我整理糖果配比销售答案.doc

2011年河南理工大学公选课数学建模课程论文论文题目： 糖果配比销售问题讨论 姓 名学号专业班级1林玉珊311019020104安管10-1班 糖果配比问题讨论摘要 此问题属于一个优化问题，即在一系列的限制条件中寻找最优的方案。所以这里我们采用优化模型，优化模型是为了使在原材料供应量受到限制的前提下使得决策的问题达到最优，在本问题中，要求我们决定各品牌果仁糖的总周利润最大的前提下，需购进杏仁、核桃仁、腰果仁和胡桃仁的数量以及各果仁糖中果仁的配比问题。我们在这篇文章里假设购进的果仁全部配制成糖果销售，对问题进行优化，然后通过糖果制作及供应量限制的条件列出条件方程，然后用Lindo软件进行求解。在问题二中，我们通过四种情况来讨论当供货量增加10%时配比是否变化以及利润如何变化等。通过这种分析，可以为决策者找到一个最优方案，无论是在淡季还是在旺季，都可以找到最优的销售配比方案。关键词优化模型、周利润、销售量、果仁配比。一、问题的重述某糖果店出售三种不同品牌的果仁糖，每个品牌含有不同比例的杏仁、核桃仁、腰果仁、胡桃仁。为了维护商店的质量信誉，每个品牌中所含有的果仁的最大、最小比例是必须满足的，如下表所示：品牌含量需求售价/ 美元/kg普通腰果仁不超过20%0.89胡桃仁不低于40%核桃仁不超过25%杏仁没有限制豪华腰果仁不超过35%1.10杏仁不低于40%核桃仁、胡桃仁没有限制蓝带腰果仁含量位于30%50%之间1.80杏仁不低于30%核桃仁、胡桃仁没有限制每周商店从供应商处能够得到的每类果仁的最大数量和售价如下表：售价/ 美元/kg每周最大供应量/ kg杏仁0.452000核桃仁0.554000腰果仁0.705000胡桃仁0.5030001） 商店希望确定每周购进杏仁、核桃仁、腰果仁、胡桃仁的数量，使周利润最大，建立数学模型，帮助该商店管理人员解决果仁混合的问题。2） 若在圣诞周，豪华和蓝带品牌的销售量会增加，这时商店会让果仁供应量增加10%,试问在这种情况下混合配比是否改变，圣诞周利润会改变多少？请分情况说明。二、模型假设1.糖果店配置的各种品牌的糖果都能及时售出；2.糖果的售价和果仁的进价都保持不变；3.糖果店严格按照标准配置各种品牌的糖果；4.糖果店存在足够的周转资金用于进货；5.果仁供货商能及时足量供货。三、符号说明设P、H、L分别表示普通、豪华、蓝带品牌糖果的量，Py表示普通品牌的糖果中腰果仁的含量，Pu表示普通品牌的糖果中胡桃仁的含量，Pe表示普通品牌的糖果中核桃仁的含量，Px表示普通品牌的糖果中杏仁的含量，Hy表示豪华品牌的糖果中腰果仁的含量，Hu表示豪华品牌的糖果中胡桃仁的含量，He表示豪华品牌的糖果中核桃仁的含量，Hx表示豪华品牌的糖果中杏仁的含量，Ly表示蓝带品牌的糖果中腰果仁的含量，Lu表示蓝带品牌的糖果中胡桃仁的含量，Le表示蓝带品牌的糖果中核桃仁的含量，Lx表示蓝带品牌的糖果中杏仁的含量。4、 模型的建立与求解（1） 问题一1.1问题分析这个优化问题的目标是使商店的周利润最大，要做的决策是进行果仁混合配比，即每周分别购进多少杏仁、核桃仁、腰果仁、胡桃仁，该决策受到2个条件的限制：糖果中各果仁的含量、原料每周最大供应量。1.2模型的建立根据糖果配比要求有Py20%P，Pu40%P，Pe25%P，Hy35%H，Hx40%H，30%LLy50%L，Lx30%L这里Py+Pu+Pe+Px=PHy+Hu+He+Hx=HLy+Lu+Le+Lx=L将此三个等式依次代入上述不等式，并整理得到0.8Py-0.2Pu-0.2Pe-0.2Px00.4Py-0.6Pu+0.4Pe+0.4Px0-0.25Py-0.25Pu+0.75Pe-0.25Px00.65Hy-0.35Hu-0.35He-0.35Hx00.4Hy+0.4Hu+0.4He-0.6Hx0-0.7Ly+0.3Lu+0.3Le+0.3Lx00.5Ly-0.5Lu-0.5Le-0.5Lx00.3Ly+0.3Lu+0.3Le-0.7Lx0根据原料每周最大供应量，有Px+Hx+Lx2000Pe+He+Le4000Py+Hy+Ly5000Pu+Hu+Lu3000在约束条件中共有12个变量，为计算和叙述方便，分别用x1，x12表示。令x1=Py，x2=Pu，x3=Pe，x4=Pxx5=Hy，x6=Hu，x7=He，x8=Hxx9=Ly，x10=Lu，x11=Le，x12=Lx由此约束条件可以表示为：0.8x1-0.2x2-0.2x3-0.2x400.4x1-0.6x2+0.4x3+0.4x40-0.25x1-0.25x2+0.75x3-0.25x400.65x5-0.35x6-0.35x7-0.35x800.4x5+0.4x6+0.4x7-0.6x80-0.7x9+0.3x10+0.3x11+0.3x1200.5x9-0.5x10-0.5x11-0.5x1200.3x9+0.3x10+0.3x11-0.7x120x4+x8+x122000x3+x7+x114000x1+x5+x95000x2+x6+x103000x1，x120我们的目的是使周利润最大，即糖果的售价减去果仁的进价为最大。