数据模型与决策结课作业

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。

数据模型与决策习题与参考答案

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 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 ）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。

数据模型与决策作业

《数据模型与决策》作业题1、某装配车间共有装配工人200人，某日对其日装配工件数进行统计，分组资料如表1，根据资料计算：（1）算术平均数；（2）标准差；（3）标准差系数；（4）偏态系数和峰度系数。

表1 日装配工件数资料日装配工件数（件）工人数（人）4—6 6—8 8—10 10—12 12—14 25 40 85 35 15 合计2002、对来自三个不同地区的蔬菜的品质进行检验，结果如表2所示。

试在05.0=α的显著性水平下，确定蔬菜的品质与产地是否存在依赖关系。

（列联分析）表2 对三个产地的蔬菜质检结果质量等级产地一等品重量（公斤）二等品重量（公斤）三等品重量（公斤）合计甲 60 40 100 200 乙 50 80 70 200 丙 70 80 50 200 合计1802002206003、某企业采用四种不同的方式包装其产品，采用随机抽样的方式选择了5个销售区域对每种包装方式做了采样，得到如表3中所示的数据，试以05.0=α的显著性水平作出统计决策。

（方差分析）表3 不同包装方式的销售情况包装方式销售区域方式一方式二方式三方式四 1 66 32 46 25 2 45 16 32 15 3 78 66 12 30 4 25 23 45 16 5344650404、已知我国1990年～1999年的货运量y 、工业总产值x 1、农业总产值x 2资料如表4所示：表4 1990年～1999年我国货运量、工业总产值和农业总产值资料年份货运量（万吨）工业总产值（亿元）农业总产值（亿元）1990 970602 23924 7662.1 1991 985793 26625 8157.0 1992 1045899 34599 9084.7 1993 1115771 48402 10995.5 1994 1180273 70176 15750.5 1995 1234810 91894 20340.9 1996 1296200 99595 22353.7 1997 1278087 113733 23788.4 1998 1267200 119048 24541.9 1999129265012611124519.1要求计算：（1）相关系数（2）二元线性回归方程（3）进行各种统计显著性检验。

数据模型与决策--作业大全详解

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。

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）。

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。

数据模型与决策课

数据模型与决策课程大作业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》练习题及答案

《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∑==n1i ixn1x =37.1岁样本方差)X n X (1-n 1)X (X 1-n 1s 2n 1i 2i2n 1i i 2-=-=∑∑===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)X X (21m 1)2n ()2n (++==(35+36)/2=35.5岁极差=-=1)()n (X X R 63-21=42岁众数=0m 35岁（ii ）样本分为7组、且组距相同的列表数据、直方图如下所示样本均值i k1i f Xi n 1X ∑===36.3岁样本方差)X n f X (1-n 1f )X (X 1-n 1s 2k 1i i 2i i2k 1i i 2-=-=∑∑===174.3724 样本中位数810230730f F 2n i I m -+=-+==34.375岁众数=--⨯-+=---+=+448248730f f 2f f f iI m 1m 1-m m 1-m m 033.5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：解：样本均值i k1i f Xi n 1X ∑===1566.667元样本标准差)X n f X (1-n 1f )X (X 1-n 1s 2k 1i i2i i 2k 1i i -=-=∑∑===398.1751元样本中位数在累计74人的那一组，m=1500元；众数1500m 0=元。

数据模型与决策--作业大全

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。

MBA《数据、模型与决策》作业(案例)

《数据、模型与决策》案例1企业背景SGT特殊钢铁公司是我国西部地区最大、西北地区唯一的百万吨资源型特殊钢生产基地，是国家级创新型企业、国家军工产品配套企业。

经过多年的发展，通过传统产业升级改造和优势产业发展壮大，已形成年产焦炭75万吨、普通钢400万吨、特殊钢400万吨的综合生产能力，成为集“特钢制造、煤炭焦化”为一体的资源综合开发钢铁联合企业。

SGT公司钢铁制造始终坚持以“打造西部重要的特种钢生产基地”为己任，牢固树立品牌意识，顺应新型特殊钢材料发展趋势，努力提升工艺技术及装备水平，拥有“高炉—转炉—精炼—连铸—连轧”五位一体的长流程生产线和“电炉（兑铁水）冶炼—精炼—连铸（模铸）—连轧（半连轧）”的短流程生产线。

