《数据模型与决策》练习题及答案
数据模型与决策 第1章 - baidu
P14
第1章：P15第8题（a , c）
回忆一下下面的生产模型
max=10x
s.t.
5x≤40
x≥0
假设现在公司考虑生产第二张产品，它的单位利润是5美元，单位制造用时为2小时。
用y表示产品2的产量。
a.考虑同时生产两种产品，请写出数学模型。
c. 画出该模型的输入输出流程图。
a. 答:max 10x+5y
s.t.
5x+2y≤40
x≥0；y≥0
b.
第1章：P15第14题
Eastman出版公司正在考虑出版一本关于电子数据表在商业上的应用的平板教科书。
估计原稿准备、教科书设计以及生产设备的固定成本是80000美元，每本书可变的生产和原料成本估计是3美元，对书的需求估计是4000册。
该公司想以每本20美元的价格卖给大学和学院的书店。
a.盈亏平衡点是多少？
b.需求是4000册时，预期利润或损失时多少？
c.需求是4000册时，为达到盈亏平衡，该公司必须索取的最低单价是多少？
答：a.max 17x-80000
s.t. x≥0
盈亏平衡点是 4706本
b.将x=4000带入求解 17*4000-80000=-12000，所以亏本12000元
c.设最低单价为￥，（￥-3）*4000-80000=0,求得，￥=23元/本。
数据模型与决策(运筹学)课后习题和案例答案(6)
CHAPTER 7NETWORK OPTIMIZATION PROBLEMS Review Questions7.1-1 A supply node is a node where the net amount of flow generated is a fixed positive number.A demand node is a node where the net amount of flow generated is a fixed negativenumber. A transshipment node is a node where the net amount of flow generated is fixed at zero.7.1-2 The maximum amount of flow allowed through an arc is referred to as the capacity of thatarc.7.1-3 The objective is to minimize the total cost of sending the available supply through thenetwork to satisfy the given demand.7.1-4 The feasible solutions property is necessary. It states that a minimum cost flow problemwill have a feasible solution if and only if the sum of the supplies from its supply nodesequals the sum of the demands at its demand nodes.7.1-5 As long as all its supplies and demands have integer values, any minimum cost flowproblem with feasible solutions is guaranteed to have an optimal solution with integervalues for all its flow quantities.7.1-6 Network simplex method.7.1-7 Applications of minimum cost flow problems include operation of a distribution network,solid waste management, operation of a supply network, coordinating product mixes atplants, and cash flow management.7.1-8 Transportation problems, assignment problems, transshipment problems, maximum flowproblems, and shortest path problems are special types of minimum cost flow problems. 7.2-1 One of the company’s most important distribution centers (Los Angeles) urgently needs anincreased flow of shipments from the company.7.2-2 Auto replacement parts are flowing through the network from the company’s main factoryin Europe to its distribution center in LA.7.2-3 The objective is to maximize the flow of replacement parts from the factory to the LAdistribution center.7.3-1 Rather than minimizing the cost of the flow, the objective is to find a flow plan thatmaximizes the amount flowing through the network from the source to the sink.7.3-2 The source is the node at which all flow through the network originates. The sink is thenode at which all flow through the network terminates. At the source, all arcs point awayfrom the node. At the sink, all arcs point into the node.7.3-3 The amount is measured by either the amount leaving the source or the amount entering thesink.7.3-4 1. Whereas supply nodes have fixed supplies and demand nodes have fixed demands, thesource and sink do not.2. Whereas the number of supply nodes and the number of demand nodes in a minimumcost flow problem may be more than one, there can be only one source and only onesink in a standard maximum flow problem.7.3-5 Applications of maximum flow problems include maximizing the flow through adistribution network, maximizing the flow through a supply network, maximizing the flow of oil through a system of pipelines, maximizing the flow of water through a system ofaqueducts, and maximizing the flow of vehicles through a transportation network.7.4-1 The origin is the fire station and the destination is the farm community.7.4-2 Flow can go in either direction between the nodes connected by links as opposed to onlyone direction with an arc.7.4-3 The origin now is the one supply node, with a supply of one. The destination now is theone demand node, with a demand of one.7.4-4 The length of a link can measure distance, cost, or time.7.4-5 Sarah wants to minimize her total cost of purchasing, operating, and maintaining the carsover her four years of college.7.4-6 When “real travel” through a network can end at more that one node, a dummy destinationneeds to be added so that the network will have just a single destination.7.4-7 Quick’s management must consider trade-offs between time and cost in making its finaldecision.7.5-1 The nodes are given, but the links need to be designed.7.5-2 A state-of-the-art fiber-optic network is being designed.7.5-3 A tree is a network that does not have any paths that begin and end at the same nodewithout backtracking. A spanning tree is a tree that provides a path between every pair of nodes. A minimum spanning tree is the spanning tree that minimizes total cost.7.5-4 The number of links in a spanning tree always is one less than the number of nodes.Furthermore, each node is directly connected by a single link to at least one other node. 7.5-5 To design a network so that there is a path between every pair of nodes at the minimumpossible cost.7.5-6 No, it is not a special type of a minimum cost flow problem.7.5-7 A greedy algorithm will solve a minimum spanning tree problem.17.5-8 Applications of minimum spanning tree problems include design of telecommunicationnetworks, design of a lightly used transportation network, design of a network of high- voltage power lines, design of a network of wiring on electrical equipment, and design of a network of pipelines.Problems7.1a)b)c)1[40] 6 S17 4[-30] D1 [-40] D2 [60] 5 8S2 6[-30] D37.2a)supply nodestransshipment nodesdemand nodesb)[200] P1560 [150]425 [125][0] W1505[150]490 [100]470 [100][-150]RO1[-200]RO2P2 [300]c)510 [175]600 [200][0] W2390 [125]410[150] 440[75]RO3[-150]7.3a)supply nodestransshipment nodesdemand nodesV1W1F1V2V3W2 F21P1W1RO1RO2P2W2RO3[-50] SE3000[20][0]BN5700[40][0]HA[50]BE 4000 6300[40][30] [0][0]NY2000[60]2400[20]3400[10] 4200[80][0]5900[60]5400[40]6800[50]RO[0]BO[0]2500[70]2900[50]b)c)7.4a)LA 3100 NO 6100 LI 3200 ST[-130] [70] [30] [40] [130]1[70]11b)c) The total shipping cost is $2,187,000.7.5a)[0][0] 5900RONY[60] 5400[0] 2900 [50]4200 [80][0] [40] 6800 [50]BO[0] 2500LA 3100 NO 6100 LI 3200 ST [-130][70][30] [40][130]b)c)SEBNHABERONYNY(80) [80] (50) [60](30)[40] ROBO (40)(50) [50] (70)[70]11d)e)f) $1,618,000 + $583,000 = $2,201,000 which is higher than the total in Problem 7.5 ($2,187,000). 7.6LA(70) NO[50](30)LI (30) ST[70][30] [40]There are only two arcs into LA, with a combined capacity of 150 (80 + 70). Because ofthis bottleneck, it is not possible to ship any more than 150 from ST to LA. Since 150 actually are being shipped in this solution, it must be optimal. 