数据模型与决策习题解答

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数据模型与决策练习题含答案

1、某企业目前的损益状况如在下：销售收入（1000件×10元/件） 10 000销售成本：变动成本（1000件×6元/件） 6 000固定成本 2 000销售和管理费（全部固定） 1 000利润 1 000（1）假设企业按国家规定普调工资，使单位变动成本增加4％，固定成本增加1％，结果将会导致利润下降。

为了抵销这种影响企业有两个应对措施：一是提高价格5％，而提价会使销量减少10％；二是增加产量20％，为使这些产品能销售出去，要追加500元广告费。

请做出选择，哪一个方案更有利？（2）假设企业欲使利润增加50％，即达到1 500元，可以从哪几个方面着手，采取相应的措施。

2、某企业每月固定制造成本1 000元，固定销售费100元，固定管理费150元；单位变动制造成本6元，单位变动销售费0.70元，单位变动管理费0.30元；该企业生产一种产品，单价10元，所得税税率50％；本月计划产销600件产品，问预期利润是多少?如拟实现净利500元，应产销多少件产品?3、某企业生产甲、乙、丙三种产品，固定成本500000元，有关资料见下表（单位：元）：要求：（1）计算各产品的边际贡献；（2）计算加权平均边际贡献率；（3）根据加权平均边际贡献率计算预期税前利润。

4、某企业每年耗用某种材料3 600千克，单位存储成本为2元，一次订货成本25元。

则经济订货批量、每年最佳订货次数、最佳订货周期、与批量有关的存货总成本是多少？5．有10个同类企业的生产性固定资产年平均价值和工业总产值资料如下：（1）说明两变量之间的相关方向；（2）建立直线回归方程；（3）估计生产性固定资产（自变量）为1100万元时总产值（因变量）的可能值。

6、某商店的成本费用本期发生额如表所示，采用账户分析法进行成本估计。

首先，对每个项目进行研究，根据固定成本和变动成本的定义及特点结合企业具体情况来判断，确定它们属于哪一类成本。

例如，商品成本和利息与商店业务量关系密切，基本上属于变动成本；福利费、租金、保险、修理费、水电费、折旧等基本上与业务量无关，视为固定成本。

数据模型与决策试题及参考答案

数据模型与决策试题及参考答案本文为《数据模型与决策》复，共分为五个填空题。

1.已知成年男子的身高服从正态分布N(167.48,6.092)，随机调查100位成年男子的身高，那么，这100位男子身高的平均数服从的分布是N(167.48,0.609)。

2.某高校想了解大学生每个月的消费情况，随机抽取了100名大学生，算得平均月消费额为1488元，标准差是2240元。

根据正态分布的“68-95-99”法则，该高校大学生每个月的消费额的95%估计区间为[1040,1936]。

3.从遗传规律看，一个产妇生男生女的概率是一样的，都是50%，但也有个人的特殊情况。

假设某人前一胎是女孩，那么她的下一胎也是女孩的概率为0.55；如果某人前一胎是男孩，那么她的下一胎还是男孩的概率为0.48.已知___第一胎是女孩，那么她的第三胎生男孩的概率是0.4653.4.调查发现，一个刚参加工作的MBA毕业生在顶级管理咨询公司的初始年薪可以用均值为9万美元和标准差是2万美元的正态分布来表示，那么一个这样的毕业生初始年薪超过9万美元的概率是0.5.5.结合生活实际，判断两个量之间的相关系数大概有多大？比如问您孩子身高与父母身高的的相关系数可能是0.6.1.孩子与父母的身高存在相关性，这个相关性可以用相关系数来衡量。