建成了极具特色的“功勋牌”特钢产品体系，产品涵盖轴承钢、模合结钢、碳工钢、工具钢等钢种，产品广泛应用于汽车、工程制造、机械制造、石油、军工、航空、铁路运输、新能源、新基建等多种行业。

公司整体生产工艺及技术在行业中处于领先水平，特别是“十二五”期间，公司对工艺装备进行了全面升级改造，先后建成110吨Consteel电炉、410* 530mm三机三流大方坯连铸机、精品特钢大棒材生产线、精品特钢小棒材生产线。

尤其是精品特钢大、小棒材生产线采用了当今世界顶尖工艺技术及装备，达到国际钢铁工业的先进水平。

公司目前已拥有五十余项特殊钢生产的专有技术，其中主要有钢包、连铸耐材材质控制技术，分钢种脱氧工艺与技术，分钢种渣系控制技术，残余元素与有害元素的控制技术，非金属夹杂物形态控制技术，硫化物夹杂弥散细化技术，含硫钢纯净化技术，含S，Al，B钢连铸技术，炉前化学成分精确控制技术，炉前淬透性（DI值）动态控制技术，模铸、连铸碳偏析及凝固组织控制技术，在线正火轧制技术，在线超快冷技术，在线控轧控冷一热机轧制（TMCP）技术，大规格热轧材高精度轧制技术，小规格热轧材高精度轧制技术，热轧材非标规格轧制技术，银亮材剥、碾光表面光洁度控制技术，低中碳钢低硬度球化退火技术，钢材零缺陷无损探伤检测技术，电炉、转炉全铁冶炼技术，电炉、转炉底吹搅拌，高品质特殊钢精品生产技术集成，富氮合金、高氮中间合金及氮化硅锰冶炼螺纹钢技术，小粒度石灰石在电炉、转炉直接炼钢技术，高品质系列钢锭缺陷控制技术，马氏体不锈钢大方坯连铸技术，电弧炉炼钢复合吹炼技术等。

数据模型与决策课程总结

数据模型与决策课程总结学习总结（期中论文）我们所用的教材叫做《数据、模型与决策》，我记得老师第一天给我们上课就提到过一些基本的概念以及思想，例如“什么是管理”；“什么是模型”；“如何对实际问题简化”等等。

在这其中我认为非常重要的有以下几点：首先，管理的最初根源是因为资源是有限的。

如何将有限的资源进行合理配制、优化从而达到最大的效益是我们应该要去注意的问题。

其次，数学问题是有最优解的，当我们给定了一个确定的数学问题我们能够得到一个确定的解，但当我们在研究一个给定的现实管理问题的时候，我们是很难去找到一个最优解的，甚至可以说，管理问题是没有最优解的。

（这不同于我们平时所做的运筹学等问题，因为我们平时所做的问题都已经经过了很多的化简，已经把现实管理问题进行了抽象，与其说那些问题是一个管理问题不如说它们是数学问题）这是因为现实中的管理问题比较复杂，具有很强的不确定性，我们只能是抓住主要矛盾，暂且不考虑次要矛盾。

（当然了，当我们已经解决了主要矛盾之后我们可以开始考虑次要矛盾，因为这个时候次要矛盾已经上升为主要矛盾了。

）所以我们去寻找的是管理问题的满意解而不是最优解。

这两点在后面的学习建模中得到了很好的验证。

我们之前的学习大多是倾向于解决一个数学问题而不是一个管理问题。

这一门课之所以在大三才开设我认为有其道理，在没有掌握基本的数学基本知识之前，我们是不可能很好地解决管理问题的，因为我们解决一个管理问题是先将其转化为一个可以解决的数学问题。

但是并不是说我们掌握了高数、运筹学等知识就能顾很好的解决管理问题，因为如何把现实存在复杂的管理问题转化成为我们可以解决的数学问题正是这门课的核心内容之一。

以企业的生产计划安排作为例子，总结一下应用现行规划建模的步骤：●我们的问题是什么？（如何安排生产）如何组合不同产品的生产、生产的种类。

●我们能做什么？（不同产品的生产数量）明确决策变量，也就是管理中可以人为设定的要素。

●确定决策的准则（利润最大化、成本最小化、社会责任最大化）根据决策变量写出目标函数。

MBA数据、模型课堂大作业

《数据、模型与决策》案例案例1 北方化工厂月生产计划安排（0.7+0.3）一. 问题提出根据经营现状和目标，合理制定生产计划并有效组织生产，是一个企业提高效益的核心。