7.7[-50] SE3000 [20] [0] BN 5700 [40][0] HA[50] BE4000 6300[40][0] NY2000 [60] 2400 [20][30] [0]5900RO [60]17.8 a) SourcesTransshipment Nodes Sinkb)7.9 a)AKR1[75]A [60]R2[65] [40][50][60] [45]D [120] [70]B[55]E[190]T [45][80] [70][70]R3CF[130][90]SE PT KC SL ATCHTXNOMES S F F CAb)Oil Fields Refineries Distribution CentersTXNOPTCACHATAKSEKCME c)SLSFTX[11][7] NO[5][9] PT[8] [2][5] CA [4] [7] [8] [7] [4] [6][8] CH [7][5][9] [4] ATAK [3][6][6][12] SE KC[8][9][4][8] [7] [12] [11]MESL [9]SF[15][7]d)3Shortest path: Fire Station – C – E – F – Farming Community 7.11 a)A70D40 60O60 5010 B 20 C5540 10 T50E801c)Shortest route: Origin – A – B – D – Destinationd)Yese)Yes7.12a)31,00018,000 21,00001238,000 10,000 12,000b)17.13a) Times play the role of distances.B 2 2 G5ACE 1 31 1b)7.14D F1. C---D: Cost = 14.E---G: Cost = 5E---F: Cost = 1 *choose arbitrarilyD---A: Cost = 4 2.E---G: Cost = 5 E---B: Cost = 7 E---B: Cost = 7 F---G: Cost = 7 E---C: Cost = 4 C---A: Cost = 5F---G: Cost = 7C---B: Cost = 2 *lowestF---C: Cost = 3 *lowest5.E---G: Cost = 5 F---D: Cost = 4 D---A: Cost = 43. E---G: Cost = 5 B---A: Cost = 2 *lowestE---B: Cost = 7 F---G: Cost = 7 F---G: Cost = 7 C---A: Cost = 5F---D: Cost = 46.E---G: Cost = 5 *lowestC---D: Cost = 1 *lowestF---G: Cost = 7C---A: Cost = 5C---B: Cost = 2Total = $14 million7.151. B---C: Cost = 1 *lowest 4. B---E: Cost = 72. B---A: Cost = 4 C---F: Cost = 4 *lowestB---E: Cost = 7 C---E: Cost = 5C---A: Cost = 6 D---F: Cost = 5C---D: Cost = 2 *lowest 5. B---E: Cost = 7C---F: Cost = 4 C---E: Cost = 5C---E: Cost = 5 F---E: Cost = 1 *lowest3. B---A: Cost = 4 *lowest F---G: Cost = 8B---E: Cost = 7 6. E---G: Cost = 6 *lowestC---A: Cost = 6 F---G: Cost = 8C---F: Cost = 4C---E: Cost = 5D---A: Cost = 5 Total = $18,000D---F: Cost = 57.16B 34 2E HA D 2 G I K3C F 12J34B41E6A C41G2 FD1. F---G: Cost = 1 *lowest 6. D---A: Cost = 62. F---C: Cost = 6 D---B: Cost = 5F---D: Cost = 5 D---C: Cost = 4F---I: Cost = 2 *lowest E---B: Cost = 3 *lowestF---J: Cost = 5 F---C: Cost = 6G---D: Cost = 2 F---J: Cost = 5G---E: Cost = 2 H---K: Cost = 7G---H: Cost = 2 I---K: Cost = 8G---I: Cost = 5 I---J: Cost = 33. F---C: Cost = 6 7. B---A: Cost = 4F---D: Cost = 5 D---A: Cost = 6F---J: Cost = 5 D---C: Cost = 4G---D: Cost = 2 *lowest F---C: Cost = 6G---E: Cost = 2 F---J: Cost = 5G---H: Cost = 2 H---K: Cost = 7I---H: Cost = 2 I---K: Cost = 8I---K: Cost = 8 I---J: Cost = 3 *lowestI---J: Cost = 3 8. B---A: Cost = 4 *lowest4. D---A: Cost = 6 D---A: Cost = 6D---B: Cost = 5 D---C: Cost = 4D---E: Cost = 2 *lowest F---C: Cost = 6D---C: Cost = 4 H---K: Cost = 7F---C: Cost = 6 I---K: Cost = 8F---J: Cost = 5 J---K: Cost = 4G---E: Cost = 2 9. A---C: Cost = 3 *lowestG---H: Cost = 2 D---C: Cost = 4I---H: Cost = 2 F---C: Cost = 6I---K: Cost = 8 H---K: Cost = 7I---J: Cost = 3 I---K: Cost = 85. D---A: Cost = 6 J---K: Cost = 4D---B: Cost = 5 10. H---K: Cost = 7D---C: Cost = 4 I---K: Cost = 8E---B: Cost = 3 J---K: Cost = 4 *lowestE---H: Cost = 4F---C: Cost = 6F---J: Cost = 5G---H: Cost = 2 *lowest Total = $26 millionI---H: Cost = 2I---K: Cost = 8I---J: Cost = 37.17a) The company wants a path between each pair of nodes (groves) that minimizes cost(length of road).b)7---8 : Distance = 0.57---6 : Distance = 0.66---5 : Distance = 0.95---1 : Distance = 0.75---4 : Distance = 0.78---3 : Distance = 1.03---2 : Distance = 0.9Total = 5.3 miles7.18a) The bank wants a path between each pair of nodes (offices) that minimizes cost(distance).b) B1---B5 : Distance = 50B5---B3 : Distance = 80B1---B2 : Distance = 100B2---M : Distance = 70B2---B4 : Distance = 120Total = 420 milesHamburgBostonRotterdamSt. PetersburgNapoliMoscowA IRFIELD SLondonJacksonvilleBerlin RostovIstanbulCases7.1a) The network showing the different routes troops and supplies may follow to reach the Russian Federation appears below.PORTSb)The President is only concerned about how to most quickly move troops and suppliesfrom the United States to the three strategic Russian cities. Obviously, the best way to achieve this goal is to find the fastest connection between the US and the three cities.We therefore need to find the shortest path between the US cities and each of the three Russian cities.The President only cares about the time it takes to get the troops and supplies to Russia.It does not matter how great a distance the troops and supplies cover. Therefore we define the arc length between two nodes in the network to be the time it takes to travel between the respective cities. For example, the distance between Boston and London equals 6,200 km. The mode of transportation between the cities is a Starlifter traveling at a speed of 400 miles per hour * 1.609 km per mile = 643.6 km per hour. The time is takes to bring troops and supplies from Boston to London equals 6,200 km / 643.6 km per hour = 9.6333 hours. Using this approach we can compute the time of travel along all arcs in the network.By simple inspection and common sense it is apparent that the fastest transportation involves using only airplanes. We therefore can restrict ourselves to only those arcs in the network where the mode of transportation is air travel. We can omit the three port cities and all arcs entering and leaving these nodes.The following six spreadsheets find the shortest path between each US city (Boston and Jacksonville) and each Russian city (St. Petersburg, Moscow, and Rostov).The spreadsheets contain the following formulas:Comparing all six solutions we see that the shortest path from the US to Saint Petersburg is Boston → London → Saint Petersburg with a total travel time of 12.71 hours. The shortest path from the US to Moscow is Boston → London → Moscow with a total travel time of 13.21 hours. The shortest path from the US to Rostov is Boston →Berlin → Rostov with a total travel time of 13.95 hours. The following network diagram highlights these shortest paths.-1c)The President must satisfy each Russian city’s military requirements at minimum cost.Therefore, this problem can be solved as a minimum-cost network flow problem. The two nodes representing US cities are supply nodes with a supply of 500 each (wemeasure all weights in 1000 tons). The three nodes representing Saint Petersburg, Moscow, and Rostov are demand nodes with demands of –320, -440, and –240,respectively. All nodes representing European airfields and ports are transshipment nodes. We measure the flow along the arcs in 1000 tons. For some arcs, capacityconstraints are given. All arcs from the European ports into Saint Petersburg have zero capacity. All truck routes from the European ports into Rostov have a transportation limit of 2,500*16 = 40,000 tons. Since we measure the arc flows in 1000 tons, the corresponding arc capacities equal 40. An analogous computation yields arc capacities of 30 for both the arcs connecting the nodes London and Berlin to Rostov. For all other nodes we determine natural arc capacities based on the supplies and demands at the nodes. We define the unit costs along the arcs in the network in $1000 per 1000 tons (or, equivalently, $/ton). For example, the cost of transporting 1 ton of material from Boston to Hamburg equals $30,000 / 240 = $125, so the costs of transporting 1000 tons from Boston to Hamburg equals $125,000.The objective is to satisfy all demands in the network at minimum cost. The following spreadsheet shows the entire linear programming model.HamburgBoston Rotterdam St.Petersburg+500-320Napoli Moscow A IRF IELDSLondon -440Jacksonville Berlin Rostov+500-240Istanbul The total cost of the operation