数据模型与决策--作业大全详解

合集下载

数据模型与决策课程大作业(完整资料).doc

【最新整理，下载后即可编辑】数据模型与决策课程大作业以我国汽油消费量为因变量，乘用车销量、城镇化率和90#汽油吨价与城镇居民人均可支配收入的比值为自变量时行回归（数据为年度时间序列数据）。

试根据得到部分输出结果，回答下列问题：1）“模型汇总表”中的R方和标准估计的误差是多少？2）写出此回归分析所对应的方程；3）将三个自变量对汽油消费量的影响程度进行说明；4）对回归分析结果进行分析和评价，指出其中存在的问题。

1）“模型汇总表”中的R方和标准估计的误差是多少？答案：R方为0.993^2=0.986 ；标准估计的误差为120910.147^（0.5）=347.722）写出此回归分析所对应的方程；答案：假设汽油消费量为Y，乘用车销量为a，城镇化率为b，90#汽油吨价/城镇居民人均可支配收入为c,则回归方程为：Y=240.534+0.00s027a+8649.895b-198.692c3）将三个自变量对汽油消费量的影响程度进行说明；乘用车销量对汽油消费量相关系数只有0.00027，数值太小，几乎没有影响，但是城镇化率对汽油消费量相关系数是8649.895，具有明显正相关，当城镇化率每提高1，汽油消费量增加8649.895。

乘用90#汽油吨价/城镇居民人均可支配收入相关系数为-198.692，呈明显负相关，即乘用90#汽油吨价/城镇居民人均可支配收入每增加1个单位，汽油消费量降低198.692个单位。

a, b, c三个自变量的sig值为0.000、0.000、0.009，在显著性水平0.01情形下，乘用车消费量对汽油消费量的影响显著为正。

（4）对回归分析结果进行分析和评价，指出其中存在的问题。

在学习完本课程之后，我们可以统计方法为特征的不确定性决策、以运筹方法为特征的策略的基本原理和一般方法为基础，结合抽样、参数估计、假设分析、回归分析等知识对我国汽油消费量影响因素进行了模拟回归，并运用软件计算出回归结果，故根据回归结果，对具体回归方程，回归准确性，自变量影响展开分析。

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.。

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)。

数据模型与决策作业答案

教材习题答案1.2 工厂每月生产A 、B 、C 三种产品 ,单件产品的原材料消耗量、设备台时的消耗量、资源限量及单件产品利润如表1－22所示．和130.试建立该问题的数学模型,使每月利润最大．【解】设x 1、x 2、x 3分别为产品A 、B 、C 的产量，则数学模型为123123123123123max 1014121.5 1.2425003 1.6 1.21400150250260310120130,,0Z x x x x x x x x x x x x x x x =++++≤⎧⎪++≤⎪⎪≤≤⎪⎨≤≤⎪⎪≤≤⎪≥⎪⎩ 1.3 建筑公司需要用6m 长的塑钢材料制作A 、B 两种型号的窗架．两种窗架所需材料规格及数量如表1－23所示：【解】设x j （j =1,2,…，14）为第j 种方案使用原材料的根数，则（1）用料最少数学模型为14112342567891036891112132347910121314min 2300322450232400232346000,1,2,,14jj j Z x x x x x x x x x x x x x x x x x x x x x x x x x x x x x j ==⎧+++≥⎪++++++≥⎪⎪++++++≥⎨⎪++++++++≥⎪⎪≥=⎩∑ 用单纯形法求解得到两个基本最优解X (1)=( 50 ,200 ,0 ,0,84 ,0,0 ,0 ,0 ,0 ,0 ,200 ,0 ,0 );Z=534 X (2)=( 0 ,200 ,100 ,0,84 ,0,0 ,0 ,0 ,0 ,0 ,150 ,0 ,0 );Z=534 （2）余料最少数学模型为134131412342567891036891112132347910121314min 0.60.30.70.40.82300322450232400232346000,1,2,,14j Z x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x j =+++++⎧+++≥⎪++++++≥⎪⎪++++++≥⎨⎪++++++++≥⎪⎪≥=⎩ 用单纯形法求解得到两个基本最优解X (1)=( 0 ,300 ,0 ,0,50 ,0,0 ,0 ,0 ,0 ,0 ,200 ,0 ,0 );Z=0，用料550根 X (2)=( 0 ,450 ,0 ,0,0 ,0,0 ,0 ,0 ,0 ,0 ,200 ,0 ,0 );Z=0，用料650根显然用料最少的方案最优。

