Data, Model and Decisions 数据、模型与决策

自考数据、模型与决策

高纲1396江苏省高等教育自学考试大纲30447 数据、模型与决策南京大学编江苏省高等教育自学考试委员会办公室I 课程性质、设置目的与要求一、《数据、模型与决策》课程的性质随着社会信息化水平的提高和科学管理意识的普遍增强，人们对如何从数据资料角度进行认识显示出越来越多的兴趣。

数据资料本身并没有什么意义，关键是采用合适的方法对其进行分析和处理，只有这样才能探索客观现象发展变化的内在规律，从而更好地服务于管理决策的需要。

《数据、模型与决策》属于数量性质的课程，侧重于讲解数据资料的搜集、描述、分析和解释，以及管理决策方法和技术方面的知识。

管理决策分为两类，一类是理性决策一类是行为决策。

数据分析与决策模型中，不论是以不确定性为特征的统计决策，还是以确定性为特征的管理科学优化决策，和以策略互动为特征的博弈决策，都可以把它们归结为理性决策范畴。

既然是理性决策，必然会要求建立某种决策准则，然后在既定的准则下通过度量来选择决策方案。

这一过程一方面要对研究的问题进行结构化处理，另一方面也需要有相应的数据资料。

前者是为了能够建立决策模型，后者则是帮助实现计算。

有鉴于此，数据与模型在决策分析中的重要意义不言而喻。

数据与模型除了共同服务于决策分析以外，两者之间也存在密切的关系。

从应用的角度，统计方法比较强调实证性做法，统计分析与决策中，没有大量的、客观准确的数据资料，统计决策分析只能停留在纯理论的状态，无法形成具体的分析结论。

管理运筹优化和博弈决策分析中，虽然不像统计分析那样，需要拥有充足的数据，但是必要的不可控因素比如模型中的有关参数，其数值资料就必须事先给以确定。

尽管现在的企业一般都积累了大量的可供开发利用的数据资料，不过由于这样那样的原因，数据资料本身总会存在不系统、不充分、不完备的情况。

因此，对于背景数据必须经过科学的编辑、处理、汇总和提炼，然后才能用于决策分析。

对此，模型起着重要的转化作用，通过模型化处理，不仅能对数据的价值结构进行改造，而且还能对决策赋以深层次的分析。

数据、模型与决策——线性规划与电子表格-精品文档64页

线性规划要确定决策变量 x1, x2, … , xn 使
得 MaximiZzec1x1 cnxn
Objective Function
subject to
and
a11x1a1nxn b1 a21x1a2nxn b2
am1x1amnxn bm
线性规划与电子表格
Assumptions of Linear Programming 线性规划的假设
Linearity 线性
Divisibility 可分性
Certainty 确定性
Nonnegativity 非负性
All Rights Reserved, Prof. Ren Jian Biao,2004
问题。调查表明，在世界500家最大的企业中，有 85% 的企业
都曾使用过线性规划解决经营管理中遇到的复杂问题。线性规
划的使用为应用者节约了数以亿万计的资金。
All Rights Reserved, Prof. Ren Jian Biao,2004
数据
Line模ar P型rogramming With S决prSeae策dssshioene2t
线性最优化建模的若干例子
线性规划与电子表格
现在，管理者面临的决策是：
利用现有的设备与可获得的原料，每月工厂生产多少刀及
多少剪可以获得最大总利润，且这个最大利润是多少？
如果再进一步，管理人员还希望知道这个能够使工厂获得
最大利润的生产计划在哪些设备或原料方面已经充分利用，哪
些还有空闲的余地。
下面，我们首先飞行此问题需要一种什么样的决策模型加
3、深水炸弹的合理爆炸深度，摧毁德军潜艇数增加400%。
4、商船如何编队，遭潜艇攻击时如何减少损失。使船只受敌机攻击时，中弹数由47%降到29% 。

数据模型与决策学习感悟

数据模型与决策学习感悟“数据、模型与决策”，看这个名字给人的感觉是既理论又实践还颇有些高深。

所谓的数据模型与决策就是管理科学的另外一种称呼方式。

管理科学，它包含了管理和科学两门课程的内容，或者说是管理的科学。

如果这个定义还是非常的模糊，那么还可以这么解释，它就是对与定量因素有关的管理问题通过应用科学的方法进行辅助管理决策制定的一门科学。

再说的通俗一些，就是将管理过程中出现的定量问题，运用科学的方法，建立相应的模型进行分析，从而为管理者提供决策的依据。

我在课程学习过程中感受到其实质内容主要是线性规划模型和概率统计（检验、估计），内容主要包括统计学和数据模型决策两部分。

我自己以前没有学过线性规划，所以感觉课程的这部分是成功的，通过课程的学习懂得了高级线性规划和应用。

统计学主要讲授数据收集方法和数据处理方法，包括抽样方法、样本分布、参数估计、置信区间、假设检验、方差分析和回归分析。

数据模型决策主要讲述线性规划内容，包括线性规划模型的建立、求解模型的软件使用。

通过该课程学习我了解和掌握数据、模型和决策的基本原理、基本方法及其在管理决策中的广泛应用，提升了计算机数量分析的应用分析能力。

统计决策的思想贯穿了企业管理的始终，对各种决策方案进行科学评估，为管理决策服务，使得企业管理者更有效合理地利用有限资源。

优胜劣汰，适者生存，这是自然界的生存法则，也是企业的生存法则。

只有那些能够成功地应付环境挑战的企业，才是得以继续生存和发展的企业。

作为企业的管理者，把握并运用好数据模型与决策的理念定会取得“运筹帷幄之中，决胜千里之外”之功效。

一、企业发展原则与战略管理企业战略管理是企业在宏观层次通过分析、预测、规划、控制等手段，充分利用本企业的人、财、物等资源，以达到优化管理，提高经济效益的目的。

随着我国经济市场化的日益加深，市场竞争日趋激烈，我国企业面临着更多的环境因素的影响与冲击。

企业要求得生存与发展，必须运筹帷幄，长远谋划，根据自身的资源来制定最优的经营战略，以战略统揽全局。

数据模型与决策概念简述

数据模型与决策概念简述数据模型与决策中理论主要有线性规划及其数学模型，线性规划的单纯行法，整数规划、运输问题、动态规划、网络计划技术、库存问题、预测与决策、博弈论等。

一、线性规划与单纯形法确定影响决策问题的变量，进行分析，做具体方案，用线性函数进行表述。

确定目标函数最大或最小。

求解。

解决现实中实际问题，诸如合理下料问题、运输问题、生产的组织与计划问题、投资证券组合问题、分派问题、生产工艺优化问题。

解决问题是先建摸，设置决策变量，选择方案，确定目标函数，确定约束条件，约束条件一般为不等式或等式，最后确定决策变量的取值范围。

决策变量因为是现实中问题，一般不能为负，且是连续的，中间会有系数和等式约束值的限定。

问题的解决分为两类，一是在条件限定下，使得某一目标达到最大化，如何安排和计划，另一类是任务确定后，如何计划和安排，用最少的人力、物力和财力去实现任务，使成本最小。

函数表现式为：∑==nj j j X C Z 1max (min)),,2,1(0),,2,1(1n j X m i b X aj i n j j ij =≥=∑=≤=≥二维线性规划问题，图解法。

在平面直角坐标系上做图，将决策变量进行绘制，根绝约束条件找出可行域，进行平移，确定最优值。

变量都非负，所以图像在第一象限内。

求解过程中会有有可行解、无可行解的情况。

可行解中有最优解（有唯一最优解或无穷多最优解）或无最优解（无界解或无可行解）。

线性规划问题的标准形式分一般式、矩阵式、向量式、化标准形式。

化标准式，（1）目标函数，目标函数一般以最大值表示，当时求最小值时，转化为最大值，minz=max （-z ）（2）约束条件，将不等式变为等式，中间加入松弛变量，当不等式为小于等于时，左端加入非负松弛变量，当不等式为大于等于时，左端减去非负松弛变量。