糖果售价为：0.89（x1+x2+x3+x4）普通品牌1.10（x5+x6+x7+x8）豪华品牌1.80（x9+x10+x11+x12）蓝带品牌果仁进价为：0.45（x4+x8+x12）杏仁0.55（x3+x7+x11）核桃仁0.70（x1+x5+x9）腰果仁0.50（x2+x6+x10）胡桃仁所以，目标函数为max z=0.89（x1+x2+x3+x4）+1.10（x5+x6+x7+x8）+1.80（x9+x10+x11+x12）-0.45（x4+x8+x12）-0.55（x3+x7+x11）-0.70（x1+x5+x9）-0.50（x2+x6+x10）=0.19x1+0.39x2+0.34x3+0.44x4+0.4x5+0.6x6+0.55x7+0.65x8+1.1x9+1.3x10+1.25x11+1.35x121.3模型求解可以用单纯形法求解此线性规划问题，在此我们采用Matlab软件求解，在Matlab命令窗口中输入如下命令：f=0.19,0.39,0.34,0.44,0.4,0.6,0.55,0.65,1.1,1.3,1.25,1.35;A=0.8,-0.2,-0.2,-0.2,zeros(1,8); 0.4,-0.6,0.4,0.4,zeros(1,8); -0.25,-0.25,0.75,-0.25,zeros(1,8); zeros(1,4),0.65,-0.35*ones(1,3),zeros(1,4); zeros(1,4),0.4*ones(1,3),-0.6,zeros(1,4); zeros(1,8),-0.7,0.3*ones(1,3); zeros(1,8),0.5,-0.5*ones(1,3); zeros(1,8),0.3*ones(1,3),-0.7; zeros(1,3),1,zeros(1,3),1,zeros(1,3),1; zeros(1,2),1,zeros(1,3),1,zeros(1,3),1,0; 1,zeros(1,3),1,zeros(1,3),1,zeros(1,3); 0,1,zeros(1,3),1,zeros(1,3),1,0,0;b=zeros(1,8),2000,4000,5000,3000;lb=zeros(12,1);ub=;x,fval=linprog(-f,A,b,lb,ub);x=x,format long;maxz=-fvalxR=x(4)+x(8)+x(12),eR=x(3)+x(7)+x(11),uR=x(2)+x(6)+x(10),yR=x(1)+x(5)+x(9)P=sum(x(1:4),H=sum(x(5:8),L=sum(x(9:12)求得Optimization terminated.x = Columns 1 through 8 1090.9 3000 1363.6 6.4125e-009 1.074e-008 1.5468e-008 3.2809e-008 4.6414e-008 Columns 9 through 12 2030.3 6.9177e-008 2636.4 2000maxz = 1.006969696933194e+004xR = 1.999999999994909e+003eR = 3.999999999729673e+003uR = 2.999999999962490e+003yR = 3.121212120747583e+003P = 5.454545453871487e+003H = 1.054311416334892e-007L = 6.666666666457737e+003结果表示：当x1，x12分别取1090.9、3000.0、1363.6、0.0、0.0、0.0、0.0、0.0、2030.3、0.0、2636.4、2000.0时，糖果店获利最大，可以获利10069.70美元。此时每周需购进杏仁、核桃仁、腰果仁、胡桃仁的数量分别约为2000.0kg、4000.0kg、3121.2kg、3000.0kg。配置的普通、豪华、蓝带三种不同品牌的果仁糖的量分别约为5454.5kg、0.0kg、6666.7kg。配置方法为：用腰果仁约1090.9kg、胡桃仁约3000.0kg、核桃仁约1363.6kg、杏仁约0.0kg配置普通品牌的果仁糖约5454.5kg；用腰果仁约0.0kg、胡桃仁约0.0kg、核桃仁约0.0kg、杏仁约0.0kg配置豪华品牌的果仁糖约0.0kg；用腰果仁约2030.3kg、胡桃仁约0.0kg、核桃仁约2636.4kg、杏仁约2000.0kg配置蓝带品牌的果仁糖约6666.7kg。求解结果可以用下表直观的显示出来。