数据模型与决策课程学习大作业.doc

数据模型与决策课程大作业以我国汽油消费量为因变量，乘用车销量、城镇化率和90#汽油吨价与城镇居民人均可支配收入的比值为自变量时行回归（数据为年度时间序列数据）。

a, b, c三个自变量的sig 值为0.000、0.000、0.009，在显著性水平0.01情形下，乘用车消费量对汽油消费量的影响显著为正。

（4）对回归分析结果进行分析和评价，指出其中存在的问题。

数据模型与决策完整

数据,模型和决策第一章决策分析一、比尔.桑普拉斯的夏季打工决策一个决策树模型及分析比尔. 桑普拉斯(bill Sampras) 在麻省理工学院的斯隆管理学院就读第一学期，已经是第三周了。

除了花在准备功课上的时间外，bill开场认真考虑有关明年夏季打工的事情，特别是该决策在几周后必须做出。

8月底，在bill飞往波士顿的途中，他坐在vanessa Parker 的旁边，并与她就双方感兴趣的问题进展了交谈。

vanessa 是一个重要的商业投资银行有关资产预算的副总裁。

在飞机到达波士顿后，vanessa坦率地告诉bill，她愿意考虑明年夏季雇佣bill的可能性，并希望在她的公司于11月中旬开场进展的夏季招聘方案时，请bill直接与她联系。

bill感觉到自己的经历和所具有的风度给vanessa留下了很深的印象〔bill曾经在一个财富500强公司的金融部门就来自税收业务的额外现金的短期投资工作过4年〕。

当bill 8月离开公司去攻读MBA时，他的老板john Mason把他叫到一边，对他许诺，到第二年夏季可以雇佣他。

夏季回到公司进展为期12个星期的打工薪水将是12000美元。

但john也告诉bill夏季工作招聘期限仅到10月底有效。

因此，bill在得到vanessa提供夏季工作的细节之前，必须决定是否承受john的工作。

vanessa已经解释，她的公司在11月中旬之前不愿意讨论夏季工作方案的细节。

如果bill回绝john的好意，bill要么承受vanessa的提供〔如果vanessa承受bill的申请〕，要么通过参加斯隆管理学院在1月和2月举办的公司夏季招聘方案中，寻找另一个夏季工作时机。

决策准那么假设bill认为所有的夏季工作时机〔为john工作，为vanessa工作和参加斯隆学院的夏季打工方案〕都将会给bill提供类似的学习、交流以及丰富经历的时机。

那么，bill判断夏季工作时机的优劣的唯一标准就是工作的薪水，以薪水越高越好。

合肥工业大学MBA数据模型与决策课后作业参考答案完整版

2014秋季MBA脱产班刘德玉学号是2014170985产品混合问题报告书为了使TJ公司合理购入这批坚果，在不考虑其他因素的在理想状态下，对公司给出了各项数据进行分析、加工得出结论。

一、建立数据模型max=1.65*x1+2*x2+2.25*x3;0.15*x1+0.2*x2+0.25*x3<=6000;0.25*x1+0.2*x2+0.15*x3<=7500;0.25*x1+0.2*x2+0.15*x3<=7500;0.1*x1+0.2*x2+0.25*x3<=6000;0.25*x1+0.2*x2+0.2*x3<=7500;x1>=10000;x2>=3000;x3>=5000;经过计算得出普通型、高级型、假日性等坚果产品的成本和最优生产组合的总利润。

如下表二、利润最大化研究通过表得出Righthand Side RangesRow Current Allowable AllowableRHS Increase Decrease2 6000.000 583.3333 610.00003 7500.000 INFINITY 250.00004 7500.000 INFINITY 250.00005 6000.000 INFINITY 875.00006 7500.000 250.0000 750.00007 10000.00 7500.000 INFINITY8 3000.000 7625.000 INFINITY9 5000.000 4692.308 5000.000其中杏仁可以多购买583磅和胡桃可以增加250磅可以在增加购买，并且普通型、高级型、假日型利润也是可以变动的，这样会使产品利润增加。