相关系数的取值范围为-1到1，绝对值越接近1表示相关性越强，绝对值越接近0表示相关性越弱。

在这个问题中，孩子与父母平均身高的相关性比较高，应该选0.9作为相关系数。

2.模拟仿真的关键步骤包括：确定仿真目标、建立仿真模型、选择仿真工具、设计实验方案、进行仿真实验、分析仿真结果、验证仿真模型。

模拟仿真是一种通过计算机模拟来研究和分析实际系统的方法，可以帮助人们更好地理解和预测系统的行为，从而提供决策支持和优化方案。

3.___某天上班路上捡到10元钱属于小概率事件。

小概率事件是指在一次试验中，出现的概率很小的事件。

通常认为，小概率事件的概率小于等于0.05.在这个问题中，其他选项中抛硬币的结果全是正面的概率都大于0.05，因此不属于小概率事件。

数据模型与决策习题与参考答案

《数据模型与决策》复习题及参考答案第一章绪言一、填空题1．运筹学的主要研究对象是各种有组织系统的管理问题，经营活动。

2．运筹学的核心是运用数学方法研究各种系统的优化途径及方案，为决策者提供科学决策的依据。

3．模型是一件实际事物或现实情况的代表或抽象。

4、通常对问题中变量值的限制称为约束条件，它可以表示成一个等式或不等式的集合。

5．运筹学研究和解决问题的基础是最优化技术，并强调系统整体优化功能。

运筹学研究和解决问题的效果具有连续性。

6．运筹学用系统的观点研究功能之间的关系。

7．运筹学研究和解决问题的优势是应用各学科交叉的方法，具有典型综合应用特性。

8．运筹学的发展趋势是进一步依赖于_计算机的应用和发展。

9．运筹学解决问题时首先要观察待决策问题所处的环境。

10．用运筹学分析与解决问题，是一个科学决策的过程。

11.运筹学的主要目的在于求得一个合理运用人力、物力和财力的最佳方案。

12．运筹学中所使用的模型是数学模型。

用运筹学解决问题的核心是建立数学模型，并对模型求解。

13用运筹学解决问题时，要分析，定议待决策的问题。

14．运筹学的系统特征之一是用系统的观点研究功能关系。

15.数学模型中，“s· t ”表示约束。

16．建立数学模型时，需要回答的问题有性能的客观量度，可控制因素，不可控因素。

17．运筹学的主要研究对象是各种有组织系统的管理问题及经营活动。

二、单选题1. 建立数学模型时，考虑可以由决策者控制的因素是（ A ）A．销售数量B．销售价格C．顾客的需求D．竞争价格2．我们可以通过（C）来验证模型最优解。

A．观察B．应用C．实验D．调查3．建立运筹学模型的过程不包括（ A ）阶段。

A．观察环境B．数据分析C．模型设计4. 建立模型的一个基本理由是去揭晓那些重要的或有关的（D．模型实施B）A 数量B变量C约束条件D目标函数5.模型中要求变量取值（ D ）A可正B可负C非正D非负6. 运筹学研究和解决问题的效果具有（A）A连续性B整体性C阶段性D再生性7.运筹学运用数学方法分析与解决问题，以达到系统的最优目标。