特别是对于一个化工厂而言，由于其原料品种多，生产工艺复杂，原材料和产成品存储费用较高，并有一定的危险性，对其生产计划作出合理安排就显得尤为重要。

现要求我们对北方化工厂的生产计划做出合理安排。

二. 有关数据1．生产概况北方化工厂现在有职工120人，其中生产工人105名。

主要设备是2套提取生产线，每套生产线容量为800kg，至少需要10人看管。

该工厂每天24小时连续生产，节假日不停机。

原料投入到成品出线平均需要10小时，成品率约为60％，该厂只有4吨卡车1辆，可供原材料运输。

2．产品结构及有关资料该厂目前的产品可分为5类，多有原料15种，根据厂方提供的资料，经整理得表1。

3．供销情况①根据现有运出条件，原料3从外地购入，每月只能购入1车。

②根据前几个月的销售情况，产品1和产品3应占总产量的70％，产品2的产量最好不要超过总产量的5％，产品1的产量不要低于产品3与产品4的产量之和。

问题：a．请作该厂的月生产计划，使得该厂的总利润最高。

b．找出阻碍该厂提高生产能力的瓶颈问题，提出解决办法。

案例2 石华建设监理公司监理工程师配置问题（0.8+0.2）石华建设监理公司（国家甲级），侧重国家大中型项目的监理，仅在河北省石家庄市就正在监理七项工程，总投资均在5000万元以上。