equals $412.867 million. The entire supply for SaintPetersburg is supplied from Jacksonville via London. The entire supply for Moscow is supplied from Boston via Hamburg. Of the 240 (= 240,000 tons) demanded by Rostov, 60 are shipped from Boston via Istanbul, 150 are shipped from Jacksonville viaIstanbul, and 30 are shipped from Jacksonville via London. The paths used to shipsupplies to Saint Petersburg, Moscow, and Rostov are highlighted on the followingnetwork diagram.PORTSd)Now the President wants to maximize the amount of cargo transported from the US tothe Russian cities. In other words, the President wants to maximize the flow from the two US cities to the three Russian cities. All the nodes representing the European ports and airfields are once again transshipment nodes. The flow along an arc is againmeasured in thousands of tons. The new restrictions can be transformed into arccapacities using the same approach that was used in part (c). The objective is now to maximize the combined flow into the three Russian cities.The linear programming spreadsheet model describing the maximum flow problem appears as follows.The spreadsheet shows all the amounts that are shipped between the various cities. The total supply for Saint Petersburg, Moscow, and Rostov equals 225,000 tons, 104,800 tons, and 192,400 tons, respectively. The following network diagram highlights the paths used to ship supplies between the US and the Russian Federation.PORTSHamburgBoston Rotterdam St.Petersburg+282.2 -225NapoliMoscowAIRFIELDS-104.8LondonJacksonvilleBerlin Rostov +240 -192.4Istanbule)The creation of the new communications network is a minimum spanning tree problem.As usual, a greedy algorithm solves this type of problem.Arcs are added to the network in the following order (one of several optimal solutions):Rostov - Orenburg 120Ufa - Orenburg 75Saratov - Orenburg 95Saratov - Samara 100Samara - Kazan 95Ufa – Yekaterinburg 125Perm – Yekaterinburg 857.2a) There are three supply nodes – the Yen node, the Rupiah node, and the Ringgit node.There is one demand node – the US$ node. Below, we draw the network originatingfrom only the Yen supply node to illustrate the overall design of the network. In thisnetwork, we exclude both the Rupiah and Ringgit nodes for simplicity.b)Since all transaction limits are given in the equivalent of $1000 we define the flowvariables as the amount in thousands of dollars that Jake converts from one currencyinto another one. His total holdings in Yen, Rupiah, and Ringgit are equivalent to $9.6million, $1.68 million, and $5.6 million, respectively (as calculated in cells I16:K18 inthe spreadsheet). So, the supplies at the supply nodes Yen, Rupiah, and Ringgit are -$9.6 million, -$1.68 million, and -$5.6 million, respectively. The demand at the onlydemand node US$ equals $16.88 million (the sum of the outflows from the sourcenodes). The transaction limits are capacity constraints for all arcs leaving from thenodes Yen, Rupiah, and Ringgit. The unit cost for every arc is given by the transactioncost for the currency conversion.Jake should convert the equivalent of $2 million from Yen to each US$, Can$, Euro, and Pound. He should convert $1.6 million from Yen to Peso. Moreover, he should convert the equivalent of $200,000 from Rupiah to each US$, Can$, and Peso, $1 million from Rupiah to Euro, and $80,000 from Rupiah to Pound. Furthermore, Jake should convert the equivalent of $1.1 million from Ringgit to US$, $2.5 million from Ringgit to Euro, and $1 million from Ringgit to each Pound and Peso. Finally, he should convert all the money he converted into Can$, Euro, Pound, and Peso directly into US$. Specifically, he needs to convert into US$ the equivalent of $2.2 million, $5.5 million, $3.08 million, and $2.8 million Can$, Euro, Pound, and Peso, respectively. Assuming Jake pays for the total transaction costs of $83,380 directly from his American bank accounts he will have $16,880,000 dollars to invest in the US.c)We eliminate all capacity restrictions on the arcs.Jake should convert the entire holdings in Japan from Yen into Pounds and then into US$, the entire holdings in Indonesia from Rupiah into Can$ and then into US$, and the entire holdings in Malaysia from Ringgit into Euro and then into US$. Without the capacity limits the transaction costs are reduced to $67,480.d)We multiply all unit cost for Rupiah by 6.The optimal routing for the money doesn't change, but the total transaction costs are now increased to $92,680.e)In the described crisis situation the currency exchange rates might change every minute.Jake should carefully check the exchange rates again when he performs thetransactions.The European economies might be more insulated from the Asian financial collapse than the US economy. To impress his boss Jake might want to explore other investment opportunities in safer European economies that provide higher rates of return than US bonds.。
数据、模型与决策(运筹学)课后习题和案例答案013
CHAPTER 13FORECASTINGReview Questions13.1-1 Substantially underestimating demand is likely to lead to many lost sales, unhappycustomers, and perhaps allowing the competition to gain the upper hand in the marketplace. Significantly overestimating the demand is very costly due to excessive inventory costs, forced price reductions, unneeded production or storage capacity, and lost opportunity to market more profitable goods.13.1-2 A forecast of the demand for spare parts is needed to provide good maintenanceservice.13.1-3 In cases where the yield of a production process is less than 100%, it is useful toforecast the production yield in order to determine an appropriate value of reject allowance and, consequently, the appropriate size of the production run.13.1-4 Statistical models to forecast economic trends are commonly called econometricmodels.13.1-5 Providing too few agents leads to unhappy customers, lost calls, and perhaps lostbusiness. Too many agents cause excessive personnel costs.13.2-1 The company mails catalogs to its customers and prospective customers severaltimes per year, as well as publishing mini-catalogs in computer magazines. They then take orders for products over the phone at the company’s call center.13.2-2 Customers who receive a busy signal or are on hold too long may not call back andbusiness may be lost. If too many agents are on duty there may be idle time, which wastes money because of labor costs.13.2-3 The manager of the call center is Lydia Weigelt. Her current major frustration is thateach time she has used her procedure for setting staffing levels for the upcoming quarter, based on her forecast of the call volume, the forecast usually has turned out to be considerably off.13.2-4 Assume that each quarter’s cal l volume will be the same as for the preceding quarter,except for adding 25% for quarter 4.13.2-5 The average forecasting error is commonly called MAD, which stands for MeanAbsolute Deviation. Its formula is MAD = (Sum of forecasting errors) / (Number of forecasts)13.2-6 MSE is the mean square error. Its formula is (Sum of square of