MidTerm 2020 PDF

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Practice Problems – End Term

I. Champion manufactures winter fleece jackets for sale in the United States. Demand for jackets during the season is normally distributed with a mean of 20,000 and a standard deviation of 10,000. Each jacket sells for $60 and costs $30 to produce. Any left over jackets at the end of the season are currently sold for $25 at the year end clearance sale. Holding jackets until the year end sale adds another $5 to their cost. A recent recruit has suggested shipping leftover jackets to South America for sale in the winter there rather than running a clearance. Each jacket will fetch a price of $35 in South America and all jackets sent there are likely to sell. Shipping costs add $5 to the cost of any jacket sold in South America. Would you recommend the South American option? How will this decision impact production decisions at Champion? SOLUTION:

Salvage value Cost of understocking Cost of overstocking Outputs Optimal cycle service level Optimal production size

$20 $30 $10

$25 $30 $5

0.7500 26,745

0.8571 30,676

Clearly the South American option generates higher service level (Cu/(Cu+Co)). Hence this is preferred.

II. Weekly demand for Hewlett Packard printers at a Sam's Club store is normally distributed with a mean of 250 and a standard deviation of 150. The store manager continuously monitors inventory and currently orders a thousand printers each time the inventory drops to 600 printers. Hewlett Packard currently takes two weeks to fill an order. How much safety inventory does the store carry? What cycle service level does Sam's Club achieve as a result of this policy? SOLUTION: Safety Inventory = 100, Cycle Service Level = 68% (normdist(600,500,212,1), demand during lead time follows a normal distribution with mean 250*LeadTime = 250*2 = 500 and standard deviation of 150*sqrt(2) = 212

III. Wholemark is an Internet order business that sells one popular New Year’s greeting card once a year. The cost of the paper on which the card is printed is $0.05 per card, and the cost of printing is $0.15 per card. The company receives $2.15 per card sold. Because the cards have the current year printed on them, unsold cards have no salvage value. Its customers are from the four areas: Los Angeles, Santa Monica, Hollywood, and Pasadena. Based on past data, the number of customers from each of the four regions is normally distributed with a mean of 2,000 and a standard deviation of 500. (Assume these four are independent.) What is the optimal production quantity for the card? SOLUTION: Cu = 2.15 – .2 = 1.95 Co = .2, Cu/(Co+Cu) = 1.95/(.2+1.95) = .90697 From the standard normal table, Z-value is 1.325.

Combined demand has mean 2000*4=8000, and standard deviation = sqrt(4*5002) = 1000. Using the above, the optimal production quantity is 8000+1.325*1000 = 9325.

IV. Lakeside Bakery bakes fresh pies every morning. The daily demand for its apple pies is a random variable with (discrete) distribution, based on past experience, given by Demand 5 10 15 20 25 30 Probability 10% 20% 25% 25% 15% 5% Each apple pie costs the bakery $6.75 to make and is sold for $17.99. Unsold apple pies at the end of the day are purchased by a nearby soup kitchen for 99 cents each. Assume no goodwill cost. a. If the company decided to bake 15 apple pies each day, what would be its expected profit? b. Based on the demand distribution given, how many apple pies should the company bake each day to maximize its expected profit? SOLUTION: Cu = 17.99 – 6.75 = 11.24, Co = 6.75 – 0.99 = 5.76 a. If demand is 5 (probability 0.1), we will sell 5 apple pies and have 10 leftover. If demand is 10 (probability 0.2), we will sell 10 apple pies and have 5 leftover. If demand is at least 15 (probability 0.7), we will sell all the 15 apple pies. The expected profit is 0.1(5*11.24 – 10*5.76) + 0.2(10*11.24 – 5*5.76) + 0.7(15*11.24) = $134.60 b. Cu = 17.99 – 6.75 = 11.24, Co = 6.75 – 0.99 = 5.76 Increase Q so long as P(demand =11.24/(11.24+5.76) = .663 The optimal quantity of apple pies is 20.

