西工大作业集-运筹学作业集
运筹学作业_解的题目(发布版)
运筹学作业一、 有如下线性规划问题 Max f(x)=2x1 +3x2 x 1 + 2x 2 ≤ 8 4x 1 ≤ 16 4x 2 ≤ 12 x 1 ,x 2 ≥ 0 1） 用图解法求最优解； 解：线性规划问题为： Max f(x)=2x 1 +3x 2x 1 + 2x 2 ≤ 8 ① 4x 1 ≤ 16 ② 4x 2 ≤ 12 ③ x 1 ≥ 0 ④ x 2 ≥ 0 ⑤1、 以x 1为横坐标、x 2为纵坐标，建立平面坐标系。
然后在该平面坐标系上画出各个约束条件，包括非负约束条件。
（见下图1.1）0 1 2 3 4 5 6 7 8543211x图1.12、 图1.1所示的凸多边形OABCD 即为给定线性规划问题的可行域。
3、 将目标函数f(x)=2x 1 + 3x 2，写成x 2 = -2/3x 1 + 1/3f(x) ⑥令f(x)=0，则上式变为x 2 = -2/3x 1，对应直线见下图1.2的⑥。
图1.24、 观察图1.2将直线⑥平行移动至与凸多边形OABCD 的顶点C 相切时，对应直线⑦，目标函数f(x)取得最大值，此时得最优解，为：x 1 =4，x 2 =2（即C 点的坐标值），目标函数值f(x)=14。
即图1.3所示图1.32） 写出所有基本可行解，并指出它在图解法图中的位置； 解：基本可行解如下：(0,0)、(0,3) 、(2,3) 、(4,2) 、(4,0)分别对应图解法图1.1中凸多边形OABCD 的五个顶点O 、A 、B 、C 、D 。
3） 用QSB 软件求最优解，对C 1,C 2,b 1,b 2进行灵敏度分析，打印出计算结果。
解：1、 QSB 软件求出最优解：灵敏度分析：q 1=1.5，资源1的影子价格为1.5，资源1无剩余； q 2=0.5，资源2的影子价格为0.5，资源2无剩余； q 3=0，资源3的影子价格为0，资源3有剩余； q 1最大，资源1最紧缺。
C 1、C 2：从上面最优解可看到，C 1、C 2为当前值2、3时，x 1、x 2均投入生产，此时保持最优解结构不变的C 1、C 2取值范围为C 1≥1.5（C 2不变时），0≤C 2≤4（C 1不变时）。
西工大17秋《运筹学》在线作业
------------------------------------------------------------------------------------------------------------------------------ 西工大17秋《运筹学》在线作业一、单选题：1.（单选题）使用人工变量法求解极大化线性规划问题时，当所有的检验数，在基变量中仍含有非零的人工变量，表明该线性规划问题（）。
(满分A有唯一的最优解B有无穷多个最优解C无可行解D为无界解正确答案:——C——2.（单选题）以下不属于运用运筹学进行决策的步骤的是（）。
(满分A观察待决策问题所处的环境B分析定义待决策的问题并拟定模型C提出解并验证其合理性D进行灵敏度分析正确答案:——D——3.（单选题）以下各项中不属于运输问题的求解程序的是（）。
(满分A分析实际问题，绘制运输图B用单纯刑法求得初始运输方案C计算空格的改进指数D根据改进指数判断是否已得最优解正确答案:——C——4.（单选题）在解运输问题时，若调整路线已确定，则调整运量应为（）。
(满分 A负号格的最小运量B负号格的最大运量C正号格的最小运量D正号格的最大运量正确答案:————5.（单选题）若运输问题在有条件的总供应量大于总需要量时，（）。
(满分A不能求解B不存在可行解C虚设一个需求点再求解D虚设一个供应点再求解正确答案:————6.（单选题）按决策的可靠程度将决策分类中，不包括（）。
(满分:)A确定型决策B风险型决策C单项决策D不确定型决策正确答案:————7.（单选题）在求解运输问题的过程中运用到下列哪些方法（）。
(满分:)A西北角法B位势法C闭回路法------------------------------------------------------------------------------------------------------------------------------ D以上都是正确答案:————8.（单选题）关于整数规划的分类，下列描述错误的是（）。
西工大18春运筹学在线作业辅导资料
西工大18春《运筹学》在线作业1、C2、A3、D4、C5、C一、单选题共15题，60分1、用割平面法求解整数规划是，构造的割平面只能切去（）。
A整数可行解B整数解最优解C非整数解D无法确定正确答案是：C2、实际应用中遇到各种非标准形式的指派问题时，通常的处理方法是（）。
A先转化为标准形式，然后用匈牙利解法求解B用匈牙利算法求解C用割平面法求解D用分枝定界法求解正确答案是：A3、若运输问题在有条件的总供应量大于总需要量时，（）。
A不能求解B不存在可行解C虚设一个需求点再求解D虚设一个供应点再求解正确答案是：D4、下列说法正确的为（）。
A如果线性规划的原问题存在可行解，则其对偶问题也一定存在可行解B如果线性规划的对偶问题无可行解，则原问题也一定无可行解C在互为对偶的一对原问题与对偶问题中，不管原问题是求极大或极小，原问题可行解的目标函数值都一定不超过其对偶问题可行解的目标函数D如果线性规划问题原问题有无界解，那么其对偶问题必定无可行解正确答案是：C5、以下各项中不属于运输问题的求解程序的是（）。
A分析实际问题，绘制运输图B用单纯刑法求得初始运输方案C计算空格的改进指数D根据改进指数判断是否已得最优解正确答案是：C6、在产销平衡运输问题的数学模型中，约束条件的关系是（）。
A=B< =C> =D< =，=，>=都有正确答案是：A7、一般的指派问题不包括（）。
A最小化指派问题B人数和事数不等的指派问题C一个人可做几件事的指派问题D某事一定不能由某人做的指派问题正确答案是：A8、在解运输问题时，若调整路线已确定，则调整运量应为（）。
A负号格的最小运量B负号格的最大运量C正号格的最小运量D正号格的最大运量正确答案是：A9、在不确定的条件下进行决策，下列哪个条件是不必须具备的（）。
A确定各种自然状态可能出现的概率值B具有一个明确的决策目标C可拟定出两个以上的可行方案D可以预测或估计出不同的可行方案在不同的自然状态下的收益值正确答案是：A10、在用单纯形法求解线性规划问题时，下列说法错误的是（）。
大工22春《运筹学》在线作业123答案
大工22春《运筹学》在线作业1-00001试卷总分:100 得分:100一、单选题(共 5 道试题,共40 分)1.线性规划的四条基本假设不包括( )。
A.比例性B.连续性C.确定性D.发散性答案:D2.下列有关运筹学的说法不正确的为( )。
A.是管理学的简称B.涉及到应用数学、形式科学、经济学、管理学等学科C.采用数学建模、统计学和计算方法等来求解复杂问题，以达到最优和近似最优的解决方案D.利用科学的管理方法，为管理人员达到管理目标提供决策支持答案:A3.数学规划的研究对象为( )。
A.数值最优化问题B.最短路问题C.整数规划问题D.最大流问题答案:A4.( )是解决多目标决策的定量分析的数学规划方法。
A.线性规划B.非线性规划C.目标规划D.整数规划答案:C5.数学规划模型的三个要素不包括( )。
A.决策变量B.目标函数C.约束条件D.最优解答案:D二、判断题(共15 道试题,共60 分)6.线性规划基本假设中的可加性是指线性规划中所有目标函数和约束函数中的系数都是确定的常数,不含随机因素。
答案:错误7.家庭中的存储储备品,工厂储备原材料,商店存储商品等都是存储问题。
答案:正确8.线性规划的求解方法包括图解法、单纯形法、椭球法、内点法等。
答案:正确9.运筹学是运用数学方法,对需要进行管理的问题统筹规划,为决策机构进行决策时提供以数量化为基础的科学方法。
答案:正确10.整数规划问题中的整数变量可以分为一般离散型整数变量和连续型整数变量。
答案:错误11.决策变量、目标函数和约束条件是数学规划模型的三个要素,若目标函数和约束条件均为线性的数学规划问题称为非线性规划。
答案:错误12.数学规划的应用极为普遍,它的理论和方法已经渗透到自然科学、社会科学和工程技术中。
答案:正确13.线性规划基本假设中的连续性要求每个决策变量在目标函数和约束函数中,其贡献与决策变量的值存在直接比例性。
答案:错误14.线性规划可行域的顶点定是最优解。
运筹学习题集二
运筹学习题集二运筹学习题集二习题一1.1 用法求解下列线性规划问题并指出各问题是具有唯一最优解、无穷多最优解、无界解或无可行解。
(1) min z ＝6x1＋4x2 (2) max z ＝4x1＋8x2 st. 2x1＋x2≥1 st. 2x1＋2x2≤103x1＋4x2≥1.5 －x1＋x2≥8x1, x2≥0 x1, x2≥0(3) max z ＝x1＋x2 (4) max z ＝3x1－2x2 st. 8x1＋6x2≥24 st. x1＋x2≤14x1＋6x2≥－12 2x1＋2x2≥42x2≥4 x1, x2≥0x1, x2≥0(5) max z＝3x1＋9x2 (6) max z ＝3x1＋4x2 st. x1＋3x2≤22 st. －x1＋2x2≤8－x1＋x2≤4 x1＋2x2≤12x2≤6 2x1＋x2≤162x1－5x2≤0 x1, x2≥0x1, x2≥01.2. 在下列线性规划问题中找出所有基本解指出哪些是基本可行解并分别代入目标函数比较找出最优解。
(1) max z ＝3x1＋5x2 (2) min z ＝4x1＋12x2＋18x3 st. x1 ＋x3 ＝4 st. x1 ＋3x3－x4 ＝32x2 ＋x4 ＝12 2x2＋2x3 －x5＝53x1＋2x2 ＋x5 ＝18 xj ≥0 （j＝1, (5)xj ≥0 （j＝1, (5)1.3. 分别用法和单纯形法求解下列线性规划问题并对照指出单纯形法迭代的每一步相当于法可行域中的哪一个顶点。
(1) max z ＝10x1＋5x2st. 3x1＋4x2≤95x1＋2x2≤8x1, x2≥0(2) max z ＝100x1＋200x2st. x1＋x2≤500x1 ≤2002x1＋6x2≤1200x1, x2≥01.4. 分别用大M法和两阶段法求解下列线性规划问题并指出问题的解属于哪一类：(1) max z ＝4x1＋5x2＋x3 (2) max z ＝2x1＋x2＋x3 st. 3x1＋2x2＋x3≥18 st. 4x1＋2x2＋2x3≥42x1＋x2 ≤4 2x1＋4x2 ≤20x1＋x2－x3＝5 4x1＋8x2＋2x3≤16xj ≥0 （j＝1,2,3）xj ≥0 （j＝1,2,3）(3) max z ＝x1＋x2 (4) max z ＝x1＋2x2＋3x3－x4 st. 8x1＋6x2≥24 st. x1＋2x2＋3x3＝154x1＋6x2≥－12 2x1＋x2＋5x3＝202x2≥4 x1＋2x2＋x3＋x4＝10x1, x2≥0 xj ≥0 （j＝1, (4)(5) max z ＝4x1＋6x2 (6) max z ＝5x1＋3x2＋6x3 st. 2x1＋4x2 ≤180 st. x1＋2x2＋x3≤183x1＋2x2 ≤150 2x1＋x2＋3x3≤16x1＋x2＝57 x1＋x2＋x3＝10x2≥22 x1, x2≥0x3无约束x1, x2≥01.5 线性规划问题max z＝CXAX＝bX≥0如X*是该问题的最优解又λ0为某一常数分别讨论下列情况时最优解的变化：（1）目标函数变为max z＝λCX；（2）目标函数变为maxz＝（C＋λ）X；（3）目标函数变为max z＝X约束条件变为AX＝λb。
西工大作业集-经济数学基础课程练习册上
