运筹学本科版答案
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运筹学本科版答案
【篇一:运筹学课后习题答案】
xt>1.用xj(j=1.2…5)分别代表5中饲料的采购数,线性规划模型: minz?0.2x1?0.7x2?0.4x3?0.3x4?0.8x5st.3x1?2x2?x3?6x4
+18x5?700x1?0.5x2?0.2x3+2x4?x5?30
0.5x1?x2?0.2x3+2x4?0.8x5?100
2.解:设x1x2x3x4x5x6x表示在第i个时期初开始工作的护士人数,z表示所需的总人数,则
minz?x1?x2?x3?x4?x5?x6st.x1?x6?60x?x2?701
x2?x3?60x3?x4?50x4?x5?20x5?x6?30
xj(j?1,2,3,4,5,6)?0
3.解:设用i=1,2,3分别表示商品a,b,c,j=1,2,3分别代表前,中,后舱,xij表示装于j舱的i种商品的数量,z表示总运费
收入则:
maxz?1000(x11?x12?x13)?700(x21?x22?x23)?600(x31?x32?x3 3)st.x11?x12?x13?600x21?x22?x23?1000
x31?x32?x33?80010x11?5x21?7x31?4001
0x12?5x22?7x32?540010x13?5x23?7x33?1500
8x11?6x21?5x31?20008x12?6x22?5x32?3000
8x13?6x23?5x33?15008x?6x21?5x3111?0.15
8x12?6x22?5x328x?6x23?5x33
13?0.15
8x12?6x22?5x328x?6x21?5x31
11?0.1
8x13?6x23?5x33xij?0(i?1,2.3.j?1,2,3)
xi(i?1,2.3.4.5.6)?0
5. (1)
z = 4
(2)
maxz?x1?x2
st.6x1?10x2?120x1?x2?705?x1?10
解:如图:由图可得:
x?(10,6);z
*
t
*
3?x2?8
?16
*
即该问题具有唯一最优解x
?
(10,6)
t
(3)
无可行解
(4)
maxz?5x1?6x2st.2x1?x2?2?2x1?3x2?2 x1,x2?0
如图:
由图知,该问题具有无界解。
6(1)
maxz?3x1?4x2?2x3?5xst.-2x1?x
2
4
?5x
4
?0x
5
?0x
6
?2x
3
?x
44
4
?x4?2
x1?x2?x3?2x-2x1?3x2?x3?x
?2x4+x5?14?x4-x6?6
6
(2)
x1,x2,x3,x4,x4,x5,x
?0
maxz?2x1?2x2?3x3?3xst.x1?x 2
3
?0x
4
?x3?x3?4
2x1?x2?x3?x3+x4?6 x1,x2,x3,x3,x
4
?0?12
?8??3?
310
6?40
300
0??
20=(p1 p2 p3 p4 p5 p6) ?
0?1??
7.1)系数矩阵a:c6
b1?p1p2
3
?20种组合
可b1构成基。
23?8
3
310
???54?;∴00
求b1的基本解,
?12
?8??3?
310
6?40
9????10=??
?0???
010
001
?
?16/3
?-7/6?? 0
(b,b)=
∴ y1=(0,16/3,-7/6,0,0,0)t
同理y2=(0,10,0,-7,0,0)t y3=(0, 3,0,0,7/2,0)t y4=(7/4,-4,0,0,0,21/4)t y5=(0,0,-5/2,8,0,0)t y6=(0,0,3/2,0,8,0)ty7=(1,0,-1/2,0,0,3)t y8=(0,0,0,3,5,0)t y9=(5/4,0,0,-2,0,15/4)t
y10=(0, 3,-7/6,0,0,0)t y11=(0,0,-5/2,8,0,0)t y12=(0,0,-5/2,3,5,0)ty13=(4/3,0,0,0,2,3/4)t y14=(0,10,0,-7,0,0)t y15=(0, 3,0,0,7/3,0)t
y16=(0,0,3/2,0,8,0)t
基可行解:(每个x值都大于0),(y3,y6,y8,y12,y13,
y15,y16)最优解:(y3,y6, y15,y16) zmax=3
[p2 p3 p4],[p2 p3 p5],[p3 p4 p5],[p2 p4 p5]为奇异,∴只有16个基。
c4?6
2
1
011
b?2??5?0?1?2???3
8.基的定义
3
∴x1 x2 x3所对应的列向量可以构成基
011
6?
?3?4??
b 由 x1 x2 x3 列向量构成 =
n 由非基变量对应的向量构成 =
?1
?2???3
011
6
1??1