forecasting errors) /(Number of forecasts).13.2-7 A time series is a series of observations over time of some quantity of interest.13.3-1 In general, the seasonal factor for any period of a year measures how that periodcompares to the overall average for an entire year.13.3-2 Seasonally adjusted call volume = (Actual call volume) / (Seasonal factor).13.3-3 Actual forecast = (Seasonal factor)(Seasonally adjusted forecast)13.3-4 The last-value forecasting method sometimes is called the naive method becausestatisticians consider it naive to use just a sample size of one when additional relevant data are available.13.3-5 Conditions affecting the CCW call volume were changing significantly over the pastthree years.13.3-6 Rather than using old data that may no longer be relevant, this method averages thedata for only the most recent periods.13.3-7 This method modifies the moving-average method by placing the greatest weighton the last value in the time series and then progressively smaller weights on the older values.13.3-8 A small value is appropriate if conditions are remaining relatively stable. A largervalue is needed if significant changes in the conditions are occurring relatively frequently.13.3-9 Forecast = α(Last Value) + (1 –α)(Last forecast). Estimated trend is added to thisformula when using exponential smoothing with trend.13.3-10 T he one big factor that drives total sales up or down is whether there are any hotnew products being offered.13.4-1 CB Predictor uses the raw data to provide the best fit for all these inputs as well asthe forecasts.13.4-2 Each piece of data should have only a 5% chance of falling below the lower line and a5% chance of rising above the upper line.13.5-1 The next value that will occur in a time series is a random variable.13.5-2 The goal of time series forecasting methods is to estimate the mean of theunderlying probability distribution of the next value of the time series as closely as possible.13.5-3 No, the probability distribution is not the same for every quarter.13.5-4 Each of the forecasting methods, except for the last-value method, placed at leastsome weight on the observations from Year 1 to estimate the mean for each quarter in Year 2. These observations, however, provide a poor basis for estimating the mean of the Year 2 distribution.13.5-5 A time series is said to be stable if its underlying probability distribution usuallyremains the same from one time period to the next. A time series is unstable if both frequent and sizable shifts in the distribution tend to occur.13.5-6 Since sales drive call volume, the forecasting process should begin by forecastingsales.13.5-7 The major components are the relatively stable market base of numerous small-niche products and each of a few major new products.13.6-1 Causal forecasting obtains a forecast of the quantity of interest by relating it directlyto one or more other quantities that drive the quantity of interest.13.6-2 The dependent variable is call volume and the independent variable is sales.13.6-3 When doing causal forecasting with a single independent variable, linear regressioninvolves approximating the relationship between the dependent variable and the independent variable by a straight line.13.6-4 In general, the equation for the linear regression line has the form y = a + bx. Ifthere is more than one independent variable, then this regression equation has a term, a constant times the variable, added on the right-hand side for each of these variables.13.6-5 The procedure used to obtain a and b is called the method of least squares.13.6-6 The new procedure gives a MAD value of only 120 compared with the old MADvalue of 400 with the 25% rule.13.7-1 Statistical forecasting methods cannot be used if no data are available, or if the dataare not representative of current conditions.13.7-2 Even when good data are available, some managers prefer a judgmental methodinstead of a formal statistical method. In many other cases, a combination of the two may be used.13.7-3 The jury of executive opinion method involves a small group of high-level managerswho pool their best judgment to collectively make a forecast rather than just the opinion of a single manager.13.7-4 The sales force composite method begins with each salesperson providing anestimate of what sales will be in his or her region.13.7-5 A consumer market survey is helpful for designing new products and then indeveloping the initial forecasts of their sales. It is also helpful for planning a marketing campaign.13.7-6 The Delphi method normally is used only at the highest levels of a corporation orgovernment to develop long-range forecasts of broad trends.13.8-1 Generally speaking, judgmental forecasting methods are somewhat more widelyused than statistical methods.13.8-2 Among the judgmental methods, the most popular is a jury of executive opinion.Manager’s opinion is a close second.13.8-3 The survey indicates that the moving-average method and linear regression are themost widely used statistical forecasting methods.Problems13.1 a) Forecast = last value = 39b) Forecast = average of all data to date = (5 + 17 + 29 + 41 + 39) / 5 = 131 / 5 =26c) Forecast = average of last 3 values = (29 + 41 + 39) / 3 = 109 / 3 = 36d) It appears as if demand is rising so the average forecasting method seemsinappropriate because it uses older, out-of-date data.13.2 a) Forecast = last value = 13b) Forecast = average of all data to date = (15 + 18 + 12 + 17 + 13) / 5 = 75 / 5 =15c) Forecast = average of last 3 values = (12 + 17 + 13) / 3 = 42 / 3 = 14d) The averaging method seems best since all five months of data are relevant indetermining the forecast of sales for next month and the data appears relativelystable.13.3MAD = (Sum of forecasting errors) / (Number of forecasts) = (18 + 15 + 8 + 19) / 4 = 60 / 4 = 15 MSE = (Sum of squares of forecasting errors) / (Number of forecasts) = (182 + 152 + 82 + 192) / 4 = 974 / 4 = 243.513.4 a) Method 1 MAD = (258 + 499 + 560 + 809 + 609) / 5 = 2,735 / 5 = 547Method 2 MAD = (374 + 471 + 293 + 906 + 396) / 5 = 2,440 / 5 = 488Method 1 MSE = (2582 + 4992 + 5602 + 8092 + 6092) / 5 = 1,654,527 / 5 = 330,905Method 2 MSE = (3742 + 4712 + 2932 + 9062 + 3962) / 5 = 1,425,218 / 5 = 285,044Method 2 gives a lower MAD and MSE.b) She can use the older data to calculate more forecasting errors and compareMAD for a longer time span. She can also use the older data to forecast theprevious five months to see how the methods compare. This may make her feelmore comfortable with her decision.13.5 a)b)c)d)13.6 a)b)This progression indicatesthat the state’s economy is improving with the unemployment rate decreasing from 8% to 7% (seasonally adjusted) over the four quarters.13.7 a)b) Seasonally adjusted value for Y3(Q4)=28/1.04=27,Actual forecast for Y4(Q1) = (27)(0.84) = 23.c) Y4(Q1) = 23 as shown in partb Seasonally adjusted value for Y4(Q1) = 23 / 0.84 = 27 Actual forecast for Y4(Q2) = (27)(0.92) = 25Seasonally adjusted value for Y4(Q2) = 25 / 0.92 = 27 Actual forecast for Y4(Q3) = (27)(1.20) = 33Seasonally