华东交通大学：《数学模型》课程教学资源（讲义）原油的采购与加工程序与报告

方法 )求 max4.8x11+4.8x21+5.6x12+5.6x22-10x xII+x12-x0 0.4x120.6x22>0 X0 X<1000

方法一 （一）求 R1 max 4.8x11+4.8x21+5.6x12+5.6x22-10x st x11+x12-x0 0.4x12-0.6x22>0 x0 0.4x12-0.6x22>0 x>500 x<1000

48x11+4.8x21+5.6x12+5.6x22-8x-r2=1000 LP OPTIMUM FOUND AT STEP OBJECTIVE FUNCTION VALUE 6000.000 VARIABLE VALUE REDUCED COST 0.000000 0.000000 0.400000 X12 1500.000000 0.000000 X22 1000.000000 0.000000 1000.000000 0.000000 R2 5000.000000 0.000000 ROW SLACK OR SURPLUS DUAL PRICES 0.000000 0.000000 3456)8 -6.400000 0.000000 -6.000000 500.000000 0.000000 0.000000 0.000000 0.000000 0.000000 NO. ITERATIONS (三)求R3 max48x11+48x21+5.6x12+5.6X22-6x Xl1+x12-x0 0.4x120.6x22>0 X>1000 X<1500 48x11+4.8x21+5.6x12+5.6X22-6x3=3000

4.8x11+4.8x21+5.6x12+5.6x22-8x-r2=1000 end LP OPTIMUM FOUND AT STEP 7 OBJECTIVE FUNCTION VALUE 1) 6000.000 VARIABLE VALUE 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-x0 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 VALUE 8000.000 VARIABLE VALUE REDUCED COST 0.000000 0.000000 X21 1.400000 0.000000 1000.000000 0.000000 1000.000000 0.000000 5000.000000 0.000000 ROW SLACK OR SURPLUS DUAL PRICES 0.00 6.000000 0.000000 -2.400000 0.000000 -1.000000 0.000000 0.000000 NO. ITERATIONS= 方法二 model max=4.8*x11+4.8*x21+5.6*12+5.6*x22-10*x1-8*x2-6*x3; x10 0.4★x12-0.6*x22>0 x1+X2+x3-X=0 x1-500)★x2=0; x2-500)*x3=0;

LP OPTIMUM FOUND AT STEP 0 OBJECTIVE FUNCTION VALUE 1) 8000.000 VARIABLE VALUE 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; x10; 0.4*x12-0.6*x22>0; x1+x2+x3-x=0; (x1-500)*x2=0; (x2-500)*x3=0;

End Local optimal solution found at iteration Objective value 4800.000 Variable educed c。st X11 0.000000 x21 X12 0.000000 0.000000 X22 0.000000 0.4000000 0.4000000 0.000000 0.000000 X 0.000000 0.000000 R。WS1 ack or Surp⊥us Dual Price 500.0000 0.000000 500.0000 0.000000 0.000000 9.600000 500.0000 0.000000 0.000000 9.600000 0.000000 10.00000 0.000000 0.7200000E-02 我们加上x>10后 max=4.8*x11+4.8*x21+5.6*12+5.6*x22-10*x1-8*x2-6*x3; x20 x1-500)★x2=0; x2-500)*x3=0;

End 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; x10; 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 21 Objective value 5000.000 Variable Value Reduced cost X11 0.000000 0.8997907 500.0000 0.000000 500.0000 0.000000 0.000000 1000.000 0.000000 R。WS1 ack or surp⊥us Dual price 5000.000 1.00000 0.000000 0.000000 500.0000 0.000000 0.000000 7.000419 3.499372 -3.501046 7.000419 -0.5999163E-02 -0.7363828E+10 990.0000 0.000000 方法三 max4.8x11+4.8x21+5.6x12+5.6x22-10x1-8x2-6x3 x21+x220

x>10; end Local optimal solution found at iteration: 21 Objective value: 5000.000 Variable Value Reduced Cost X11 0.000000 0.000000 X21 0.000000 0.8997907 X12 1500.000 0.000000 X22 1000.000 0.000000 X1 500.0000 0.000000 X2 500.0000 0.000000 X3 0.000000 0.000000 X 1000.000 0.000000 Row Slack or Surplus Dual Price 1 5000.000 1.000000 2 0.000000 0.000000 3 0.000000 0.000000 4 500.0000 0.000000 5 0.000000 7.000419 6 0.000000 3.499372 7 0.000000 -4.400837 8 0.000000 -3.501046 9 0.000000 -7.000419 10 0.000000 -0.5999163E-02 11 0.000000 -0.7363828E+10 12 990.0000 0.000000 方法三 max 4.8x11+4.8x21+5.6x12+5.6x22-10x1-8x2-6x3 st x11+x12-x0

0.4x120.6x22>0 x1-500y1<0 x2-500y2<0 X3-500y3<0 int y2 ENUMERATION COMPLETE BRANCHES= 2 PIVOTS= 35 LAST INTEGER SOLUTION IS THE BEST FOUND RE-INSTALLING BEST SOLUTION OBJECTIVE FUNCTION VALUE 5000.000 VARIABLE VALUE REDUCED COST 1.000000 0.000000 Y2 1.000000 2200.000000 Y3 1.000000 1200.000000 X11 0.800000 X21 0.800000 1500.000000 0.000000 1000.000000 XI 500.000000 0.000000 X2 500.00000 0.000000 0.000000 0.400000 1000.000000 0.000000 ROW SLACK OR SURPLUS DUAL PRICES 0.000000 550 0.000000 567 0.000000 0.000000 0.000000 -5.600000 0.000000 4.400000

0.4x12-0.6x22>0 x1+x2+x3-x=0 500y2-x1<0 x1-500y1<0 500y3-x2<0 x2-500y2<0 x3-500y3<0 end int y1 int y2 int y3 ENUMERATION COMPLETE. BRANCHES= 2 PIVOTS= 35 LAST INTEGER SOLUTION IS THE BEST FOUND RE-INSTALLING BEST SOLUTION... OBJECTIVE FUNCTION VALUE 1) 5000.000 VARIABLE VALUE REDUCED COST Y1 1.000000 0.000000 Y2 1.000000 2200.000000 Y3 1.000000 1200.000000 X11 0.000000 0.800000 X21 0.000000 0.800000 X12 1500.000000 0.000000 X22 1000.000000 0.000000 X1 500.000000 0.000000 X2 500.000000 0.000000 X3 0.000000 0.400000 X 1000.000000 0.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 5.600000 3) 0.000000 5.600000 4) 0.000000 0.000000 5) 0.000000 0.000000 6) 0.000000 -5.600000 7) 0.000000 4.400000

8) 0.000000 0.000000 0.000000 2.400000 10) 0.000000 500.000000 0.000000 NO. ITERATIONS= BRANCHES 2 DETERM E 1000E 0

8) 0.000000 0.000000 9) 0.000000 2.400000 10) 0.000000 0.000000 11) 500.000000 0.000000 NO. ITERATIONS= 39 BRANCHES= 2 DETERM.= 1.000E 0