数据-模型与决策练习题含答案

请做出选择，哪一个方案更有利？（2）假设企业欲使利润增加50％，即达到1 500元，可以从哪几个方面着手，采取相应的措施。

4、某企业每年耗用某种材料3 600千克，单位存储成本为2元，一次订货成本25元。

6、某商店的成本费用本期发生额如表所示，采用账户分析法进行成本估计。

首先，对每个项目进行研究，根据固定成本和变动成本的定义及特点结合企业具体情况来判断，确定它们属于哪一类成本。

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》复习题及参考答案

《数据模型与决策》复习题及参考答案《数据模型与决策》复习题及参考答案第一章绪言一、填空题1．运筹学的主要研究对象是各种有组织系统的管理问题，经营活动。

2．运筹学的核心是运用数学方法研究各种系统的优化途径及方案，为决策者提供科学决策的依据。

3．模型是一件实际事物或现实情况的代表或抽象。

4、通常对问题中变量值的限制称为约束条件，它可以表示成一个等式或不等式的集合。

5．运筹学研究和解决问题的基础是最优化技术，并强调系统整体优化功能。

运筹学研究和解决问题的效果具有连续性。

6．运筹学用系统的观点研究功能之间的关系。

7．运筹学研究和解决问题的优势是应用各学科交叉的方法，具有典型综合应用特性。

8．运筹学的发展趋势是进一步依赖于_计算机的应用和发展。

9．运筹学解决问题时首先要观察待决策问题所处的环境。

10．用运筹学分析与解决问题，是一个科学决策的过程。

11.运筹学的主要目的在于求得一个合理运用人力、物力和财力的最佳方案。

12．运筹学中所使用的模型是数学模型。

用运筹学解决问题的核心是建立数学模型，并对模型求解。

13用运筹学解决问题时，要分析，定议待决策的问题。

14．运筹学的系统特征之一是用系统的观点研究功能关系。

15.数学模型中，“s〃t”表示约束。

16．建立数学模型时，需要回答的问题有性能的客观量度，可控制因素，不可控因素。

17．运筹学的主要研究对象是各种有组织系统的管理问题及经营活动。

二、单选题1.建立数学模型时，考虑可以决策者控制的因素是第 1 页共40页A．销售数量B．销售价格C．顾客的需求D．竞争价格 2．我们可以通过来验证模型最优解。

A．观察B．应用C．实验D．调查 3．建立运筹学模型的过程不包括阶段。

A．观察环境B．数据分析C．模型设计D．模型实施 4.建立模型的一个基本理是去揭晓那些重要的或有关的 A数量B变量 C 约束条件 D 目标函数5.模型中要求变量取值A可正B可负C非正D非负 6.运筹学研究和解决问题的效果具有A 连续性B 整体性C 阶段性D 再生性7.运筹学运用数学方法分析与解决问题，以达到系统的最优目标。

数据模型与决策练习题含答案

请做出选择，哪一个方案更有利？（2）假设企业欲使利润增加50％，即达到1 500元，可以从哪几个方面着手，采取相应的措施。

4、某企业每年耗用某种材料3 600千克，单位存储成本为2元，一次订货成本25元。

6、某商店的成本费用本期发生额如表所示，采用账户分析法进行成本估计。

首先，对每个项目进行研究，根据固定成本和变动成本的定义及特点结合企业具体情况来判断，确定它们属于哪一类成本。

《数据模型与决策》复习题及参考答案

答：运筹学的系统特征可以概括为以下四点：一、用系统的观点研究功能关系
二、应用各学科交叉的方法三、采用计划方法四、为进一步研究揭露新
问题。
5、线性规划数学模型具备哪几个要素？答：（1）.求一组决策变量Xi或xj
的值（i=1,2,…m j=1,2…n）使目标函数达到极大或极小；（2）.表示约束条件的数学式都是线性等式或不等式；（3）.表示问题最优化指标的目标函数都是决策变量的线性函数
17.求解线性规划问题可能的结果有无解，有唯一最优解，有无穷多个最优解。
18.如果某个约束条件是“w”情形，若化为标准形式，需要引入一松弛变量。
19.如果某个变量X为自由变量，则应引进两个非负变量XjXj,同时令X=
20.表达线性规划的简式中目标函数为max(min)Z=ECjX』。
二、单选题
1.如果一个线性规划问题有n个变量，m个约束方程(m<n)，系数矩阵的数
A.观察B.应用C.实验D.调查
3•建立运筹学模型的过程不包括（A）阶段。
A.观察环境B.数据分析C.模型设计D.模型实施
4.建立模型的一个基本理由是去揭晓那些重要的或有关的（B）
A数量B变量C约束条件D目标函数
5.模型中要求变量取值（D）
A可正B可负C非正D非负
6.运筹学研究和解决问题的效果具有（A）
的集合。
5•运筹学研究和解决问题的基础是最优化技术，并强调系统整体优化功能。运筹学研究和解决问题的效果具有连续性。
6•运筹学用系统的观点研究功能之间的关系。
7•运筹学研究和解决问题的优势是应用各学科交叉的方法，具有典型综合应用
特性。
8•运筹学的发展趋势是进一步依赖于_计算机的应用和发展。
9•运筹学解决问题时首先要观察待决策问题所处的环境。