第一章测试题一.问答题1:求下列函数的定义域：(1)y = e21-x(2)y=2-4x (3)y =x1+ ln(x + 1) (4)y=⎪⎪⎩⎪⎪⎨⎧≤<-=<≤-21112cos2xxxxx，，，答案(1)由,得定义域为(2)由，得定义域为(3)由,得定义域为(4)由在定义域为[一2,1) U (1,2]2:求下列函数的定义域并作图:(1)y=1-x(2)y=⎪⎩⎪⎨⎧≤<+=<<--111122xxxxx，，，答案(1)定义域为(2)定义域为(一1，1]3:设f(x)=⎪⎩⎪⎨⎧>≤-1,sin112xxxx，，求f(1)，f(-21)，f(2π)，f(π23-)答案4:试判断下列函数对中函数f(x)和g(x)是否相同，并说明理由(1)f(x)=4lnx，g(x)=lnx4(2)f(x)=cosx，g(x)=x2sin1-(3)f(x)=x+1，g(x)=11x2--x答案(1)不相同，因为(2)不相同，因为，对应规则不同(3)不相同，因为5:设f(x)=a x+b，求)()(xfhxfy-+=∆答案6:设xx f 1)(=，求)()(x f h x f y -+=∆ 答案7:判断下列函数的奇偶性：(1)2xx e e y --=(2)32x x y -=(3))1ln(2x x y ++= (4)xxx y -+=11ln 答案 (1)因，故函数为奇函数(2)因故函数为非奇非偶函数(3)因故函数为奇函数(4)因，故函数为偶函数8:指出下列函数是由哪些函数复合而成的 (1)x y cos = (2)21x e y +=(3))1ln(1x e y +=答案 (1)由复合而成(2)由复合而成 (3)由复合而成9:设)(31)1(22x f xx x x f ，求++=+ 答案令则 故10:)(cos 1)(2x f x x x f ，求设-= 答案11:求下列函数的反函数及反函数的定义域 (1)13+=x y(2)1)2lg(++=x y 答案(1)(2)二.综合题1:一无盖的长方体木箱，容积为1 m 3，髙为2 m ，设底面一边的长为x m ，试将木箱的表面积表示为x 的函数。
[西北工业大学]《运筹学》(202110)[69164]
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二、判断题 1.运输问题解的情况有四种：无可行解；无界解；唯一最优解；无穷多最优解。 答案：错误
2.在线性规划的图解法中，基可行解一定可以在顶点得到。 答案：正确
3.42求极大值的目标值是各分枝的上界 答案：正确
18.动态规划的基本方程是将一个多阶段的决策问题转化为一系列具有递推关系的单阶段的决策问 题。 答案：正确
19.图解法和单纯形法虽然求解的形式不同，但从几何上理解，两者是一致的。 答案：正确
20.基本解对应的基是可行基X当非负时为基本可行解，对应的基叫可行基 答案：错误
21.运输问题的表上作业法实质上就是求解运输问题的单纯形法。 答案：正确
5.有m个产地n个销地的平衡运输问题模型具有特征 A.有mn个变量m+n个约束m+n-1个基变量 B.有m+n个变量mn个约束 C.有mn个变量m+n1约束 D.有m+n1个基变量，mnmn1个非基变量 答案：A
6.12则() A.无可行解 B.有唯一最优解 C.有无界解 D.有多重解 答案：A
7.5有6个产地4个销地的平衡运输问题模型具有特征 A.有10个变量24个约束 B.有24个变量10个约束 C.有24个变量9个约束
22.若线性规划的原问题有无穷多最优解，则其对偶问题也一定具有无穷多最优解。 答案：正确
23.44原问题求最大值，第i个约束是“”约束，则第i个对偶变量yi0 答案：正确
24.求最小值问题的目标函数值是各分枝函数值的下界 减去一个数后最优解不变 答案：错误
3.1线性规划具有唯一最优解是指 A.最优表中存在常数项为零 B.最优表中非基变量检验数全部非零 C.最优表中存在非基变量的检验数为零 D.可行解集合有界 答案：B
运筹学大作业
运筹大作业
文件说明
rosenbrock.m：函数的matlab符号函数实现。
naive_opt.m：负梯度法
F_R.m：Fletcher-Reeves共轭梯度法
P_R.m：Polak-Ribiere共轭梯度法
B_S.m：Beale-Sorenson共轭梯度法
程序说明
四个程序遵循相同的架构。
对于符号函数，matlab有gradient函数可以求符号函数的梯度，norm函数可以求符号函数的范数。
使用sub可以将符号函数转为数值函数。
之后严格遵循算法流程代入值求解即可。
调用方法
四个程序都不带参数。
直接调用即可获得图像，返回值为double的数值解。
求解结果
负梯度法
共迭代377次，最后结果为(0.999999999906938, 0.999999999790556)（数值解）
Fletcher-Reeves共轭梯度法
（第三个点因为其log为-inf因此无法被画出）
共迭代2次，最后结果为(1.000000000000001, 1.000000000000001)
Polak-Ribiere共轭梯度法
（第三个点因为其log为-inf因此无法被画出）
共迭代2次，最后结果为(1.000000000000001, 1.000000000000001)
Beale-Sorenson共轭梯度法
（第三个点因为其log为-inf因此无法被画出）
共迭代2次，最后结果为(1.000000000000001, 1.000000000000001)。
《运筹学》_习题_线性规划部分练习题及_答案
一、思考题1. 什么是线性规划模型，在模型中各系数的经济意义是什么？2. 线性规划问题的一般形式有何特征？3. 建立一个实际问题的数学模型一般要几步？4. 两个变量的线性规划问题的图解法的一般步骤是什么？5. 求解线性规划问题时可能出现几种结果，那种结果反映建模时有错误？6. 什么是线性规划的标准型，如何把一个耳非标准形式的线性规划问题转化成标准形式。
7. 试述线性规划问题的可行解、基础解、基础可行解、最优解、最优基础解的概念及它们之间的相互关系。
8 试述单纯形法的计算步骤，如何在单纯形表上判别问题具有唯一最优解、有无穷多个最优解、无界解或无可行解。
9. 在什么样的情况下采用人工变量法，人工变量法包括哪两种解法？10 .大M 法中，M 的作用是什么？对最小化问题，在目标函数中人工变量的系数取什么？最大化问题呢？ 11. 什么是单纯形法的两阶段法？两阶段法的第一段是为了解决什么问题？在怎样的情况下，继续第二阶段？二、判断下列说法是否正确。
1. 线性规划问题的最优解一定在可行域的顶点达到。
2. 线性规划的可行解集是凸集。
3. 如果一个线性规划问题有两个不同的最优解，则它有无穷多个最优解。
4. 线性规划模型中增加一个约束条件，可行域的范围一般将缩小，减少一个约束条件，可行域的范围一般将扩大。
5. 线性规划问题的每一个基本解对应可行域的一个顶点。
6.如果一个线性规划问题有可行解，那么它必有最优解。
CT i A 07. 用单纯形法求解标准形式（求最小值）的线性规划问题时，与J对应的变量都可以被选作换入变量。
8 单纯形法计算中，如不按最小非负比值原则选出换出变量，则在下一个解中至少有一个基变量的值是负的。
9.单纯形法计算中，选取最大正检验数 k 对应的变量 x k 作为换入变量，可使目标函数值得到最快的减少。
10 . 一旦一个人工变量在迭代中变为非基变量后，该变量及相应列的数字可以从单纯形表中删除，而不影响计算结果。
运筹学作业答案
运筹学作业答案《运筹学》作业答案作业一一、是非题：下列各题，你认为正确的打在每小题后的括号内打“√”，错的打“×”。
：1. 图解法与单纯形法虽然求解的形式不同，但从几何上理解，两者是一致的。
（√ ）2. 线性规划问题的每一个基解对应可行解域的一个顶点。
（╳ ）3. 如果线性规划问题存在最优解，则最优解一定可以在可行解域的顶点上获得。
（√ ）4. 用单纯形法求解Max 型的线性规划问题时，检验数Rj ＞0对应的变量都可以被选作入基变量。
（√ ）5. 单纯形法计算中，如果不按最小比值规划选出基变量，则在下一个解中至少有一个基变量的值为负。
（√ ）6. 线性规划问题的可行解如为最优解，则该可行解一定是基可行解。
（╳ ）7. 若线性规划问题具有可行解，且可行解域有界，则该线性规划问题最多具有有限个数的最优解。
（╳ ）8. 对一个有n 个变量，m 个约束的标准型线性规划问题，其可行域的顶点数恰好为m nC个。
（╳）9. 一旦一个人工变量在迭代中变为非基变量后，该变量及相应列的数字可以从单纯形表中删除，而不影响计算结果。
（√）10. 求Max 型的单纯形法的迭代过程是从一个可行解转换到目标函数值更大的另一个可行解。
（√）二、线性规划建模题：1.某公司一营业部每天需从A 、B 两仓库提货用于销售，需提取的商品有：甲商品不少于240件，乙商品不少于80台，丙商品不少于120吨。
已知：从A 仓库每部汽车每天能运回营业部甲商品4件，乙商品2台，丙商品6吨，运费200元/每部；从B 仓库每部汽车每天能运回营业部甲商品7件，乙商品2台，丙商品2吨，运费160元/每部。
问：为满足销售量需要，营业部每天应发往A 、B 两仓库各多少部汽车，并使总运费最少？解：设营业部每天应发往A 、B 两仓库各x 1，x 2部汽车,则有：12121212min 200160472402280621200(1,2)j W x x x x x x x x x j =++≥??+≥??+≥??≥=?2．现有一家公司准备制定一个广告宣传计划来宣传开发的新产品，以使尽可能多的未来顾客特别是女顾客得知。
运筹学1至6章习题参考答案
，运筹学1至6章习题参考答案第1章 线性规划工厂每月生产A 、B 、C 三种产品 ,单件产品的原材料消耗量、设备台时的消耗量、资源限量及单件产品利润如表1－23所示．310和130.试建立该问题的数学模型,使每月利润最大．【解】设x 1、x 2、x 3分别为产品A 、B 、C 的产量，则数学模型为123123123123123max 1014121.5 1.2425003 1.6 1.21400150250260310120130,,0Z x x x x x x x x x x x x x x x =++++≤⎧⎪++≤⎪⎪≤≤⎪⎨≤≤⎪⎪≤≤⎪≥⎪⎩ 建筑公司需要用5m 长的塑钢材料制作A 、B 两种型号的窗架．两种窗架所需材料规格及数量如表1－24所示：【解设x j （j =1,2,…，10）为第j 种方案使用原材料的根数，则 （1）用料最少数学模型为10112342567368947910min 28002120026002239000,1,2,,10jj j Z x x x x x x x x x x x x x x x x x x j ==⎧+++≥⎪+++≥⎪⎪+++≥⎨⎪+++≥⎪⎪≥=⎩∑ （2）余料最少数学模型为（2345681012342567368947910min 0.50.50.528002120026002239000,1,2,,10j Z x x x x x x x x x x x x x x x x x x x x x x x x j =++++++⎧+++≥⎪+++≥⎪⎪+++≥⎨⎪+++≥⎪⎪≥=⎩某企业需要制定1～6月份产品A 的生产与销售计划。
已知产品A 每月底交货，市场需求没有限制，由于仓库容量有限，仓库最多库存产品A1000件，1月初仓库库存200件。
1～6月份产品A 的单件成本与售价如表1－25所示。
（2）当1月初库存量为零并且要求6月底需要库存200件时，模型如何变化。
西工大作业集-运筹学作业集
第一章测试题1:某工厂用三种原料Q1,Q2 ,Q3生产三种产品P1,P2,P3。
已知单位产品所需原料数量如下表所示，试制订出利润最大的生产计划。