adjusted value for Y4(Q3) = 33/1.20 = 27Actual forecast for Y4(Q4) = (27)(1.04) = 28d)13.8 Forecast = 2,083 – (1,945 / 4) + (1,977 / 4) = 2,09113.9 Forecast = 782 – (805 / 3) + (793 / 3) = 77813.10 Forecast = 1,551 – (1,632 / 10) + (1,532 / 10) = 1,54113.11 Forecast(α) = α(last value) + (1 –α)(last forecast)Forecast(0.1) = (0.1)(792) + (1 –0.1)(782) = 783 Forecast(0.3) = (0.3)(792) + (1 –0.3)(782) = 785 Forecast(0.5) = (0.5)(792) + (1 – 0.5)(782) = 78713.12 Forecast(α) = α(last value) + (1 –α)(last forecast)Forecast(0.1) = (0.1)(1,973) + (1 –0.1)(2,083) = 2,072 Forecast(0.3) = (0.3)(1,973) + (1 –0.3)(2,083) = 2,050 Forecast(0.5) = (0.5)(1,973) + (1 – 0.5)(2,083) = 2,02813.13 a) Forecast(year 1) = initial estimate = 5000Forecast(year 2) = α(last value) + (1 –α)(last forecast)= (0.25)(4,600) + (1 –0.25)(5,000) = 4,900 Forecast(year 3) = (0.25)(5,300) + (1 – 0.25)(4,900) = 5,000b) MAD = (400 + 400 + 1,000) / 3 = 600MSE = (4002 + 4002 + 1,0002) / 3 = 440,000c) Forecast(next year) = (0.25)(6,000) + (1 – 0.25)(5,000) = 5,25013.14 Forecast = α(last value) + (1 –α)(last forecast) + Estimated trendEstimated trend = β(Latest trend) + (1 –β)(Latest estimate of trend) Latest trend = α(Last value – Next-to-last value) + (1 –α)(Last forecast – Next-to-last forecast)Forecast(year 1) = Initial average + Initial trend = 3,900 + 700 = 4,600Forecast (year 2) = (0.25)(4,600) + (1 –0.25)(4,600)+(0.25)[(0.25)(4,600 –3900) + (1 –0.25)(4,600 –3,900)] + (1 –0.25)(700) = 5,300Forecast (year 3) = (0.25)(5,300) + (1 – 0.25)(5,300) + (0.25)[(0.25)(5,300 – 4,600) + (1 – 0.25)(5,300 – 4,600)]+(1 – 0.25)(700) = 6,00013.15 Forecast = α(last value) + (1 –α)(last forecast) + Estimated trendEstimated trend = β(Latest trend) + (1 –β)(Latest estimate of trend) Latest trend = α(Last value – Next-to-last value) + (1 –α)(Last forecast – Next-to-last forecast)Forecast = (0.2)(550) + (1 – 0.2)(540) + (0.3)[(0.2)(550 – 535) + (1 – 0.2)(540 –530)] + (1 – 0.3)(10) = 55213.16 Forecast = α(last value) + (1 –α)(last forecast) + Estimated trendEstimated trend = β(Latest trend) + (1 –β)(Latest estimate of trend) Latest trend = α(Last value – Next-to-last value) + (1 –α)(Last forecast – Next-to-last forecast)Forecast = (0.1)(4,935) + (1 – 0.1)(4,975) + (0.2)[(0.1)(4,935 – 4,655) + (1 – 0.1) (4,975 – 4720)] + (1 – 0.2)(240) = 5,21513.17 a) Since sales are relatively stable, the averaging method would be appropriate forforecasting future sales. This method uses a larger sample size than the last-valuemethod, which should make it more accurate and since the older data is stillrelevant, it should not be excluded, as would be the case in the moving-averagemethod.b)c)d)e) Considering the MAD values (5.2, 3.0, and 3.9, respectively), the averagingmethod is the best one to use.f) Considering the MSE values (30.6, 11.1, and 17.4, respectively), the averagingmethod is the best one to use.g) Unless there is reason to believe that sales will not continue to be relatively stable,the averaging method should be the most accurate in the future as well.13.18 Using the template for exponential smoothing, with an initial estimate of 24, thefollowing forecast errors were obtained for various values of the smoothing constant α:use.13.19 a) Answers will vary. Averaging or Moving Average appear to do a better job thanLast Value.b) For Last Value, a change in April will only affect the May forecast.For Averaging, a change in April will affect all forecasts after April.For Moving Average, a change in April will affect the May, June, and July forecast.c) Answers will vary. Averaging or Moving Average appear to do a slightly better jobthan Last Value.d) Answers will vary. Averaging or Moving Average appear to do a slightly better jobthan Last Value.13.20 a) Since the sales level is shifting significantly from month to month, and there is noconsistent trend, the last-value method seems like it will perform well. Theaveraging method will not do as well because it places too much weight on olddata. The moving-average method will be better than the averaging method butwill lag any short-term trends. The exponential smoothing method will also lagtrends by placing too much weight on old data. Exponential smoothing withtrend will likely not do well because the trend is not consistent.b)Comparing MAD values (5.3, 10.0, and 8.1, respectively), the last-value method is the best to use of these three options.Comparing MSE values (36.2, 131.4, and 84.3, respectively), the last-value method is the best to use of these three options.c) Using the template for exponential smoothing, with an initial estimate of 120, thefollowing forecast errors were obtained for various values of the smoothingconstant α:constant is appropriate.d) Using the template for exponential smoothing with trend, using initial estimates of120 for the average value and 10 for the trend, the following forecast errors wereobtained for various values of the smoothing constants α and β:constants is appropriate.e) Management should use the last-value method to forecast sales. Using thismethod the forecast for January of the new year will be 166. Exponentialsmoothing with trend with high smoothing constants (e.g., α = 0.5 and β = 0.5)also works well. With this method, the forecast for January of the new year will be165.13.21 a) Shift in total sales may be due to the release of new products on top of a stableproduct base, as was seen in the CCW case study.b) Forecasting might be improved by breaking down total sales into stable and newproducts. Exponential smoothing with a relatively small smoothing constant canbe used for the stable product base. Exponential smoothing with trend, with arelatively large smoothing constant, can be used for forecasting sales of each newproduct.c) Managerial judgment is needed to provide the initial estimate of anticipated salesin the first month for new products. In addition, a manger should check theexponential smoothing forecasts and make any adjustments that may benecessary based on knowledge of the marketplace.13.22 a) Answers will vary. Last value seems to do the best, with exponential smoothingwith trend a close second.b) For last value, a change in April will only affect the May forecast.For averaging, a change in April will affect all forecasts after April.For moving average, a change in April will affect the May, June, and July forecast.For exponential smoothing, a change in April will affect all forecasts after April.For exponential smoothing with trend, a change in April will affect all forecastsafter April.c) Answers will vary. last value or exponential smoothing seem to do better than theaveraging or moving average.d) Answers will vary. last value or exponential smoothing seem to do better than theaveraging or moving average.13.23 a) Using the template for exponential smoothing, with an initial estimate of 50, thefollowing MAD values were obtained for various values of the smoothing constantα:Choose αb) Using the template for exponential smoothing, with an initial estimate of 50, thefollowing MAD values were obtained for various