因为根据运输情况最大杏仁才可以增加583磅，在买1000磅，这种情况情况下就会有库存，从利益的考虑，不用在购买。

三、不用满足订单不用满足订单的情况下，公司只能生产普通型和高级型。

《数据模型决策》复习(作业)题

《数据模型决策》复习(作业)题《数据模型决策》复习（作业）题二、分析、建模题1、（广告策划）一家广告公试司想在电视、广播及杂志做广告，其目的是尽可能多地招徕顾客。

下面是市场调查结果：这家公司希望广告费用不超过800（千元），还要求：（1）至少有二百万妇女收看广告；（2）电视广告费用不超过500（千元）；（3）电视广告白天至少播出3次，最佳时间至少播出2次；（4）通过广播、杂志做的广告各重复5到10次。

试建立该问题的数学模型，并用软件求解。

解：设变量X1, X 2, X 3, X 4为白天、最佳时间、无线电广播、杂志次数目标函数maxZ=400 X1+900X2+500 X 3+200 X 4约束条件s.t40 X 1+75 X 2+30 X 3+15 X 4≤80040X1+400X2+200X3+100X4≥80040X1+75X2≤500X1≥3X2≥2,X3≥5X3≤10X4≥5X4≤10X i≥0 i=1,2,3,4软件求解2、（指派问题）分配甲、乙、丙、丁四人分别去完成A、B、C、D 四项工作。

已知每人完成各项工作的时间如下表所示。

规定每项工作只能由一人去单独完成，每个人最多承担一项工作。

如何分配工作，使完成四项工作总的耗时为最少？建立线性规划数学模型（不求解）。

解：设变量X11,X12,X13,X14为甲参加1，2，3，4工作，X X22,X23,X24为乙参加1，2，3，4工作，21,X31,X32,X33,X34为丙参加1，2，3，4工作，X41,X42,X43,X44为丁参加1，2，3，4工作目标函数maXZ=10X11+5X12+15X13,+20X14 +2X21+10X22+5X23+15X24+3X31+15X32+14X33+13X34 +15X41+2X42+7X43+6X44约束条件s.tX11+X12+X13, +X14=1X21+X22+X23+X24=1X31+X32+X33+X34=1X41+X42+X43+X44=1X i,j≥0 i=1,2,3,4 j=1,2,3,4软件求解3、昼夜运营的公交线路每天各时间区段内所需要的司机和乘务员人数如下表：设司机和乘务员分别在各时间区段一开始时上班，并连续工作8小时，问该公交线路至少配备多少名司机和乘务人员。

1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。

P45.1.21.2N ewtowne有一副珍贵的油画，并希望被拍卖。

有三个竞争者想得到该幅油画。

第一个竞拍者将于星期一出价，第二个竞拍者将于星期二出价，而第三个竞拍者将于星期三出价。

每个竞拍者必须在当天作出接受或拒绝的决定。

如果三个竞拍者都被拒绝，那个该油画将被标价90万美元出售。

Newtowne 拍卖行的主任对拍卖计算的概率结果列在表1.5中。

例如拍卖人的估计第二个拍卖人出价200万美元的概率p=0.9.（a）对接受拍卖者的决策问题构造决策树。

1、买家1：如果出价300万，就接受，如果出价200万，就拒绝；2、买家2：如果出价400万，就接受，如果出价200万，也接受。

接受买家1200 200200接受买家22002002000.50.9接受买家3买家1出价200万买家2出价200万0.7100 21买家3出价100万100100 0220020010100拒绝买家390拒绝买家290900190接受买家30.3400买家3出价400万400400拒绝买家1104000220拒绝买家3909090接受买家24004004000.1接受买家3买家2出价400万0.71001买家3出价100万100100040010100260拒绝买家390拒绝买家290900190接受买家30.3400买家3出价400万40040010400拒绝买家3909090接受买家1300 300300接受买家22002002000.50.9接受买家3买家1出价300万买家2出价200万0.7100 11买家3出价100万100100 0300020010100拒绝买家390拒绝买家290900190接受买家30.3400买家3出价400万400400拒绝买家1104000220拒绝买家3909090接受买家24004004000.1接受买家3买家2出价400万0.71001买家3出价100万100100040010100拒绝买家390拒绝买家290900190接受买家30.3400买家3出价400万40040010400拒绝买家39090902.9在美国有55万人感染HIV病毒。