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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：29.625 18.000 8.625 18.5009.250 79.375 1.250 14.00010.000 8.750 24.250 35.25032.250 53.375 11.500 9.37534.000 8.000 7.625 33.62516.500 11.375 48.375 9.00037.000 37.875 21.625 19.37529.625 16.625 52.000 9.25043.250 28.500 30.375 31.12538.000 38.875 18.000 33.500（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1.25, 7.625, 8, 8.625, 8.75, 9, 9.25, 9.25, 9.375, 10, 11.375, 11.5, 14, 16.5, 16.625, 18, 18, 18.5, 19.375, 21.625, 24.25, 28.5, 29.625, 29.625, 30.375, 31.125, 32.25, 33.5, 33.625, 34, 35.25, 37, 37.875, 38, 38.875, 43.25, 48.375, 52, 53.375, 79.375，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。

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87.5%，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 ≤≤⋯≤≤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。

从茎叶图和箱线图可以看出其分布特征：中间（上下四分位数部分）比较集中，但是最大值是奇异点。

数据分布明显不对称，右拖尾比较长。

（4）现用原始数据计算常用的描述性统计量样本均值：421875.2540/875.1016401401===∑=i ixx样本方差：196.26340391240122=⎪⎪⎭⎫ ⎝⎛⨯-=∑=x x s i i样本标准差：2233.16196.2634039124012==⎪⎪⎭⎫ ⎝⎛⨯-=∑=x x s i i 用分组数据计算常用的描述性统计量：97561='∑=k kk x f，33925612='∑=k kk x f样本均值：375.2440/97540161=='=∑=k kk x fx样本方差：4968.260403912612=⎪⎪⎭⎫ ⎝⎛⨯-'=∑=x x f s k kk 样本标准差：1399.164968.260403912612==⎪⎪⎭⎫ ⎝⎛⨯-'=∑=x x f s k k k 与用原始数据计算的结果差别不大。

此外，可以用Excel 中的数据分析直接进行描述性统计分析，结果如下：补充习题：1. 测量血压14次,记录收缩压,得样本如下:121,123,119,130,125,115,128,126,109,112,120,126,125,125 求样本均值，样本方差，样本中位数，众数和极差。

2．根据列表数据3. 调查30个中学生英语成绩,得样本如下：54, 66, 69, 69, 72,75, 77, 75, 76, 79, 76,77, 78, 79,81, 81, 85, 87, 83, 84,89, 86,89, 89, 92, 95,96,96, 98, 99把样本分为5组，组距为10，且最小组的下限为50，作出列表数据和直方图补充习题答案1. 测量血压14次,记录收缩压,得样本如下:121,123,119,130,125,115,128,126,109,112,120,126,125,125 求样本均值，样本方差，样本中位数，众数和极差。