由于工程开工的时间不同，多工程工期之间相互搭接，具有较长的连续性，2012年监理的工程量与2011年监理的工程量大致相同。

每项工程安排多少监理工程师进驻工地，一般是根据工程的投资，建筑规模，使用功能，施工的形象进度，施工阶段来决定的。

监理工程师的配置数量是随之而变化的。

由于监理工程师从事的专业不同，他们每人承担的工作量也是不等的。

有的专业一个工地就需要三人以上，而有的专业一人则可以兼管三个以上的工地。

数据模型与决策作业

数据模型与决策作业一、Cropain公司基建部问题1、首先将60组数据单独列出，找到因变量Y（earn）和自变量X（size、p15、inc、nrest、price）,数据如下所示：打开EXCEL表格-工具-数据分析-回归-确定-Y值区域为EARN 列，X值区域为SIZE到PRICE列，点标志-确定，生成数据如下：SUMMARY OUTPUT回归统计Multiple R 0.92532R Square 0.856217Adjusted RSquare 0.842903标准误差36.20023观测值60方差分析df SS MS F SignificanceF回归分析 5 421397.1 84279.41 64.313 1.64E-21残差54 70764.67 1310.457总计59 492161.7回归方程如下：Y(earn)=0.77size+0.04p15+8.78inc+1.41nrest-2.69price-35 3.82多重共线性检验如下：回到EXCEL表格-工具-数据分析-相关系数-确定,选定区域为从EARN到PRICE的所有列,点击标志在第一行,确定,生成相关性系数.2、将Y变量（earn）和X变量（从size到price）粘贴到MINITAB中，统计-回归-逐步回归-响应（earn），预测变量（从size到price）-确定，得出数据如下：逐步回归: EARN 与 SIZE, P15, INC, NREST, PRICE入选用 Alpha: 0.15 删除用 Alpha: 0.15响应为 5 个自变量上的 EARN，N = 50步骤 1 2 3 4 5常量 -0.6348 -354.7460 -421.2498 -412.5582 -379.4336P15 0.0459 0.0416 0.0397 0.0436 0.0432T 值 5.81 6.35 7.20 10.00 10.49P 值 0.000 0.000 0.000 0.000 0.000INC 11.7 12.8 9.7 9.7T 值 4.93 6.37 5.85 6.19P 值 0.000 0.000 0.000 0.000NREST 1.33 1.48 1.46T 值 4.56 6.43 6.70P 值 0.000 0.000 0.000SIZE 0.61 0.63T 值 5.51 6.04P 值 0.000 0.000PRICE -2.01T 值 -2.55P 值 0.014S 68.0 55.8 46.8 36.5 34.5R-Sq 41.26 61.28 73.33 84.09 86.14R-Sq（调整） 40.04 59.63 71.59 82.67 84.56Mallows Cp 140.4 78.9 42.6 10.5 6.0五个变量的回归方程如下：Earn=0.0432p15+9.7inc+1.46nrest+0.63size-2.01price-379.43;数据中51到60相关数据如下：各组实际利润率=earn/k,因而51到60的每组实际利润率如下：各组的预测利润=变量回归方程上各项*相关各项的数据举例：第51店的预测利润为Earn=0.0432*2800+9.7*32.1+1.46*26+0.63*146-2.0111.6-379 .43=159.524;因而各组利润为：各组预测利润率、与实际利润率比较为：60组数据的相关系数和回归方程如下所示：逐步回归: EARN 与 SIZE, P15, INC, NREST, PRICE 入选用 Alpha: 0.15 删除用 Alpha: 0.15响应为 5 个自变量上的 EARN，N = 60步骤 1 2 3 4 5 常量 -3.083 -103.061 -145.274 -399.009 -353.820P15 0.0482 0.0501 0.0485 0.0444 0.0442T 值 6.15 7.95 9.10 10.10 10.95P 值 0.000 0.000 0.000 0.000 0.000SIZE 0.798 0.852 0.754 0.772T 值 5.73 7.23 7.73 8.61P 值 0.000 0.000 0.000 0.000NREST 1.39 1.45 1.41T 值 4.93 6.34 6.71P 值 0.000 0.000 0.000INC 8.8 8.8T 值 5.44 5.91P 值 0.000 0.000PRICE -2.69T 值 -3.37P 值 0.001S 71.7 57.6 48.5 39.5 36.2R-Sq 39.47 61.60 73.22 82.59 85.62R-Sq（调整） 38.42 60.26 71.79 81.33 84.29Mallows Cp 171.3 90.2 48.6 15.4 6.0Y（earn）=0.0442p15+0.772size+1.41nrest+8.8inc-2.69price-353.820 根据60组数据的预测回归方程对未来10组数据进行预测如下：二、菲拉托伊·里尤尼蒂纺织厂问题首先,把要求的相关决策变量清空,如下所示：根据题意，本题给出了各工厂生产机器的单位时间和月度最大使用时间，和各工厂生产四种产品的单位成本和运输成本，要求的是在满足各品种需求量的基础上各工厂如何生产总成本最低的问题，根据题意，首先建立目标函数：COST OF PRODUCTIONCOST OF TRANSPORTATION将各共厂两项成本相加可得：目标函数为各工厂各品种生产的成本与各工厂生产产品的乘积。