（3）变量，变量无非负约束，对变量进行转换。

（4）右端项系数，右端项必须非负，在等式变换时进行变形。

数据,模型,和决策

第一章（管理科学简介）Ｐ５（１）管理科学介绍管理科学本质：是对与定量因素有关的管理问题通过应用科学的方法进行辅助管理决策制定的一门学科．管理科学发展过程：快速发展开始于２０世纪四五十年代起初的动力来自于第二次世界大战另一个里程碑是１９４７年丹捷格发明单纯形罚更大的推动作用的是计算机革命的爆发管理决策：管理者考虑管理科学对定量因素进行分析得出的结果后，再考虑管理科学以外的众多无形因素，然后根据其最佳判断做出决策管理科学小组系统和考察时步骤：定义问题与收集数据——构件数学模型——从模型中形成对于一个问题进行求解的基于计算机的程序——测试模型并在必要时进行修正——应用模型分析问题以及提出管理建议——帮助实施被管理者采纳的小组建议课后问题：１．管理科学什么时候有了快速发展？快速发展开始于２０世纪四五十年代２．商学院以外还广泛使用的对管理科学学科的叫法：运筹学３．管理科学研究提供给管理者什么？对问题涉及的定量因素进行分析并向开明的管理者提出建议４．管理科学以哪些领域作为基础？科学领域：数学，计算机社会领域：经济学５．什么是决策支持系统？辅助管理决策制定的交互式基于计算机的系统６．与管理问题有关的一般定量因素有哪些？生产数量，收入，成本，资源Ｐ１１（２）一个例子：盈亏平衡分析步骤：分析问题——建立模型——敏感性分析，电子表格模型提供上述三者了方便的途径如果预测销售数量＜盈亏平衡点，Ｑ＝０预测销售数量＞盈亏平衡点，Ｑ＝预测销售数量敏感性分析目的：研究如果一个估计值发生了变化，将会给模型带来什么样的变化Ｍｉｎ（ａ，ｂ）：取ａ，ｂ中的最小值Ｉｆ（Ａ，ｂ，ｃ）：如果表达式Ａ为真，则值为ｂ，否则为ｃ第二章（线性规划：基本概念）Ｐ３１（３）在电子表格上建立恩德公司问题的模型１．开始在电子表格上建立线性规划模型时需要回答的三个问题：要做出的决策是什么？在做出这些决策上有哪些约束条件？这些决策的全部绩效测度是什么？２．以下各个单元格的作用数据单元格：显示数据的单元格可变单元格：需要做出决策的单元格输出单元格：依赖于可变单元格的输出结果的单元格目标单元格：在生产率做出决策时目标值定为尽可能大的特殊单元格３．该案例中每个输出单元格（包括目标单元格）的Ｅｘｃｅｌ等式的形式：可以表达为一个ＳＵＭＰＲＯＤＵＣＴ函数，这里的每一项是一个数据单元格和可变单元格的乘积Ｐ３３（４）电子表格的数学模型１．电子表格模型与代数模型相同的初始步骤：收集相关数据确定要做出的决策确定对这些决策的约束条件确定为这些决策的完全绩效测度把约束条件和绩效测度的口头描述转化为数据和决策表示的定量表达式２．用代数形式建立线性规划模型时，模型中需要引入代数符号来表示哪几类数量？用来表示绩效测度与决策４．模型的一个可行解是什么意思？决策变量的任何一个取值Ｐ４１（５）求两问题变量的图解法１．图解法能用来求解带有几个决策变量的线性规划问题？只有两个２．什么是约束边界线？形成一个约束条件所允许的边界的直线什么是约束边界方程？形成一个约束条件所允许的边界的直线的方程可行域？比所有约束边界线更靠近原点的那些点成为可行解，可行解所在的区域成为可行域会作图求解Ｐ４４（６）应用Ｅｘｃｅｌ求解线性规划问题(solver)１．用来输入目标单元格和可变单元格地址的对话框是什么？目标单元格：set target cell 可变单元格：by changing cells２．具体化模型的函数约束的对话框是什么？ｓｕｂｊｅｃｔｔｏｔｈｅＣｏｎｓｔｒａｉｎｔｓ３．在ｓｏｌｖｅｒ中，哪些选项一般需要选定以求得一个线性规划模型？采用线性模型，假定非负Ｐ４８（７）一个最小化的例子——利博公司广告组合问题３. 利博公司的目标？在达到市场份额的前提下，确定最低的总成本并决定要在每种媒体上做多少钱的广告４．在电子表格中设置目标单元格和可变单元格的基本原理是什么？目标单元格：？可变单元格：？Ｐ５０（８）管理视角的线性规划１．管理部门一般对线性规划研究的技术细节设计深么？不深，没有必要２．一般问题有两个以上决策变量看，那么研究两个决策变量问题的图解法的意义？实际意义中没有价值，但对于传达线性规划设计确定约束边界和使目标值往尽可能大的这一方向移动的这一基本观念有很大价值３．开明的管理者关于线性规划应该知道哪些事项？需要知道线性规划是什么的一个良好直觉对线性规划的适用性和作用有一个正确的评价使得在合适的时候鼓励应用能够区分能胜任和以次充好的线性规划研究理解如何解释研究成果第三章（电子表格建模的艺术）（多简答和选择）Ｐ７６（２）电子表格建模程序的概述１．当你不知道从哪里开始时，帮助你开始建立电子表格模型的方法是什么？设想一下的目标手工进行一些计算建立一个小的电子表格２．手工计算可从那两方面帮助你？首先，它能帮助理清输出单元格公式的形式其次，它可以帮助检验表格３．描述一下组织和编排电子表格的一个有用方法计划设想一下你的目标手工进行一些计算建立一个电子表格建模先建立一个小模型测试利用不同的测试数据分析模型的逻辑关系，将模型扩展为完整的模型分析评估建议的解和／或利用ｓｏｌｖｅｒ优化４．哪些数值应被输入数据单元格以测试模型？试着输入一些我知道输出单元格结果的数值５．单元格绝对坐标：当被填入其它单元格时，坐标不会改变的坐标，如＆Ｅ＆１１单元格相对坐标：公式中单元格或者范围的坐标通常是基于他们相对含有公式的单元格的位置Ｐ８１（３）建立一个好的电子表格模型的几个原则１．模型的哪个部分最先输入电子表格？在建立电子表格之前，先输入和仔细编排所有数据２．数据应包含在公式中，还是被单独输入数据单元格？单独３．区域名称是如何使公式和模型在Ｓｏｌｖｅｒ对话框中更易于理解？如何选择区域名称？？１：用区域名称取代单元格地址写入公式中，使得公式更容易说明？２：选择“插入”菜单中“名称／定义”，然后输入一个名称，获得区域名称４．区分数据单元格，可变单元格，输出单元格，目标单元格有哪些方法？对不同类型的单元格使用不同的边框，单元格阴影５．在电子表格中完整的表达一个约束条件需要多少单元格？３Ｐ８６（４）调试电子表格模型１．调试电子表格模型的第一个步骤是什么？在你预知输出单元格正确结果的情况下，将不同的数值输入可变单元格，然后观察模型的计算结果是否和预期结果一致２．如何输出单元格在数值和公式中的切换？ｐｃ上同时按ｃｏｎｔｒｏｌ和～键（Ｍａｃ上同时按Ｃｏｍｍａｎｄ和～键）３．对于一个给定的单元格，哪一个Ｅｘｃｅｌ工具可以用来追踪其从属单元格或引用单元格？“工具”——“审核／追踪从属单元格”，会显示出箭头，以观察单元格之间的联系建立一个好的电子表格所需原则：●首先输入数据●组织和清楚标识数据●每个数据输入唯一的单元格●将数据与公式分离●保持简单化●使用区域名称●使用相对和绝对坐标简化公式的复制●使用边框、阴影和颜色来区分单元格类型●在电子表格中显示整个模型第四章（线性规划：建模与应用）Ｐ９７（１）案例研究——超级食品公司的广告混合问题４．在评价使用线性规划来表示该实际问题的准确性时，要做出的假设条件有哪些？允许有分数解包括目标单元格和可变单元格都可以用ＳＵＭＰＲＯＤＵＣＴ函数以数据单元格和可变单元格表示（有时候只是可变单元格的加总）Ｐ１０６（２）资源分配问题１．资源分配问题的共性在线性规划模型中每一个函数限制均为资源限制，并且，每一种资源都可以表现为如下的形式：使用的资源数量＜＝可用的资源数量２．资源限制的形式如何？使用的数量＜＝可获得的数量３．为解决资源分配问题，必须收集哪三类数据？每种资源的可供量每一种活动所需要的各种资源的数量，对于每一种资源约活动的组合，单位活动所消耗的资源量必须首先估计出来每一种活动对总的绩效测度的单位贡献Ｐ１１３（３）成本收益平衡问题１．资源分配问题与成本收益平衡问题在管理目标上的差异是什么？资源分配问题：各种资源是受限制的因素（包括财务资源），问题的目标是（根据特定的总绩效测度）最有效的利用各种资源成本收益问题：管理层采取更为主动的姿态，他们指定哪些收益必须实现（不管如何使用资源），并且要以最低的成本实现所指明的收益２．成本收益平衡问题的共性是什么？所有的函数约束均为收益约束，并具有如下的形式：完成的水平＞＝最低可接受的水平３．收益限制的形式如何？完成的水平＞＝最低可接受的水平４．为解决成本收益平衡问题必须收集的三类数据包括哪些？每种收益的最低可接受水平（管理决策）每一种活动对每一种收益的贡献（单位活动的贡献）每种活动的单位成本Ｐ１１７（４）网络配送问题１．为什么这类问题为网络配送问题？这类问题通过配送网络能以最小的成本完成货物的配送，所以称之为网络配送问题２．网络配送问题的共性是什么？确定需求的约束，提供的数量＝需要的数量３．确定需求的约束与资源约束和收益约束的区别是什么？确定需求的约束：提供的数量＝需要的数量资源约束：使用的资源数量＜＝可用的资源数量收益约束：完成的水平＞＝最低可接受的水平Ｐ１２９（７）管理视角的建模１．为什么ｗｈａｔ－ｉｆ分析是线性规划的研究中非常重要的一个组成部分？尽管可能使用许多变异的模型，但是对于一个特定版本的模型，一次只能求得一个解，但是在求得一个解以后，管理层会有很多问题：如果模型的参数估计有误怎么办？如果做出不同的似是而非的假设，问题的将会如何变化？如果管理方面所要求的某一选项没有被考虑在内，会产生怎样的结果？Ｗｈａｔ－ｉｆ分析有助于解决上述等相关问题Ｐ１３１（８）线性规划应用经典回顾１．比较三种线性规划的应用，注意各种类型的问题应该使用哪一类型的线性规划模型（资源分配、成本收益平衡、网络配送以及混合问题）第五章（线性规划的Ｗｈａｔ－ｉｆ分析）Ｐ１５８（３）只有一个目标函数系数变动１．目标函数系数允许变化范围的含义是什么？能使最优解保持不变的目标函数系数的变化范围称为目标函数系数允许变化范围２．如果目标系数的估计值不是实际值，并且不在允许变化范围之内，会有怎样的影响？最优解不正确３．在Ｅｘｃｅｌ的灵敏度分析报告中，目标函数系数一栏该如何解释？允许增加值和允许减少值一栏又该如何解释？目标函数系数一栏：目标函数系数的现值允许增加值和允许减少值一栏：是这些系数在最优的范围内，允许增加和减少的量（１Ｅ＋３０）：１０的三十次方的缩写，表示无穷大Ｐ１６５（４）目标函数系数同时变动的影响１．目标系数变动百分比法则中，变动的百分比指什么？各个变动系数占该系数允许变化范围允许变动量的百分比之和（有方向）２．在百分百法则中，如果变动的百分比之和不超过100%，最初的最优解将如何？不会改变３．在百分百法则中，如果变动的百分比之和超过100%，是否就意味着最初的最优解已经不再是最优解？不能确定最优解是否改变Ｐ１７２（５）单个约束条件变化的影响１．为什么要研究函数的约束条件的变化带来的影响？因为在建模时,还不能得到模型的这些参数的精确值更重要的是:这些常数往往不是由外界决定而是由管理层的政策决策决定的２．为什么函数约束的右端值可能改变？这些常数往往不是由外界决定而是由管理层的政策决策决定的，因此，在建模并求解之后管理者想要知道如果改变这些决策是否会提高最终的收益３．影子价格的含义是什么？约束常数增加微小的量1，使得目标函数增加的量４．用电子表格如何找到影子价格？用Ｓｏｌｖｅｒ表格呢？用灵敏度报告呢？电子表格：改变某一约束条件的值，重新按下Solver键，尝试在约束条件变化范围内找出每单位约束条件变化引起的目标函数值的变化即为影子价格Ｓｏｌｖｅｒ表格：？灵敏度报告:Shadow price栏５．为什么管理者会对影子价格感兴趣？管理者可以用影子价格评价，在影子价格的有效域内幅度不大的改变工作时间的各种决策６．影子价格是否也同样适用于减少函数约束右端值的数值的情况？是７．影子价格０对管理者来说是什么意思？该影子价格对应的约束条件在其变化范围内对目前的最优解没有影响８．为什么管理层会对可行域感兴趣？？１７６（６）约束右端值同时变动的情形１．为什么要研究约束条件同时发生变化的情况？经常会出现需要我们考虑约束条件同时变动的情况。