解：排序：109 112 115 119 120 121 123 125 125 125 126 126128130均值：1nii xx n==∑= 121.71方差：222211()11nniii i x x xnxs n n ==--==--∑∑= 37.76中位数：12202n nx x m -+== 124众数：m e = 125 极差：R=x n -x 1= 212．根据列表数据解:样本均值：11ki ii x x f n ==∑= 31.59091样本方差：222211()11kni ii i i i x x f x f nxs n n ==--==--∑∑=25.32468样本标准差：s =3调查30个中学生英语成绩,得样本如下：54, 66, 69, 69, 72,75, 77, 75, 76, 79, 76,77, 78, 79,81, 81, 85, 87, 83, 84,89, 86,89, 89, 92, 95,96,96, 98, 99把样本分为5组，组距为10，且最小组的下限为50，作出列表数据和直方图解：列表第四章习题(p118)21.下面的10个数据是来自一个正态总体的样本数据： 10，8，16，12，15，6，5，14，13，9 （1）总体均值的点估计是多少？（2）总体标准差的点估计是多少？（3）总体均值99%的置信区间是多少？解：（1）总体均值的点估计8.10108101ˆ=⨯==x μ （2）总体标准差的点估计()7947.34.148.10101296911091ˆ221012==⨯-=⎪⎪⎭⎫ ⎝⎛⨯-==∑=x x s i i σ （3）这是正态总体方差未知的条件下，总体均值μ的区间估计问题99.01=-α，01.0=α，2498.3)9()1(005.02/==-t n t α总体均值99%的置信区间为：⎪⎪⎭⎫⎝⎛-+--n s n t x n s n t x )1(,)1(2/2/αα）（6997.14,9.6107947.32498.38.10,107947.32498.38.10=⎪⎪⎭⎫ ⎝⎛⨯+⨯-= 第五章习题(p154)7.某一问题的零假设和备择假设分别如下：25:0≥μH 25:1<μH当某个样本容量为100，总体标准差为12时，对下面每一个样本的结果，都采用显著性水平05.0=α计算检验统计量的值，并得出相应的结论。

（1）0.221=x 。

（2）5.232=x 。

（3）8.223=x 。

（4）0.244=x 。

解：这是总体分布未知，大样本前提下，总体均值的单边检验问题。

故，可以用大样本情况下单个总体均值的检验。

提出原假设与备择假设：25:0≥μH 25:1<μH 选择检验统计量nx z /25σ-=，当0H 成立时，)1,0(~//25N nx nx z σμσ-≤-=给定显著性水平05.0=α，645.105.0==z z α，拒绝域645.105.0-=-≤z z （1）0.221=x ，645.15.2100/122522/25-<-=-=-=nx z σ，拒绝0H 。

接受1H ，即不能认为25≥μ。

（2）5.231=x ，645.125.1100/12255.23/25->-=-=-=nx z σ，接受0H 。

即认为25≥μ。

（3）8.221=x ，645.183333.1100/12258.22/25-<-=-=-=nx z σ，拒绝0H 。

接受1H ，即不能认为25≥μ。

（4）0.241=x ，645.18333.0100/122524/25->-=-=-=nx z σ，接受0H 。

即认为25≥μ。

12.有一项研究要作的假设检验是： 20:0=μH 20:1≠μH某个样本有6个数据，他们分别是：20，18，19，16，17，18。

根据这6个数据，分别回答以下问题：（1）它们的均值和标准差各是多少？（2）当显著性水平05.0=α时，拒绝规则是什么？（3）计算检验统计量t 的值。

（4）根据以上信息，你所得出的结论是什么？解：说明：本题是小样本，应该有总体服从正态分布),(2σμN 的假定。

（1）由样本数据得6=n ，10861=∑=i ix，1954612=∑=i ix样本均值：186/1086161===∑=i ixx ；样本方差：2)1861954(51651226122=⨯-=⎪⎪⎭⎫ ⎝⎛⨯-=∑=x x s i i样本标准差：4142.126512612==⎪⎪⎭⎫ ⎝⎛⨯-=∑=x x s i i （2）在总体服从正态分布的假定之下，这是正态总体方差未知的条件下，总体均值的双边检验问题，用t 检验。

提出原假设与备择假设：20:0=μH 20:1≠μH 选择检验统计量：ns x t /20-=，当原假设0H 成立时，)1(~//20--=-=n t ns x ns x t μ当显著性水平05.0=α时，5706.2)5()1(025.02/==-t n t α，因此：拒绝域为： 5706.2)5(025.0=≥t t（3）计算检验统计量t 的值236.255/22018/20-=-=-=-=ns x t（4）5706.2236.2/20<=-=ns x t ，接受0H 。