V. You are a newsvendor selling the San Pedro Times every morning. Before you get to work, you go to the printer and buy the day’s paper for $0.25 a copy. You sell a copy of the San Pedro Times for $1.00. Daily demand is distributed normally with mean = 250 and standard deviation = 50. At the end of each morning, any leftover copies are worthless and they go to a recycle bin. a. How many copies of the San Pedro Times should you buy each morning? b. Based on part (a), what is the probability that you will run out of stock? SOLUTION: Co = $0.25, Cu = $0.75 a) Service level = ($0.75)/($0.75 + 0.25) = 0.75. Z-value for 75% is 0.67. Q = 250 + 0.67 * 50 = 283.5 b) 25%

VI. Annual demand for a product is 13,000 units; weekly demand is 250 units with a standard deviation of 40 units. The time from ordering to receipt is four weeks. To provide a 98% service probability, what must the reorder point (ROP) be? Suppose the production manager is told to reduce the safety stock of this item by 100 units. If this is done, what will the new service probability be? 2 2 SOLUTION:  L  L  4(40) = 80 units 98% CSL (Cycle Service Level)  From standard normal distribution, z = 2.05

ROP  d L  z L = 250(4) + (2.05)80 = 1000 + 164 = 1164 If safety stock is reduced by 100 units, then ss = 64 units. ss  z L ,

z

ss

L



64 80 = .80

From standard normal distribution, z = .80, service probability is 79%

VII. Demand for vacuum cleaners at Crow’s Rental is shown in the following table. Machines are rented by the day only. Profit on the vacuum cleaners is $10 per day. Crow’s has stocked 4 vacuum cleaners. Demand Frequency 0 0.3 1 0.2 2 0.2 3 0.15 4 0.1 5 0.05 1.00 Assuming Crow’s stocking decision is optimal, what is the implied range of overstocking cost per machine? SOLUTION: Determine the implied range of excess cost per machine: Q*=4 Cu = $10 Co = unknown For 4 machines to be optimal, the critical ratio (Service ratio) must be ≥ .85 and ≤ .95. Step 1: Set SL = .85 and solve for Co:

(round to two decimals) Step 2: Set SL = .95 and solve for Ce:

(round to two decimals) Conclusion: Implied range of excess cost: $.53 ≤ ≤ $1.76.

VIII. Two independent methods of forecasting based on judgment and experience have been prepared each month for the past 10 months. The forecasts and actual sales are as follows: Month Sales Forecast 1 Error Forecast 2 Error 1 ……………. 770 771 -1 769 1 2 ……………. 789 785 4 787 2 3 ……………. 794 790 4 792 2 4 ……………. 780 784 -4 798 -18 5 ……………. 768 770 -2 774 -6 6 ……………. 772 768 4 770 2 7 ……………. 760 761 -1 759 1

8 ……………. 775 771 4 775 9 …………… 786 784 2 788 10 …………… 790 788 2 (12) 788 Compute a tracking signal for the 10th month for each forecast. Use performance of the forecast processes. No need of drawing chart.

0 -2 2 (-16) 4 limits to investigate the

SOLUTION: MAD1=2.8, MAD2=3.6 TS=12/2.8=4.3 TS2=-16/3.6=-4.44

IX. Using the appropriate control chart, determine two sigma control limits for the following case: “An inspector found an average of 3.9 scratches in the exterior paint of each of the automobiles being prepared for shipment to dealers”. SOLUTION: c-chart, c-bar=3.9, UCL=3.9+2*sqrt(3.9), LCL=3.9-2*sqrt(3.9)

MCQ1. Suggest one measure that Widget.com could use to see how well their production processes for existing, mature products meet customer requirements. [0.5] a. UCL & LCL b. UTL & LTL c. Process Capability Index* d. OEE

MCQ2. Consider the following potential quality problems and answer where a  -chart can be used. a. Wine served in a restaurant sometimes is served too warm while at other times it is served too cold.* b. A surgeon in a hospital follows the hygiene procedures in place on most days, but not all days. c. A passenger travelling with an airline might be seated at a seat with a defective audio system. d. An underwriter in a bank might sometimes accidentally approve loans to consumers that are not creditworthy.

MCQ3. Suppose control limits for a process are set at 3 standard deviations from the mean. If the process is in control, the probability of observing a sample outside the control limits is independent of the sigma capability of the process. a. TRUE* b. FALSE

MCQ4. An assembly process has twenty successive stages, each with 3-σ capability. The company has come up with a new design that will require only ten assembly stages. The capability at each stage still remains 3- σ. The overall assembly process capability with the new design will be a. Higher than the original design.* b. Same as the original design. c. Lower than the original design.