数据、模型与决策(运筹学)课后习题和案例答案009

CHAPTER 9INTEGER PROGRAMMING Review Questions9.1-1 In some applications, such as assigning people, machines, or vehicles, decisionvariables will make sense only if they have integer values.9.1-2 Integer programming has the additional restriction that some or all of the decisionvariables must have integer values.9.1-3 The divisibility assumption of linear programming is a basic assumption that allowsthe decision variables to have any values, including fractional values, that satisfy the functional and nonnegativity constraints.9.1-4 The LP relaxation of an integer programming problem is the linear programmingproblem obtained by deleting from the current integer programming problem the constraints that require the decision variables to have integer values.9.1-5 Rather than stopping at the last instant that the straight edge still passes through thefeasible region, we now stop at the last instant that the straight edge passes through an integer point that lies within the feasible region.9.1-6 No, rounding cannot be relied on to find an optimal solution, or even a good feasibleinteger solution.9.1-7 Pure integer programming problems are those where all the decision variables mustbe integers. Mixed integer programming problems only require some of the variables to have integer values.9.1-8 Binary integer programming problems are those where all the decision variablesrestricted to integer values are further restricted to be binary variables.9.2-1 The decisions are 1) whether to build a factory in Los Angeles, 2) whether to build afactory in San Francisco, 3) whether to build a warehouse in Los Angeles, and 4) whether to build a warehouse in San Francisco.9.2-2 Binary decision variables are appropriate because there are only two alternatives,choose yes or choose no.9.2-3 The objective is to find the feasible combination of investments that maximizes thetotal net present value.9.2-4 The mutually exclusive alternatives are to build a warehouse in Los Angeles or build awarehouse in San Francisco. The form of the resulting constraint is that the sum of these variables must be less than or equal to 1 (x3 + x4≤ 1).9.2-5 The contingent decisions are the decisions to build a warehouse. The forms of theseconstraints are x3≤ x1 and x4≤x2.9.2-6 The amount of capital being made available to these investments ($10 million) is amanagerial decision on which sensitivity analysis needs to be performed.9.3-1 A value of 1 is assigned for choosing yes and a value of 0 is assigned for choosing no.9.3-2 Yes-or-no decisions for capital budgeting with fixed investments are whether or notto make a certain fixed investment.9.3-3 Yes-or-no decisions for site selections are whether or not a certain site should beselected for the location of a certain new facility.9.3-4 When designing a production and distribution network, yes-or-no decisions likeshould a certain plant remain open, should a certain site be selected for a new plant, should a certain distribution center remain open, should a certain site be selected fora new distribution center, and should a certain distribution center be assigned toserve a certain market area might arise.9.3-5 Should a certain route be selected for one of the trucks.9.3-6 It is estimated that China is saving about $6.4 billion over the 15 years.9.3-7 The form of each yes-or-no decision is should a certain asset be sold in a certaintime period.9.3-8 The airline industry uses BIP for fleet assignment problems and crew schedulingproblems.9.4-1 A binary decision variable is a binary variable that represents a yes-or-no decision.An auxiliary binary variable is an additional binary variable that is introduced into the model, not to represent a yes-or-no decision, but simply to help formulate the model as a BIP problem.9.4-2 The net profit is no longer directly proportional to the number of units produced so alinear programming formulation is no longer valid.9.4-3 An auxiliary binary variable can be introduced for a setup cost and can be defined as1 if the setup is performed to initiate the production of a certain product and 0 if thesetup is not performed.9.4-4 Mutually exclusive products exist when at most one product can be chosen forproduction due to competition for the same customers.9.4-5 An auxiliary binary variable can be defined as 1 if the product can be produced and 0if the product cannot be produced.9.4-6 An either-or constraint arises because the products are to be produced at eitherPlant 3 or Plant 4, not both.9.4-7 An auxiliary binary variable can be defined as 1 if the first constraint must hold and 0if the second constraint must hold.9.5-1 Restriction 1 is similar to the restriction imposed in Variation 2 except that it involvesmore products and choices.9.5-2 The constraint y1+ y2+ y3≤ 2 forces choosing at most two of the possible newproducts.9.5-3 It is not possible to write a legitimate objective function because profit is notproportional to the number of TV spots allocated to that product.9.5-4 The groups of mutually exclusive alternative in Example 2 are x1 = 0, 1, 2, or 3, x2 = 0,1, 2, or 3, and x3 = 0,1,2, or 3.9.5-5 The mathematical form of the constraint is x1 + x4 + x7 + x10≥ 1. This constraint saysthat sequence 1, 4, 7, and 10 include a necessary flight and that one of the sequences must be chosen to ensure that a crew covers the flight.Problems9.1 a) Let T= the number of tow bars to produceS= the number of stabilizer bars to produce Maximize Profit = $130T+ $150Ssubject to 3.2T+ 2.4S≤ 16 hours2T+ 3S≤15 hours and T≥ 0, S≥ 0 T, S are integers.b) Optimal solution: (T,S) = (0,5). Profit = $750.c)1 2 3 4 5 6 7 8 9A B C D E FTow Bars Stabilizer BarsUnit P rofit$130$150Hours HoursUsed Available M achine 1 3.2 2.412<=16 M achine 22315<=15Tow Bars Stabilizer Bars Total P rofit Units P roduced05$750Hours Used P er Unit P roduced9.2 a)1 2 3 4 5 6 7 8 9 10 11 12 13A B C D E FM odel A M odel B(high speed)(low er speed)Unit Cost$6,000$4,000Total CapacityCapacity Needed Capacity20,00010,00080,000>=75,000 M odel A M odel B(high speed)(low er speed)Total Total Cost P urchase246$28,000 >=>=1M in Needed6Copies per Dayb) Let A= the number of Model A (high-speed) copiers to buyB= the number of Model B (lower-speed) copiers to buy Minimize Cost = $6,000A+ $4,000Bsubject to A+ B≥ 6 copiersA≥ 1 copier20,000A+ 10,000B≥ 75,000 copies/day and A≥ 0, B≥ 0 A, B are integers.c) Optimal solution: (A,B) = (2,4). Cost = $28,000.9.3 a) Optimal solution: (x1, x2) = (2, 3). Profit = 13.b) The optimal solution to the LP-relaxation is (x1, x2) = (2.6, 1.6). Profit = 14.6.Rounded to the nearest integer, (x1, x2) = (3, 2). This is not feasible since it violatesthe third constraint.RoundedFeasible? Constraint Violated PSolution(3,2) No 3rd-(3,1) No 2nd & 3rd-(2,2) Yes - 12(2,1) Yes - 11None of these is optimal for the integer programming model. Two are notfeasible and the other two have lower values of Profit.9.4 a) Optimal solution: (x1, x2) = (2, 3). Profit = 680.b) The optimal solution to the LP-relaxation is (x1, x2) = (2.67, 1.33). Profit = 693.33.Rounded to the nearest integer, (x1, x2) = (3, 1). This is not feasible since it violatesthe second and third constraint.Rounded Solution Feasible? Constraint Violated P(3,1) No 2nd & 3rd-(3,2) No 2nd-(2,2) Yes - 600(2,1) Yes - 520None of these is optimal for the integer programming model. Two are notfeasible and the other two have lower values of Profit.9.5 a)1 2 3 4 5 6 7 8 9 10 11 12A B C D E F GLong-Range M edium-R ange Short-RangeJets Jets JetsAnnual P rofit ($m illion) 4.23 2.3Resource ResourceResource Used P er Unit P roduced Used Available Budget6750351498<=1500M aintenance Capacity 1.667 1.333139.333<=40 P ilot Crew s11130<=30Long-Range M edium-R ange Short-Range Total AnnualJets Jets Jets P rofit ($m illion) P urchase1401695.6b) Let L= the number of long-range jets to purchaseM= the number of medium-range jets to purchaseS= the number of short-range jets to purchase Maximize Annual Profit ($millions) = 4.2L+ 3M+ 2.3Ssubject to 67L+ 50M+ 35S≤ 1,500 ($million)(5/3)L+ (4/3)M+ S≤ 40 (maintenan ce capacity)L+ M+ S≤ 30 (pilot crews) and L≥ 0, M≥ 0, S≥ 0 L, M, S are integers.9.6 a) Let x ij= tons of gravel hauled from pit i to site j(for i= N, S; j= 1, 2, 3)y ij = the number of trucks hauling from pit i to site j (for i = N, S; j = 1, 2, 3) Minimize Cost = $130x N1+ $160x N2+ $150x N3+ $180x S1+ $150x S2+ $160x S3+$50y N1+ $50y N2+ $50y N3+ $50y S1+ $50y S2+ $50y S3 subject to x N1+ x N2+ x N3≤ 18 tons (supply at North Pit)x S1+ x S2+ x S3≤ 14 tons (supply at South Pit)x N1+ x S1= 10 tons (demand at Site 1)x N2+ x S2= 5 tons (demand at Site 2)x N3+ x S3= 10 tons (demand at Site 3)x ij≤ 5y ij(for i= N, S; j= 1, 2, 3) (max 5 tons per truck) and x ij≥ 0, y ij≥ 0, y ij are integers (for i = N, S; j = 1, 2, 3)b)9.7 a) Let F LA= 1 if build a factory in Los Angeles; 0 otherwiseF SF= 1 if build a factory in San Francisco; 0 otherwiseF SD= 1 if build a factory in San Diego; 0 otherwiseW LA= 1 if build a warehouse in Los Angeles; 0 otherwiseW SF= 1 if build a warehouse in San Francisco; 0 otherwiseW SD= 1 if build a warehouse in San Diego; 0 otherwise Maximize NPV ($million) = 9F LA+ 5F SF+ 7F SD+ 6W LA+ 4W SF+ 5W SDsubject to 6F LA+ 3F SF+ 4F SD+ 5W LA+ 2W SF+ 3W SD≤ $10 million (Capital)W LA+ W SF+ W SD≤ 1 warehouseW LA≤ F LA(warehouse only if factory)W SF≤ F SFW SD≤ F SD and F LA, F SF, F SD, W LA, W SF, W SD are binary variables.b)9.8 See the articles in Interfaces.9.9 a) Let E M= 1 if Eve does the marketing; 0 otherwiseE C= 1 if Eve does the cooking; 0 otherwiseE D= 1 if Eve does the dishwashing; 0 otherwiseE L= 1 if Eve does the laundry; 0 otherwiseS M= 1 if Steven does the marketing; 0 otherwiseS C= 1 if Steven does the cooking; 0 otherwiseS D= 1 if Steven does the dishwashing; 0 otherwiseS L= 1 if Steven does the laundry; 0 otherwise Minimize Time (hours) = 4.5E M+ 7.8E C+ 3.6E D+ 2.9E L+4.9S M+ 7.2S C+ 4.3S D+ 3.1S Lsubject to E M+ E C+ E D+ E L= 2 (each person does 2 tasks)S M+ S C+ S D+ S L= 2E M+ S M= 1 (each task is done by 1 person)E C+ S C= 1E D+ S D= 1E L+ S L= 1and E M, E C, E D, E L, S M, S C, S D, S L are binary variables.b)9.10 a) Let x1= 1 if invest in project 1; 0 otherwisex2= 1 if invest in project 2; 0 otherwisex3= 1 if invest in project 3; 0 otherwisex4= 1 if invest in project 4; 0 otherwisex5= 1 if invest in project 5; 0 otherwise Maximize NPV ($million) = 1x1+ 1.8x2+ 1.6x3+ 0.8x4+ 1.4x5subject to 6x1+ 12x2+ 10x3+ 4x4+ 8x5≤ 20 ($million capital available)and x1, x2, x3, x4, x5 are binary variables.b)c)12 13 14 15 16 17 18 19 20 21 22 23A B C D E F G Capital Total Available Undertake?P rofit ($m illion)P roject 1P roject 2P roject 3P roject 4P roject 5($m illion) 10110 3.4 1601010 2.6 1810011 3.2 2010110 3.4 2200111 3.8 24101014 2611001 4.2 2810111 4.8 301101159.11 a)b)18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34A B C D E F G H I Capital Total Available Undertake Investm ent Opportunity?P rofit ($m illion)1234567($m illion) 101000141 80101000032 90011000134 100101000141 110101001045 120101010148 130101011052 140100111056 150101011161 160101011161 170100111165 180100111165 190100111165 2001001111659.12Mutually exclusive alternatives: Each swimmer can only swim one stroke.Each stroke can only be swum by one swimmer.9.139.149.159.16An alternative optimal solution is to produce 3 planes for customer 1 and 2 planes for customer 2.9.179.18 a) Let y ij = 1 if x i = j ; 0 otherwise (for i = 1, 2; and j = 1, 2, 3)Maximize Profit = 3y 11 + 8y 12 + 9y 13 + 9y 21 + 24y 22 + 9y 23 subject to y 11 + y 12 + y 13 ≤ 1 (x i can only take on one value) y 21 + y 22 + y 13 ≤ 1 (y 11 + 2y 12 + 3y 13) + (y 21 + 2y 22 + 3y 23) ≤ 3 andy ij are binary variables (for i = 1, 2; and j = 1, 2, 3)b)c) Optimal Solution (x 1,x 2) = (1, 2). Profit = 27.9.19The constraints in C11:E13 are mutually exclusive alternative (at each stage, exactly one arc is used). The constraints in D6:I8 are contingent decisions (a route can leavea node only if a route enters the node).9.209.21The six equality constraints (total stations = 2; one station assigned to each tract) correspond to mutually exclusive alternatives. In addition, there are the following contingent decision constraints: each tract can only be assigned to a station location if there is a station at that location (D21:D25 ≤ B21:B25; E21:E25 ≤ B21:B25; F21:F25 ≤ B21:B25; G21:G25 ≤ B21:B25; H21:H25 ≤ B21:B25).9.22 a) Let x i= 1 if a station is located in tract i; 0 otherwise (for i= 1, 2, 3, 4, 5)Minimize Cost ($thousand) = 200x1+ 250x2+ 400x3+ 300x4+ 500x5subject to x1+ x3+ x5≥ 1 (stations within 15 minutes of tract 1)x1+ x2+ x4≥ 1 (stations within 15 minutes of tract 2)x2+ x3+ x5≥ 1 (stations within 15 minutes of tract 3)x2+ x3+ x4+ x5≥ 1 (stations within 15 minutes of tract 4)x1+ x3+ x4+ x5≥ 1 (stations within 15 minutes of tract 5) and x i are binary variables (for i = 1, 2, 3, 4, 5).b)Cases9.1 a) With this approach, we need to formulate an integer program for each monthand optimize each month individually.In the first month, Emily does not buy any servers since none of the departmentsimplement the intranet in the first month.In the second month she must buy computers to ensure that the Sales Department can start the intranet. Emily can formulate her decision problem as an integer problem (the servers purchased must be integer. Her objective