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Some questions of forecasting

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School
IIT Kanpur**We aren't endorsed by this school
Course
GENERAL ASSESSMENT
Subject
Economics
Date
Dec 23, 2024
Pages
9
Uploaded by MasterArt15889
Good Good Good Good Good GoodEconometrics and Mathematical EconomicsShanghai University of Finance and Economics8 pag.Document shared on https://www.docsity.com/en/good-good-good-good-good-good-1/7057274/
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1/ 8Problem 1.Here are the actual tabulated demands for an item for a nine-month period (January through September). Your supervisor wants to test two forecasting methods to see which method was better over this period.MonthActual DemandMonthActual DemandJanuary110June180February130July140March150August130April170September140May160a.Forecast April through September using a three-month moving average.b.Use simple exponential smoothing with an alpha of 0.3 to estimate April through September, using the average of January through March as the initial forecast input (i.e., ??) to help forecast for April.c.Use MAD to decide which method produced the better forecast over the six-month period.Answer: For the exponential smoothing forecast we need a beginning forecast for March and an . For the beginning forecast use the average of the first three periods and select = 0.3. Other choices will produce different answers.MonthDemand3-Mo.MAAbsolute DeviationExponential SmoothingAbsolute DeviationJanuary110February130March150130.00April17013040136.0034.00May16015010146.2013.80June18016020150.3429.66July14017030159.2419.24August13016030153.4723.47September14015010146.436.43MAD23.321.1Based upon MAD, the exponential smoothing model appears to be the best.Problem 2.Historical demand for a product is as follows.Month Actual Demand Month Actual Demand April 60 July 60 May 55 August 80 June 75 September 75 Document shared on https://www.docsity.com/en/good-good-good-good-good-good-1/7057274/
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2/ 8a.Using a simple four-month moving average, calculate a forecast for October.b.Using single exponential smoothing with ? = 0.2and a September forecast = 65, calculate a forecast for October.c.Using simple linear regression, calculate the trend line for the historical data. Let us say the ?axis is April = 1, May = 2, and so on, while the ?axis is demand.d.Calculate a forecast for October using your regression formula.Answer:a.?October=(75 + 80 + 60 + 75)/4 = 72.5b.?October=?September+ ?(?September− ?September) =65 +0.2(75 – 65) = 67.0c.y= 405/6 = 67.5x= 21/6 = 3.5b = 222)5.3(6915.67)5.3(61485−−=−−xnxyxnxy= 3.86a = xby−= 67.5 – 3.86(3.5) = 54.00Y = a + bx = 54.0 + 3.86xd.?October=54.00 + 3.86×7 = 81.01 Problem 3.Zeus Computer Chips, Inc., used to have major contracts to produce the Centrino-type chips. The market has been declining during the past three years because of the dual-core chips, which it cannot produce, so Zeus has the unpleasant task of forecasting next year. The task is unpleasant because the firm has not been able to find replacement chips for its product line. Here is demand over the past 12 quarters:2017 2018 2019 Quarter 1 4,800 Quarter 1 3,500 Quarter 1 3,200 Quarter 2 3,500 Quarter 2 2,700 Quarter 2 2,100 Quarter 3 4,300 Quarter 3 3,500 Quarter 3 2,700 Quarter 4 3,000 Quarter 4 2,400 Quarter 4 1,700 Use the decomposition technique (seasonal factor with linear trend) to forecast the sales of four quarters of 2020.Answer:Document shared on https://www.docsity.com/en/good-good-good-good-good-good-1/7057274/
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3/ 8Forecast: Quarter1 = 2488.28, Quarter2 = 1646.54, Quarter3 = 1894.04, Quarter4 = 1152.97 Problem 4.A company currently using an inspection process in its material receiving department is trying to install an overall cost reduction program. One possible reduction is the elimination of one inspection position. This position tests material that has a defective content on the average of 0.04. By inspecting all items, the inspector is able to remove all defects. The inspector can inspect 50 units per hour. The hourly rate including fringe benefits for the position is $9. If the inspection position is eliminated, defects will go into product assembly and will have to be replaced later at a cost of $10 each when they are detected in final product testing.a.Should this inspection position be eliminated?b.What is the cost to inspect each unit?c.Is there benefit (or loss) from the current inspection process? How much?Answer: Defective average rate= 0.04, inspection rate = 50 per hour, cost of inspector = $9 per hour, and repair cost is $10 each.a.Calculation Cost per hour No inspection 0.04 * (50) * 10 $20 Inspection $9 Therefore, it is cheaper to inspect in this case.b.Cost per unit for inspection = 9/50 = $0.18. c.Benefit from the current inspection process isHourly: Cost of no inspection – Cost of inspection = 20 -9 = $11,orPer unit: Average cost of quality – Average Cost of inspection = 0.04*10 -0.18 = $0.22.Document shared on https://www.docsity.com/en/good-good-good-good-good-good-1/7057274/
