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1、一、选择填空1.2.3.4.5.6.7.8.9.10.二、判断正误1.2.3.4.5.6.7.8.9.10.三、将下列问题化为标准型1.[解]令,,在约束1中引入非负的松弛变量,约束2两边同乘以-1。整理得:即:2.MinZ=-x1+5x2-2x3s.t.x1+x2-x3≤62x1-x2+3x3≥5x1+x2=10x1≥0,x2≤0,x3符号不限[解]首先,令对变量x3进行处理,令x3=x’3-x4;再令x’2=-x2。然后对目标函数和约束条件进行标准化。MaxZ=x1+5x2+2x3-2x4s.t.x1-x2-x3
2、+x4+x5=62x1+x2+3x3-3x4-x6=5x1-x2=10x1,x2,x3,x4,x5,x6≥0四、用图解法求解下列线性规1.minZ=-x1+2x2s.t.x1-x2≥-2x1+2x2≤6x1,x2≥0E-2-1012345678x1x2654321x1-x2≥-2x1+2x2≤6[解]画图如下:根据上图,最优解为X*=(x1,x2)T=(6,0)T,最优值为-6。2.MaxZ=-x1+2x2s.t.x1-x2≥-2x1+2x2≤6x1,x2≥0E-2-1012345678x1x2654321x1-x2
3、≥-2x1+2x2≤6[解]画图如下:根据上图,最优解为,最优值为。一、用单纯形法求解下列线性规划1.MaxZ=3x1+5x2s.t.x1≤42x2≤123x1+2x2≤18x1,x2≥0[解]首先,标准化后线性规划如下:(1)MaxZ=3x1+5x2+0x3+0x4+0x5s.t.x1+x3=42x2+x4=123x1+2x2+x5=18x1,x2,x3,x4,x5,x6≥0再用表格单纯形法求解如下:(一、按照第一个正检验数对应非基变量进基的方法)CBXBcjbxj35000x1x2x3x4x5θj000x3x4x
4、5412181010002010320014/118/3-Z035000300x1x4x54126101000201002-30112/26/2-Z-1205-300305x1x4x2463101000031-101-3/201/24/16/3-Z-27009/20-5/2305x1x3x2226100-1/31/30011/3-1/30101/20-Z-36000-3/2-1(二、按照最大正检验数对应非基变量进基的方法)CBXBcjbxj35000x1x2x3x4x5θj000x3x4x54121810100020
5、103200112/218/2-Z0350000x34101004/150x2x5660101/20300-116/3-Z-12300-5/20053x3x2x12620011/3-1/30101/20100-1/31/3-Z-3600-3-3/2-1因此,最优解为X*=(2,6,2,0,0)T,最优值为Zmax=36。2.MaxZ=2x1-x2+x3s.t.3x1+x2+x3≤60x1-x2+2x3≤10x1+x2-x3≤20x1,x2,x3≥0[解]首先,标准化后线性规划如下:(1)MaxZ=2x1-x2+x3s
6、.t.3x1+x2+x3+x4=60x1-x2+2x3+x5=10x1+x2-x3+x6=20x1,x2,x3,x4,x5,x6≥0再用表格单纯形法求解如下:CBXBcjbxj2-11000x1x2x3x4x5x6θj000x4x5x66010203111001-1201011-100160/310/120/1-Z02-11000020x4x1x630101004-51-301-1201002-30-1130/410/2-Z-2001-30-2002-1x4x1x2101550011-1-2101/201/21/20
7、1-3/20-1/21/2-Z-2500-3/20-3/2-1/2因此,最优解为X*=(15,5,0,10,0,0)T,最优值为Zmax=25。一、表格单纯形法计算题1.(2)初始线性规划模型如下:MaxZ=5x1+20x2+25x3s.t.2x1+x2≤402x2+x3≤303x1-1/2x3≤15x1,x2,x3≥0(3)用单纯形法求出最优解及相应的最优值。[解](按照最大检验数对应非基变量进基的方法)CBXBcjbxj52025000x1x2x3x4x5x6θj000x4x5x64030152101000210
8、1030-1/200130/1-Z0520250000250x4x3x640303021010002101031001/2140/230/3-Z-7505-3000-2500255x4x3x120301001/301-1/3-2/302101011/3001/61/3-Z-8000-95/300-155/6-5/3一、用大M法和两阶段法求解下列