答案该问题的数学模型为2:用图解法求解下列LP问题,并指出各问题是否具有唯一最优解，无穷多最优解,无界解或无可行解。
答案(1)有唯一最优解Z=3，x1=1/2，x2=0(见图(a))(2)无可行解(见图(b))(3)有无穷多最优解(4)无可行解3:将下列LP问题变换成标准型，并列出单纯形表:答案(1)令，再引入变量x6，x7，原问题化为标准型初始单纯形表为(2)令，再引入变量x7，x8，x9，则原问题的标准形式为初始单纯形表为4:某线性规划问题的约束条件是问变量x2, x4而所对应的列向量A1，A2是否构成可行基?如果是，写出B，N，并求出B所对应的基本可行解。
答案可以构成，B对应的基本可行解为5:对于下面的线性规划问题，以B = (A2，A3,A6)为基写出对应的典式。
答案6:用单纯形法求解下列LP问题。
答案(1)首先将问题化为标准形式此时取基变量x3 ,x4 ,x5,然后进行如下迭代由上表可得时最优值为16，时最优值仍为16。
由于可行解都是凸集，凸集中的两相异的最优解，的连线上的点都是最优解，即都为最优解,其最优值为16。
(2)首先将问题化为标准型初始表取基变量为x3，x4，单纯形表如下从初始的单纯形表可以看出，检验数为正数的第二列中不含有正数,故此问题无最优解。
7:用两阶段法求解下列LP问题:答案(1)因为系数矩阵中已包含一个单位向量,所以第一阶段的只要增加两个人工变量x5，x6，所得辅助问题为进行如下表的迭送按单纯形法迭代关于目标函数G的检验数都不是正数，因此辅助问题LP的最优解为X =(0,8/5,18/5,0,41/5,0)T,其最优值G = 41/5 > 0，所以原问题无解。
8:写出下列LP问题的对偶问题。
答案对偶问题为9:用对偶单纯形法求解下面LP问题答案引进剩余变量x4，x5≥0，将不等式约束化为等式约束对其应用单纯形迭代如下此时b>0,故原问题的最优解为X = (2/13，7/13,0,0)T,其最优值为9/13。
大学运筹学大作业【精选】
1.1管理部门的目标是什么?
我们认为企业是以‘利润最大化’作为企业管理目标。利润从一定程度上反映出 企业经济效益的高低和对社会贡献的大小。同时，利润是企业补充资本，扩大经 营规模，实现再生产的源泉。因此，以追逐利润最大化作为企业管理的目标。 而以利润最大化为企业管理目标也有许多优点。首先，它反应了当期经营活动中 投入与产出对照的结果，在一定程度上体现了企业经济效益的高低。利润最大化 对企业的投资者、债权人和经营者和职工都是有益的，有利于企业整体经济效益 的提高。再而，利润这个指标在实际利用方面比较简便。利润额直观、明确，容 易计算，便于分解落实，大多数职工都能理解。人人都有逐利的心理，以利润最 大化为管理目标能使员工等受益者更卖力地工作，更有激情，提高了企业效率， 使企业上下一派生机勃勃。 为了达到‘利润最大化’的管理目标，我们可以通过进行经济核算，加 强管理，改进技术，提高劳动生产率，降低产品成本等方式实现目 标。
目标函数的确定
用收益减去人工成本，再减去原料成本，可得目标函数： maxZ={(4-2.52)*(X1+X4)/18+(4.5-
3.18)*(X2+X5)/20+(3.8-1.95)*(X3+X6)/25}{(X1+X2+X3)*9.32/100+(X4+X5+X6)*5.18/100}; 化简整理得：
约束3 1
0
0
1
0
0
0
<=
14400 000
约束4 0
1
0
0
1
0
53333 3.3
<=
10000 00
约束5 0
0
1
0
129319-运筹学-作业-HW_12++Filatoir+Case
Filatoi Riuniti The Italian textile industry The northern regions ofItaly are the heartlands of the Italian textile industry,providing textile products for many great Italian (and non-Italian) fashion houses.The industry's history reaches back to Roman times when Biella (a town in thenorthwestern Alps) was a center for wool production. Most of today's companieswere founded at the turn of the century and have grown and thrived despite aseries twentieth century catastrophes, including the great Depression (whichcrushed the silk industry), World War II, and the flash floods of 1968 (whichdestroyed many of Biella's mills). The families that run the wool-producingcompanies, many of whose ancestors had worked the land for centuries, havealways come back with great competitive spirit.Sales in the entire Italian textile and apparel industry were about $50 billion in 1994 with exports of $23 billion, according to Italy's National Statistics Institute. The entire textile sector employs over 700,000 people. Italy's largest export markets in the textiles sector are Germany, France, Great Britain, the U.S., and Japan. In 1994, over 1,200 companies exported textile products to the U.S. totaling about $625 million (over 6% of Italy's overall textile exports).The major Italian textile companies are fighting hard to stay at the forefront of the world textile market, and the battles are numerous. Competition from east Asia, a tug-of-war with China over silk and cashmere exports, lack of sufficient qualified employees, and a fast-moving fashion industry (which requires the mills to create and deliver rapidly) are among the new challenges facing the industry. In the face of these challenges, many Italian textile firms are committed to making massive investments in state-of-the-art machinery and in research into new finishings.Italian manufacturers are confident that Italian textile producers will always have an edge on the competition because of the high concentration of dyeing, finishing, and printing specialists i n Italy. “Italy’s textile sector is unique because there is this microcosm of small companies who are very creative and always up-to-date on the latest trends. They provide a constant stimulus and an endless source of ideas for manufacturers. In the end, this means continued creativity and flexibility of the industry”, says industry spokesperson Enrico Cerruti. “The trump card for the Italian textile industry is its cultural tradition and human resources… You can copy technology, you can copy design, but you can't duplicate the people here. It takes 100 years to build that and it's our greatest advantage.”How cotton yarn is producedPrior to the Industrial Revolution textiles were spun by hand using aspinning wheel (Figure 1).Today most commercial yarns are produced in textile mills, and althoughthe tools and techniques used are different from hand spinning, most of theprocesses are still the same. Most yarns are spun from staple fibers madeusing one of three systems: the Cotton Process, the Woolen Process, or theWorsted Process. These processes vary only slightly from each other butthey all include the three basic processes of preparation, spinning andfinishing. The Cotton Process is used to spin cotton fibers between 3/4" and 2" inlength. Synthetic fibers can also be blended with the cotton to form blendedyarns. Likewise the Worsted and Woolen Processes are used to spin wool fibers and wool blends. The