values of the smoothing constantα:Choose αc) Using the template for exponential smoothing, with an initial estimate of 50, thefollowing MAD values were obtained for various values of the smoothing constantα:13.24 a)b)Forecast = 51.c) Forecast = 54.13.25 a) Using the template for exponential smoothing with trend, with an initial estimatesof 50 for the average and 2 for the trend and α = 0.2, the following MAD values were obtained for various values of the smoothing constant β:Choose β = 0.1b) Using the template for exponential smoothing with trend, with an initial estimatesof 50 for the average and 2 for the trend and α = 0.2, the following MAD valueswere obtained for various values of the smoothing constant β:Choose β = 0.1c) Using the template for exponential smoothing with trend, with an initial estimatesof 50 for the average and 2 for the trend and α = 0.2, the following MAD valueswere obtained for various values of the smoothing constant β:13.26 a)b)0.582. Forecast = 74.c) = 0.999. Forecast = 79.13.27 a) The time series is not stable enough for the moving-average method. Thereappears to be an upward trend.b)c)d)e) Based on the MAD and MSE values, exponential smoothing with trend should beused in the future.β = 0.999.f)For exponential smoothing, the forecasts typically lie below the demands.For exponential smoothing with trend, the forecasts are at about the same level as demand (perhaps slightly above).This would indicate that exponential smoothing with trend is the best method to usehereafter.13.2913.30 a)factors:b)c) Winter = (49)(0.550) = 27Spring = (49)(1.027) = 50Summer = (49)(1.519) = 74Fall = (49)(0.904) = 44d)e)f)g) The exponential smoothing method results in the lowest MAD value (1.42) and thelowest MSE value (2.75).13.31 a)b)c)d)e)f)g) The last-value method with seasonality has the lowest MAD and MSE value. Usingthis method, the forecast for Q1 is 23 houses.h) Forecast(Q2) = (27)(0.92) = 25Forecast(Q3) = (27)(1.2) = 32Forecast(Q4) = (27)(1.04) = 2813.32 a)b) The moving-average method with seasonality has the lowest MAD value. Using13.33 a)b)c)d) Exponential smoothing with trend should be used.e) The best values for the smoothing constants are α = 0.3, β = 0.3, and γ = 0.001.C28:C38 below.13.34 a)b)c)d)e) Moving average results in the best MAD value (13.30) and the best MSE value(249.09).f)MAD = 14.17g) Moving average performed better than the average of all three so it should beused next year.h) The best method is exponential smoothing with seasonality and trend, using13.35 a)••••••••••0100200300400500600012345678910S a l e sMonthb)c)••••••••••0100200300400500600012345678910S a l e sMonthd) y = 410.33 + (17.63)(11) = 604 e) y = 410.33 + (17.63)(20) = 763f) The average growth in sales per month is 17.63.13.36 a)•••01000200030004000500060000123A p p l i c a t i o n sYearb)•••01000200030004000500060000123A p p l i c a t i o n sYearc)d) y (year 4) = 3,900+ (700)(4) = 6,700 y (year 5) = 3,900 + (700)(5) = 7,400 y (year 6) = 3,900 + (700)(6) = 8,100 y (year 7) = 3,900 + (700)(7) =8,800y (year 8) = 3,900 + (700)(8) = 9,500e) It does not make sense to use the forecast obtained earlier of 9,500. Therelationship between the variable has changed and, thus, the linear regression that was used is no longer appropriate.f)•••••••0100020003000400050006000700001234567A p p l i c a t i o n sYeary =5,229 +92.9x y =5,229+(92.9)(8)=5,971the forecast that it provides for year 8 is not likely to be accurate. It does not make sense to continue to use a linear regression line when changing conditions cause a large shift in the underlying trend in the data.g)Causal forecasting takes all the data into account, even the data from before changing conditions cause a shift. Exponential smoothing with trend adjusts to shifts in the underlying trend by placing more emphasis on the recent data.13.37 a)••••••••••50100150200250300350400450500012345678910A n n u a l D e m a n dYearb)c)••••••••••50100150200250300350400450500012345678910A n n u a l D e m a n dYeard) y = 380 + (8.15)(11) = 470 e) y = 380 = (8.15)(15) = 503f) The average growth per year is 8.15 tons.13.38 a) The amount of advertising is the independent variable and sales is the dependentvariable.b)•••••0510*******100200300400500S a l e s (t h o u s a n d s o f p a s s e n g e r s )Amount of Advertising ($1,000s)c)•••••0510*******100200300400500S a l e s (t h o u s a n d s o f p a s s e n g e r s )Amount of Advertising ($1,000s)d) y = 8.71 + (0.031)(300) = 18,000 passengers e) 22 = 8.71 + (0.031)(x ) x = $429,000f) An increase of 31 passengers can be attained.13.39 a) If the sales change from 16 to 19 when the amount of advertising is 225, then thelinear regression line shifts below this point (the line actually shifts up, but not as much as the data point has shifted up).b) If the sales change from 23 to 26 when the amount of advertising is 450, then the linear regression line shifts below this point (the line actually shifts up, but not as much as the data point has shifted up).c) If the sales change from 20 to 23 when the amount of advertising is 350, then the linear regression line shifts below this point (the line actually shifts up, but not as much as the data point has shifted up).13.40 a) The number of flying hours is the independent variable and the number of wingflaps needed is the dependent variable.b)••••••024*********100200W i n g F l a p s N e e d e dFlying Hours (thousands)c)d)••••••024*********100200W i n g F l a p s N e e d e dFlying Hours (thousands)e) y = -3.38 + (0.093)(150) = 11f) y = -3.38 + (0.093)(200) = 1513.41 Joe should use the linear regression line y = –9.95 + 0.10x to develop a forecast forCase13.1 a) We need to forecast the call volume for each day separately.1) To obtain the seasonally adjusted call volume for the past 13 weeks, we firsthave to determine the seasonal factors. Because call volumes follow seasonalpatterns within the week, we have to calculate a seasonal factor for Monday,Tuesday, Wednesday, Thursday, and Friday. We use the Template for SeasonalFactors. The 0 values for holidays should not factor into the average. Leaving themblank (rather than 0) accomplishes this. (A blank value does not factor into theAVERAGE function in Excel that is used to calculate the seasonal values.) Using thistemplate (shown on the following page, the seasonal factors for Monday, Tuesday,Wednesday, Thursday, and Friday are 1.238, 1.131, 0.999, 0.850, and 0.762,respectively.2) To forecast the call volume for the next week using the last-value forecasting method, we need to use the Last Value with Seasonality template. To forecast the next week, we need only start with the last Friday value since the Last Value method only looks at the previous day.The forecasted call volume for the next week is 5,045 calls: 1,254 calls are received on Monday, 1,148 calls are received on Tuesday, 1,012 calls are received on Wednesday, 860 calls are received on Thursday, and 771 calls are received on Friday.3) To forecast the call volume for the next week using the averaging forecasting method, we need to