所有这些人中，27.5万人是吸毒者，其余的人是非吸毒者。

美国总人口为2.5亿。

在美国有10000万人吸毒。

HIV感染的标准血液检测并不总是准确的。

某人感染HIV,检测HIV为肯定的概率是0.99.某人没有感染HIV，检测HIV为否定的概率也是0.99。

回答下列问题，清晰的说明你需要作出的任何假设。

（A）假设随机选择一个人进行HIV标准血液测试，测试结果是肯定的。

这个人感染HIV的概率是多少？你的答案令人吃惊吗？（B）假设随机选择一个吸毒者进行HIV标准血液测试，测试结果是肯定的。

这个人感染HIV的概率是多少？第一问：答：设：P（x）为随机抽取一个人为HIV感染者的概率；P（y）为从美国人中随机抽取一个人检测HIV为肯定的概率。

那么：假设随机选择一个人进行HIV标准血液测试，测试结果是肯定的，这个人感染HIV的概率：P（X｜Y）= P（Y｜X）P（X）/P（Y）P（Y｜X）=0.99P（X）=550000/250000000*100%=0.0022P（Y）= P（X）*0.99+(1-P（x）)*0.01=0.012156因此：假设随机选择一个人进行HIV标准血液测试，测试结果是肯定的，这个人感染HIV的概率P（X｜Y）为17.92%。

第二问：答：设P（X）为随机抽取一个吸毒者为HIV感染者的概率；P（Y）为从吸毒者中随机抽取一个人检测HIV为肯定的概率。

那么假设随机选择一个吸毒者进行HIV标准血液测试，结果是肯定的，这个人感染HIV的概率表示为：P（X｜Y）= P（Y｜X）P（X）/P（Y）P（Y｜X）=0.99P（X）=275000/10000000*100%=0.0275那么假设随机选择一个吸毒者进行HIV标准血液测试，结果是肯定的，这个人感染HIV的概率P（X｜Y）为74.59%。

2.16在一个小型造船厂每月制造的木质航海船的树木是一个随机变量，它服从下表中所给出的概率分布。

4800美元。

（A）计算每月制造船的费用的均值和标准离差。

（B）制造航海船的月费用的均值和标准离差是多少。

（C）如果每月的固定费用从3万每月增加到5.3万美元，在问题（B）中，答案会怎样变化？请仅利用（B）中计算的结果，重新计算答案。

（D）如果每支船的建造费用从4800美元增加到7000美元，但每月的固定费用仍是3万美元，在问题(B)中，你的答案会如何变化？请仅利用(A)和(B)中计算的结果，重新计算你的答案。

答案：均值=2×0.25+3×0.20+4×0.30+5×0.25+6×0.05+7×0.05=3（1）此教授退休金购买的基金为Z=30%X + 70%Y。

由于X~N(0.07,0.02)，Y~N(0.13,0.08)E(X)=0.07，E(Y)=0.13。

因此E（Z）= 30%E(X) + 70%E(Y)=0.021 + 0.091=0.112 （2）教授退休金年收益率标准离差σzσz 2=(0.3σx)2+(0.7σy)2+2×0.3×0.7×σx×σy×CORR（X，Y）将相关数值代入σz2=0.000036+0.003136-0.0002688σz2=0.0029032σz=0.054（3）教授退休金年收益率的分布服从正态分布Z~N （0.112 ，0.054）（4）教授年收益在10%和15%之间的概率P设K为服从一个均值μz=0.112和标准差σz =0.054的正态分布那么：P（0.1≤K≤0.15）=P（Z≤（0.15-μz ）/σz）- P（Z≤（0.1-μz）/σz）将相关数值代入公式：P（0.1≤K≤0.15）=P（Z≤（0.15-0.112）/0.054）- P（Z≤（0.1-0.112）/0.054）= P（Z≤0.704）- P（Z≤-0.22）检查表A.1 在表中得到数字：P（0.1≤K≤0.15）=0.758-0.4129=0.3451因此，教授年收益在10%和15%之间的概率为34.51%。