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4/ 8For Problem 5 and 6, we assume the desired quality level is 3-Sigma.Problem 5.A metal fabricator produces connecting rods with an outer diameter that has a 1±0.01-inch specification. A machine operator takes several sample measurements over time and determines the sample mean outer diameter to be 1.002 inches with a standard deviation of 0.003 inch. Calculate the process capability index for this example.Answer: 333.1,889.min)003(.399.002.1,)003(.3002.101.1min3,3min=−−=−−=LTLXXUTLCpk= 0.889 Problem 6. C-Spec, Inc., is attempting to determine whether an existing machine is capable of milling an engine part that has a key specification of4±0.003-inches. After a trial run on this machine, C-Spec has determined that the machine has a sample mean of 4.001 inches with a standard deviation of 0.002 inch. Calculate the ???for this machine.Answer:667,.333.min)002(.3997.3001.4,)002(.3001.4003.4min3,3min=−−=−−=LSLXXUSLCpk= 0.333Problem 7. The state and local police departments are trying to analyze crime rates so they can shift their patrols from decreasing-rate areas to where rates are increasing. The city and county have been geographically segmented into areas containing 5,000 residences. The police recognize that not all crimes and offenses are reported: People do not want to become involved, consider the offense too small to report, are too embarrassed to make a police report, or do not take the time, among other reasons. Every month, because of this, the police are contacting by phone a random sample of 1,000 of the 5,000 residences for data on crime. (Respondents are guaranteed anonymity.) Here are the data collected for the past 12 months for one area.Month Crime Incidence Sample Size Crime Rate January 7 1,000 0.007 February 9 1,000 0.009 March 7 1,000 0.007 April 7 1,000 0.007 May 7 1,000 0.007 Document shared on https://www.docsity.com/en/good-good-good-good-good-good-1/7057274/
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5/ 8June 9 1,000 0.009 July 7 1,000 0.007 August 10 1,000 0.010 September 8 1,000 0.008 October 11 1,000 0.011 November 10 1,000 0.010 December 8 1,000 0.008 Construct a?-chart for 95 percent confidence (i.e., the number of standard deviations is 1.96) and plot each of the months. If the next three months show crime residences in this area asJanuary = 10(out of 1,000 sampled)February = 12(out of 1,000 sampled)March = 11(out of 1,000 sampled)What comments can you make regarding the crime rate?Answer:?̅ =10012×1000= .008333,??= √?̅(1−?̅)?= √.008333×(1−.008333)1000= .00287,UCL = ?̅ + 1.96??= .008333 + 1.96 × .00287 = .01396,LCL = ?̅ − 1.96??= .008333 − 1.96 × .00287 = .00270.Document shared on https://www.docsity.com/en/good-good-good-good-good-good-1/7057274/
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6/ 8The process is in control. Therefore, it can be stated that the crime rate has not increased. However, there appears to be a gradual increase in the crime rate. Problem 8.The following table contains the measurements of the key length dimension from a fuel injector. These samples of size five were taken at one-hour intervals.Observations Sample Number 1 2 3 4 5 1 .486 .499 .493 .511 .481 2 .499 .506 .516 .494 .529 3 .496 .500 .515 .488 .521 4 .495 .506 .483 .487 .489 5 .472 .502 .526 .469 .481 6 .473 .495 .507 .493 .506 7 .495 .512 .490 .471 .504 8 .525 .501 .498 .474 .485 9 .497 .501 .517 .506 .516 10 .495 .505 .516 .511 .497 11 .495 .482 .468 .492 .492 12 .483 .459 .526 .506 .522 13 .521 .512 .493 .525 .510 14 .487 .521 .507 .501 .500 15 .493 .516 .499 .511 .513 16 .473 .506 .479 .480 .523 17 .477 .485 .513 .484 .496 18 .515 .493 .493 .485 .475 19 .511 .536 .486 .497 .491 20 .509 .490 .470 .504 .512 Construct a three-sigma ?̅-chart and ?-chart (using the form of ?2, ?3, ?4provided in the slides) for the length of the fuel injector. What can say about the process?Answer:Sample number12345MeanRange1.486.499.493.511.481.494.0302.499.506.516.494.529.509.0353.496.500.515.488.521.504.0334.495.506.483.487.489.492.0235.472.502.526.469.481.490.0576.473.495.507.493.506.495.0347.495.512.490.471.504.494.0418.525.501.498.474.485.497.051Document shared on https://www.docsity.com/en/good-good-good-good-good-good-1/7057274/
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7/ 89.497.501.517.506.516.507.02010.495.505.516.511.497.505.02111.495.482.468.492.492.486.02712.483.459.526.506.522.499.06713.521.512.493.525.510.512.03214.487.521.507.501.500.503.03415.493.516.499.511.513.506.02316.473.506.479.480.523.492.05017.477.485.513.484.496.491.03618.515.493.493.485.475.492.04019.511.536.486.497.491.504.05020.509.490.470.504.512.497.042=X.499, R= .037 Control limits for ?̅-chart: UCL = ?+ ?2?= .499 + .58×.037 = .520, LCL =?− ?2?= .499 − .58×.037 = .478. Control limits for ?-chart: UCL = ?4?= 2.11×.037 = .078,LCL = ?3?= 0×.037 = 0.00.Document shared on https://www.docsity.com/en/good-good-good-good-good-good-1/7057274/
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8/ 8Process is in statistical control.Document shared on https://www.docsity.com/en/good-good-good-good-good-good-1/7057274/
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