is to minimize the purchase cost. She has to satisfy to constraints. She cannot spend more than $9500 (she still has her entire budget for the first two months since she didn't buy any computers in the first month) and the computer(s) must support at least 60 employees. She solves her integer programming problem using the Excel solver.5A B C D EUnit Cost=(1-Discount)*OriginalCost=(1-Discount)*OriginalCost=(1-Discount)*OriginalCost=(1-Discount)*OriginalCostRange N ame Cells Budget B8:E8 BudgetAvailable F8 BudgetSpent H8Discount B4:E4 OriginalCost B3:E3 ServersP urchased B12:E12 Support B8:E8 SupportNeeded H8 TotalCost H12 TotalSupport F8 UnitCost B4:E56789101112FTotalSupport=SUM P ROD UCT(Support,ServersP urchased)BudgetSpent=SUM P ROD UCT(B12:E12,ServersP urchased)141516HTotalCost=SUM P ROD UCT(UnitCost,ServersP urchased)Note, that there is a second optimal solution to this integer programming problem. For the same amount of money Emily could buy two standard PC's that would also support 60 employees. However, since Emily knows that she needs to support more employees in the near future, she decides to buy the enhanced PCsince it supports more users.For the third month Emily needs to support 260 users. Since she has already computing power to support 80 users, she now needs to figure out how to support additional 180 users at minimum cost. She can disregard the constraint that the Manufacturing Department needs one of the three larger servers, since she already bought such a server in the previous month. Her task leads her to the following integer programming problem and solution.Emily decides to buy one SGI Workstation in month 3. The network is now able to support 280 users.In the fourth month Emily needs to support a total of 290 users. Since she has already computing power to support 280 users, she now needs to figure out how to support additional 10 users at minimum cost. This task leads her to the following integer programming problem:Emily decides to buy a standard PC in the fourth month. The network is now able to support 310 users.Finally, in the fifth and last month Emily needs to support the entire company witha total of 365 users. Since she has already computing power to support 310 users,she now needs to figure out how to support additional 55 users at minimum cost.This task leads her to the following integer programming problem and solution.Emily decides to buy another enhanced PC in the fifth month. (Note that again she could have also bought two standard PC's, but clearly the enhanced PC provides more room for the workload of the system to grow.) The entire network of CommuniCorp consists now of 1 standard PC, 2 enhanced PC's and 1 SGI workstation and it is able to support 390 users. The total purchase cost for this network is $22,500.b) Due to the budget restriction and discount in the first two months Emily needs todistinguish between the computers she buys in those early months and in the later months. Therefore, Emily uses two variables for each server type.Emily essentially faces four constraints. First, she must support the 60 users in the sales department in the second month. She realizes that, since she no longer buys the computers sequentially after the second month, that it suffices to include only the constraint on the network to support the all users in the entire company. This second constraint requires her to support a total of 365 users. The third constraint requires her to buy at least one of the three large servers. Finally, Emily has to make sure that she stays within her budget during the second month.5A B CMonth 2 Cost=(1-Month2Discount)*Month3to5Cost=(1-Month2Discount)*Month3to5Cost678910111213F Total Support=SUM P R ODUCT(M onth2Support,M onth2P urchases)=SUM P R ODUCT(M onth3to5Support,TotalP urchases)Budget Spent=SUM P R ODUCT(M onth2Budget,M onth2P urchases)171819G HM onth 2 C ost =SU M P R OD U C T(M onth2C ost,M onth2P urchases)M onth 3-5 C ost =SU M P R OD U C T(M onth3to5C ost,M onth3to5P urchases)Total C ost =H 17+H 18Emily should purchase a discounted SGI workstation in the second month, and another regular priced one in the third month. The total purchase cost is $19,000.c) Emily's second method in part (b) finds the cost for the best overall purchase policy. The method in part (a) only finds the best purchase policy for the given month, ignoring the fact that the decision in a particular month has an impact on later decisions. The method in (a) is very short-sighted and thus yields a worse result that the method in part (b).d) Installing the intranet will incur a number of other costs. These costs include:Training cost,Labor cost for network installation,Additional hardware cost for cabling, network interface cards, necessary hubs, etc.,Salary and benefits for a network administrator and web master,Cost for establishing or outsourcing help desk support.e) The intranet and the local area network are complete departures from the waybusiness has been done in the past. The departments may therefore beconcerned that the new technology will eliminate jobs. For example, in the pastthe manufacturing department has produced a greater number of pagers thancustomers have ordered. Fewer employees may be needed when themanufacturing department begins producing only enough pagers to meet orders.The departments may also become territorial about data and procedures, fearingthat another department will encroach on their business. Finally, the departmentsmay be concerned about the security of their data when sending it over thenetwork.9.2 a) We want to maximize the number of pieces displayed in the exhibit. For eachpiece, we therefore need to decide whether or not we should display the piece.Each piece becomes a binary decision variable. The decision variable is assigned1 if we want to display the piece and assigned 0 if we do not want to display thepiece.We group our constraints into four categories – the artistic constraints imposedby Ash, the personal constraints imposed by Ash, the constraints imposed byCeleste, and the cost constraint. We now step through each of these constraintcategories.Artistic Constraints Imposed by AshAsh imposes the following constraints that depend upon the type of art that isdisplayed. The constraints are as follows:1. Ash wants to include only one collage. We have four collages available:“Wasted Resources” by Norm Marson, “Consumerism” by Angie Oldman, “MyNamesake” by Ziggy Lite, and “Narcissism” by Ziggy Lite. A constraint forces us toinclude exactly one of these four pieces (D36=D38 in the spreadsheet model thatfollows).2. Ash wants at least one wire-mesh sculpture displayed if a computer-generated drawing is displayed. We have three wire-mesh sculptures available and two computer-generated drawings available. Thus, if we include either one or two computer-generated drawings, we have to include at least one wire-mesh sculpture. Therefore, we constrain the total number of wire-mesh sculptures (total) to be at least (1/2) time the total number of computer-generated drawings (L40 ≥ N40).3. Ash wants at least one computer-generated drawing displayed if a wire-mesh sculpture is displayed. We have two computer-generated drawings available and three wire-mesh sculptures available. Thus, if we include one, two, or three wire-mesh sculptures, we have to include either one or two computer-generated drawings. Therefore, we constraint the total number of wire-mesh sculptures (total) to be at least (1/3) times the total number of computer-generated drawings (L41 ≥ N41).4. Ash wants at least one photo-realistic painting displayed. We have three photo-realistic paintings available: “Storefront Window” by David Lyman, “Harley” by David Lyman, and “Rick” by Rick Rawls. At least one of these three paintings has to be displayed (G36 ≥ G38).5. Ash wants at least one cubist painting displayed. We have three cubist paintings available: “Rick II” by Rick Rawls, “Study of a Violin” by Helen Row, and “Study of a Fruit Bowl” by Helen Row. At least one of these three paintings has t o be displayed (H36 ≥ H38).6. Ash wants at least one expressionist painting displayed. We have only one expressionist painting available: “Rick III” by Rick Rawls. This painting has to be displayed (I36 ≥ I38).7. Ash wants at least one watercolor painting displayed. We have six watercolor paintings available: “Serenity” by Candy Tate, “Calm Before the Storm” by Candy Tate, “All That Glitters” by Ash Briggs, “The Rock” by Ash Briggs, “Winding Road” by Ash Briggs, and “Dreams Come True” by Ash Br iggs. At least one of these six paintings has to be displayed (J36 ≥ J38).8. Ash wants at least one oil painting displayed. We have five oil paintings available: “Void” by Robert Bayer, “Sun” by Robert Bayer, “Beyond” by Bill Reynolds, “Pioneers” by Bill Reynolds, and “Living Land” by Bear Canton. At least one of these five paintings has to be displayed (K36 ≥ K38).9. Finally, Ash wants the number of paintings to be no greater than twice the number of other art forms. We have 18 paintings available and 16 other art forms available. We classify the followi ng pieces as paintings: “Serenity,” “Calm Before the Storm,” “Void,” “Sun,” “Storefront Window,” “Harley,” “Rick,” “Rick II,” “Rick III,” “Beyond,” “Pioneers,” “Living Land,” “Study of a Violin,” “Study of a Fruit Bowl,” “All That Glitters,” “The Rock,” “Winding Road,” and “Dreams Come True.” The total number of these paintings that we display has to be less than or equal to twice the total number of other art forms we display (L42 ≤ N42).Personal Constraints Imposed by Ash 1. Ash wants all of his own paintings included in the exhibit, so we must include “All That Glitters,” “The Rock,” “Winding Road,” and “Dreams Come True.” (In the spreadsheet model, we constraint the total number of Ash paintings to equal 4: N36=N38.)2. Ash wants all of Candy Tate’s work included in the exhibit, so we must include “Serenity” and “Calm Before the Storm.” (In the spreadsheet model, we constrain the total number of Candy Tate works to equal 2: O36=O38.)3. Ash wants to include at least one piece from David Lyman, so we have to include one or more of the pieces “Storefront Window” and “Harley”(P36 ≥ P38).4. Ash wants to include at least one piece from Rick Rawls, so we have to include one or more of the pieces “Rick,” “Rick II,” and “Rick III” (Q36 ≥ Q38)5. Ash wants to display as many pieces from David Lyman as from Rick Rawls. Therefore we constrain the total number of David Lyman works to equal the total number of Rick Rawls works (L43 = N43).6. Finally, Ash wants at most one piece from Ziggy Lite displayed. We can therefore include no more than one of “My Namesake” and “Narcissism”(R36 ≤ R38).Constraints Imposed by Celeste 1. Celeste wants to include at least one piece from a female artist for every two pieces included from a male artist. We have 11 pieces by female artists available: “Chaos Reigns” by Rita Losky, “Who Has Control?” by Rita Losky, “Domestication” by Rita Losky, “Innocence” by Rita Losky, “Serenity” by Candy Tate, “Calm Before the Storm” by Candy Tate, “Consumerism” by Angie Oldman, “Reflection” by Angie Oldman, “Trojan Victory” by Angie Oldman, “Study of a Violin” by Helen Row, and “Study of a Fruit Bowl” by Helen Row. The total number of these pieces has to be greater-than-or-equal-to (1/2) times the total number of pieces by male artists (L44 ≥ N44).2. Celeste wants at least one of the pieces “Aging Earth” and “Wasted Resources” displayed in order to advance environmentalism (V36 ≥ V38).3. Celeste wants to include at least one piece by Bear Canton, so we must include one or more o f the pieces “Wisdom,” “Superior Powers,” and “Living Land” to advance Native American rights (W36 ≥ W38).4. Celeste wants to include one or more of the pieces “Chaos Reigns,” “Who Has Control,” “Beyond,” and “Pioneers” to advance science (X36 ≥ X38).5. Celeste knows that the museum only has enough floor space for four sculptures. We have six sculptures available: “Perfection” by Colin Zweibell, “Burden” by Colin Zweibell, “The Great Equalizer” by Colin Zweibell, “Aging Earth” by Norm Marson, “Reflection” by Angie Oldman, and “Trojan Victory” by Angie Oldman. We can only include a maximum of four of these six sculptures (Y36 ≤ Y38).6. Celeste also knows that the museum only has enough wall space for 20 paintings, collages, and drawings. We have 28 paintings, collages, and drawings available: “Chaos Reigns,” “Who Has Control,” “Domestication,” “Innocence,” “Wasted Resources,” “Serenity,” “Calm Before the Storm,” “Void,” “Sun,” “Storefront Window,” “Harley,” “Consumerism,” “Rick,” “Rick II,” “Rick III,” “Beyond,” “Pioneers,” “Wisdom,” “Superior Powers,” “Living Land,” “Study of a Violin,” “Study of a Fruit Bowl,” “My Namesake,” “Narcissism,” “All That Glitters,” “The Rock,” “Winding Road,” and “Dreams Come True.” We can only include a maximum of 20 of these 28 wall pieces (Z36 ≤ Z38).7. Finally, Celeste wants “Narcissism” displayed if “Reflection” is displayed. So if the decision variable for “Reflection” is 1, the decision variable for “Narcissism” must also be 1. However, the decision variable for “Narcissism” can still be 1 even if the decision variable for “Reflection” is 0 (L45 ≥ N45).Cost Constraint The cost of all of the pieces displayed has to be less than or equal to $4 million (C36 ≤ C38).。

