50 0.0610.10 5.00 0.040

0.00 0.02 0.04 0.06 0.08 0.10 0.120.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

OC-Curve for the sampling plan, n=50, C = 1

Pd (% of defectives in the lot)

P(Ac

cept

ing

lot)

(d)Pd Pa AOQ

0.00 1.000 0.000000.01 0.910 0.008190.02 0.736 0.013250.03 0.558 0.015070.04 0.406 0.014620.05 0.288 0.012940.06 0.199 0.010750.07 0.137 0.008600.08 0.092 0.006620.09 0.061 0.004940.10 0.040 0.00360

AOQL = 0.01507

19.N = 5000 nPd = 6n = 200 Pa = 0.446 <-- from Poisson tablec = 5 AOQ = .03(.446)(5000-200)/5000

Pd = 3% AOQ =0.012844

8 i.e. = 1.28%

0.00 0.02 0.04 0.06 0.08 0.10 0.120.00000

0.00500

0.01000

0.01500

0.02000AOQ

PdPerc

ent d

efec

tive

acce

pted