数据、模型与决策课后答案17

CD CHAPTER 17 SOLUTION CONCEPTS FOR LINEAR PROGRAMMINGReview Questions17.1-1 A corner point is a point that lies at a corner of the feasible region.17.1-2 The corner point with the best value of the objective function is the optimal solution.17.1-3 The simplex method has a way of quickly getting to the best corner point and detecting that thispoint is optimal, so it stops without needing to evaluate the rest of the corner points.17.1-4 No. It can have one optimal solution, an infinite number of optimal solutions, or no optimalsolution.17.1-5 Yes, and every point of the line segment between these two corner points will be an optimalsolution.17.1-6 The constraints of a linear programming problem can be so restrictive that it is impossible for asolution to satisfy all of the constraints simultaneously.17.1-7 If some necessary constraint were not included in the linear programming model, it is possible tohave no limit on the best objective function value.17.2-1 The best corner point is always an optimal solution.17.2-2 The big advantage of searching for an optimal solution simply by finding the best corner point isthat it tremendously reduces the number of solutions that need to be considered.17.2-3 When the simplex method is ready to move from the current corner point to the next one, theadjacent corner points are candidates to be the next one.17.2-4 If none of the adjacent corner points are better than the current corner point, then the currentcorner point is an optimal solution.17.2-5 If the simultaneous solution of a set of constraint boundary equations is feasible then that solutionis a corner point.17.2-6 Two corner points are adjacent to each other if they share all but one of the same constraintboundaries.17.2-7 The simplex method examines corner points in an order such that each one is adjacent to thepreceding one. This greatly streamlines the algebra.17.2-8 A problem with 20 decision variable and 30 functional constraints might have more than a billioncorner points.17.2-9 The simplex method can solve problems that the enumeration-of-corner-points method cannotbecause it is able to reach and identify the optimal corner point for a large problem after examining only a small fraction of all the corner points.17.3-1 The simplex method focuses solely on solutions that are corner points.17.3-2 Each iteration of the simplex method consists of a prescribed series of steps for moving from thecurrent corner point to a new corner point.17.3-3 Management science algorithms typically are iterative algorithms.17.3-4 Whenever possible, the initialization step of the simplex method chooses the origin (0,0) to be theinitial corner point to be examined.17.3-5 When the simplex method finishes examining a corner point it then gathers information about theadjacent corner points. It identifies the rates of improvement in the objective function along theedges that lead to the adjacent corner points.17.3-6 The optimality test consists simply of checking whether any of the edges give a positive rate ofimprovement in the objective function. If none do, then the current point is optimal.17.4-1 A problem with several thousand functional constraints and many thousand decision variables isnot considered unusually large for a fast computer.17.4-2 The software package needs to help formulate, input, and modify the model. In addition it shouldanalyze solutions from the model and present results in the language of management.17.5-1 Narenda Karmarkar.17.5-2 Today, the more powerful software packages include at least one interior-point algorithm alongwith the simplex method.17.5-3 Interior-point algorithms shoot through the interior of the feasible region toward an optimalsolution instead of taking a less direct path around the boundary of the feasible region.17.5-4 The computer time per iteration for an interior-point algorithm is many times longer than for thesimplex method.17.5-5 For fairly small problems, the number of iterations needed by an interior-point algorithm and thesimplex method tend to be somewhat comparable. For large problems, interior-point algorithms do not need many more iterations while the simplex method does. The reason for the largedifference is the difference in the paths followed.17.5-6 The simplex method is very well suited for what-if analysis while the interior-point approachcurrently has limited capability in this area.17.5-7 The limited capability for performing what-if analysis can be overcome by switching over to thesimplex method.17.5-8 Switching over begins by identifying a corner point that is very close to the final trial solution.Problems17.1Optimal Solution: (A 1, A 2) = (5, 5) and Profit = $15,000.17.2Optimal Solution: (TV , PM ) = (4, 3) and Cost = $12.17.3b & e) Optimal Solution: (x1 ,x2 ) = (3, 3) and Profit = $150.Corner Points: (0,0), (4.5,0), (3,3), (1.667,4), and (0,4).c)Optimal Solution: (x1, x2 ) = (3,3) and Profit = $150. 