cotton process method employs the following steps:Figure 1 - Old spinning wheel Filatoi RiunitiSpinningPreparation FinishingFigure 2 - The main steps of the cotton yarn production1. Preparation:A.Opening & Loosening. Upon arrival at the mill the cotton bails are opened and loosened. This helpsseparate and clean the cotton before it is fed into carding machines. Impurities such as rocks, coins, and other solid objects (there are stories about bullets found in the raw cotton!) are removed.B.Carding. Carding machines further loosen and separate the cotton fibers by passing the fibers betweentwo rotating metal drums covered with wire needles. This aligns the cotton in a thin web of parallel fibers, which is formed into a rope-like strand called a sliver. The sliver is collected in a sliver can in preparation for roving. For high quality yarns the sliver is combed after carding to make the fibers more parallel and to remove smaller fibers.C.Drawing and Roving. Slivers are then drawn out, blending the fibers and making them more parallel.No twist is added to the sliver during drawing, but several slivers can be blended together. Slivers can go through multiple drawings for further parallelization and blending. Drawn out slivers are then fed to the roving frame where they are drawn further while a slight twist is added. The roving strands are collected in cans and fed to the spinning machine.2. Spinning and Winding. The spinning machine draws out the roving strand, adds twist and winds theyarn onto bobbins. The slivers, which are relatively short segments of cotton, are twisted against each other into a long, continuous, strong yarn. Multiple bobbins of yarn are then wound onto larger spools called cheeses. Now the yarn is ready for texturing and dying and finally weaving into fabric.3. Re-spooling/texturizing. Some finer qualities of thread require an additional step of spooling or passagein a gas-saturated room. This further step is necessary to chemically eradicate remaining impurities or to impart aesthetic and/or functional properties to the yarn.Filatoi RiunitiFilatoi Riuniti is a family-owned spinning mill located in Piemonte, a region in northwest Italy. Filatoi Riuniti produces cotton yarn, which is one the raw materials (together with wool) that is used to produce the fabrics that are then cut and sewn by Italian stylists into the worldwide famous Italian clothes.Filatoi Riuniti was founded at the beginning of the 19th century by an Italian entrepreneur who built a spinning machine and began to produce cotton yarn on a low scale in the barn in his backyard. The company grew tremendously in the period 1880-1920. After World War I, Filatoi Riuniti benefited from the development of certain machine tool industries and replaced and otherwise modernized most of their machine capacity. Unfortunately, after World War II, Italy had to rebuild its economy almost from scratch; Filatoi Riuniti struggled to remain solvent and it was many years before they had the capital needed to replenish and then expand their production facilities.Filatoi Riuniti’s current situationFor the past twenty years Filatoi Riuniti had followed a strategy of rapid expansion to increase market share. As the Italian fashion industry grew in both stature and revenues, demand for Filatoi Riuniti’s products increased, and management expected this pattern to continue. Filatoi Riuniti invested aggressively in new machine capacity to meet the demands of its current clients as well as to serve anticipated new customers. Filatoi Riunitibegan to experience financial problems in the early 1990’s, and had to halt this expansion strategy due to an economic recession in Italy that was accompanied by a strengthening of the Italian lira:“Recession has hit the European textile industry from all sides…cotton yarn weavers are having anexceptionally hard time”… “Some companies collapsed only to re-emerge shrunk to a third of theirinitia l size” (Daily News Record, March 29, 1993). “The '80s…were very good in both sales andprofitability, but there are market cycles and now we're facing a descending cycle…The decrease inoverall consumption of apparel, new competition, market problems in the U.S. and the strength ofthe lira against the deutsche mark and yen present uncertain signs for the future.” (Daily NewsRecord, December 5, 1990).In the face of disappointing demand in the early 1990’s, it might seem obvious that Filatoi Riuniti shoul d have adjusted its production and its production capacity downward. Such down-sizing is not so easy to do in Italy’s unionized industries. Italian trade unions achieved substantial gains in power and stature during the 1970’s and 1980’s. As a result, it was (and still is) very difficult for a company to lay off employees solely due to a fall in sales, unless the company could prove to the government that it was in severe financial distress. If the company could prove such financial distress, they could lay off employees (who would then be