use the Averaging with Seasonality template.The forecasted call volume for the next week is 4,712 calls: 1,171 calls are received on Monday, 1,071 calls are received on Tuesday, 945 calls are received on Wednesday, 804 calls are received on Thursday, and 721 calls are received onFriday.4) To forecast the call volume for the next week using the moving-average forecasting method, we need to use the Moving Averaging with Seasonality template. Since only the past 5 days are used in the forecast, we start with Monday of the last week to forecast through Friday of the next week.The forecasted call volume for the next week is 4,124 calls: 985 calls are received on Monday, 914 calls are received on Tuesday, 835 calls are received on Wednesday, 732 calls are received on Thursday, and 658 calls are received on Friday.5) To forecast the call volume for the next week using the exponential smoothing forecasting method, we need to use the Exponential with Seasonality template. We start with the initial estimate of 1,125 calls (the average number of calls on non-holidays during the previous 13 weeks).The forecasted call volume for the next week is 4,322 calls: 1,074 calls are received on Monday, 982 calls are received on Tuesday, 867 calls are received onWednesday, 737 calls are received on Thursday, and 661 calls are received on Friday.b) To obtain the mean absolute deviation for each forecasting method, we simplyneed to subtract the true call volume from the forecasted call volume for each day in the sixth week. We then need to take the absolute value of the five differences.Finally, we need to take the average of these five absolute values to obtain the mean absolute deviation.1) The spreadsheet for the calculation of the mean absolute deviation for the last-value forecasting method follows.This method is the least effective of the four methods because this method depends heavily upon the average seasonality factors. If the average seasonality factors are not the true seasonality factors for week 6, a large error will appear because the average seasonality factors are used to transform the Friday call volume in week 5 to forecasts for all call volumes in week 6. We calculated in part(a) that the call volume for Friday is 0.762 times lower than the overall average callvolume. In week 6, however, the call volume for Friday is only 0.83 times lower than the average call volume over the week. Also, we calculated that the call volume for Monday is 1.34 times higher than the overall average call volume. In Week 6, however, the call volume for Monday is only 1.21 times higher than the average call volume over the week. These differences introduce a large error.。
MBA数据模型与决策考卷及答案
MBA数据模型与决策考卷及答案一、选择题（每题1分，共5分）A. 线性模型B. 非线性模型C. 网络模型D. 层次分析法模型A. 期望收益B. 折现率C. 净现值D. 敏感性分析A. 敏感性分析B. 概率树C. 决策树D. 蒙特卡洛模拟A. 目标函数为线性函数B. 约束条件为非线性函数C. 变量之间存在相关性D. 变量取值范围为整数A. ExcelB. SPSSC. MATLABD. AutoCAD二、判断题（每题1分，共5分）1. 数据模型只能用于定量分析，不能用于定性分析。
（）2. 在决策过程中，确定性决策的风险一定低于不确定性决策。
（）3. 敏感性分析可以找出影响项目收益的关键因素。
（）4. 多目标规划问题中，各个目标函数之间一定是相互矛盾的。
（）5. 网络计划技术（PERT）是一种确定型网络图。
（）三、填空题（每题1分，共5分）1. 数据模型的三个基本要素是变量、______和关系。
2. 决策树分析中，节点分为______节点和______节点。
3. 在线性规划问题中，目标函数和约束条件均为______函数。
4. 概率树分析是一种______分析工具，适用于评估项目风险。
5. 数据挖掘的五个基本步骤包括：数据准备、______、数据挖掘、结果评估和______。
四、简答题（每题2分，共10分）1. 简述蒙特卡洛模拟的基本原理。
2. 什么是网络计划技术（PERT）？它有哪些优点？3. 简述线性规划在企业管理中的应用。
4. 如何运用决策树分析解决实际问题？5. 数据挖掘技术在市场营销中的作用是什么？五、应用题（每题2分，共10分）1. 某企业生产两种产品，产品A的利润为50元/件，产品B的利润为80元/件。
生产一件产品A需要2小时，生产一件产品B需要3小时。
企业每月共有240小时的生产能力，请问如何安排生产计划，使得总利润最大化？2. 某项目有三种投资方案，方案一的投资额为100万元，收益率为10%；方案二的收益率为12%，投资额为150万元；方案三的投资额为200万元，收益率为15%。
作业题(数据模型与决策)
《数据模型与决策》课程作业（2014春秋MBA周末班）：一、生产轮班人员的双向选择问题解：1）建立运输模型假设以24名工人为产地，4名组长为销地，24名普通员工与4位组长之间的相互满意度值为运输单价，每名工人到一个小组为产量，每个小组需要的工人数为销量，列下表：解一：即：第一组：1、3、4、9、15、23；第二组：2、6、7、8、10、20；第三组：5、11、12、13、14、16；第四组：17、18、19、21、22、24；解二：即：第一组：1、2、4、9、15、23；第二组：3、6、7、8、10、20； 第三组：5、11、12、13、14、16；第四组：17、18、19、21、22、24； 2）建立0-1整数规划模型：令x ij = 1(指派第 i 工人去j 组长小组工作时)或0（指第 i 工人不去j 组长小组工作工作时)。
这样可以表示为一个0－1整数规划问题： 设C ij 为第i 员工与第j 组长之间的相互满意度值 则minZ=∑∑==24141i j j ij Xi Cs.t.{∑=411jjx=1....∑=4124jjx=1{6 2411=∑=ii x6 2412=∑=ii x6 2413=∑=ii x6 2414=∑=ii xx ij = 1—0，（i=1,2,3,……，24；j=1,2,3,4）二、证券营业网点设置问题解：建立0—1模型令x i =1（指在该地建立营业网点）或0（指在该地不建立营业网点）。
这样可以表示为一个0－1整数规划问题：投资额b j ；利润额c j ；市场平均份额r j 均为原题目中表格内的数据。
maxZ=∑∑==201201j i ij x cs.t.{∑∑==201201j i i j x b ≤220000000∑∑==201201j i i j x r ≤10∑=201i i x ≤124321x x x x +++≥31312111098765x x x x x x x x x ++++++++≥420191817161514x x x x x x x ++++++≤54∗(4321x x x x +++)+3∗(1312111098765x x x x x x x x x ++++++++)+2∗(20191817161514x x x x x x x ++++++)≤40x i =1—0；(i=1,2,3,……20)。
2016.04数据模型与决策(解答)
‘2016年4月江苏省高等教育自学考试数据、模型与决策试卷（30447）本试卷分为两部分，满分一百分，考试时间150分钟。
第一部分为选择题1页至2页。
应考者必须按照试题顺序在“答题卡”上按要求填涂，答在试卷上无效。
第二部分为非选择题3页至4页，共2页。
应考者必须按照试题顺序在“答题卡”作答，答在试卷上无效。
第一部分选择题（共10分）一、单项选择题（本大题共10小题，每小题1分，共10分）在每小题列出的四个选项中只有一个是符合题目要求的，请将其选出并将“答题卡”的相应代码涂黑。
错涂、多涂或未涂均无分。
对时间序列移动平均，得到的新时间序列比原时间序列的项数少，如6期移动平均少（D ）A.8项B.16项C.7项D.6项2.对事物进行分类或分组的到的结果，就是（ B ）A.定量资料B.定类资料C.调查资料D.测量资料3.对二人有限零和博弈，A为局中人I的赢得矩阵，则局中人II的赢得矩阵是（D ）A.AB.2AC.1/AD.-A4.如果排队系统的顾客到达数为泊松流，则顾客到达间隔服从（D ）A.指数分布B.定长分布C.爱尔朗分布D.泊松分布5．对于一个线性规划问题，下列说法正确的是（C ）A.一定能找到最优解B.有可行解必有最优解C.最优解同时也是最基本可行解D.一定有可行解6.C控制图主要用于检查（ B ）A.少数关键性要素B.单位产品的缺陷数C.系统性原因D.异常波动7．贝尔实验室工程师休哈特博士发明的，运用统计方法确定管理界限，并用于管理监控的一种图表叫做（B ）A.鱼刺图B.控制图C.怕雷特图D.直方图8.在下面的几个决策准则中，属于风险型决策准则是（D ）A.小中取大准则B.最小后悔准则C.等可能性准则D.最大可能性准则9.把每一年各月（季）观察值直接与上一年同一月份（季）的观察值进行对比，计算年距环比发展速度，以消除季节变动和长期趋势的影响，能够粗略地描述时间序列的（ D ）A.指数变化B.趋势变动C.季节变动D.循环变动10.多元线性回归分析中，拟合优度系数R2是（ B ）A.因变量数目的递增函数B. 自变量数目的递增函数C. 修正系数的递增函数D. 模型真实性的递增函数第二部分非选择题（共90分）二、填空题（本大题共10小题，每小题1分，共10分）11.检验两个总体变异程度的大小，需要用到的分布是F分布12.衡量产品的加工过程中的稳定性可以使用工序能力指数13.改变自变量的回归系数，可以引进虚拟变量14.时间序列的各项观察值，叫做水平15.增长量与基期水平相比得到增长速度16.决策人不能控制的客观存在的环境或条件叫做自然状态17.加工过程出现条件变化时反映在直方图上可能会呈孤岛型18.出现次数最多的观察值叫做众数19.到达加油站加油的车辆数其分布刻画可以用泊松分布20.线性规划问题中，通常将资源对目标函数的边际影响叫做对偶解三、名词解释题（本大题共5小题，每小题3分，共15分）21.队长：P298 队长是排队系统中的顾客数，由排队等候的顾客数和正在接受服务的顾客数两部分组成。
MBA数据模型与决策试卷与解答
数据模型与决策试卷一、建模计算分析题（30分）下面两个题目任选一个做即可1艾伦是一个个人投资者，有70000美元可用于不同的投资。
不同投资的年回报率分别为：政府债券8.5%，存款5%，短期国库券6.5%，增长股基金13%。
所有的投资都要满1年后才对收益进行评估。
然而，每项投资都会有不同的风险，因此建议进行多元投资。
艾伦想知道每项投资各位多少可以使收益最大化。
下面的方针可以使投资具有多样性，从而降低投资者的风险：（1）对政府债券的投资比例不要超过全部投资的20%。
（2）在存款方面的投资不要超过其他3种投资的总和。
（3）在存款和短期国库券方面的投资至少要占全部投资的30%（4）为了投资安全，在存款和短期国库券方面的投资与在政府债券和增长股基金方面的投资比例至少是1.2:1。
（5）艾伦想把70000美元全部用来投资。
如果全部投资不再正好是70000美元，其他条件不变，则模型应该如何变化？2某投资咨询公司，为大量的客户管理高达1.2亿元的资金。
公司运用一个很有价值的模型，为每个客户安排投资量，分别投资在股票增长基金、收入基金和货币市场基金。
为了保证客户投资的多元化，公司对这三种投资的数额加以限制。
一般来说，投资在股票方面的资金应该占总投资的20%~40%，投资在收入基金方面的资金应该确保在20%~50%，货币市场方面的投资至少应该占30%。
此外，公司还尝试着引入了风险承受能力指数，以迎合不同投资者的需求。
如该公司的一位新客户希望投资800000元。
对其风险承受能力进行评估得出其风险指数为0.05。
公司的风险分析人员计算出，股票市场的风险指数是0.10，收入基金的风险指数是0.07，货币市场的风险指数是0.01，整个投资的风险指数是各项投资占投资的百分率与其风险指数乘积的代数和。
此外该公司预测股票基金的年收益是18%，收入基金的收益率是12.5%，货币市场基金的收益率是7.5%。
现在基于以上信息，公司应该如何安排这位客户的投资呢？建立线性规划模型，求出使总收益最大的解，并根据模型写出管理报告。
数据模型重点习题及答案
数据模型与决策14版重点习题与答案第4章16题：芬古森造纸公司生产用于法器、台式计算器和收银机的卷纸。
这些卷纸每卷长度为200英尺，宽度可为1.5,2.5和3.5英尺。
生产过程只能提供200英尺长和10英尺宽的卷纸。
所以，公司必须剪切卷纸以满足所需的宽度。
7种剪切方案以及每种方案造成的浪费如下表所示。
A．若公司希望使用的10英寸卷纸最少，则每一种方案应剪切多少个10英寸卷纸？总欧冠那个需要多少个？最后浪费了多少（英寸）？B．若公司希望造成的浪费最少，每一种方案应剪切多少个10另存卷纸？总共需要多少？最后浪费了多少？C．A问题和b问题有什么不同？在这个案例里，你偏好哪一种目标？请加以解释。