17.4b & e) Optimal Solution: (x1, x2 ) = (3, 5) and Profit = $1,900.Corner Points: (0,0), (6,0), (3,5), (1.5,6), and (0,6).c)Optimal Solution: (x1, x2 ) = (3, 5) and Profit = $1,900.17.5 a) Objective Function: Profit = $400x1 + $400x2Optimal Solution: (x1, x2 ) = (2, 6) and Profit = $3,200Objective Function: Profit = $500x1 + $300x2Optimal Solution: (x1, x2 ) = (4, 3) and Profit = $2,900Objective Function: Profit = $300x1– $100x2Optimal Solution: (x1, x2 ) = (4, 0) and Profit = $1,200 Objective Function: Profit = –$100x1 + $500x2Optimal Solution: (x1, x2 ) = (0, 6) and Profit = $3,000Objective Function: Profit = –$100x1– $100x2Optimal Solution: (x1, x2 ) = (0, 0) and Profit = $0 b) Objective Function: Profit = $400x1 + $400x2Optimal Solution: (x1, x2 ) = (2, 6) and Profit = $3,200Objective Function: Profit = $500x1 + $300x2Optimal Solution: (x1, x2 ) = (4, 3) and Profit = $2,900Objective Function: Profit = $300x1– $100x2Optimal Solution: (x1, x2 ) = (4, 0) and Profit = $1,200Objective Function: Profit = –$100x1 + $500x2Optimal Solution: (x1, x2 ) = (0,6) and Profit = $3,000Objective Function: Profit = –$100x1– $100x2Optimal Solution: (x1, x2 ) = (0, 0) and Profit = $017.6 a) True. (Example: Maximize Profit = –x1 + 4x2 )b) True. (Example: Maximize Profit = –x1 + 3x2 )c) False. (Example: Maximize Profit = –x1–x2 )17.7 a) x1≤ 6x2≤ 3–x1 + 3x2≤ 6b)c) Objective Function: Profit = –x1 + 5x2Optimal Solution: (x1, x2 ) = (3, 3) and Profit = $12d) Objective Function: Profit = –x1 + 2x2Optimal Solution: (x1, x2 ) = (0, 2) and Profit = $417.8 a)Adjustable CellsFinal Reduced Obj ectiv e Allow able Allow able Cell Name Value Cost Coefficient Increase Decrease$B$11Number of Units Activity 1 1.2502008$C$11Number of Units Activity 2 2.50030200 ConstraintsFinal Shadow Constraint Allow able Allow able Cell Name Value Price R.H. Side Increase Decrease$D$5Resource A Used16.250201E+30 3.75$D$6Resource B Used30 3.3333330 2.64713$D$7Resource C Used15015 1.6667 2.1429b) The sensitivity report indicates that the problem has other optimal solutions because theallowable increase of Activity 1 and the allowable decrease of Activity 2 are 0. An alternativeoptimal solution is shown below (obtained by adjusting the unit profit for Activity 2 to 29.99).c) The other optimal solutions will be located on the line segment connecting the two optimalsolutions found in parts a and b.d) Optimal Solution: (x 1, x 2 ) = (2.857, 1.429), (1.25, 2.5) and all points on the connecting line.Profit = $100 million.17.9a)Adjustable CellsFinal Reduced Obj ectiv e Allow able Allow able Cell NameValue Cost CoefficientIncrease Decrease$B$10Number of Units Activity 115.000.005004000$C$10Number of Units Activity 215.000.00300133.3333Constraints Final Shadow ConstraintAllow ableAllow able Cell NameValue Price R.H. Side Increase Decrease $D$5Resource A Used 30003006083.3333$D$6Resource B Used 2405024042.857140$D$7Resource C Used30004501E+30150b) The sensitivity report indicates that the problem has other optimal solutions because theallowable decrease of Activity 1 and the allowable increase of Activity 2 are 0. An alternative optimal solution is shown below (obtained by adjusting the profit for activity 2 to $300.01).c) The other optimal solutions will be located on the line segment connecting the two optimalsolutions found in parts a) and b).d) Optimal Solution: (x1, x2 ) = (15, 15), (2.5, 35.833) and all points on the connecting line.Profit = $12,000.17.10 Feasible Region:Case 1 (c2 = 0):If c1 > 0, the objective increases as x1 increases, so the optimal solution is (x1, x2) = (5.5, 0).If c2 < 0, the opposite is true, so the optimal solution is (x1, x2) = (0,0), (0, 1) and the connecting line.(Note if c1 = 0, every feasible point gives the optimal solution, as 0x1 + 0x2 = 0).Case 2 (c2 > 0): (slope of objective function line is –(c1 / c2))If –(c1 / c2) > 1/2 (or equivalently, c1/c2 < –1/2), then the optimal solution is (x1, x2) = (0, 1).If –(c1 / c2) < -2, then the optimal solution is (x1, x2) = (5.5, 0).If 1/2 > -(c1 / c2) > -2, then the optimal solution is (x1, x2) = (4, 3).