paid a minimal wage by the government). However, companies are very hesitant to take this course of action, because being labeled as “financially distressed” makes it almost impossible to buy from suppliers on account or to borrow money at low interest rates.Filatoi Riuniti’s management chose not to lay off any employees in the early 1990’s. The resulting cash-flow problems severely limited the funds Filatoi Riuniti could use to increase their Spinning capacity (the second step). This was most unfortunate, since Filatoi Riuniti had previously expanded their Preparation machine capacity (first step) and Finishing capacity (third step), and their Spinning capacity needed to be expanded to fully balance their production facilities.Out-sourcing Spinning ProductionWith the recent upturn in the Italian economy, demandfor Filatoi Riuniti’s production is strong again, butFilatoi Riuniti’s Spinning machine capacity isinsufficient to meet its production orders, and theydecided six months ago to outsource part of theSpinning production to six local family-owned spinningmills named Ambrosi, Bresciani, Castri, De Blasi,Estensi, and Giuliani (all six mills operate primarily forlocal markets such as Filatoi Riuniti). Filatoi Riuniticurrently processes the raw material (combed cotton),and then sends part of the combed cotton to these sixspinning mills for the spinning step. The semi-finishedproduct is then returned to Filatoi Riu niti’s productionFigure 3 - Industrial spinning machinesite where it is finished and stored until delivery tocustomers. The local mills charge higher prices for spinning finer yarns and for this reason Filatoi Riuniti has decided to spin as much as possible of the finer yarn sizes entirely in-house and to outsource only the spinning of low-end (coarser) yarns. Last month, for example, Filatoi Riuniti faced a total demand of 104,500 Kg of cotton and they outsourced 31,530 Kg of the spinning of the low-end sizes. Exhibit 1 shows the production schedule and the prices charged by the six mills for February.Milan Consulting Group Inc.Last fall, Giorgio Armeni was named the new CEO of Filatoi Riuniti. Faced with a myriad of challenging issues at Filatoi Riuniti and sensing that Filatoi Riuniti’s internal management team was not up to the task of radical internal re-engineering, he decided to hire Milan Consulting Group Inc. (MCG) to help address some of the company’s ongoing problems. MCG’s team was composed of Maurizio Donato, a junior partne r, and RobertoBenello and Sofia Cominetti, two young associates in their first engagements. The three consultants spent four days at Filatoi Riuniti touring the production facilities, reviewing operations, interviewing managers, and studying mounds of data. After a weekend of hard thinking back in Milan, Maurizio Donato started the next Monday’s project meeting with his two young associates with the words: “Our first priority is to find ways to reduce the costs that our client faces which are jeopardizing the future of the factory and the jobs of the almost 200 employees. Our goal is to come up with smart, workable ideas, e subito!” He then outlined what he thought were four areas for obvious cost reduction:1. reducing machine down-time through improved inspection strategies,2. differential scheduling of machine maintenance,3. different management of overtime on production shifts, and4. improved outsourcing strategies from the six local spinning mills.Roberto and Sofia immediately went to work on the analysis of these four proposals. They found that the total expected cost savings from the first three proposals would probably amount to roughly $200,000 per year. This was not a particularly large sum of money in comparison to Filatoi Riuniti’s sales of $15 mill ion (although it easily justified their consulting fees!). However, when they started to work on the fourth proposal, they immediately saw the potential for very large savings through more optimal outsourcing strategies with the six local spinning mills.Optimizing the Outsourcing of Spinning ProductionFilatoi Riuniti produces four different sizes of yarn (coarse, medium, fine and extra fine) and there is an autonomous demand for each of the four sizes. Filatoi Riuniti can prepare enough raw cotton to meet their total demand, but they lack sufficient machine capacity to spin the four sizes of yarn, and as discussed above, they have been outsourcing part of the spinning production to six local mills: Ambrosi, Bresciani, Castri, De Blasi, Estensi and Giulia ni. Exhibit 1 shows February’s spinning production schedule. Of the total demand of 104,500 Kg of yarn, 31,530 Kg were outsourced to the six local mills. All of the spinning of Coarse yarn was outsourced (29,000 Kg), and 11% of the spinning of Medium yarn was outsourced. The exhibit also shows the prices charged by the six mills to Filatoi Riuniti. The head of the purchasing department at Filatoi Riuniti said “We spin the finer sizes in-house and outsource the rest of