MCQ5. A customer wants delivery to be ensured between 10 am and 2 pm. Your truck leaves the factory at 4 am and the time taken to reach the customer is normally distributed with an average of 8 hrs and a standard deviation of 1 hr. (i) What is the process capability ratio of the process? a. 0.67* b. 0.50 c. 1.00

d. 2.00 e. None of the above (ii) What should be the target mean and standard deviation to improve the ability of the delivery process to be a 6-σ process ? Answer within the space provided below. 12noon, s.d.=1/3

X. K-Log produces cereals that are sold in boxes labeled to contain 390 grams. If the cereal content is below 390 grams, K-Log may invite FDA scrutiny. Filling much more than 390 grams costs the company since it essentially means giving away more of the product. Accordingly, KLog has set specification limits between 390 and 410 grams for the weight of cereal boxes. Currently the boxes are filled automatically by a filling machine and they have an average weight of 405 g with a standard deviation of 4 g. What process targets (in terms of mean and standard deviation of the filling process) are needed for the filling machine to have a six-sigma capability? Solution: We need to center the process at the middle of the specs, i.e., at 400 gms. To yield six sigma quality the standard deviation must be (410-390)/(2*6) = 1.67 gms

XI. A process has a mean surface roughness of 3.56 and standard deviation of 0.35. The specification is to have a roughness of less than or equal to 4 (USL). a) Find the capability of the process. b) Estimate the percent defective if the process is normally distributed and it is given that NORMSDIST(1.2571)=0.896

MCQ6. A machine produces drive shafts. Manufacturing currently takes samples of 10 shafts every half an hour to check if the process is in control. Control limits have been set at  3 standard deviations of the sample means and are set at 10  30.001 cm. A suggestion calls for control limits to be tightened to  2 standard deviations, i.e., 10  20.001 cm. As a result the proportion of shafts produced that are defective (outside specification limits) will (i) (ii) (iii)

Increase Decrease Remain unchanged*

MCQ7. Which one of the following is not an example of “Appraisal” costs? a. b. c. d.

Overtime to cover production losses* In process inspection costs Inspection Laboratory costs Periodic inspection

MCQ8. A workstation in a computer plant assembles and packs computer mice. Demand for the dual button model for a specific brand runs approximately 60 mice per 8 hour shift. Average replenishment lead time is 2 hours. Each container holds 5 mice. If the company wants a 15% safety stock factor, determine the appropriate number of kanbans.

a. 3 b. 28 c. 4* d. 27 e. none of the above

XII. The Good Chocolate company makes a variety of chocolte candies, including a 12-ounce chocolate bar (340gms) and a box of six 1-ounce chocolate bars (170gms). a. Specifications for the 12-ounce bar are 330gms and 350gms. What is the largest standard deviation (in grams) that the machine that fills the bar moulds can have and still run a four sigma process if the average fill is 340 gms? The process is centered so we can use Cp.

Cp 

USL  LSL 350  330  6 6

For a 4 sigma process Cp = 4/3=1.33 and solve for σ.

1.33(6σ) = 20 (1.33)6σ = 20 7.98σ = 20 σ = 20/7.98 σ = 2.506 (round to three decimals) b. The machine that fills the bar models for the 1-ounce (1 ounce=28.33 grams) bars has a standard deviation of .80 grams per bar. The filling machine is set to deliver an average of 1.01 ounces per bar, Specifications for the six-bar box are 160 to 180 grams. Hint: The variance for the box is equal to six times the bar variance. What is the process capability score? Round to a maximum of three decimals: 1 ounce = 28.33 grams. Bar variance = (.80)2 = 0.64 Box variance = 6 * 0.64 = 3.84 Box standard deviation = √ Average box weight = 6 *1.01 = 6.06 ounces * 28.33 = 171.68 grams for the six-bar box.