(Note, if -(c1 / c2) = 1/2 or -2, then the optimal solution is at (4, 3) and (0, 1) or (5.5, 0),respectively, along with the connecting line.)Case 3 (c2 < 0): (slope of objective function line is still -(c1 / c2), but the objective function value increases as the line is shifted down)If –(c1 / c2) > 0 (i.e., c1 > 0), then the optimal solution is (x1, x2) = (5.5, 0).If –(c1 / c2) < 0 (i.e., c1 < 0), then the optimal solution is (x1, x2) = (0, 0).(Note, if –(c1 / c2) = 0, then the optimal solution is at (5.5, 0), (0, 0) and the connecting line.)17.11Solver could not find a feasible solution.c)x 117.12Solver could not find a feasible solution.c)17.13 a)b) Yes. Optimal Solution: (x1, x2 ) = (0, 10) and Profit = 10.c) No. The objective function value is maximized by sliding the objective function line to the right.This can be done forever, so there is no optimal solution.d) No, solutions exist that will make the profit arbitrarily large. This usually occurs when aconstraint is left out of the model.e)The Solver message was that the Set Cell values do not converge. There is no optimal solution, because a better solution can always be found.17.14 a)b) No. The objective function value is maximized by sliding the objective function line up. Thiscan be done forever, so there is no optimal solution.c) Yes. Optimal Solution: (x 1, x 2 ) = (10, 0) and Profit = 10.d) No, solutions exist that will make Z arbitrarily large. This usually occurs when a constraint isleft out of the model.e)The Solver message was that the Set Cell values do not converge. There is no optimal solution, because a better solution can always be found.17.15 a)b) Corner Point (0, 0): x2 = 0, and x1 = 0(0, 2): x1 = 0, and x2 = 2(1, 2): x1 + x2 = 3, and x1 = 2(2, 1): x2= 2, and x1 + x2 = 3(0, 2): x1 = 2, and x2 = 0c) Corner Points (x1, x2) = (0, 0), (2, 0), (2, 1), (1, 2), (0, 2)d) Corner Point (0, 0): (0, 2) and (2, 0) are adjacent(0, 2): (0, 0) and (1, 2) are adjacent(1, 2): (0, 2) and (2, 1) are adjacent(2, 1): (2, 0) and (1, 2) are adjacent(2, 0): (0, 0) and (2, 1) are adjacente) Corner Points (0, 0) and (0, 2) share x1 = 0(0, 2) and (1, 2) share x2 = 2(1, 2) and (2, 1) share x1 + x2 = 3(2, 1) and (2, 0) share x1 = 2(2, 0) and (0, 0) share x2 = 017.16 a) Optimal Solution: (x1, x2 ) = (0.667, 0.667) and Profit = $6,000.Note: corner points will be called A, B, C, D, E, and F going clockwise from (0,1).b) Corner Point A: F and B are adjacentB: A and C are adjacentC: B and D are adjacentD: C and E are adjacentE: D and F are adjacentF: E and A are adjacent17.17 a) Optimal Solution: (x1, x2 ) = (2, 2) and Profit = 10.b)c)d)Optimal Solution: (x1,x2 ) = (2,2) and Profit = 10.e)* the next corner point is (0, 3), which has already been checked 17.18 a) Optimal Solution: (x1, x2 ) = (2,2) and Profit = 6.b)c)d)Optimal Solution: (x1,x2 ) = (2,2) and Profit = 6.e)* the next corner point is (4, 0), which has already been checked17.19 a) Optimal Solution: (x1, x2) = (3, 4) and Profit = 17.b)c)d)Optimal Solution: (x1, x2) = (3, 4) and Profit = 17.e)* the next corner point is (0, 5), which has already been checked17.2017.2117.2217.2317.24 a) True (see solution concept number 6).b) False, there can be an infinite number of optimal solutions, such as the set of solutions along aline segment between two corner points.c) True, if the objective function line is parallel to the constraint boundary line that connects thetwo.17.25 a) If the feasible region is unbounded then there may be no optimal solution.b) An optimal solution may contain all points on a line segment between two corner points.c) If an adjacent corner point has an equal objective function value then all the points on theconnecting line segment will also be optimal.17.26 a) The problem may not have an optimal solution.b) The optimality test checks whether the current corner point is optimal. The iterative step onlymoves to a new corner point.c) The simplex method only chooses the origin as the initial corner point when it is a feasiblepoint.d) One of the adjacent points is likely to be better, not necessarily optimal.e) The simplex method only identifies the rate of improvement, not all the adjacent corner points.17.27b)Optimal Solution: (x1, x2 ) = (4, 3) and Profit = $11 million.d)* the next corner point is (0,5) which has already been checked17.28a)Corner Points: (0,0), (4,0), and (0,4).b)* the next corner point is (0, 4) which has already been checkedc)17.29 a) Optimal Solution: (x1, x2 ) = (3,3) and Profit = 6.b) Corner points: (0, 0), (4.5, 0), (0, 4.5), and (3, 3).c) Examine (4.5, 0) first.* the next corner point is (0,4.5) which has already been checkedExamine (0,4.5) first.* the next corner point is (4.5, 0) which has already been checkedd)e)17.30 a)b)a)17.31b)。