the work. We outsource each yarn size to the lowest-price mill and then meet demand with the next-lowest-price mill.” Roberto and Sofia thought that this outsourcing strategy could easily lead to sub-optimal outsourcing decisions, since outsourcing decisions were optimized only one at a time as opposed to optimizing all outsourcing simultaneously. In order to analyze the potential savings from optimizing the outsourcing of spinning production, they started to work with the client to identify the decision variables, the constraints, and the objective function to optimize.Decision variables. Given the amount of each yarn size that Filatoi Riuniti needs to deliver to meet demand, the problem was how to allocate spinning production (both at Filatoi Riuniti and at the six local mills) in order to minimize costs. The decision variables of the optimization model are denoted X ij, which represents the amount of yarn of size i that the company j would be assigned to produce. In this context, i=1,2,3 and 4 means “extra fine”, “fine”, “medium” and “coarse”, respectively. Similarly, j=A, B, C, D, E, F, G mean Ambrosi, Bresciani, Castri, De Blasi, Estensi, Filatoi Riuniti, and Giuliani. Each X ij must of course be non-negative because none of the mills can produce negative amounts of spun yarn!Variable Costs of Production. Roberto and Sofia knew the prices charged to Filatoi Riuniti by the six local mills (see Exhibit 1). For internal purposes, they also needed to know Filatoi Riuniti’s internal production costs in order to determine how much of each yarn size should optimally be produced internally versus externally. After a couple of days spent with the plant managers and the chief accountant, they came up with a fair estimate of the production cost for each of the four yarn sizes. See the table “Cost of Product” in Exhibit 2. The two blanks in the table indicate that Ambrosi and De Blasi cannot produce extra fine yarn.Transportation costs. The yarn that is spun by the six local mills needs to be transported from Filatoi Riuniti to the mills for spinning and then be transported back to the production plant of Filatoi Riuniti in order to refineit and store it prior to delivery to customers. Sofia realized that they needed to obtain accurate data on transportation costs. One of the operations m anagers explained to her: “We have an agreement with a local truck company which takes care of all the transportation. The contract with the truck company is very simple. They charge a fixed amount per kilometer per unit volume”. Each product has a diff erent density and therefore takes up a different volume per Kg. One Kg of finer product is more dense and so is less expensive to transport on a per Kg basis. Of course, each local mill is located at a different distance from Filatoi Riuniti. Armed with the contract with the truck company, a road map with the location of the six local mills, and product specification data, Sofia was able to estimate the transportation cost per Kg of each product for all the local mills. These numbers are shown in the ta ble “Cost of Transportation” in Exhibit 2. For example, it costs $0.01 per Kg per Km to transport fine yarn, and the round trip distance from Filatoi Riuniti to the Giuliani mill is 2⨯25=50 Km: therefore, the table shows that it costs (0.01⨯50)=$0.50 to transport one Kg of fine yarn to Giuliani and back.Resource consumption. Another important task was to understand the actual spinning machine production capacity of the six local mills and of Filatoi Riuniti itself. During the time spent with the plant manager, Roberto learned that production capacity is measured in machine hours per month and each product size requires a different amount of machine hours per Kg of product. He spent some more time with the plant engineer trying to estimate the six local mills’ capacity in terms of machine hours per month and their production rate in terms of hours needed to produce one Kg of a given product size. Because each mill has different types of machines in different configurations, the number of machine hours required to produce one Kg of product differs among the mills. After a full day of work and very many telephone calls, fax and email messages, Roberto and the plant engineer produced a table containing the production capacity and production rate per product for each of the six mills plus Filatoi Riuniti itself. These capacity and production rate numbers are shown in the two tables “Machine Hours Required for Production” and “Production Capacity” in Exhibit 2. For example, at the Bresciani mill, it takes 0.70 hours to produce one Kg of extra fine yarn and there are at total of 3,000 machine hours per month available.Product Demand. After talking to the marketing and sales manager at Filatoi Riuniti, Sofia estimated the demand for the four spun yarn sizes for March, which is shown in the table “Demand to Meet” in Exhibit 2. Armed with all of this data, Roberto and Sofia felt that they had enough