表3 最佳糖果配比（单位：kg）腰果仁胡桃仁核桃仁杏仁待售普通1090.93000.01363.60.05454.5豪华0.00.00.00.00.0蓝带2030.30.02636.42000.06666.7进购3121.23000.04000.02000.0在上述糖果配置方案中，糖果店每销售1kg普通品牌的果仁糖获利为（0.895454.5-0.701090.9-0.503000.0-0.551363.6-0.450.0）/5454.50.338美元糖果店每销售1kg蓝带品牌的果仁糖获利为（1.86666.7-0.703121.2-0.503000.0-0.554000.0-0.452000.0）6666.70.782美元在上述配置方案中，糖果店每进购1kg腰果仁可获利（0.891090.9+1.802030.3-0.703121.2）3121.20.782美元糖果店每进购1kg胡桃仁可获利（0.893000.0-0.503000.0）3000.0=0.49美元糖果店每进购1kg核桃仁可获利（0.891363.6+1.802636.4-0.554000.0）4000.00.940美元糖果店每进购1kg杏仁可获利（1.802000.0-0.452000.0）2000.0=1.35美元在此最优配比方案中，杏仁、核桃仁、胡桃仁都是按每周的最大供应量进货，而尽管果仁店每进购1kg腰果仁的获利比每进购1kg胡桃仁的获利多0.782-0.45=0.332美元，但是，腰果仁并未按每周的最大供应量进货，这是要求商店要严格的按照标准（配比要求）配置果仁糖。同理，在此方案中，糖果店每销售1kg的蓝带品牌的果仁糖比每销售1kg普通品牌的果仁糖可以多获利0.782-0.338=0.444美元。但是，此糖果店并没有只配置蓝带一种品牌的糖果，因为糖果店要严格按照标准配置，并且要充分利用资源，于是，在众多约束条件下，糖果店为追求获利最大，才形成了上述配置糖果的方案。问题二若在圣诞周，豪华和蓝带品牌的销售量会增加，这时商店会让果仁供应量增加10%,试问在这种情况下混合配比是否改变，圣诞周利润会改变多少？请分情况说明。由于问题（1）所求得的糖果的最优配置方案中豪华品牌的配置量为零，而糖果店的配置标准并未改变，与此同时，配置的所有糖果均能及时售出，所以，尽管圣诞周豪华和蓝带品牌的销售量会增加，若所有果仁供应量均增加10%，但是糖果店糖果店糖果的混合配比并不会改变，只是配比量在问题（1）的最优方案的基础上各增加10%，圣诞周的利润相应增加10%，变为11076.67美元。我们将上述求解命令中的b=zeros(1,8),2000,4000,5000,3000;改为b=zeros(1,8),2200,4400,5500,3300;即可求得结果。若是某种果仁的供应量增加10%，则商店糖果的混合配比方案会有所改变。下面我们分情况来讨论。.若只是杏仁供应量增加10%，我们可以将上述求解命令中的b=zeros(1,8),2000,4000,5000,3000;改为b=zeros(1,8),2200,4000,5000,3000;再进行求解。求解结果为x = 1.0e+003 * Columns 1 through 5 1.09090909090908 3.00000000000000 1.36363636363636 0.00000000000000 0.00000000000000 Columns 6 through 10 0.00000000000000 0.00000000000000 0.00000000000000 2.49696969696969 0.00000000000000 Columns 11 through 12 2.63636363636363 2.20000000000000maxz = 1.085303030303029e+004xR = 2.199999999999999e+003eR = 3.999999999999988e+003uR = 2.999999999999998e+003yR = 3.587878787878778e+003P = 5.454545454545435e+003H = 3.020533797989380e-012L = 7.333333333333327e+003.若只是核桃仁供应量增加10%，我们可以将上述求解命令中的b=zeros(1,8),2000,4000,5000,3000;改为b=zeros(1,8),2000,4400,5000,3000;再进行求解。求解结果为x = 1.0e+003 * Columns 1 through 5 1.09090909090909 3.00000000000000 1.36363636363636 0.00000000000000 0.00000000000000 Columns 6 through 10 0.00000000000000 0.00000000000000 0.00000000000000 2.00000000000000 0.00000000000000 Columns 11 through 12 2.66666666666665 2.00000000000000maxz = 1.007424242424241e+004xR = 2.000000000000000e+003eR = 4.030303030303014e+003uR = 2.999999999999998e+003yR = 3.090909090909090e+003P = 5.454545454545440e+003H = 4.504576832222785e-012L = 6.666666666666657e+003.