管理科学-数据模型与决策简介

Data, Model and Decisions 数据、模型与决策
Introduction to Management Science 管理科学简介
What is Data, Model and Decisions 数据模型与决策是什么
管理者信息提供模型
结论
决策
执行
反馈
结果
管理者在组织内制定决策，数据、模型与决策的目的是在科学、符合逻辑和合理的基础上制定决策。内容主要是管理科学和统计学。
Fixed Cost = $50,000 (if Q > 0) Variable Cost = $400 Q Total Cost =
0, if Q = 0 $50,000 + $400 Q, if Q > 0
Profit:
Profit = Total revenue – Total cost
为管理科学实践者颁发的最负盛名的奖项是弗兰茨·厄德曼(Franz Edelman) 奖。这些奖项授予全世界年度管理科学的最佳应用。评奖活动由国际运筹学会、管理科学协会、管理科学实践学会联合发起。
Theory of Quantitative Analysis 管理定量分析理论
解决方法
线性规划整数规划非线性规划多目标规划预测网络分析决策分析库存模型模拟
实际问题
Many real world examples 许多实际问题举例
▪ Breakeven point Analysis 盈亏平衡分析 ▪ Optimum loan plan of the bank银行的最优贷款计划 ▪Portfolio selection 投资方案的选择 ▪ Optimal payback plan for the investment 最优还款计划

