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NO. ITERATIONS= 6
计算lindo截屏
2.1a:
对偶问题为： maxz=2y1+3y2+5y3 s.t.
y1+2y2+y3≤2 3y3+y2+4y3≤2 4y1+3y2+3y3=4 y1≥0, y 2≤0,y3无约束
因为原问题的对偶问题的对偶问题仍是原问题，因此本问题的对偶问题的对偶问题为：
minz=2x1+2x2+4x3 s.t.
x1+3x2+4x3≥2 2x1+x2+3x3≤3 x1+4x2+3x3=5
x1,x2≥0,x3无约束
81页2.12
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a)设x1,x2,x3分别为A，B，C产品数量
maxz=3x1+x2+4x3 s.t.
6x1+3x2+5x3≤45 3x1+4x2+5x3≤30 x1,x2,x3≥0 用lomdo求解为
LP OPTIMUM FOUND AT STEP 2
OBJECTIVE FUNCTION VALUE
1) 27.00000
VARIABLE VALUE REDUCED COST X1 5.000000 0.000000 X2 0.000000 2.000000 X3 3.000000 0.000000 X1,X2,X3 0.000000 0.000000
ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 0.200000 3) 0.000000 0.600000 4) 0.000000 0.000000
NO. ITERATIONS= 2
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最大生产计划为A生产5个单位，C生产3个单位
b)
LP OPTIMUM FOUND AT STEP 2
OBJECTIVE FUNCTION VALUE
1) 27.00000
VARIABLE VALUE REDUCED COST X1 5.000000 0.000000 X2 0.000000 2.000000 X3 3.000000 0.000000 X1,X2,X3 0.000000 0.000000
ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 0.200000 3) 0.000000 0.600000 4) 0.000000 0.000000
NO. ITERATIONS= 2
RANGES IN WHICH THE BASIS IS UNCHANGED:
OBJ COEFFICIENT RANGES
VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE X1 3.000000 1.800000 0.600000 X2 1.000000 2.000000 INFINITY X3 4.000000 1.000000 1.500000 X1,X2,X3 0.000000 0.000000 INFINITY
RIGHTHAND SIDE RANGES
ROW CURRENT ALLOWABLE ALLOWABLE RHS INCREASE DECREASE 2 45.000000 15.000000 15.000000 3 30.000000 15.000000 7.500000 4 0.000000 0.000000 INFINITY
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可知A产品的利润变化范围【6. 8,2.4】,上述计划不变。
c)
设x4为产品D的数量
maxz=3x1+x2+4x3+3x4 s.t.
6x1+3x2+5x3+8x4≤45 3x1+4x2+5x3+2x4≤30 x1,x2,x3 ,x4≥0 用lomdo求解为
LP OPTIMUM FOUND AT STEP 0
OBJECTIVE FUNCTION VALUE
1) 27.50000
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VARIABLE VALUE REDUCED COST X1 0.000000 0.100000 X2 0.000000 1.966667 X3 5.000000 0.000000 X4 2.500000 0.000000
ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 0.233333 3) 0.000000 0.566667
NO. ITERATIONS= 0
安排生产D有利，新最有生产计划为x1=x2=0,x3=5,x4=2.5，利润为27.5
d)
maxz=3x1+x2+4x3-0.4y s.t.
6x1+3x2+5x3≤45 3x1+4x2+5x3-y≤30 x1,x2,x3,y≥0 用lomdo求解为
LP OPTIMUM FOUND AT STEP 0
OBJECTIVE FUNCTION VALUE
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