若只是腰果仁供应量增加10%，我们可以将上述求解命令中的b=zeros(1,8),2000,4000,5000,3000;改为b=zeros(1,8),2000,4000,5500,3000;再进行求解。求解结果为x = 1.0e+003 * Columns 1 through 5 1.09090909072738 2.99999999986968 1.36363636335176 0.00000000000660 0.00000000001083 Columns 6 through 10 0.00000000001661 0.00000000003405 0.00000000004840 2.03030303010031 0.00000000007319 Columns 11 through 12 2.63636363633557 1.99999999993962maxz = 1.006969696933781e+004xR = 1.999999999994627e+003eR = 3.999999999721375e+003uR = 2.999999999959489e+003yR = 3.121212120838514e+003P = 5.454545453955418e+003H = 1.098942748775471e-007L = 6.666666666448693e+003.若只是胡桃仁供应量增加10%，我们可以将上述求解命令中的b=zeros(1,8),2000,4000,5000,3000;改为b=zeros(1,8),2000,4000,5000,3300;再进行求解。求解结果为x = 1.0e+003 * Columns 1 through 5 1.19999999991264 3.29999999994905 1.49999999993180 0.00000000000483 0.00000000000916 Columns 6 through 10 0.00000000001005 0.00000000002092 0.00000000003269 2.16666666661367 0.00000000001713 Columns 11 through 12 2.49999999992652 1.99999999995854maxz = 1.023333333313440e+004xR = 1.999999999996060e+003eR = 3.999999999879236e+003uR = 3.299999999976228e+003yR = 3.366666666535466e+003P = 5.999999999798311e+003H = 7.281526530041396e-008L = 6.666666666515863e+003.若果仁供应量增加的10%（即1400kg）全为杏仁，我们可以将上述求解命令中的b=zeros(1,8),2000,4000,5000,3000;改为b=zeros(1,8),3400,4000,5000,3000;再进行求解。求解结果为x = 1.0e+003 * Columns 1 through 5 0.65382340326988 2.56004823311234 0.85279503028443 0.00000000000000 0.00000000000000 Columns 6 through 10 0.00000000000000 0.00000000000000 0.00000000000000 4.34617659673012 0.43995176688764 Columns 11 through 12 3.14720496971555 3.39999999999999maxz = 1.528933333333330e+004xR = 3.399999999999999e+003eR = 3.999999999999985e+003uR = 2.999999999999988e+003yR = 4.999999999999997e+003P = 4.066666666666650e+003H = 1.045164780900433e-011L = 1.133333333333331e+004.若果仁供应量增加的10%（即1400kg）全为核桃仁，我们可以将上述求解命令中的b=zeros(1,8),2000,4000,5000,3000;改为b=zeros(1,8),2000,5400,5000,3000;再进行求解。求解结果为x = 1.0e+003 * Columns 1 through 5 1.09090908675965 2.99999999626337 1.36363635953805 0.00000000044224 0.00000000057218 Columns 6 through 10 0.00000000082154 0.00000000246899 0.00000000321955 1.99999999569704 0.00000000086917 Columns 11 through 12 2.66666665955821 1.99999999600291maxz = 1.007424240708538e+004xR = 1.999999999664705e+003eR = 4.030303021565261e+003uR = 2.999999997954069e+003yR = 3.090909083028862e+003P = 5.454545443003307e+003H = 7.082254933473905e-006L = 6.666666652127335e+003.