什么样的情况下，另一种目标更有吸引力？答案：A：根据题目要求，定义决策变量如下：X1=方案1中应剪切10英寸卷的个数X2=方案2中应剪切10英寸卷的个数X3=方案3中应剪切10英寸卷的个数X4=方案4中应剪切10英寸卷的个数X5=方案5中应剪切10英寸卷的个数X6=方案6中应剪切10英寸卷的个数X7=方案7中应剪切10英寸卷的个数如果建模目标是希望使用10英寸的卷纸数量最少，那么目标函数是：Min X1+X2+X3+X4+X5+X6+X7根据已知信息，已知模型的约束条件为：6X1+ 2X3+X5+ X6+ 4X7 ≥10004X2+ X4 +3X5+ 2X6≥20002X3+2X4 + X6+ X7 ≥4000X1，X2，X3，X4，X5，X6，X7≥0对上述模型用LINGO进行求解，结果如下：综上可得，总共需要10英寸卷纸的数量为2125个，其中每种方案所需数量为：X1=0，X2=125，X3=500，X4=1500，X5=0，X6=0，X7=0；最后浪费的尺寸为：X1+0X2+0X3+0.5X4+X5+0X6+0.5X7=750英寸；b.目标函数是：Min X1+0X2+0X3+0.5X4+X5+0X6+0.5X7=X1+0.5X4+X5+0.5X7根据已知信息，已知模型的约束条件为：6X1+ 2X3+X5+ X6+ X7 ≥10004X2+ X4 +3X5+ 2X6≥20002X3+2X4 + X6+ X7 ≥4000X1，X2，X3，X4，X5，X6，X7≥0对上述模型用LINGO求解得，无浪费，总共需要10英寸卷纸的数量为2500个，但是1.5英寸的规格多生产了3000个；第4章17题弗朗德克公司制造、组装和改造仓库和分销中心使用装卸装备。
数据模型决策复习题集.docx
《数据模型决策》作业集课程英文名称：Data, Model and DecisionK某新产品在两个城市的试销资料如下，试比较两城市的市场稳定情况。
月份销售量（万件）甲城市乙城市1110702986539584499120598916100100797688987391021401010112011102110121001592、某科技产品核心元件的耐用时数服从均值为10万小时的正态分布，现随机抽取10件新工艺条件下的产品作耐用性能测试，测得其平均耐用时数为10.77万小时，标准差为0.5197万小时，能否以0.5%的显著性水平，判断该元件在新工艺条件下已得到明显提高？如果有明显提高，其概率是多少？4、企业有以下情爭・1）欲了解本企业彙?导产品的市场前景；2）能提供本企业该产品至少5年的相关统计资料；3）欲重点开拓一、二个城市作为目标市场。
请以定量分析为着重点，应该从哪几个方面入手开展工作，就目前所学的统计学方法设想怎样应用？请思考O试分析该企业原材料费用总额的变动受产量、单耗、原材料价格的影响程度。
2）如果生产该产品需要两种原材料，请在上述资料的基础上，自行再假定一个原材料单耗和原材料价格资料，其他条件不变，并在此基础上分析该企业原材料费用总额的变动受产量、单耗、原材料价格的影响程度。
说明：不一定要具体计算，但是要写出你的做题思路和步骤。
2、试自行设立生活开支比重数据，计算相关系数，要求所计算的相关系戮为负相关。
并且在显著性水平为5%下对相关系数进行检验。
此题最和具体计算一下，看一看结果。
3、拟设立企业文化优劣判定指标，请从概念、设立原则、假设条件、判定指标及指标解释、指标计算方法等方面进行讨论。
案例研究o逮合固囹份翁銀唸參创迢裁颐窓一、设计议题1、湖北省省辖市战略产业选择评价指标体系2、生活质量评价指标体系3、世界观评价指标体系4、企业绩效评价指标体系5^人才评价指标体系6、和谐社会评价指标体系7、企业经营危机识别指标体系二、要求1、以小组为单位形成文稿，若小组成员超过10人，则必须有两个以上的设计报告2、内容至少要包括：前言、指标体系的设计依据或者基本思路、指标体系的主要内容、主要指标的定义或者范畴、指标体系的计算方法和分析方法、指标体系的模拟数据、其他需要说明的问题3、指定专人利用多媒体发言，发言稿要交。
数据模型与决策习题解答
第二章习题(P46)14.某天40只普通股票的收盘价（单位：元/股）如下：（1）构建频数分布*。
（2）分组，并绘制直方图，说明股价的规律。
（3）绘制茎叶图*、箱线图，说明其分布特征。
（4）计算描述统计量，利用你的计算结果，对普通股价进行解释。
解：（1）将数据按照从小到大的顺序排列, , 8, , , 9, , , , 10, , , 14, , , 18, 18, , , , , , , , , , , , , 34, , 37, , 38, , , , 52, , ，结合（2）建立频数分布。
（2）将数据分为6组，组距为10。
分组结果以及频数分布表。
为了方便分组数据样本均值与样本方差的计算，将基础计算结果也列入下表。
区间组频数累计频数组中值组频数×组中值组频数×组中值×组中值,0[99 54522510)[1019 1515022501020,)[524 25125312520),30[1135 353851347530,)4040[237 45904050 50,)50[ 340 6018010800 ,)合计4097533925根据频数分布与累积频数分布，画出频率分布直方图与累积频率分布的直方图。
频率分布直方图从频率直方图和累计频率直方图可以看出股价的规律。
股价分布10元以下、10—20元、30—40元占到60%，股价在40元以下占%，分布不服从正态分布等等。
累积频率分布直方图（3）将原始数据四舍五入取到整数。
1，8 ，8 ，9 ，9 ，9 ，9 ，9 ，9 ，10 ，11 ，12 ，14 ，17 ，17 ，18 ，18 ，19 ，19 ，22 ，24 ，29 ，30 ，30 ，30 ，31 ，32 ，34 ，34 ，34 ，35 ，37 ，38 ，38 ，39 ，43 ，48 ，52 ，53 ，79以10位数为茎个位数为叶，绘制茎叶图如下茎（十位数）叶（个位数及其小数） 0 9 1 9 2 249 3 89 4 38 5 23 6 7 9由数据整理，按照从小到大的准许排列为：)40()39()2()1(x x x x ≤≤⋯≤≤频率0.050.10.150.20.250.30—1010—2020—3030—4040—5050及以上累计频率0.20.40.60.811.20—1010—2020—3030—4040—5050及以上最小值25.1)1(=x ，下四分位数()03125.11375.11431041341)11()10(=⨯+⨯=⨯+=x x Q l ，中位数()9375.22225.24625.2121)21()20(=+=+=x x M e ，上四分位数())30()29(341x x Q u +⨯= 3125.3425.35413443=⨯+⨯=，最大值375.79)40(=x ，四分位数间距28125.2313=-=Q Q IQR ，375.792344.695.1)40(3=<=+x IQR Q ,因此可以做出箱线图为：茎叶图的外部轮廓反映了样本数据的分布状况。
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《管理统计学》习题解答 (20XX年秋MBA周末二班，邢广杰，学号：) 第3章 描述性统计量 (一) P53 第1题 抽查某系30个教工，年龄如下所示： 61，54，57，53，56，40，38，33，33，45，28，22，23，23，24，22，21，45，42，36，36，35，28，25，37，35，42，35，63，21 （i）求样本均值、样本方差、样本中位数、极差、众数； （ii）把样本分为7组，且组距相同。作出列表数据和直方图； （iii）根据分组数据求样本均值、样本方差、样本中位数和众数。
解：
（i）样本均值n1iixn1x=37.1岁 样本方差)XnX(1-n1)X(X1-n1s2n1i2i2n1ii2=189.33448 把样本按大小顺序排列：21，21，22，22，23，23，24，25，28，28，33，33，35，35，35，36，36，37，38，40，42，42，45，45，53，54，56，57，61，63
样本中位数)XX(21m1)2n()2n(=(35+36)/2=35.5岁
极差1)()n(XXR63-21=42岁 众数0m35岁 （ii）样本分为7组、且组距相同的列表数据、直方图如下所示 教工年龄分组(岁) 教工分组频数(if) 组中值（x）
累计频
数
(16,23] 6 19.5 6 (23,30] 4 26.5 10 (30,37] 8 33.5 18 (37,44] 4 40.5 22 (44,51] 2 47.5 24 (51,58] 4 54.5 28 (58,65] 2 61.5 30 （iii）根据分组数据求样本均值、样本方差、样本中位数和众数。
样本均值ik1ifXin1X=36.3岁
教职工岁数分组频数图0246810
23303744515865教职工岁数
频数频数样本方差)XnfX(1-n1f)X(X1-n1s2k1ii2ii2k1ii2=174.3724 样本中位数810230730fF2niIm=34.375岁 众数448248730ff2fffiIm1m1-mm1-mm033.5岁 (二)P53 第2题 某单位统计了不同级别的员工的月工资水平资料如下： 月工资（元） 800 1000 1200 1500 1900 2000 2400 员工数（人） 5 8 25 36 24 16 6 累计频数 5 13 38 74 98 114 120 求样本均值、样本标准差、样本中位数和众数。
解：
样本均值ik1ifXin1X=1566.667元
样本标准差)XnfX(1-n1f)X(X1-n1s2k1ii2ii2k1ii=398.1751元 样本中位数在累计74人的那一组，m=1500元； 众数1500m0元。 第7章 参数统计推断 (一)某种零件的厚度服从正态分布，从该批产品中随机抽取9件，测得平均长度为23.4mm。已知总体的标准差0.15mm,试估计该批零件厚度的区间范围，给定置信水平为95﹪。
解：本题中，n=9，4.23x，0.15，05.0，96.1u
20.05
故的置信水平为95%的区间估计为：
)96.1915.04.23()unx(20.05=(23.302，23.498)（mm）
(二) 某灯泡厂生产一种新型灯泡，为了了解灯泡的使用寿命，在生产线上随机抽取9只灯泡进行测试，得到下列数据（小时）：5100，5100，5400，5260，5400，5100，5320，5180，4940。若灯泡的使用寿命服从正态分布，现以95﹪的可靠性估计该批新型灯泡平均使用寿命的区间范围。
解：本题中，n=9，)49405180532052602540035100(
9
1x=5200，05.0，
96.1u20.05；这是方差未知时小样本正态总体均值的区间估计问题，
根据题中数据，有：n1i2i)x(x1n1s=156.5248 故的置信水平为95%的区间估计为： )306.29156.52485200())8(9156.52485200())1(tnsx(025.02tn=(5079.68
5，5320.315)（小时） (三) 某厂生产一种耐高温的零件，根据质量管理资料，在以往一段时间里，零件抗热的平均温度是1250℃，零件抗热温度的标准差是150℃。在最近生产的一批零件中，随机测试了100个零件，其平均抗热温度为1200℃。该厂能否认为最近生产的这批零件仍然符合产品质量要求，而承担的生产者风险为0.05。
解：这是一个正态总体、
2
已知时的均值双侧假设检验问题
构造原假设0H和对立假设1H为 1250:H 1250:H10
当20.050unx时接收0H，否则拒绝0H 本题中，0=1250，150，1200x，n100，05.0，96.1u20.05 计算得：96.1333.310015012501200，故样本落入拒绝域，因而拒绝原假设0H，即在
05.0的生产者风险下，认为最近生产的这批零件不符合产品质量要求。
(四)阀门厂的零件需要钻孔，要求孔径10cm，孔径过大过小的零件都不合格。为了测试钻孔机是否正常，随机抽取了100件钻孔的零件进行检验，测得cmX6.9，cms1。给定05.0，检验钻孔机的操作是否正常。
解：这是一个非正态总体、
2
未知、的大样本均值双侧假设检验问题
构造原假设0H和对立假设1H为 10:H 10:H10
当20.050unsx时接收0H，否则拒绝0H 本题中，0=10，s1，6.9x，n100，05.0，96.1u20.05 计算得：96.141001106.9，故样本落入拒绝域，因而拒绝原假设0H，即在水平05.0
（置信度为95%）下，认为钻孔机的操作不正常。
(五)某日用化工厂用一种设备生产香皂，其厚度要求为5cm。今欲了解设备的工作性能是否良好，随机抽取10块香皂，测得平均厚度为5.3cm，标准差为0.3cm，试分别以0.01，0.05的显著性水平检验设备的工作性能是否合乎要求。
解：这是一个非正态总体、
2
未知、的小样本均值双侧假设检验问题
构造原假设0H和对立假设1H为 5:H 5:H10
当)1n(tnsx20时接收0H，否则拒绝0H 本题中，0=5，s0.3，3.5x，n10。 当10.0时，24984.3)9(t)1n(t0.0052
计算得：24984.31623.3100.353.5，故样本未落入拒绝域，因而无法拒绝原假设0H，
即在10.0时，认为生产香皂的设备工作性能正常。 当50.0时，26216.2)9(t)1n(t0.0252，而26216.21623.3，故样本落入拒绝域，
从而拒绝原假设0H，即在50.0时，认为生产香皂的设备工作性能不正常。 (六) P147 第1题 某市环保局对空气污染物质24小时的最大容许量为942/mg，在该城市中随机选取的测量点来检测24小时的污染物质量。数据为 82，97，94，95，81，91，80，87，96，77（2/mg） 设污染物质量服从正态分布，求该市24小时污染物质量的95％区间估计，据此数据，你认为污染物质是否超标？
解：本题中，n=10，88)77968780191859499782(
10
1x，05.0，
96.1u20.05；这是方差未知时小样本正态总体均值的区间估计问题，
根据题中数据，有：n1i2i)x(x1n1s=7.5277 故的置信水平为95%的区间估计为： )2622.2107.527788())9(107.527788())1(tnsx(025.02tn=(82.615，93.385)
（2/mg） 这是一个正态总体、2未知、的小样本均值单侧假设检验问题 构造原假设0H和对立假设1H为 2120/94:H /94:Hmgmg
当)1n(tnsx0时接收0H，否则拒绝0H 本题中，0=94，s7.5277，88x，n10。 当50.0时，833.1)9(t)1n(t0.05
计算得：)1n(t39.297.52779488nsx0
，故样本未落入拒绝域，因而无法拒绝原
假设0H，即在50.0时，认为污染物质未超标。
第9章 相关分析和回归分析 P195 第1题(i)、(ii) 企业希望了解每周的广告费与销售额之间的关系，记录了如下数据（单位：万元）： 广告费x 40 25 35 45 30 28 40 24 32 28 销售额y 395 350 380 430 370 380 420 330 350 360 （i）求回归直线； （ii）求广告费和销售额之间的相关系数；
解：（ii）
5.9002450.15.1824)yny)(xnx(yxnyxSSSr2n1i2i2n1i2in1iiiyyxxxy
=0.906375
广告费与销售额正相关，即广告费多，销售额也大 （i）4.05351.4505.1824SSbxxxy^，243.957.320535.45.376xbya^^（截距）
回归直线方程为：0535x.495.243xbay^^ 第10章 时间序列 P219第3题（只用加法模型求解） 某地区连续三年半经济活力指数如下：
年 1 2 3 4 季 1 2 3 4 1 2 3 4 1 2 3 4 1 2 指数 100 91 89 108 104 93 90 112 109 97 94 115 110 100 （i）求直线趋势方程； （ii）用加法模型求季节因子； （iii）用4点滑动平均模型求曲线趋势，并根据该曲线对加法模型求季节因子； （iv）用加法模型预测第四年三、四季度的经济活力指数。（预报）