数据、模型与决策(运筹学)课后习题和案例答案001

CHAPTER 1INTRODUCTIONReview Questions1.1-1 The rapid development of the discipline began in the 1940’s and 1950’s.1.1-2 The traditional name given to the discipline is operations research.1.1-3 A management science study provides an analysis and recommendations, based onthe quantitative factors involved in the problem, as input to the managers.1.1-4 Management science is based strongly on some scientific fields, includingmathematics and computer science. It also draws upon the social sciences, especially economics.1.1-5 A decision support system is an interactive computer-based system that aidsmanagerial decision-making. The system draws current data from databases or management information systems and then solves the various versions of the model specified by the manager.1.1-6 Many managerial problems revolve around such quantitative factors as productionquantities, revenues, costs, the amounts available of needed resources, etc.1.2-1 The production and sales volume needs to exceed the break-even point to make itworthwhile to introduce a product.1.2-2 The number of clocks produced cannot be less than 0, nor should it exceed thenumber that can be sold. Also, the objective is to make the decision that maximizes the company’s profit.1.2-3 The purpose of sensitivity analysis is to check the effect on the recommendations of amodel if the estimates turn out to be wrong.1.2-4 Simply enter a variety of new values and see what happens.1.2-5 The MIN(a, b) function gives the minimum of a and b.1.2-6 The IF(a, b, c) function returns b if a is true, otherwise it returns c.1.3-1 These applications typically resulted in annual savings in the millions of dollars. Problems1.1 Answers will vary.1.2 Answers will vary.1.3 If Q units are produced per month, thenMonthly Profit = $0 {if Q= 0} and –$20,000 + ($20 –$10)Q{if Q> 0}.Break-even point = $20,000 / ($20 – $10) = 2000,so it will be profitable to produce if Q > 2000.1.4 a) $40,000b) $15.c) $15.1.5 a) Let Q be the number of units produced and sold. ThenMonthly Profit = $0 {if Q= 0} and –$500,000 + ($35 – $15)Q{if Q> 0}.Break-even point = $500,000 / ($35 – $15) = 25,000.b) Let Q be the number produced and s the number that can be sold. ThenProfit = [0 if Q = 0] and [-$500,000 + $35 * MIN(Q, s) – $15Q if Q > 0].c)d) Q≤ s.1.6 a) $150,000b) $733.33c) $566.671.7 a)b) Break-even point = ($50,000) / ($700 – $500) = 250.c) Maximize Profit = $0 {if Q= 0} and –$50,000 + $200Q{if Q> 0}subject to 0 ≤ Q≤ s.d)e)1.8 a) Jennifer must decide how much to ship from each plant (A and B) to each retailoutlet (1 and 2). Let xˆj = amount to ship from plant i (for i = A, B) to each retailoutlet j (for j = 1, 2).b) Shipping Cost = $700x A1 + $400x A2 + $800x B1 + $600x B2c) x A1 + x A2 = 30; x B1 + x B2≤ 500; x A1 + x B1 = 40; x A2 + x B2≥ 25; all x ij≥ 0.d) Minimize Shipping Cost = $700x A1+ $400x A2+ $800x B1+ $600x B2subject to x A1+ x A2= 30x B1+ x B2≤ 500x A1+ x B1= 40x A2+ x B2≥ 25 andall x ij≥ 0e) Jennifer should ship all of Retail Outlet 2’s 25 units from Plant A because it is $200cheaper than from Plant B. Retail Outlet 1 should get all it can from Plant A (5units) because it is $100 cheaper than from Plant B. The remaining 35 units shouldcome from Plant B. The decision variables would be x A1 = 5, x A2 = 25, x B1 = 35, x B2 =0.1.9 a)b) Break-even point = $1,000,000 / ($2,000 - $1,600) = 2,500.1.10 a)b) The make option appears to be better ($100,000 profit for the make option vs.$75,000 profit for the buy option).c) Q= number of grandfather clocks to produce for sale.Make Option: Profit = $0 {if Q= 0} and –$50,000 + $500Q{if Q> 0}.Buy Option: Profit = ($900 –$650)Q= $250QIncremental profit from choosing make option rather than buy option = $0 {if Q=0} and= –$50,000 + $500Q–$250Q= –$50,000 + $250Q{if Q> 0}.Mathematical model: Now interpret Q as the number to produce with the make option. The model is to find the value of Q so as toMaximize Incremental Profit = $0 {if Q=0} and= –$50,000 + $500Q–$250Q= –$50,000 + $250Q{if Q> 0}subject to Q≤ s(sales forecast)Q≥ 0.d)e) Make-option cost = $50,000 + $400QBuy-option cost = $650Q Break-even point = $50,000 / ($650 – $400) = 200 units.。

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1500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）：产品甲乙丙要求：（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：单价1009095单位变动成本807580销量12000150008000企业编号12345678910合计生产性固定资产价值(万元)3189102004094155023141210102212256525工业总产值（万元）52410196388159139286051516121916249801（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。

数据模型与决策课程大作业

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》习题解答 (2)

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。

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