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线性规划习题及答案 线性规划问题的解题步骤

解决简单线性规划问题的方法是图解法,即借助直线(线性目标函数看作斜率确定的一族平行直线)与平面区域(可行域)有交点时,直线在y轴上的截距的最大值或最小值求解,它的步骤如下:

(1)设出未知数,确定目标函数。

(2)确定线性约束条件,并在直角坐标系中画出对应的平面区域,即可行域。

(3)由目标函数变形为,所以求z的最值可看成是求直线在y轴上截距的最值(其中a、b是常数,z随x、y的变化而变化)。

(4)作平行线:将直线平移(即作的平行线),使直线与可行域有交点,且观察在可行域中使最大(或最?。┦彼牡?,求出该点的坐标。

(5)求出最优解:将(4)中求出的坐标代入目标函数,从而求出z的最大(?。┲?。

扩展资料:

线性规划基本概念:

(1)可行解:把满足约束条件的一组决策变量值称为该线性规划问题的可行解。

(2)可行解集/可行解域:满足约束条件的可行解的全体称为可行解集,在平面上,所有可行解的点的集合称为可行解域。

(3)最优解:在可行解集中,使目标函数达到最优值的可行解称为最优解。

参考资料:

百度百科-线性规划

线性规划习题

同学,这是最基本的线性规划问题,可以用基本的“单纯形法”求解,网上应该有相应的教程的,我的图片里列出了我亲自笔算的详细表格,最终的x1=2,x2=4,x3=0目标函数最大值为22

线性规划问题

该问题的关键词有两个,一是“最小值”,二是“无穷多个”。这样lz就能明白为什么参考答案是正确的解法了。望lz再细心体会下这两关键词。不明白的话再说吧,这样提高最大。

ps:给个提示,如果选择直线l的话,是没取得u的最小值的。

------------------------------------------------------

呵呵,既然如此,我就再说详细点。

由题意,u的意义就是直线kx+y-u=0在y轴上的截距,按照lz原来的那张图上标的l(图没了),是经过直线AB与y轴交点的l的截距小呢还是将l进行平移移到经过A点的直线l'在y轴上的截距???显然是后者吧,最优解就是l'与可行域(三角形ABC)的交点。所以我说直线l没有达到最优解

线性规划问题

下面是最小费用的两组解,对应的最小费用为1008元:

{{小巴个数,所跑次数,限载人数,费用},{大巴个数,所跑次数,限载人数,费用},{限载总人数,总费用}}

{{3,2,96,288},{4,3,384,720},{480,1008}},

{{2,3,96,288},{4,3,384,720},{480,1008}},

下面是所有满足情况的解(不排除有些重复解):

