原油的采购与加工程序与报告
- 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
方法一
(一)求R1
max 4.8x11+4.8x21+5.6x12+5.6x22-10x
st
x11+x12-x<500
x21+x22<1000
0.5x11-0.5x21>0
0.4x12-0.6x22>0
x<500
end
LP OPTIMUM FOUND AT STEP 4
OBJECTIVE FUNCTION V ALUE
1) 4800.000
V ARIABLE V ALUE REDUCED COST X11 500.000000 0.000000
X21 500.000000 0.000000
X12 0.000000 0.266667
X22 0.000000 0.000000
X 0.000000 0.400000
ROW SLACK OR SURPLUS DUAL PRICES
2) 0.000000 9.600000
3) 500.000000 0.000000
4) 0.000000 -9.600000
5) 0.000000 -9.333333
6) 500.000000 0.000000
NO. ITERATIONS= 4
(二)求R2
max 4.8x11+4.8x21+5.6x12+5.6x22-8x
st
x11+x12-x<500
x21+x22<1000
0.5x11-0.5x21>0
0.4x12-0.6x22>0
x>500
x<1000
4.8x11+4.8x21+
5.6x12+5.6x22-8x-r2=1000
end
LP OPTIMUM FOUND AT STEP 7
OBJECTIVE FUNCTION V ALUE
1) 6000.000
V ARIABLE V ALUE REDUCED COST X11 0.000000 0.000000
X21 0.000000 0.400000
X12 1500.000000 0.000000
X22 1000.000000 0.000000
X 1000.000000 0.000000
R2 5000.000000 0.000000
ROW SLACK OR SURPLUS DUAL PRICES
2) 0.000000 8.000000
3) 0.000000 2.000000
4) 0.000000 -6.400000
5) 0.000000 -6.000000
6) 500.000000 0.000000
7) 0.000000 0.000000
8) 0.000000 0.000000
NO. ITERATIONS= 7
(三)求R3
max 4.8x11+4.8x21+5.6x12+5.6x22-6x
st
x11+x12-x<500
x21+x22<1000
0.5x11-0.5x21>0
0.4x12-0.6x22>0
x>1000
x<1500
4.8x11+4.8x21+
5.6x12+5.6x22-6x-r3=3000
End
LP OPTIMUM FOUND AT STEP 0
OBJECTIVE FUNCTION V ALUE
1) 8000.000
V ARIABLE V ALUE REDUCED COST
X11 0.000000 0.000000
X21 0.000000 1.400000
X12 1500.000000 0.000000
X22 1000.000000 0.000000
X 1000.000000 0.000000
R3 5000.000000 0.000000
ROW SLACK OR SURPLUS DUAL PRICES
2) 0.000000 6.000000
3) 0.000000 5.000000
4) 0.000000 -2.400000
5) 0.000000 -1.000000
6) 0.000000 0.000000
7) 500.000000 0.000000
8) 0.000000 0.000000
NO. ITERATIONS= 0
方法二
model:
max=4.8*x11+4.8*x21+5.6*x12+5.6*x22-10*x1-8*x2-6*x3;
x1<500;
x2<500;
x3<500;
x11+x12 x21+x22<1000; 0.5*x11-0.5*x21>0; 0.4*x12-0.6*x22>0; x1+x2+x3-x=0; (x1-500)*x2=0; (x2-500)*x3=0; Local optimal solution found at iteration: 9 Objective value: 4800.000 Variable Value Reduced Cost X11 500.0000 0.000000 X21 500.0000 0.000000 X12 0.000000 0.000000 X22 0.000000 0.4000000 X1 0.000000 0.4000000 X2 0.000000 0.000000 X3 0.000000 0.000000 X 0.000000 0.000000 Row Slack or Surplus Dual Price 1 4800.000 1.000000 2 500.0000 0.000000 3 500.0000 0.000000 4 500.0000 0.000000 5 0.000000 9.600000 6 500.0000 0.000000 7 0.000000 -9.600000 8 0.000000 -10.00000 9 0.000000 -9.600000 10 0.000000 -0.3200000E-02 11 0.000000 -0.7200000E-02 我们加上x>10后 model: max=4.8*x11+4.8*x21+5.6*x12+5.6*x22-10*x1-8*x2-6*x3; x1<500; x2<500; x3<500; x11+x12 x21+x22<1000; 0.5*x11-0.5*x21>0; 0.4*x12-0.6*x22>0; x1+x2+x3-x=0; (x1-500)*x2=0; (x2-500)*x3=0;