若果仁供应量增加的10%（即1400kg）全为腰果仁，我们可以将上述求解命令中的b=zeros(1,8),2000,4000,5000,3000;改为b=zeros(1,8),2000,4000,6400,3000;再进行求解。求解结果为x = 1.0e+003 * Columns 1 through 5 1.09090909074948 2.99999999989478 1.36363636341602 0.00000000000849 0.00000000001378 Columns 6 through 10 0.00000000001905 0.00000000004074 0.00000000005893 2.03030303028766 0.00000000003854 Columns 11 through 12 2.63636363614622 1.99999999992557maxz = 1.006969696929303e+004xR = 1.999999999992986e+003eR = 3.999999999602981e+003uR = 2.999999999952373e+003yR = 3.121212121050924e+003P = 5.454545454068771e+003H = 1.325048549314246e-007L = 6.666666666397990e+003.若果仁供应量增加的10%（即1400kg）全为胡桃仁，我们可以将上述求解命令中的b=zeros(1,8),2000,4000,5000,3000;改为b=zeros(1,8),2000,4000,5000,4400;再进行求解。求解结果为x = 1.0e+003 * Columns 1 through 5 1.59999999999922 4.39999999999954 1.99999999999924 0.00000000000003 0.00000000000005 Columns 6 through 10 0.00000000000006 0.00000000000009 0.00000000000017 2.66666666666646 0.00000000000026 Columns 11 through 12 1.99999999999937 1.99999999999978maxz = 1.083333333333201e+004xR = 1.999999999999978e+003eR = 3.999999999998696e+003uR = 4.399999999999866e+003yR = 4.266666666665731e+003P = 7.999999999998026e+003H = 3.706669060433271e-010L = 6.666666666665875e+003对于更复杂的情况，在此我们不予讨论。对于上述各种情况求得的结果，与问题（1）类似，我们不做更多的解释。五、模型的评价1.本模型没有考虑糖果销售情况，只是假设所有配制糖果均能完全销售，然而在实际问题中考虑到商店的实际情况，包括平时员工工资，销售时长，营业税以及糖果多样性对商店销售情况的影响，单纯进行的一个优化，用销售额与进价之差表示利润。具有一定的局限性。2.我们可以一眼看出，无论果仁的供应量如何变化，在现有配置标准下，豪华品牌的糖果的配置量几乎都为0。因此上述所有方案都有一个明显的缺陷，即它不能满足部分顾客对豪华品牌果仁糖的需求，会因此而失去部分客源，这是果仁店商家所不愿看到的。所以我们需要对上述各方案进行改进。六、模型的改进改进方法有：适当调整豪华品牌果仁糖的价格；适当改变豪华果仁糖的配比；在原有标准和价格下，配置少量、适量豪华品牌的果仁糖。7、 参考文献【1】谢金星 薛毅，优化建模与LINDO/LINGO软件，清华大学出版社，2005年07月【2】 运筹学教材编写组编，运筹学.本科版，北京：清华大学出版社，2005.9八、附录1.