{{小巴个数,所跑次数,限载人数,费用},{大巴个数,所跑次数,限载人数,费用},{限载总人数,总费用}}

{{6,1,96,288},{4,3,384,720},{480,1008}},

{{2,3,96,288},{4,3,384,720},{480,1008}},

{{2,4,128,384},{4,3,384,720},{512,1104}},

{{2,5,160,480},{4,3,384,720},{544,1200}},

{{3,2,96,288},{4,3,384,720},{480,1008}},

{{3,3,144,432},{4,3,384,720},{528,1152}},

{{3,4,192,576},{3,3,288,540},{480,1116}},

{{3,4,192,576},{4,3,384,720},{576,1296}},

{{3,5,240,720},{3,3,288,540},{528,1260}},

{{3,5,240,720},{4,2,256,480},{496,1200}},

{{4,2,128,384},{4,3,384,720},{512,1104}},

{{4,3,192,576},{3,3,288,540},{480,1116}},

{{4,3,192,576},{4,3,384,720},{576,1296}},

{{4,4,256,768},{3,3,288,540},{544,1308}},

{{4,4,256,768},{4,2,256,480},{512,1248}},

{{4,5,320,960},{2,3,192,360},{512,1320}},

{{4,5,320,960},{3,2,192,360},{512,1320}},

{{4,5,320,960},{4,2,256,480},{576,1440}},

{{5,2,160,480},{4,3,384,720},{544,1200}},

{{5,3,240,720},{3,3,288,540},{528,1260}},

{{5,3,240,720},{4,2,256,480},{496,1200}},

{{5,4,320,960},{2,3,192,360},{512,1320}},

{{5,4,320,960},{3,2,192,360},{512,1320}},

{{5,4,320,960},{4,2,256,480},{576,1440}},

{{5,5,400,1200},{1,3,96,180},{496,1380}},

{{5,5,400,1200},{2,2,128,240},{528,1440}},

{{5,5,400,1200},{3,1,96,180},{496,1380}},

{{5,5,400,1200},{4,1,128,240},{528,1440}},

{{6,1,96,288},{4,3,384,720},{480,1008}},

{{6,2,192,576},{3,3,288,540},{480,1116}},

{{6,2,192,576},{4,3,384,720},{576,1296}},

{{6,3,288,864},{2,3,192,360},{480,1224}},

{{6,3,288,864},{3,2,192,360},{480,1224}},

{{6,3,288,864},{4,2,256,480},{544,1344}},

{{6,4,384,1152},{1,3,96,180},{480,1332}},

{{6,4,384,1152},{2,2,128,240},{512,1392}},

{{6,4,384,1152},{3,1,96,180},{480,1332}},

{{6,4,384,1152},{4,1,128,240},{512,1392}},

{{6,5,480,1440},{0,0,0,0},{480,1440}},

{{6,5,480,1440},{1,0,0,0},{480,1440}},

{{6,5,480,1440},{2,0,0,0},{480,1440}},

{{6,5,480,1440},{3,0,0,0},{480,1440}},

{{6,5,480,1440},{4,0,0,0},{480,1440}},

{{7,1,112,336},{4,3,384,720},{496,1056}},

{{7,2,224,672},{3,3,288,540},{512,1212}},

{{7,2,224,672},{4,2,256,480},{480,1152}},

{{7,3,336,1008},{2,3,192,360},{528,1368}},

{{7,3,336,1008},{3,2,192,360},{528,1368}},

{{7,3,336,1008},{4,2,256,480},{592,1488}},

{{7,4,448,1344},{1,1,32,60},{480,1404}},

{{7,4,448,1344},{2,1,64,120},{512,1464}},

{{7,4,448,1344},{3,1,96,180},{544,1524}},

{{7,4,448,1344},{4,1,128,240},{576,1584}},

{{7,5,560,1680},{0,0,0,0},{560,1680}},

{{7,5,560,1680},{1,0,0,0},{560,1680}},

{{7,5,560,1680},{2,0,0,0},{560,1680}},

{{7,5,560,1680},{3,0,0,0},{560,1680}},

{{7,5,560,1680},{4,0,0,0},{560,1680}}

下面是按照车费由小到大排序的结果:

{{小巴个数,所跑次数,限载人数,费用},{大巴个数,所跑次数,限载人数,费用},{限载总人数,总费用}}

{{3,2,96,288},{4,3,384,720},{480,1008}},

{{2,3,96,288},{4,3,384,720},{480,1008}},

{{7,1,112,336},{4,3,384,720},{496,1056}},

{{4,2,128,384},{4,3,384,720},{512,1104}},

{{2,4,128,384},{4,3,384,720},{512,1104}},

{{6,2,192,576},{3,3,288,540},{480,1116}},

{{4,3,192,576},{3,3,288,540},{480,1116}},

{{3,4,192,576},{3,3,288,540},{480,1116}},

{{7,2,224,672},{4,2,256,480},{480,1152}},

{{3,3,144,432},{4,3,384,720},{528,1152}},

{{5,3,240,720},{4,2,256,480},{496,1200}},

{{5,2,160,480},{4,3,384,720},{544,1200}},

{{3,5,240,720},{4,2,256,480},{496,1200}},

{{2,5,160,480},{4,3,384,720},{544,1200}},

{{7,2,224,672},{3,3,288,540},{512,1212}},

{{6,3,288,864},{3,2,192,360},{480,1224}},

{{6,3,288,864},{2,3,192,360},{480,1224}},

{{4,4,256,768},{4,2,256,480},{512,1248}},

{{5,3,240,720},{3,3,288,540},{528,1260}},

{{3,5,240,720},{3,3,288,540},{528,1260}},

{{6,2,192,576},{4,3,384,720},{576,1296}},

{{4,3,192,576},{4,3,384,720},{576,1296}},

{{3,4,192,576},{4,3,384,720},{576,1296}},

{{4,4,256,768},{3,3,288,540},{544,1308}},

{{5,4,320,960},{3,2,192,360},{512,1320}},

{{5,4,320,960},{2,3,192,360},{512,1320}},

{{4,5,320,960},{3,2,192,360},{512,1320}},

{{4,5,320,960},{2,3,192,360},{512,1320}},

{{6,4,384,1152},{3,1,96,180},{480,1332}},

{{6,4,384,1152},{1,3,96,180},{480,1332}},

{{6,3,288,864},{4,2,256,480},{544,1344}},

{{7,3,336,1008},{3,2,192,360},{528,1368}},

{{7,3,336,1008},{2,3,192,360},{528,1368}},

{{5,5,400,1200},{3,1,96,180},{496,1380}},

{{5,5,400,1200},{1,3,96,180},{496,1380}},

{{6,4,384,1152},{4,1,128,240},{512,1392}},

{{6,4,384,1152},{2,2,128,240},{512,1392}},

{{7,4,448,1344},{1,1,32,60},{480,1404}},

{{6,5,480,1440},{4,0,0,0},{480,1440}},

{{6,5,480,1440},{3,0,0,0},{480,1440}},

{{6,5,480,1440},{2,0,0,0},{480,1440}},

{{6,5,480,1440},{1,0,0,0},{480,1440}},

{{6,5,480,1440},{0,0,0,0},{480,1440}},

{{5,5,400,1200},{4,1,128,240},{528,1440}},

{{5,5,400,1200},{2,2,128,240},{528,1440}},

{{5,4,320,960},{4,2,256,480},{576,1440}},

{{4,5,320,960},{4,2,256,480},{576,1440}},

{{7,4,448,1344},{2,1,64,120},{512,1464}},

{{7,3,336,1008},{4,2,256,480},{592,1488}},

{{7,4,448,1344},{3,1,96,180},{544,1524}},

{{7,4,448,1344},{4,1,128,240},{576,1584}},

{{7,5,560,1680},{4,0,0,0},{560,1680}},

{{7,5,560,1680},{3,0,0,0},{560,1680}},

{{7,5,560,1680},{2,0,0,0},{560,1680}},

{{7,5,560,1680},{1,0,0,0},{560,1680}},

{{7,5,560,1680},{0,0,0,0},{560,1680}}

附上Mathematica 程序,因为程序很小, 所以没有简化程序.没有剔除重复解.

arr = {};

For[m = 0, m <= Ceiling[480/16] && m <= 7, m++,

For[p = 0, p <= Ceiling[480/16] && p <= 5, p++,

For[n = 0, n <= Ceiling[480/32] && n <= 4, n++,

For[q = 0, q <= Ceiling[480/32] && q <= 3, q++,

If[m*p*16 + n*q*32 >= 480,

arr =

Append[arr, {{m, p, 16*m*p, 48*m*p}, {n, q, 32*n*q,

60*n*q}, {m*p*16 + n*q*32, 48*m*p + 60*n*q}}];

Break[]

]

]

]

]

]

arr

Sort[arr, #1[[-1]][[-1]] < #2[[-1]][[-1]] &]

线性规划问题

就是坐标变换

设m=x+y,n=x-y.则x=(m+n)/2,y=(m-n)/2

代入A的约束条件x+y<=1,x,y>=0得

m<=1,m+n>=0,m-n>=0

你把这个区域在MON平面上画出来,不难发现面积就是1

------------------------

lz,很简单,答案错了。本题答案就是1。如果答案说什么就是什么你还跑这来问问题干吗?凡事用自己的脑子想,多方求证。不要迷信答案

线性规划问题!

首先你要明白x,y分别代表什么,这里x代表能载6吨的汽车数量,y表示能载5吨的汽车数量,理解这就知道为什么了完整的线性规划是

目标函数:z=6x+5y

约束条件是:x<=5

y<=4

x>=0 且为整数

y>=0 且为整数

线性规划问题

1.答案为8.这道题应该考虑为一个圆心为(0,0)的圆与直线X-Y+4<=0恰好相切,由点到直线的距离公式可求出圆的半径R.再由圆的方程和直线方程联立解得,X=-2,Y=2时,值最小为8.

2.答案是0<=K<=0.5 这道题应该把它转化为斜率K的问题.设Y-1/X+1=K,则点(X,Y)可以看成与点(1.-1)构成的斜率K,由图可知,当Y=1时,K最小=0.当X=1,Y=2时,K最大等于0.5

运筹学线性规划问题~~~跪求过程和答案 谢谢!

Objective value: 6.000000

Variable Value Reduced Cost

A 0.000000 -3.000000

B 1.000000 -2.000000

C 0.000000 -1.000000

D 1.000000 -4.000000

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