数据,模型,与决策

第一章（管理科学简介）Ｐ５（１）管理科学介绍管理科学本质：是对与定量因素有关的管理问题通过应用科学的方法进行辅助管理决策制定的一门学科．管理科学发展过程：快速发展开始于２０世纪四五十年代起初的动力来自于第二次世界大战另一个里程碑是１９４７年丹捷格发明单纯形罚更大的推动作用的是计算机革命的爆发管理决策：管理者考虑管理科学对定量因素进行分析得出的结果后，再考虑管理科学以外的众多无形因素，然后根据其最佳判断做出决策管理科学小组系统和考察时步骤：定义问题与收集数据——构件数学模型——从模型中形成对于一个问题进行求解的基于计算机的程序——测试模型并在必要时进行修正——应用模型分析问题以及提出管理建议——帮助实施被管理者采纳的小组建议课后问题：１．管理科学什么时候有了快速发展？快速发展开始于２０世纪四五十年代２．商学院以外还广泛使用的对管理科学学科的叫法：运筹学３．管理科学研究提供给管理者什么？对问题涉及的定量因素进行分析并向开明的管理者提出建议４．管理科学以哪些领域作为基础？科学领域：数学，计算机社会领域：经济学５．什么是决策支持系统？辅助管理决策制定的交互式基于计算机的系统６．与管理问题有关的一般定量因素有哪些？生产数量，收入，成本，资源Ｐ１１（２）一个例子：盈亏平衡分析步骤：分析问题——建立模型——敏感性分析，电子表格模型提供上述三者了方便的途径如果预测销售数量＜盈亏平衡点，Ｑ＝０预测销售数量＞盈亏平衡点，Ｑ＝预测销售数量敏感性分析目的：研究如果一个估计值发生了变化，将会给模型带来什么样的变化Ｍｉｎ（ａ，ｂ）：取ａ，ｂ中的最小值Ｉｆ（Ａ，ｂ，ｃ）：如果表达式Ａ为真，则值为ｂ，否则为ｃ第二章（线性规划：基本概念）Ｐ３１（３）在电子表格上建立恩德公司问题的模型１．开始在电子表格上建立线性规划模型时需要回答的三个问题：要做出的决策是什么？在做出这些决策上有哪些约束条件？这些决策的全部绩效测度是什么？２．以下各个单元格的作用数据单元格：显示数据的单元格可变单元格：需要做出决策的单元格输出单元格：依赖于可变单元格的输出结果的单元格目标单元格：在生产率做出决策时目标值定为尽可能大的特殊单元格３．该案例中每个输出单元格（包括目标单元格）的Ｅｘｃｅｌ等式的形式：可以表达为一个ＳＵＭＰＲＯＤＵＣＴ函数，这里的每一项是一个数据单元格和可变单元格的乘积Ｐ３３（４）电子表格的数学模型１．电子表格模型与代数模型相同的初始步骤：收集相关数据确定要做出的决策确定对这些决策的约束条件确定为这些决策的完全绩效测度把约束条件和绩效测度的口头描述转化为数据和决策表示的定量表达式２．用代数形式建立线性规划模型时，模型中需要引入代数符号来表示哪几类数量？用来表示绩效测度与决策４．模型的一个可行解是什么意思？决策变量的任何一个取值Ｐ４１（５）求两问题变量的图解法１．图解法能用来求解带有几个决策变量的线性规划问题？只有两个２．什么是约束边界线？形成一个约束条件所允许的边界的直线什么是约束边界方程？形成一个约束条件所允许的边界的直线的方程可行域？比所有约束边界线更靠近原点的那些点成为可行解，可行解所在的区域成为可行域会作图求解Ｐ４４（６）应用Ｅｘｃｅｌ求解线性规划问题(solver)１．用来输入目标单元格和可变单元格地址的对话框是什么？目标单元格：set target cell 可变单元格：by changing cells２．具体化模型的函数约束的对话框是什么？ｓｕｂｊｅｃｔｔｏｔｈｅＣｏｎｓｔｒａｉｎｔｓ３．在ｓｏｌｖｅｒ中，哪些选项一般需要选定以求得一个线性规划模型？采用线性模型，假定非负Ｐ４８（７）一个最小化的例子——利博公司广告组合问题３. 利博公司的目标？在达到市场份额的前提下，确定最低的总成本并决定要在每种媒体上做多少钱的广告４．在电子表格中设置目标单元格和可变单元格的基本原理是什么？目标单元格：？可变单元格：？Ｐ５０（８）管理视角的线性规划１．管理部门一般对线性规划研究的技术细节设计深么？不深，没有必要２．一般问题有两个以上决策变量看，那么研究两个决策变量问题的图解法的意义？实际意义中没有价值，但对于传达线性规划设计确定约束边界和使目标值往尽可能大的这一方向移动的这一基本观念有很大价值３．开明的管理者关于线性规划应该知道哪些事项？需要知道线性规划是什么的一个良好直觉对线性规划的适用性和作用有一个正确的评价使得在合适的时候鼓励应用能够区分能胜任和以次充好的线性规划研究理解如何解释研究成果第三章（电子表格建模的艺术）（多简答和选择）Ｐ７６（２）电子表格建模程序的概述１．当你不知道从哪里开始时，帮助你开始建立电子表格模型的方法是什么？设想一下的目标手工进行一些计算建立一个小的电子表格２．手工计算可从那两方面帮助你？首先，它能帮助理清输出单元格公式的形式其次，它可以帮助检验表格３．描述一下组织和编排电子表格的一个有用方法计划设想一下你的目标手工进行一些计算建立一个电子表格建模先建立一个小模型测试利用不同的测试数据分析模型的逻辑关系，将模型扩展为完整的模型分析评估建议的解和／或利用ｓｏｌｖｅｒ优化４．哪些数值应被输入数据单元格以测试模型？试着输入一些我知道输出单元格结果的数值５．单元格绝对坐标：当被填入其它单元格时，坐标不会改变的坐标，如＆Ｅ＆１１单元格相对坐标：公式中单元格或者范围的坐标通常是基于他们相对含有公式的单元格的位置Ｐ８１（３）建立一个好的电子表格模型的几个原则１．模型的哪个部分最先输入电子表格？在建立电子表格之前，先输入和仔细编排所有数据２．数据应包含在公式中，还是被单独输入数据单元格？单独３．区域名称是如何使公式和模型在Ｓｏｌｖｅｒ对话框中更易于理解？如何选择区域名称？？１：用区域名称取代单元格地址写入公式中，使得公式更容易说明？２：选择“插入”菜单中“名称／定义”，然后输入一个名称，获得区域名称４．区分数据单元格，可变单元格，输出单元格，目标单元格有哪些方法？对不同类型的单元格使用不同的边框，单元格阴影５．在电子表格中完整的表达一个约束条件需要多少单元格？３Ｐ８６（４）调试电子表格模型１．调试电子表格模型的第一个步骤是什么？在你预知输出单元格正确结果的情况下，将不同的数值输入可变单元格，然后观察模型的计算结果是否和预期结果一致２．如何输出单元格在数值和公式中的切换？ｐｃ上同时按ｃｏｎｔｒｏｌ和～键（Ｍａｃ上同时按Ｃｏｍｍａｎｄ和～键）３．对于一个给定的单元格，哪一个Ｅｘｃｅｌ工具可以用来追踪其从属单元格或引用单元格？“工具”——“审核／追踪从属单元格”，会显示出箭头，以观察单元格之间的联系建立一个好的电子表格所需原则：●首先输入数据●组织和清楚标识数据●每个数据输入唯一的单元格●将数据与公式分离●保持简单化●使用区域名称●使用相对和绝对坐标简化公式的复制●使用边框、阴影和颜色来区分单元格类型●在电子表格中显示整个模型第四章（线性规划：建模与应用）Ｐ９７（１）案例研究——超级食品公司的广告混合问题４．在评价使用线性规划来表示该实际问题的准确性时，要做出的假设条件有哪些？允许有分数解包括目标单元格和可变单元格都可以用ＳＵＭＰＲＯＤＵＣＴ函数以数据单元格和可变单元格表示（有时候只是可变单元格的加总）Ｐ１０６（２）资源分配问题１．资源分配问题的共性在线性规划模型中每一个函数限制均为资源限制，并且，每一种资源都可以表现为如下的形式：使用的资源数量＜＝可用的资源数量２．资源限制的形式如何？使用的数量＜＝可获得的数量３．为解决资源分配问题，必须收集哪三类数据？每种资源的可供量每一种活动所需要的各种资源的数量，对于每一种资源约活动的组合，单位活动所消耗的资源量必须首先估计出来每一种活动对总的绩效测度的单位贡献Ｐ１１３（３）成本收益平衡问题１．资源分配问题与成本收益平衡问题在管理目标上的差异是什么？资源分配问题：各种资源是受限制的因素（包括财务资源），问题的目标是（根据特定的总绩效测度）最有效的利用各种资源成本收益问题：管理层采取更为主动的姿态，他们指定哪些收益必须实现（不管如何使用资源），并且要以最低的成本实现所指明的收益２．成本收益平衡问题的共性是什么？所有的函数约束均为收益约束，并具有如下的形式：完成的水平＞＝最低可接受的水平３．收益限制的形式如何？完成的水平＞＝最低可接受的水平４．为解决成本收益平衡问题必须收集的三类数据包括哪些？每种收益的最低可接受水平（管理决策）每一种活动对每一种收益的贡献（单位活动的贡献）每种活动的单位成本Ｐ１１７（４）网络配送问题１．为什么这类问题为网络配送问题？这类问题通过配送网络能以最小的成本完成货物的配送，所以称之为网络配送问题２．网络配送问题的共性是什么？确定需求的约束，提供的数量＝需要的数量３．确定需求的约束与资源约束和收益约束的区别是什么？确定需求的约束：提供的数量＝需要的数量资源约束：使用的资源数量＜＝可用的资源数量收益约束：完成的水平＞＝最低可接受的水平Ｐ１２９（７）管理视角的建模１．为什么ｗｈａｔ－ｉｆ分析是线性规划的研究中非常重要的一个组成部分？尽管可能使用许多变异的模型，但是对于一个特定版本的模型，一次只能求得一个解，但是在求得一个解以后，管理层会有很多问题：如果模型的参数估计有误怎么办？如果做出不同的似是而非的假设，问题的将会如何变化？如果管理方面所要求的某一选项没有被考虑在内，会产生怎样的结果？Ｗｈａｔ－ｉｆ分析有助于解决上述等相关问题Ｐ１３１（８）线性规划应用经典回顾１．比较三种线性规划的应用，注意各种类型的问题应该使用哪一类型的线性规划模型（资源分配、成本收益平衡、网络配送以及混合问题）第五章（线性规划的Ｗｈａｔ－ｉｆ分析）Ｐ１５８（３）只有一个目标函数系数变动１．目标函数系数允许变化范围的含义是什么？能使最优解保持不变的目标函数系数的变化范围称为目标函数系数允许变化范围２．如果目标系数的估计值不是实际值，并且不在允许变化范围之内，会有怎样的影响？最优解不正确３．在Ｅｘｃｅｌ的灵敏度分析报告中，目标函数系数一栏该如何解释？允许增加值和允许减少值一栏又该如何解释？目标函数系数一栏：目标函数系数的现值允许增加值和允许减少值一栏：是这些系数在最优的范围内，允许增加和减少的量（１Ｅ＋３０）：１０的三十次方的缩写，表示无穷大Ｐ１６５（４）目标函数系数同时变动的影响１．目标系数变动百分比法则中，变动的百分比指什么？各个变动系数占该系数允许变化范围允许变动量的百分比之和（有方向）２．在百分百法则中，如果变动的百分比之和不超过100%，最初的最优解将如何？不会改变３．在百分百法则中，如果变动的百分比之和超过100%，是否就意味着最初的最优解已经不再是最优解？不能确定最优解是否改变Ｐ１７２（５）单个约束条件变化的影响１．为什么要研究函数的约束条件的变化带来的影响？因为在建模时,还不能得到模型的这些参数的精确值更重要的是:这些常数往往不是由外界决定而是由管理层的政策决策决定的２．为什么函数约束的右端值可能改变？这些常数往往不是由外界决定而是由管理层的政策决策决定的，因此，在建模并求解之后管理者想要知道如果改变这些决策是否会提高最终的收益３．影子价格的含义是什么？约束常数增加微小的量1，使得目标函数增加的量４．用电子表格如何找到影子价格？用Ｓｏｌｖｅｒ表格呢？用灵敏度报告呢？电子表格：改变某一约束条件的值，重新按下Solver键，尝试在约束条件变化范围内找出每单位约束条件变化引起的目标函数值的变化即为影子价格Ｓｏｌｖｅｒ表格：？灵敏度报告:Shadow price栏５．为什么管理者会对影子价格感兴趣？管理者可以用影子价格评价，在影子价格的有效域内幅度不大的改变工作时间的各种决策６．影子价格是否也同样适用于减少函数约束右端值的数值的情况？是７．影子价格０对管理者来说是什么意思？该影子价格对应的约束条件在其变化范围内对目前的最优解没有影响８．为什么管理层会对可行域感兴趣？？１７６（６）约束右端值同时变动的情形１．为什么要研究约束条件同时发生变化的情况？经常会出现需要我们考虑约束条件同时变动的情况。