使Matlab软件求解问题（1），在模型窗口中输入：max 0.19x1+0.39x2+0.34x3+0.44x4+0.4x5+0.6x6+0.55x7+0.65x8+1.1x9+1.3x10+1.25x11+1.35x12f=0.19,0.39,0.34,0.44,0.4,0.6,0.55,0.65,1.1,1.3,1.25,1.35;A=0.8,-0.2,-0.2,-0.2,zeros(1,8); 0.4,-0.6,0.4,0.4,zeros(1,8); -0.25,-0.25,0.75,-0.25,zeros(1,8); zeros(1,4),0.65,-0.35*ones(1,3),zeros(1,4); zeros(1,4),0.4*ones(1,3),-0.6,zeros(1,4); zeros(1,8),-0.7,0.3*ones(1,3); zeros(1,8),0.5,-0.5*ones(1,3); zeros(1,8),0.3*ones(1,3),-0.7; zeros(1,3),1,zeros(1,3),1,zeros(1,3),1; zeros(1,2),1,zeros(1,3),1,zeros(1,3),1,0; 1,zeros(1,3),1,zeros(1,3),1,zeros(1,3); 0,1,zeros(1,3),1,zeros(1,3),1,0,0;b=zeros(1,8),2000,4000,5000,3000;lb=zeros(12,1);ub=;x,fval=linprog(-f,A,b,lb,ub);x=x,format long;maxz=-fvalxR=x(4)+x(8)+x(12),eR=x(3)+x(7)+x(11),uR=x(2)+x(6)+x(10),yR=x(1)+x(5)+x(9)P=sum(x(1:4),H=sum(x(5:8),L=sum(x(9:12)求得Optimization terminated.x = Columns 1 through 8 1090.9 3000 1363.6 6.4125e-009 1.074e-008 1.5468e-008 3.2809e-008 4.6414e-008 Columns 9 through 12 2030.3 6.9177e-008 2636.4 2000maxz = 1.006969696933194e+004xR = 1.999999999994909e+003eR = 3.999999999729673e+003uR = 2.999999999962490e+003yR = 3.121212120747583e+003P = 5.454545453871487e+003H = 1.054311416334892e-007L = 6.666666666457737e+0032.使Matlab软件求解问题（2），在模型窗口中输入：.f=0.19,0.39,0.34,0.44,0.4,0.6,0.55,0.65,1.1,1.3,1.25,1.35;A=0.8,-0.2,-0.2,-0.2,zeros(1,8); 0.4,-0.6,0.4,0.4,zeros(1,8); -0.25,-0.25,0.75,-0.25,zeros(1,8); zeros(1,4),0.65,-0.35*ones(1,3),zeros(1,4); zeros(1,4),0.4*ones(1,3),-0.6,zeros(1,4); zeros(1,8),-0.7,0.3*ones(1,3); zeros(1,8),0.5,-0.5*ones(1,3); zeros(1,8),0.3*ones(1,3),-0.7; zeros(1,3),1,zeros(1,3),1,zeros(1,3),1; zeros(1,2),1,zeros(1,3),1,zeros(1,3),1,0; 1,zeros(1,3),1,zeros(1,3),1,zeros(1,3); 0,1,zeros(1,3),1,zeros(1,3),1,0,0;b=zeros(1,8),2200,4000,5000,3000;x = 1.0e+003 * Columns 1 through 5 1.09090909090908 3.00000000000000 1.36363636363636 0.00000000000000 0.00000000000000 Columns 6 through 10 0.00000000000000 0.00000000000000 0.00000000000000 2.49696969696969 0.00000000000000 Columns 11 through 12 2.63636363636363 2.20000000000000maxz = 1.085303030303029e+004xR = 2.199999999999999e+003eR = 3.999999999999988e+003uR = 2.999999999999998e+003yR = 3.587878787878778e+003P = 5.454545454545435e+003H = 3.020533797989380e-012L = 7.333333333333327e+003.f=0.19,0.39,0.34,0.44,0.4,0.6,0.55,0.65,1.1,1.3,1.25,1.35;A=0.8,-0.2,-0.2,-0.2,zeros(1,8); 