information to solve for the outsourcing production strategy that would minimize the costs of producing spun yarn.Exhibit 1Outsourcing Production Schedule and Prices Charged to Filatoi Riuniti - February PRODUCTION SCHEDULE FOR FEBRUARY PRICES CHARGED TO FILATOI RIUNITItotal104,500Exhibit 2Production schedule, costs, and constraints - MarchDECISION VARIABLES MACHINE HOURS REQUIRED FOR PRODUCTION PRODUCTIONCAPACITY(Machine hoursCOST OF PRODUCTION COST OF TRANSPORTATION($/Kg/Km)DEMAND TO MEETAssignment1. Formulate Filatoi Riuniti’s purchasing problem for the coming month (March):A.Write down the formula for the objective function of your model.B.Your model must have a capacity constraint for each local spinning mill. Write down the capacityconstraints for the Ambrosi mill, for example.C.Filatoi Riuniti must meet demand for each of the four sizes of yarn. Your model must have a constraintfor the demand for each of the four sizes of yarn. Write down the constraint for the demand for extra fine yarn, for example.2. Filatoi Riuniti should obviously consider increasing its spinning machine capacity. They could slightlyexpand the production capacity of the existing machines by renting an upgrade. This would increase their spinning production capacity by 600 hours/month. The monthly rental cost is $1,500/month. Would you recommend that they rent the upgrade?(try to answer this question without re-optimizing your model.)3. Alternatively, Filatoi Riuniti could increase its spinning machine capacity by renting another spinningmachine for the production of only medium size yarn, for a monthly rental cost of $3,000. The machine has a production capacity of 300 hours per month (the machine would run at the same rate of 0.425 hours/Kg). Suppose that the estimated production cost of running this machine is less than for Filatoi Riuniti’s existing machines, and is estimated to be $5.70/Kg (as opposed to $11.40/Kg for their existing machines according to Exhibit 2). Would you recommend that Filatoi Riuniti rent the machine?(try to answer this question without re-optimizing your model.)4. A new client is interested in purchasing up to 6,000 Kg/month of medium size yarn. What is the minimumprice that Filatoi Riuniti should quote to this new client? Would it be a fixed price per Kg? Which additional question(s) might you ask this client? (In answering this question, assume that Filatoi Riuniti has not decided to expand its spinning machine capacity, and that Filatoi Riuniti does not want to change the prices that they currently charge their existing clients.)5. Your outsourcing production strategy optimization model is based in part on the prices charged by thelocal mills to Filatoi Riuniti and on an estimate of Fila toi Riuniti’s internal production costs. The plant manager, the accounting department, and you estimate that Filatoi Riuniti’s internal production costs could vary within a 5% range of the figures shown in Exhibit 2.Would your recommendations change in the extreme cases? Why or why not?6. You estimate that the production capacity of one of your local mills, De Blasi, could vary within a 20%range of the figures shown in Exhibit 2.Would your recommendations change in the extreme cases? Why or why not?7. Suppose that you present your proposed outsourcing plan to the owners of the Ambrosi mill. Theycomplain to you that their mill cannot easily produce fine size yarn; in fact they presently can only produce medium and coarse size yarn, and they would incur substantial one-time set-up costs to ramp up for the production of fine size yarn. However, the optimal solution of the model indicates that it would be in Filatoi Riuniti’s interests for the Ambrosi mill to produce fine size yarn. The owners want to maintai n good business relations with Filatoi Riuniti, but they do not want to bear the full cost of ramping up for production of fine yarn. The contracts that Filatoi Riuniti currently has with its customers will not expire for at least other 12 months.Up to what amount would you be willing to share the one-time set-up costs for production of fine yarn with the owners of the Ambrosi mill?8. Suppose that you find out that one of the local mills, Giuliani, has the possibility of running an overtimeshift (which would double their capacity) by paying its workers only 13% more the normal wage (it is a family owned business). You know that the workers’ salaries contribute to approximately 50% of the prices that the Giuliani mill charges Filatoi Riuniti for spinning yarn. The transportation cost componentof the objective function would not change, of course. Modify the model in order to take into account this possibility and re-optimize.Does the optimal solution change? Why?[Helpful modeling hint: think of the “overtime” part of this mill as a new mill with higher product costs]。
运筹学1至6章习题参考答案