DMD数学规划-数据、模型与决策

P ≤ 16
NONNEGATIVITY: W≥0, P≥0
所求得的最优解为： W = 12（千件）， P = 9（千件）
相应的最优值为： CONTRIBUTION = ＄2,460
线性最优化模型的基本概念
目标函数问题的目标是选取决策变量的一组值，使得这些决策变量的一个函数
─ ─目标函数，取得最大值或最小值。 NBS问题的目标函数是一个成本函数：
NO；
STEEL:
1.5W+1.0P≤27
step3: 画出可行域；
确定所有对GTC问题的约束条件可行的解的集合。
step4: 确定目标函数的等值线；
CONTRIBUTION=130W+100P
step5: 确定最优解。
把等值线向目标函数值变好的方向移动，直至等值线和
➢ 解决策树的要点: 从决策树的最后一支开始算,即从右到左。
1. 对事件点,计算其期望,标于该点上。 2. 对决策点,选择其各个分支中最优的那一支,将其期望
标于该点上,其他各支划掉。 3. 当每个点都被标上数值后,决策过程完毕。
第十章最优化方法
1.线性最优化
两个线性最优化问题的例子： New Bedford公司的订购计划， GTC公司的生产计划
的利润情况。 GTC现在要根据表中数据制定有最大利润的生产计划。
钢(磅) Molding Machine(小时) Assembly Machine(小时)
需求限量(件/天)
利润($/千件)
扳手
1.5 1.0 0.3 15000 130
钳子
1.0 1.0 0.5 16000 100
可供量
27000磅/天 21000小时/天 9000小时/天

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。

数据、模型与决策(运筹学)课后习题和案例答案003

CHAPTER 3 THE ART OF MODELING WITH SPREADSHEETSReview Questions3.1-1 The long-term loan has a lower interest rate.3.1-2 The short-term loan is more flexible. They can borrow the money only in the yearsthey need it.3.1-3 End with as large a cash balance as possible at the end of the ten years after payingoff all the loans.3.2-1 Visualize where you want to finish. What should the “answer” look like?3.2-2 First, it can help clarify what formula should be entered for an output cell. Second,hand calculations help to verify the spreadsheet model.3.2-3 Sketch a layout of the spreadsheet.3.2-4 Try numbers in the changing cells for which you know what the values of the outputcells should be.3.2-5 Relative references are based upon the position relative to the cell containing theformula. Absolute references refer to a specific cell address.3.3-1 Enter the data first.3.3-2 Numbers should be entered separately from formulas in data cells.3.3-3 With range names, the formulas and Solver dialogue box contain descriptive rangenames rather than obscure cell references. Use a range name that corresponds exactly to the label on the spreadsheet.3.3-4 Borders, shading, and colors can be used to distinguish data cells, changing cells,output cells, and target cells on a spreadsheet.3.3-5 Three. One for the left-hand-side, one for the inequality sign, and one for the right-hand-side.3.4-1 Try different values for the changing cells for which you can predict the correct resultin the output cells and see if they calculate as expected.3.4-2 Control-~ on a PC (command-~ on a Mac).3.4-3 The auditing tools can be used to trace dependents or precedents for a given cell.Problems3.13.2a. The COO will need to know how many of each product to produce. Thus, the decisions are how many end tables, how many coffee tables, and how many dining room tables to produce. The objective is to maximize total profit.b. Pine wood used = (3 end tables)(8 pounds/end table)+ (3 dining room tables)(80 pounds/dining room table)= 264 pounds Labor used = (3 end tables)(1 hour/end table) + (3 dining room tables)(4 hours/dining room table) = 15 hoursc.E nd TablesCoffee TablesDining Room TablesUnit P rofitAvailableP ine Wood<=<=Units P roducedd.3.3a. Top management will need to know how much to produce in each quarter. Thus,the decisions are the production levels in quarters 1, 2, 3, and 4. The objective is to maximize the net profit.b. Ending inventory(Q1)= Starting Inventory(Q1) + Production(Q1) – Sales(Q1)= 1,000 + 5,000 – 3,000 = 3,000 Ending inventory(Q2) = Starting Inventory(Q2) + Production(Q2) – Sales(Q2)= 3,000 + 5,000 – 4,000 = 4,000 Profit from sales(Q1) = Sales(Q1) * ($20) = (3,000)($20) = $60,000 Profit from sales(Q2) = Sales(Q2) * ($20) = (4,000)($20) = $80,000 Inventory Cost(Q1) = Ending Inventory(Q1) * ($8) = (3,000)($8) = $24,000 Inventory Cost(Q2) = Ending Inventory(Q2) * ($8) = (4,000)($8) = $32,000c.Inventory Holding C ost Gross P rofit from SalesStarting M axim um Dem and/E nding Inventory Gross ProfitNet P rofitd.e.3.4a. Fairwinds needs to know how much to participate in each of the three projects, and what their ending balances will be. The decisions to be made are how much to participate in each of the three projects. The objective is to maximize the ending balance at the end of the 6 years.b. Ending Balance(Y1) = Starting Balance + Project A + Project C + Other Projects = 10 + (100%)(–4) + (50%)(–10) + 6= 7 (in $millions) Ending Balance(Y2) = Starting Balance + Project A + Project C + Other Projects = 7 + (100%)(–6) + (50%)(–7) + 6 = 3.5 (in $millions)c.Starting C ashTotalCash Flow (at full participation, $m illion)Cash Flow OtherE nding M inim um Year123456P articipationd.e.3.5a. Web Mercantile needs to know each month how many square feet to lease andfor how long. The decisions therefore are for each month how many square feet to lease for one month, for two months, for three months, etc. The objective is to minimize the overall leasing cost.b. Total Cost = (30,000 squarefeet)($190 per square foot) + (20,000 square feet)($100 per square foot)= $7.7 million.c.M onth Covered by Lease?Total Space M onth of Lease:111112222333445Leased Required Length of Lease:M onth 1M onth 2M onth 3M onth 4M onth 5Cost of Lease (per sq. ft.)Lease (sq. ft.)d.e.3.6a. Larry needs to know how many employees should work each possible shift. Therefore, the decision variables are the number of employees that work each shift. The objective is to minimize the total cost of the employees.b. Working 8am-noon: 3 FT morning + 3 PT = 6 Working noon-4pm: 3 FT morning + 2 FT afternoon + 3 PT = 8 Working 4pm-8pm: 2 FT afternoon + 4 FT evening + 3 PT = 9 Working 8pm-midning: 4 FT evening + 3 PT = 7 Total cost per day = (3+2+4 FT)(8 hours)($14/hour) + (12 PT)(4 hours)($12/hour) = $1,584.c.Full Tim eFull Tim e Full Tim e P art Tim e P art Tim e P art Tim e P art Tim e Total Total 8am 4pm 8pm -m idnight Total Total Tim e of Day 8am 4pm 8pm -m idnightd.3.7a. Al will need to know how much to invest in each possible investment each year.Thus, the decisions are how much to invest in investment A in year 1, 2, 3, and 4; how much to invest in B in year 1, 2, and 3; how much to invest in C in year 2; and how much to invest in D in year 5. The objective is to accumulate the maximum amount of money by the beginning of year 6.b. Ending Cash (Y1) = $60,000 (Starting Balance) – $20,000 (A in Y1) = $40,000Ending Cash (Y2) = $40,000 (Starting Balance) – $20,000 (B in Y2) – $20,000 (C in Y2) = $0 Ending Cash (Y3) = $0 (Starting Balance) + $20,000(1.4) (for investment A) = $28,000 Ending Cash (Y4) = $28,000 (Starting Balance) Ending Cash (Y5) = $28,000 (Starting Balance) + $20,000(1.7) (investment B) = $62,000 Ending Cash (Y6) = $62,000 (Starting Balance) + $20,000(1.9) (investment C) = $100,000c.Beginning BalanceM inim um BalanceInvestm entA A A AB B BCDE nding Minimum >=>=>=>=>=>=Dollars Investedd.e.3.8 In the poor formulation, the data are not separated from the formula—they areburied inside the equations in column C. In contrast, the spreadsheet in Figure 3.6 separates all of the data in their own cells, and then the formulas for hours used and total profit refer to these data cells.In the poor formulation, no range names are used. The spreadsheet in Figure 3.6 uses range names for UnitProfit, HoursUsed, TotalProfit, etc.The poor formulation uses no borders, shading, or colors to distinguish between cell types. The spreadsheet in Figure 3.6 uses borders and shading to distinguish the data cells, changing cells, and target cell.The poor formulation does not show the entire model on the spreadsheet. There is no indication of the constraints on the spreadsheet (they are only displayed in the Solver dialogue box). Furthermore, the right-hand-sides of the constraints are not on the spreadsheet, but buried in the Solver dialogue box. The spreadsheet in Figure 3.6 shows all of the constraints of the model in three adjacent cells on the spreadsheet.3.9 Cell F16 has –0.47 for LT Interest, rather than –LTRate*LTLoan.Cell G14 for the 2006 ST Interest uses the LT Loan amount rather than the ST Loan amount.Cell H21 for LT Payback refers to the 2006 ST Loan rather than the LT Loan to determine the payback amount.3.10 Cell G21 for the 2013 ST Interest uses LTRate instead of STRate.Cell H21 for LT Payback in 2013 as –6.649 instead of –LTLoan.Cell I15 for ST Payback in 2007 has –LTLoan instead of –E14 (LT Loan for 2006). Case3.1 a. PFS needs to know how many units of each of the four bonds to purchase, howmuch to invest in the money market, and their ending balance in the moneymarket fund each year after paying the pensions. The decisions are how manyunits of each bond to purchase, as well as the initial investment in 2003 in themoney market. The objective is to minimize the overall initial investment necessaryin 2003 in order to meet the pension payments through 2012.b. Payment received from Bond 1 (2004) = (10 thousand units) ($1,000 face value) +(10,000 units) ($1,000 face value) (0.04 coupon rate) = $10.4 million Payment received from Bond 1 (2005) = $0Payment received from Bond 2 (2004) = (10 thousand units) ($1,000 face value)(0.02 coupon) = $0.2 millionPayment received from Bond 2 (2005) = (10 thousand units) ($1,000 face value)(0.02 coupon) = $0.2 millionBalance in money market fund (2003) = $28 million (initial investment)–$8 million (pension payment)= $20 millionBalance in money market fund (2004) = $20 million (starting balance)+ $10.4 million (payment from Bond 1)+ $0.2 million (payment from Bond 2)–$12 million (pension payment)+ $1 million (money market interest)= $19.6 millionBalance in money market fund (2005) = $19.6 million (starting balance)+ $0.2 million (payment from Bond 2)–$13 million (pension payment)+ $0.98 million (money market interest)= $7.78 millionc. PFS will need to track the flow of cash from bond investments, the initialinvestment, the required pension payments, interest from the money market, and the money market balance. The decisions are the number of units to purchase of each bond. Data for the problem include the yearly cash flows from the bonds (per unit purchased), the money market rate, and the minimum required balance in the money market fund at the end of each year. A sketch of a spreadsheet model might appear as follows.3-11Money Market RateMinimum Required BalanceRequired Money Money Bond Initial P ension Market Market20030200402005020060200702008020090201002011020120Units P urchasedBond Cash Flow s (per unit)。