X  LSL 3



171.68  160  1.99 (3)(1.96)

USL  X 180 171.68   1.41 3 (3)(1.96) C pk  min{1.99,1.41}  1.41

XIII. Six Sigma Quality at Falcon Tyres Dr. Dias, vice president of quality at Falcon Tyres, was wondering how to explain the value of Six Sigma quality to the people within his organization. He wanted to push through a Six Sigma program and significantly improve the level of quality at Falcon and he needed to build a solid case for it first. Quality Issues at Falcon Given the large number of steps in tyre manufacturing, errors tend to accumulate. As a result, even if each step produced only 1% defects, at the end of 20 steps only 81.2% of the product would be of good quality. Since each factory produced about 10000 tyres per hour, such a process would result in about 1880 defective tyres per hour. Thus it was very important that each stage had a high level of quality. Another quality issue was related to settings on various machines. Over time, these settings tended to vary because of wear and tear on the machines. In such a situation, a machine would produce defective product even if the machine had correct setting. To detect such situations, Dias implemented a SPC program. At the extruder, the rubber for the AX-527 tyres had thickness specifications of 400±10 thousandth of an inch (thou). Dias and his staff analysed many samples of output from the extruder and determined that if the extruder settings were accurate, the output produced by the extruder had a thickness that was normally distributed with a mean of 400 thou and a standard deviation of 4 thou. Answer the following questions to understand the current capability of the process. 1. If the extruder setting is accurate, what proportion of the rubber extruded will be within specifications? SOLUTION: USL=410 thou, LSL=390 thou, %conforming=2*NORMDIST(410,400,4,TRUE)-1=98.76 ~ 124 defectives per hour

2. Dr. Dias has asked operators to take a sample of 10 sheets of rubber each hour from the extruder and measure the thickness of each sheet. Based on the average thickness of this sample, operators will decide whether the extrusion process is within control or not. Given that Dias plans three sigma control limits, what upper and lower control limits should he specify to the operators? SOLUTION: Standard deviation of average thickness of a sample of size 10=4/√10=1.265 3-sigma Control Limits: UCL=400+3*4/√10=403.795, LCL=400-3*4/√10=396.205

3. If a bearing is worn out, the extruder produces a mean thickness of 403 thou when the setting is 400 thou. Under this condition, what proportion of defective sheets will the extruder produce? Assuming the control limits in question 2, what is the probability that a sample taken from the extruder with worn bearings will be out of control? On average, how many hours are likely to go by before the worn bearing is detected? SOLUTION: % conforming with mean shift, NORMDIST(410,403,4,TRUE) -NORMDIST(390,403,4,TRUE)=95.94 As a result of mean shift, the extruder will now start producing 95.94% good products, increasing the defect rate to around 4% (406 defective tires per hour) % of samples within control limits=NORMDIST(403.795,403,1.265,TRUE)-NORMDIST(396.205,403,1.265,TRUE) =73.51

On average it will take (1/.2649=3.77)~4 samples before an out-of control sample appears. That means 4 hours will go by before this mean shift is discovered because a sample of size 10 is withdrawn every hour! During this period about 1600 defective tires would have been produced with mean shift!!

XIV. The manager of a restaurant open Wednesday to Saturday says that the restaurant does about 35% of its business on Friday night, 30% on Saturday night and 20% on Thursday night. What seasonal factors would describe his situation? The restaurant is open 4 days. Thursday night accounts for 0.20 of the business. Friday night accounts for 0.35 of the business. Saturday night accounts for 0.30 of the business. Wednesday night: 1.00 – 0.20 – 0.35 – 0.30 = 0.15 (15.00%) of the business. Seasonal relatives (round to two decimals): Wednesday

= 0.15 x 4 = 0.60

Thursday

= 0.20 x 4 = 0.80

Friday

= 0.35 x 4 = 1.40

Saturday

= 0.30 x 4 = 1.20

Sum of Seasonal Relatives = 0.60 + 0.80 + 1.40 + 1.20 = 4.00

XV. A department store’s quarterly sales of pyjamas over a period of 3 years are shown below: Year 1

Quarter Sales 1 88 2 129 3 88 4 72 2 1 105 2 136 3 116 4 66 3 1 105 2 108 3 108 4 68 a) Find the seasonal factors Ans: 1.003 1.255 1.05 0.693

b) Suppose the sales of pyjamas during the four quarters of the fourth year are: Quarter Sales 1 95 2 116 3 89

4 73 Use simple exponential smoothing with alpha =0.2 to find the base level forecast for the first quarter of year 5. Assume the base level forecast for the first quarter of year 4 was 99.1. Also find the final forecasts for all the four quarters of year 5. Ans. Sales 95 116 89 73

Deseasonal Sales 94.72 92.43 84.76 105.34

Forecast 99.1 98.22 97.06 94.6 96.75 So the quarterly forecasts (rounded to the next integer) for the four quarters are: 97, 121, 102, 67...