数学建模与数学实验课程设计题目与参考答案

Residual Case Order Plot10.80.60.4Residuals0.20-0.2-0.4-0.6-0.8-124681012Case Number14161820

结果为 b =

1.0229 0.0253 1.9943 1.9128 bint =

-1.2103 3.2561 -0.0318 0.0824 1.5526 2.4360 1.0444 2.7813 r =

0.1333 -0.1660 -0.0064 -0.5015 0.0705 0.4861 0.5021 -0.3210 -0.3244 0.2593

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0.0685 0.2039 0.0201 0.4033 -0.2248 -0.1838 -0.4500 -0.1362 0.4046 -0.2378 rint =

-0.3674 0.6339 -0.8015 0.4695 -0.6921 0.6792 -1.0821 0.0792 -0.6092 0.7502 -0.1303 1.1026 -0.1138 1.1181 -0.9832 0.3413 -1.0133 0.3645 -0.3016 0.8203 -0.6238 0.7608 -0.4135 0.8214 -0.6652 0.7054 -0.2008 1.0075 -0.9053 0.4558 -0.8551 0.4875 -1.0729 0.1728 -0.8044 0.5321 -0.1949 1.0040 -0.8531 0.3775 stats =

0.9226 63.5671 0.0000 0.1125

参数回归结果为对应的置信区间分别为[ -1.2103 3.2561];;[ -0.0318 0.0824];[1.5526 2.4360];[1.0444 2.7813]

r2=0.9226(越接近于1，回归效果越显著)，F= 63.5671， p=0.0000，由p<0.05, 可知回归模型

y= 1.0229-0.0253*x1+1.9943*x2+1.9128*x3 成立

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再分别求y关于单个变量x1，x2, x3的线性回归方程

y=13.5989+0.0547*x1 y= 5.6018+1.8453*x2 y= 7.1238+4.7430*x3

分别求y关于两个变量x1，x2, x3的线性回归方程

y=1.0113+0.0252*x1+2.6057*x2 y=2.0661+0.0318*x2+2.3961*x3 y=6.0781+0.0806*x1+4.0719*x3

3.优化理论中的线性规划问题---生产安排。

设每种合金品种取值xi千克 （i?1、2、3、4、5） 根据题意建立线性规划方程得：

目标费用最小

f?x?=8.6*x1?6*x2?8.9*x3?5.7*x4?8.8*x5;

x1?x2?x3?x4?x5?100;

(30*x1+10*x2+50*x3+10*x4+50*x5)?(60*x1+20*x2+20*x3+10*x4+10*x5)=1.5;?30*x1?10*x2?50*x3?10*x4?50*x5???10*x1?70*x2?30*x3?80*x4?40*x5??0.6;?60*x1?20*x2?20*x3?10*x4?10*x5???10*x1?70*x2?30*x3?80*x4?40*x5??0.4;

利用lingo求解： 程序为：

min=8.6*x1+6*x2+8.9*x3+5.7*x4+8.8*x5; x1+x2+x3+x4+x5=100;

(30*x1+10*x2+50*x3+10*x4+50*x5)/(60*x1+20*x2+20*x3+10*x4+10*x5)=1.5;

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(30*x1+10*x2+50*x3+10*x4+50*x5)/(10*x1+70*x2+30*x3+80*x4+40*x5)=0.6;

(60*x1+20*x2+20*x3+10*x4+10*x5)/(10*x1+70*x2+30*x3+80*x4+40*x5)=0.4;

求解结果：

Global optimal solution found.

Objective value: 744.4444 Objective bound: 744.4444

Infeasibilities: 0.1405409E-10 Extended solver steps: 15 Total solver iterations: 440

Variable Value Reduced Cost X1 11.11111 0.000000 X2 0.1111111E-01

X3 44.44444 0.000000 X4 44.44444 0.000000 X5 0.000000 0.1888889

Row Slack or Surplus Dual Price 1 744.4444 -1.000000 2 0.000000 -7.444444 3 0.000000 1.703704 4 0.000000 -230.0926 5 0.000000 0.000000

结果分析

当x1=11.1111 x2=0 x3=44.44444 x4=44.44444 x5=0时

取得费用最小值为 744.4444元

当铅减少1单位时总费用将减少7.4444元 当锌减少1单位时总费用将增加1.703704元 当锡减少1单位时总费用将减少230.0926元

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0.000000