1 MAX 2 X 13 X 2 ST

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线性规划模型(1) :MAX 2 X 1+3 X 2 ? ST ( 说明: 也可写成S. T. ,

线性规划模型(1) :MAX 2 X 1+3 X 2 ? ST ( 说明: 也可写成S. T. , SUCH THAT 或 SUBJECT TO 等) ? 5 X 2<15 ? 6 X 1+2 X 2<24 ? X 1+X 2<5 ? END :GO 第 9页

线性规划模型(3) LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE 1) 8. 500000 VARIABLE

线性规划模型(3) LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE 1) 8. 500000 VARIABLE X 1 X 2 VALUE 3. 500000 1. 500000 REDUCED COST 0. 000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 7. 500000 0. 000000 3) 0. 000000 0. 250000 4) 0. 000000 0. 500000 NO. ITERATIONS= 2 DO RANGE(SENSITIVITY) ANALYSIS? 第 10页

线性规划模型(2) : min 2800 x 11+2800 x 21+2800 x 31+2800 x 41+4500 x 12

线性规划模型(2) : min 2800 x 11+2800 x 21+2800 x 31+2800 x 41+4500 x 12 +4500 x 22+4500 x 32+6000 x 13+6000 x 23+7300 x 14 ? st ? x 11+x 12+x 13+x 14>15 ? x 12+x 13+x 14+x 21+x 22+x 23>10 ? x 13+x 14+x 22+x 23+x 31+x 32>20 ? x 14+x 23+x 32+x 41>12 ? end : go 第 15页

线性规划模型(2) LP OPTIMUM FOUND AT STEP 3 OBJECTIVE FUNCTION VALUE 1) 118400. 0 VARIABLE

线性规划模型(2) LP OPTIMUM FOUND AT STEP 3 OBJECTIVE FUNCTION VALUE 1) 118400. 0 VARIABLE VALUE REDUCED COST X 11 3. 000000 0. 000000 X 21 0. 000000 2800. 000000 X 31 8. 000000 0. 000000 X 41 0. 000000 1100. 000000 X 12 0. 000000 1700. 000000 X 22 0. 000000 1700. 000000 X 32 0. 000000 X 13 0. 000000 400. 000000 X 23 0. 000000 1500. 000000 X 14 12. 000000 0. 000000 第 16页

: max 3 x 1 -2 x 2+5 x 3 ? st ? x

: max 3 x 1 -2 x 2+5 x 3 ? st ? x 1+2 x 2 -x 3<2 ? x 1+4 x 2+x 3<4 ? x 1+x 2<3 ? 4 x 2+x 3<6 ? end : int x 1 : int x 2 Int 3 : int x 3 : go LINDO输出下列结果: 第 22页

LP OPTIMUM FOUND AT STEP 2 OBJECTIVE VALUE = 8. 0000 NEW INTEGER SOLUTION

LP OPTIMUM FOUND AT STEP 2 OBJECTIVE VALUE = 8. 0000 NEW INTEGER SOLUTION OF 8. 0000 AT BRANCH 0 PIVOT 2 RE-INSTALLING BEST SOLUTION. . . OBJECTIVE FUNCTION VALUE 1) 8. 000000 VARIABLE VALUE REDUCED COST X 1 1. 000000 -3. 000000 X 2 0. 000000 2. 000000 X 3 1. 000000 -5. 000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 2. 000000 0. 000000 3) 2. 000000 0. 000000 4) 2. 000000 0. 000000 5) 5. 000000 0. 000000 第 23页

: min 3 x 1+7 x 2 -x 3+x 4 ? st ? 2

: min 3 x 1+7 x 2 -x 3+x 4 ? st ? 2 x 1 -x 2+x 3 -x 4>1 ? x 1 -x 2+6 x 3+4 x 4>8 ? 5 x 1+3 x 2+x 4>5 ? end : int 4 : go LINDO输出下列结果: 第 25页

LP OPTIMUM FOUND AT STEP 4 OBJECTIVE VALUE = 2. 10526323 NEW INTEGER SOLUTION

LP OPTIMUM FOUND AT STEP 4 OBJECTIVE VALUE = 2. 10526323 NEW INTEGER SOLUTION OF 3. 0000 AT BRANCH 0 PIVOT 4 RE-INSTALLING BEST SOLUTION. . . OBJECTIVE FUNCTION VALUE 1) 3. 000000 VARIABLE VALUE REDUCED COST X 1 1. 000000 3. 000000 X 2 0. 000000 7. 000000 X 3 1. 000000 -1. 000000 X 4 1. 000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 1. 000000 0. 000000 3) 3. 000000 0. 000000 4) 1. 000000 0. 000000 第 26页

: MIN X 1+X 2+X 3+X 4+X 5+X 6 ? ST ? X 2+2

: MIN X 1+X 2+X 3+X 4+X 5+X 6 ? ST ? X 2+2 X 3+3 X 4+4 X 5+5 X 6>10000 ? 6 X 1+5 X 2+3 X 3+2 X 4+X 5>20000 ? END : GIN 6 : GO LINDO运行后输出以下结果: STATUS:OPTIMAL LP OPTIMUM FOUND AT STEP 3 OBJECTIVE VALUE = 5200. 00000 第 28页

FIX ALL VARS. (4) WITH RC > 0. 400000 E-01 NEW INTEGER SOLUTION OF

FIX ALL VARS. (4) WITH RC > 0. 400000 E-01 NEW INTEGER SOLUTION OF 5200. 00000 AT BRANCH 0 PIVOT 3 BOUND ON OPTIMUM: 5200. 000 ENUMERATION COMPLETE. BRANCHES= 0 PIVOTS= 3 LAST INTEGER SOLUTION IS THE BEST FOUND RE-INSTALLING BEST SOLUTION. . . OBJECTIVE FUNCTION VALUE 1) 5200. 000 VARIABLE VALUE REDUCED COST X 1 0. 000000 1. 000000 X 2 4000. 000000 1. 000000 X 3 0. 000000 1. 000000 X 4 0. 000000 1. 000000 X 5 0. 000000 1. 000000 X 6 1200. 000000 1. 000000 第 29页