运筹学1至6章习题参考答案第1章 线性规划1.1 工厂每月生产A 、B 、C 三种产品 ,单件产品的原材料消耗量、设备台时的消耗量、资源限量及单件产品利润如表1－23所示．310和130.试建立该问题的数学模型,使每月利润最大．【解】设x 1、x 2、x 3分别为产品A 、B 、C 的产量，则数学模型为123123123123123max 1014121.5 1.2425003 1.6 1.21400150250260310120130,,0Z x x x x x x x x x x x x x x x =++++≤⎧⎪++≤⎪⎪≤≤⎪⎨≤≤⎪⎪≤≤⎪≥⎪⎩ 1.2 建筑公司需要用5m 长的塑钢材料制作A 、B 两种型号的窗架．两种窗架所需材料规格及数量如表1－24所示：【解设x j （j =1,2,…，10）为第j 种方案使用原材料的根数，则 （1）用料最少数学模型为10112342567368947910min 28002120026002239000,1,2,,10jj j Z x x x x x x x x x x x x x x x x x x j ==⎧+++≥⎪+++≥⎪⎪+++≥⎨⎪+++≥⎪⎪≥=⎩∑L （2）余料最少数学模型为2345681012342567368947910min 0.50.50.52800212002*********0,1,2,,10j Z x x x x x x x x x x x x x x x x x x x x x x x x j =++++++⎧+++≥⎪+++≥⎪⎪+++≥⎨⎪+++≥⎪⎪≥=⎩L1.3某企业需要制定1～6月份产品A 的生产与销售计划。
已知产品A 每月底交货，市场需求没有限制，由于仓库容量有限，仓库最多库存产品A1000件，1月初仓库库存200件。
1～6月份产品A 的单件成本与售价如表1－25所示。
（2）当1月初库存量为零并且要求6月底需要库存200件时，模型如何变化。
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第一章测试题1:某工厂用三种原料Q1,Q2 ,Q3生产三种产品P1,P2,P3。
已知单位产品所需原料数量如下表所示，试制订出利润最大的生产计划。
答案该问题的数学模型为2:用图解法求解下列LP问题,并指出各问题是否具有唯一最优解，无穷多最优解,无界解或无可行解。
答案(1)有唯一最优解Z=3，x1=1/2，x2=0(见图(a))(2)无可行解(见图(b))(3)有无穷多最优解(4)无可行解3:将下列LP问题变换成标准型，并列出单纯形表:答案(1)令，再引入变量x6，x7，原问题化为标准型初始单纯形表为(2)令，再引入变量x7，x8，x9，则原问题的标准形式为初始单纯形表为4:某线性规划问题的约束条件是问变量x2, x4而所对应的列向量A1，A2是否构成可行基?如果是，写出B，N，并求出B所对应的基本可行解。
答案可以构成，B对应的基本可行解为5:对于下面的线性规划问题，以B = (A2，A3,A6)为基写出对应的典式。
答案6:用单纯形法求解下列LP问题。
答案(1)首先将问题化为标准形式此时取基变量x3 ,x4 ,x5,然后进行如下迭代由上表可得时最优值为16，时最优值仍为16。
由于可行解都是凸集，凸集中的两相异的最优解，的连线上的点都是最优解，即都为最优解,其最优值为16。
(2)首先将问题化为标准型初始表取基变量为x3，x4，单纯形表如下从初始的单纯形表可以看出，检验数为正数的第二列中不含有正数,故此问题无最优解。
7:用两阶段法求解下列LP问题:答案(1)因为系数矩阵中已包含一个单位向量,所以第一阶段的只要增加两个人工变量x5，x6，所得辅助问题为进行如下表的迭送按单纯形法迭代关于目标函数G的检验数都不是正数，因此辅助问题LP的最优解为X =(0,8/5,18/5,0,41/5,0)T,其最优值G = 41/5 > 0，所以原问题无解。
8:写出下列LP问题的对偶问题。
答案对偶问题为9:用对偶单纯形法求解下面LP问题答案引进剩余变量x4，x5≥0，将不等式约束化为等式约束对其应用单纯形迭代如下此时b>0,故原问题的最优解为X = (2/13，7/13,0,0)T,其最优值为9/13。
10:把线性规划问题记为p。
(1)用单纯形法解p；(2)写出P的对偶D；(3)写出P的对偶松紧条件，并利用它们解对偶D。
通过计算P和D的S优值，检查你的答案。
答案(1)(2)(3)略11:考虑上一题的线性规划P，在下述的每一种情况下，试用解问题P所得到的最优单纯形表继续求解。
答案(1)(2)略第二章测试题1:有一推销员，从城市v0出发，要遍访城市v1，v2，…，v n各一次，最后返回v0。
已知从v i到v j 的旅行费为c ij，问他应按怎样的次序访问这些城市，使得总旅行费用最少？试建立这个问题的数学模型。
答案对每一对城市v i,v j,指定一个变量X U。
令X U=1，若推销员决定从v i直接进人v j；否则X U=0。
则该问题的数学模型为2:用割平面法求解下列整数规划问题答案引入松弛变量x3，x4，令Z`=-Z，将原问题标准化得用单纯形法求得的最后一张单纯形表为得最优解为由于该解不是整数解，所以生成割平面。
第一行生成的个平面条件为:可进一步求得该割平面方程为引入松弛变量s1将其化为标准形式为加到最优单纯形表中,再用对偶单纯形求解得到下表得到新的最优解由于得到这个新的解仍不是整数解，所以取第二行为割平面的约束条件可进一步求得该割平面方程为引入松弛变量s2将其化为标准形式为加到最优表中，再用对偶单纯形法求解得得新的最优解为X2=(4,1)T，Z2=14。
此时得到的解为最优解故停止。
综上原问题的最优解为X=(4,1)T,Z=143:用分枝界定法求解下列问题：答案解除整数约束,用图解法或单纯形法得此问题的最优解为X0=(3。
6,2。
8)T,Z0=32。
8因为3 < x1< 4,上述问题可分解为两个子问题由图解法或单纯形法可得这两个问题的最优解为因为Z1<Z2故原问题的最优解为4:用简单隐枚举法求解下列0-1规划问题：答案(1)x=(1,0,1)T,Z=8；(2)x=(0,0,1,1,1)T,Z=6。
第三章测试题1:两个水厂A1,A2将自来水供应三个小区B1,B2,B3,每天各厂的供应量与各小区的需求量以及各水厂调运到各小区的供水单价如下表。
试用表上作法给出下表的最优供水方案。
答案先用最小元素法确定初始方案。
这种方法的基本思想是”就近供应”,即从运输问题数据表中找出最小运价c ij,选择这个最小运价在运输方案表中对应位置x ij作为数字格(基变量），在格内填写允许取得的最大数。
每次总选择剩余运价中的最小c ij所对应的x ij作为数字格，并填写允许的最大数,直至得到初始方案,如下表下面用位势法判别其是否为最优解，即λij=z ij-c ij，可得λ12=2, λ23=-5因为λ12= 2>0，则取x12为进基变量。
调整量θ= min{90，50} = 50,所以x11为出基变量。
闭回路上奇数顶点增加50,偶数定点减少50。
调整后的方案如下表对调整后的方案表进行最优解的判别如下λ11=-2<0, λ23=-3<0则所有的检验数都小于0,所以此方案为最优方案。
最优方案的总运费为Z=50X6 + 120X4 + 160X7 + 40X5 = 21002:已知运输问题的产销平衡表和单位运价如下,试用表上作业法求其最优解。
答案最优方案与检验数表如下括号内为检验数.最小运输费用为:Zmin=335. 3:下表给出了三个产地和四个销地的某种物资的供需量及产销的运价,试求出运费最少的运输方案。
答案最优方案如下表Zmin= 5X3 + 10X2 + 5X1 + 5X2 + 10X3 = 80第四章测试题1:写出下图的关联矩阵和邻矩阵。
答案关联矩阵邻接矩阵2:分别用避圈法和破圈法求解下图的支撑树。
答案略3:分别用破圈法和避圈法求下面各图的最小支撑树。
答案4:已知网络如下图所示，求从s到t的最短路径及最短距离。
答案则最短路径为(s,3,4,t),最短距离为1+4+2=7 5:求解下图中v1至各点的最短距离与最短路径。
答案V1至各点最短距离为：v2(4),v3(7),v4(11),v5(7),v6(9),v7(11),v8(18) ,v9(13),v10(9),v11(14).6:求下面网络的最大流集。
答案最大流量为20。
第五章测试题1:求解具有下述系数矩阵的最小指派问题：答案最优指派方案为最优值分别为：48，21.2:求解具有下述系数矩阵的最大化指派问题:答案最优指派方案为最优值为：363:四个人完成四项工作，由于个人的技术专长不同，他们完成四项工作任务所获得收益如下表，且规定每人只能做一项工作，试求使总受益最大的分配方案。
答案最优指派方案为所得最大收益为：maxZ=10+9+11+13=434:分配甲，乙,丙，丁四人去完成A，B，C，D，E五项任务。
每个人完成各项任务的时间如下表所示。
由于任务数多余人数，故考虑：(1)任务E必须完成,其它四项可任选三项完成；(2)其中有一人完成两项，其它每人完成一项。
试分别确定最优分配方案,使完成的总时间最少。
答案(1)由于任务数多余人数,所以霜要有一名假想的人物，因为工作E必须完成，故设完成E的时间为M(M为非常大的数）,其余的假想时间为0，则用匈牙利算法如下所以最优解为：x12=x24=x35=1，其余的x ij=0，最少时间为105h.(2)由于所有的任务，都必须由甲乙丙丁完成，所以假想的人的效率应该对每项工作而言，都是完成它的最好的人，而不能假设为0值,所以构造的效率表如下：用匈牙利方法求解,可得最佳分配方案：甲—B，乙—C和D,丙—E，丁—A，共需要 131h.第六章测试题1:某企业有三种方案可供选择：方案S1是对原厂进行扩建；方案S2是对原厂进行技术改造；方案S3是建新厂,而未来市场可能出现滞销(E1）,一般(E2）和畅销(E3）三种状态，其受益如下表(单位:万元)。
试分别安以下决策准则确定最优方案:(1)悲观准则；(2)乐观准则；(3)折衷准则（乐观系数a=0.6)；(4)后悔值准则；(5)等可能性准则。
答案(1)根据悲观准则有故最优决策方案为S2.(2)根据乐观准则有故最优决策方案为S3.(3)根据折衷准则有故最优决策方案为S3.(4)根据后悔准则有后悔准则值表如下故最优决策方案为S l.(5)等可能准则有故最优决策方案为S l.2:某企业生产一种季节性商品。
当需求量为D 时，企业生产x件商品时获得的利润为：设D有五种可能的值:1 000件，2 000件,3 000件，4 000件和5 000件，并且他们出现的概率均为0。
2。
若企业追求最大的期望利润，那么最优生产为多少件？答案该问题收益矩阵如下表则由上表可知最优选择为生产4 000件。
3:某工程队承担一座桥梁的施工任务，由于夏季多雨，需停工三个月。
在停工期间该工程队可将施工机械搬走或留在原处。
若搬走,需运费1 800元。
如留在原处，一种方案是花500元筑护堤,防止河水上涨发生髙水位的侵袭,若不筑护堤，髙水位侵袭将损失10 000元。
发生洪水时，不管是否筑护堤,都将受到60 000元的损失。
根据历史资料，该地区夏季髙水位的概率是0.25,洪水的概率是0.02。
试用决策树找出最优方案。
答案故以不搬走并筑护堤最合算。
阶段性作业一1:已知线形规划问题要求:（1)化为标准形式；(2)若用两阶段法求解，写出其辅助问题并列出第一阶段的初始单纯行表。
答案(1)令,再引入松弛变量x4 ,x5.则原模型可化为：初始单纯形表为2:用图解法或单纯形法求解下列线形规划问题：答案(1)图解法则由图可知：当x1=4,x2=2时，此问题有最优解Z=-14(2)单纯形法增加松弛变量x3,x4 ,x5 ,x6化为标准型由上表可知所有的x1=4,x2=2,x3=x4=x5=0,x6=4，则最优解为Z=14.3:求解0 - 1规划的简单隐枚举法答案则最优解为：X=(0,0,1)T,Z0=5.4:已知某一运输问题的单位运价和产销表如下表所示，请给出两个最优方案。
答案下面给出三个最优方案5:用Dijkstra算法求下图由s到t最的最短路。
答案。