P193 4.4一个制造立体声音响系统的公司宣称，其个人CD 播放机在利用碱性电池的情况下能够连续播放近8小时。

为了给出这个干劲冲天，共测试了35个利用新的碱性电池的CD 播放机，并记录播放机电池的使用时间，平均时间是8.3小时，寿标准利离差是1.2小时。

（A ）构造一个新的利用新的碱性电池的CD 播放机电池使用的平均时间的95%的致信区间。

（B ）为了估计利用新的碱性电池的CD 播放机电池使用的平均时间位于正或负10分钟范围内，以及99%的置信水平，确定所要求的样本大小。

答案：样本数大于30的为大样本。

P195 4.17在一家百货商店的两个分店，民意调查者随机地在第一个分店抽取了100个顾客，在第二个分店抽取了80个顾客，所有的调查都是在同一天进行的。

在第一个分店，平均每个顾客的消费金额是41.25美元，样本标准离差是24.25美元。

在第二个分店，平均每个顾客的消费金额是45.74美元，样本标准利差是34.76美元。

（A ）构造两个分店中每个分店每个顾客消费金额均值的一个95%的置信区间。

（B ）构造两个分店中每个顾客消费金额均值差异的一个95%的置信区间。

（1）答第一个分店每个顾客消费金额均值的一个95%的置信区间应为：⎥⎦⎤⎢⎣⎡+---n C X n C X x x x x σσ,-X 为第一个分店随机抽取顾客消费额均值，-X =41.5，样本大小为n, n x =100；同时，当βx =95% 时C x =1.96，则 σx 表示样本的标准离差σx =24.25。

将以上数值代入，则：第一个分店每个顾客消费金额均值的一个95%的置信区间应为：⎥⎦⎤⎢⎣⎡X +X -10025.2496.125.41,10025.2496.125.41[]003.46,497.36同理，第二个分店每个顾客消费金额均值的一个95%的置信区间将表示为：⎥⎦⎤⎢⎣⎡+---n C Y n C Y y y y y σσ,⎥⎦⎤⎢⎣⎡X +X -8076.3496.175.45,8076.3496.175.45[]367.53,133.38（2）答两个分店顾客消费金额均值之差的一个95%的置信区间应表示为：⎥⎥⎦⎤⎢⎢⎣⎡++-+------y yx x y y x x n n C Y X n n C Y X 2222,σσσσ 100=x n 80=y n将相关数值代入：⎥⎥⎦⎤⎢⎢⎣⎡++-+--8076.3410025.2496.175.4525.41,8076.3410025.2496.175.4525.412222[]3101.0,311.9-解：（a）对于表6.31提出的自变量，设：Y：欠税（$）X1：税前总收入($)X2：细目单A扣除部分($)X3：细目单C收入部分($)X4：细目单C部分扣除百分比(%)X5：家庭办公室指标则预测纳税人欠税的回归模型为：Y= aX1 + bX2 + cX3 + dX4 + eX5 + ε根据计算机的回归计算结果，代入系数得：Y= 0.292X1 - 0.012X2 + 0.188X3 + 104.625X4 - 3784.564X5 + 3572.406 回归统计Multiple R0.937041964R Square0.878047641Adjusted R Square0.844171986标准误差3572.406308观测值24方差分析df SS MS F Significance F回归分析5 1.65E+09 3.31E+0825.91972 1.23631E-07残差18 2.3E+0812762087总计23 1.88E+09Coefficients标准误差t Stat P-value Lower 95%Upper 95%下限 95.0%上限 95.0% Intercept-8414.7227796239.235-1.348680.194165-21522.869874693.424-21522.94693.424税前总收入($)0.2929550870.02917710.040778.39E-090.2316574330.3542530.2316570.354253细目单A扣除部分($)-0.0120617160.161062-0.074890.941129-0.3504394710.326316-0.350440.326316细目单C收入部分($)0.1877365280.167179 1.1229680.276207-0.1634932270.538966-0.163490.538966细目单C部分扣除百分比(%)104.624828443.09016 2.4280440.0258914.09575311195.153914.09575195.1539家庭办公室指标-3784.5647911827.084-2.071370.052973-7623.12520153.99562-7623.1353.99562-500050001000050000100000显然，从回归统计结果上看，这些自变量的组合对欠税预测值Y的影响并不显著。