数据、模型与决策 (14)

Session14 Forecasting
实际应用
Sales Forecasting 销售预测
预测
航空公司现在非常依赖于在收取不看重价格的商务人员旅行支付的高额票价的同时向其他人提供折扣票价以填满座位。座位的数目在不同的运费等级上如何分配的决策对利润最大化来说是关键的。美洲航空公司（American Airlines）使用对每一种票价需求的统计预测来做出这项决策的
Session14 Forecasting 预测
Data, Model and Decisions 数据、模型与决策
Session 14
Forecasting
预测
Session14 Forecasting
Session Topics

预测

Some Forecasting Applications 一些预测实际应用 Types of Forecasting 预测的类型 Demand Management 需求管理 Qualitative Forecasting Methods 定性预测方法 Quantitative Forecasting Methods 定量预测方法
39.6% 41.6% 35.4% 48
Source: Nada Sanders and Karl Mandrodt (1994) “Practitioners Continue to Rely on Judgmental Forecasting Methods Instead of Quantitative Methods,” Interfaces, vol. 24, no. 2, pp. 92-100.
想想看！
Session14 Forecasting

数据、模型与决策

表3 各年龄段的博客创建情况
年龄段
10－12 13－19 20－29 30－39
创建的博客数
55,500 2,120,000 1,630,000 241,000
频率（％）
1.35 51.45 39.56 5.85
累积频率（％）
1.35 52.80 92.35 98.20
30－49
50－59
实际问题1：资源分配问题

潘得罗索工业公司生产胶合板，根据厚度和所用木材的质量而有所不同。因为产品在一个竞争的环境中进行销售，产品的价格由市场决定。所以每个月管理层面临的一个关键问题是选择产品组合以获取尽可能多的利润。需要考虑当前生产产品必须的各种资源的可得数量。六项最重要的资源为（1）四种类型的原木（根据原木的质量区分）和（2）生产胶合板的两项关键作业的生产能力（模压作业和刨光作业）。你们公司有这样的经历吗？

热量，胖人少吃1000卡热量，他们的生活习惯没有改变。

这项研究为肥胖者提供了新的希望。
以上几个问题说明在现实生活中，不管是进行决策还是进行研究发现新结果，都离不开数据。
第一章、数据与数据展示
1.数据概述：可分为科学数据、社会数据、商业数据。（依来源与用途）分类：数值型与属性型静态数据与动态数据时间序列数据、截面数据、面板数据定类数据、定序数据、定距数据与定比数据
6. 图表完成。如果需要，可以双击图表中任何一部分进行修改。
2001－2001年一、二、三产业总值
产值（万亿元）
8.0000 数值轴主要网格线 6.0000 4.0000 2.0000 0.0000
图表区图表标题绘图区
数值轴

《数据、模型与决策》教学大纲.doc

《数据、模型与决策》教学大纲一、课程主要内容简介本课程作为MBA的一门必修课程。

各行各业的管理者都必需具备数字信息处理能力，利用数据信息得出正确的结论，并在诸多的策略中选取最优的策略。

数据分析、模型建立、策略选择是一个完整的过程。

对管理者而言，在处理问题时往往首先遇到的是数据，必须科学地、合理地在这些数据中提取他所需要的信息，或建立相应的模型，最后作出决策。

本课程是一门系统、完整、整体结构严谨、各部分紧密关联、理论与实际并重的课程，因此采用以基本理论为本、实用为主的教学指导思想。

既要求学生了解理论的内涵，掌握方法；也要求学生能学以致用，解决实际问题。

以基本理论为本，讲清来龙去脉，讲清应用背景，讲清内容要点以及使用条件，而略去比较烦难的推导证明。

以实用为主，是选择一些典型的案例或例子，运用基本理论知识给予解决。

本课程是一门理论性与实践性都比较强的课程，教学中注意循序渐进、由浅入深，理论讲解与案例讲解交叉进行。

既要避免在课堂上进行枯燥的、不完整的、不必要的数学论证，也要避免漫无目的、不得要领地去讨论实际问题。

案例讨论是本课程教学的重要部分。

选择一些有代表性的、能清晰说明一个原理或一种方法的案例，使学生能理解原理的内涵或方法的效用，同时选择一些涉及多种原理、方法和计算技巧的案例，它们可以被用来综合学生的知识，加强彼此联络，使学生学到的东西系统化、一体化。

第1章数据、模型与决策简介《数据、模型与决策》是应用分析、试验、量化的方法，对经济管理系统中人力、物力、财力等资源进行统筹安排，为决策者提供有依据的最优化方案，以实现最有效的管理。

本章主要介绍《数据、模型与决策》的学习内容和学习方法等。

第2章线性规划首先通过对大量实际问题的介绍，引入线性规划模型。

然后介绍微软的Solver在建模和求解这些问题中的作用，再讨论建立线性规划模型的用处和一些不足。

本章的主要目标是要使得学员用电子表格建立线性规划模型对实际问题进行分析的能力。

(运筹学)运输与指派问题

An Award-Winning Application 运输问题的一个获奖应用
P&G重新设计制造和配送体系：90’S 成百上千个供应商 50多个产品类别超过60个的工厂 15个配送中心超过1000个的顾客群体
An Award-Winning Application 运输问题的一个获奖应用
+ 15($388) + 85($685)
＝ $165,595
P&T公司的运输问题
贝林翰尤基尼艾尔贝李需求
萨克拉门托
$464 $352
$995
80
盐湖城
$513 $416 $682
65
赖皮特城奥尔巴古
$654 $690 $388
70
$867 $791 $685
85
供应
75 125 100
运输问题是一种线性规划问题
令xij = 从第i个罐头加工厂运送到第j个仓库的车数最小化成本＝$464x11 + $513x12 + $654x13 + $867x14
+ $352x21 + $416x22+ $690x23 + $791x24 + $995x31 + $682x32 + $388x33 + $685x34
$37
$18
$32
$48
$29
$59
$51
$35
Shipment Plant 1 Plant 2 Plant 3
Total
Customer 1 Customer 2 Customer 3 Customer 4 Production

Session5网络最优化问题

Session5 Network Optimization Problems 网络最优化问题
Network representation 网络表述
80 units produced F1 W1 60 units needed
DC
70 units produced
F2
W2
90 units needed
Session5 Network Optimization Problems 网络最优化问题
Minimum Cost Network Flow Model 最小费用流问题
最小费用流问题的构成：
节点(nodes)（供应点、需求点、转运点）
弧（arcs）
目标: 通过网络满足需求提供供应，
最小化流的总成本
一家折扣连锁零售店，现在和以前是如何使用微型计算机去处理一个最小费用流问题。应用中公司力图使得从供应商到加工中心，再从加工中心到零售店的商流最优。其中的一些网络有超过 20,000条弧。
All Rights Reserved, Prof. Ren Jian Biao,2004
Session5 Network Optimization Problems 网络最优化问题
All Rights Reserved, Prof. Ren Jian Biao,2004
Session5 Network Optimization Problems
案例研究
BMZ Case Study BMZ案例研究
RO [ 60 ] NY [ 80 ]
网络最优化问题
[ 50 ]
[ 40 ] BO [ 70 ] ST
经典应用
Planning Vehicle Replacement at Phillips Petroleum 飞利浦石油的运输工具替换计划