0.4,-0.6,0.4,0.4,zeros(1,8); -0.25,-0.25,0.75,-0.25,zeros(1,8); zeros(1,4),0.65,-0.35*ones(1,3),zeros(1,4); zeros(1,4),0.4*ones(1,3),-0.6,zeros(1,4); zeros(1,8),-0.7,0.3*ones(1,3); zeros(1,8),0.5,-0.5*ones(1,3); zeros(1,8),0.3*ones(1,3),-0.7; zeros(1,3),1,zeros(1,3),1,zeros(1,3),1; zeros(1,2),1,zeros(1,3),1,zeros(1,3),1,0; 1,zeros(1,3),1,zeros(1,3),1,zeros(1,3); 0,1,zeros(1,3),1,zeros(1,3),1,0,0;b=zeros(1,8),2000,4400,5000,3000;x = 1.0e+003 * Columns 1 through 5 1.09090909090909 3.00000000000000 1.36363636363636 0.00000000000000 0.00000000000000 Columns 6 through 10 0.00000000000000 0.00000000000000 0.00000000000000 2.00000000000000 0.00000000000000 Columns 11 through 12 2.66666666666665 2.00000000000000maxz = 1.007424242424241e+004xR = 2.000000000000000e+003eR = 4.030303030303014e+003uR = 2.999999999999998e+003yR = 3.090909090909090e+003P = 5.454545454545440e+003H = 4.504576832222785e-012L = 6.666666666666657e+003.f=0.19,0.39,0.34,0.44,0.4,0.6,0.55,0.65,1.1,1.3,1.25,1.35;A=0.8,-0.2,-0.2,-0.2,zeros(1,8); 0.4,-0.6,0.4,0.4,zeros(1,8); -0.25,-0.25,0.75,-0.25,zeros(1,8); zeros(1,4),0.65,-0.35*ones(1,3),zeros(1,4); zeros(1,4),0.4*ones(1,3),-0.6,zeros(1,4); zeros(1,8),-0.7,0.3*ones(1,3); zeros(1,8),0.5,-0.5*ones(1,3); zeros(1,8),0.3*ones(1,3),-0.7; zeros(1,3),1,zeros(1,3),1,zeros(1,3),1; zeros(1,2),1,zeros(1,3),1,zeros(1,3),1,0; 1,zeros(1,3),1,zeros(1,3),1,zeros(1,3); 0,1,zeros(1,3),1,zeros(1,3),1,0,0;b=zeros(1,8),2000,4000,5500,3000;x = 1.0e+003 * Columns 1 through 5 1.09090909072738 2.99999999986968 1.36363636335176 0.00000000000660 0.00000000001083 Columns 6 through 10 0.00000000001661 0.00000000003405 0.00000000004840 2.03030303010031 0.00000000007319 Columns 11 through 12 2.63636363633557 1.99999999993962maxz = 1.006969696933781e+004xR = 1.999999999994627e+003eR = 3.999999999721375e+003uR = 2.999999999959489e+003yR = 3.121212120838514e+003P = 5.454545453955418e+003H = 1.098942748775471e-007L = 6.666666666448693e+003.f=0.19,0.39,0.34,0.44,0.4,0.6,0.55,0.65,1.1,1.3,1.25,1.35;A=0.8,-0.2,-0.2,-0.2,zeros(1,8); 0.4,-0.6,0.4,0.4,zeros(1,8); -0.25,-0.25,0.75,-0.25,zeros(1,8); zeros(1,4),0.65,-0.35*ones(1,3),zeros(1,4); zeros(1,4),0.4*ones(1,3),-0.6,zeros(1,4); zeros(1,8),-0.7,0.3*ones(1,3); zeros(1,8),0.5,-0.5*ones(1,3); zeros(1,8),0.3*ones(1,3),-0.7; zeros(1,3),1,zer