47页1.1b
用图解法找不到满足所有约束条件的公共范围，所以该问题无可行解47页1.1d
无界解
1.2（b）
约束方程的系数矩阵A= 1 2 3 4
( )
2 1 1 2
P1 P2 P3 P4
最优解A=(0 1/2 2 0)T和(0 0 1 1)T
49页13题
设Xij为第i月租j个月的面积
minz=2800x11+2800x21+2800x31+2800x41+4500x12+4500x22+4500x32+6000x13
+6000x23+7300x14
s.t.
x11+x12+x13+x14≥15
x12+x13+x14+x21+x22+x23≥10
x13+x14+x22+x23+x31+x32≥20
x14+x23+x32+x41≥12
Xij≥0
用excel求解为：
用LINDO求解：
LP OPTIMUM FOUND A T STEP 3
OBJECTIVE FUNCTION V ALUE
1) 118400.0
VARIABLE V ALUE REDUCED COST Z 0.000000 1.000000
X11 3.000000 0.000000
X21 0.000000 2800.000000
X31 8.000000 0.000000
X41 0.000000 1100.000000
X12 0.000000 1700.000000
X22 0.000000 1700.000000
X32 0.000000 0.000000
X13 0.000000 400.000000
X23 0.000000 1500.000000
X14 12.000000 0.000000
ROW SLACK OR SURPLUS DUAL PRICES
2) 0.000000 -2800.000000
3) 2.000000 0.000000
4) 0.000000 -2800.000000
5) 0.000000 -1700.000000
NO. ITERATIONS= 3
答若使所费租借费用最小，需第一个月租一个月租期300平方米，租四个月租期1200平方米，第三个月租一个月租期800平方米，
50页14题
设a1，a2，a3, a4, a5分别为在A1, A2, B1, B2, B3加工的Ⅰ产品数量，b1，b2，b3分别为在A1, A2, B1加工的Ⅱ产品数量，c1为在A2，B2上加工的Ⅲ产品数量。则目标函数为‘
maxz= (1.25-0.25)( a1+a2+a3)+( 2-0.35) b3+( 2.8-0.5)c1 -0.05 (a1+b1)-
0.03 (a2+b2+c1)- 0.06 (a3+b3)-0.11(a4+c1)-0.05a5
=0. 95a1+0. 97a2+0. 94a3+1.5b3+2.1c1-0.05b1-0.11a4-0.05a5
s.t.
5a1+10b1≤6000
7a2+b2+12c1≤10000
6a3+8a3≤4000
4a4+11c1≤7000
7a5≤4000
a1+a2-a3-a4-a5=0
b1+b2-b3=0
a1，a2，a3, a4, a5, b1，b2，b3, c1≥0
用lindo求解得：
LP OPTIMUM FOUND AT STEP 6
OBJECTIVE FUNCTION V ALUE
1) 16342.29
V ARIABLE V ALUE REDUCED COST
A1 1200.000000 0.000000
A2 0.000000 9.640000
A3 285.714294 0.000000
B3 10000.000000 0.000000
C1 0.000000 15.900000
B1 0.000000 0.230000
A4 342.857147 0.000000
A5 571.428589 0.000000
B2 10000.000000 0.000000
ROW SLACK OR SURPLUS DUAL PRICES
2) 0.000000 0.168000
3) 0.000000 1.500000
4) 0.000000 0.075000
5) 5628.571289 0.000000
6) 0.000000 0.008571
7) 0.000000 0.110000
8) 0.000000 -1.500000
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
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 V ALUE
1) 27.00000
V ARIABLE V ALUE 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
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 V ALUE
1) 27.00000
V ARIABLE V ALUE 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 V ARIABLE 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
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 V ALUE
1) 27.50000
V ARIABLE V ALUE 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 V ALUE