某工厂调查了解市场情况,估计在今后四个月内,市场对其产品的需求量如下表所示。
时期(月) | 需要量(产品单位) |
1 | 2 |
2 | 3 |
3 | 2 |
4 | 4 |
已知:对每个月来讲,生产一批产品的固定成本费为3 (千元),若不生产,则为零。每生产单位产品的成本费为1(千元)。同时,在任何一个月内,生产能力所允许的最大生产批量为不超过6个单位。又知每单位产品的库存费用为每月0.5(千元),同时要求在第一个月开始之初, 及在第四个月末,均无产品库存。问:在满足上述条件下,该厂应如何安排各个时期的生产与库存,使所花的总成本费用最低?
设第k月的需求量为Nk(k=1,2,3,4)
状态变量Xk:第k月初的库存量,X1=X5=0,0≤Xk≤Nk+…+N4
决策变量Uk:第k月的生产量,max{0,Nk-Xk}≤Uk≤min{6,Nk+…+N4 – Xk}
状态转移方程:Xk+1 = Uk + Xk – Nk
第k月的成本Vk = 0.5*(Xk – Nk) Uk=0
3 + Uk + 0.5*(Uk + Xk – Nk) Uk≠0
设Fk(Xk)是由第k月初的库存量Xk开始到第4月份结束这段时间的最优成本
则Fk(Xk) = min{Vk + Fk+1(Xk+1)} 1≤k≤4
= min{ 3 + Uk + 0.5*(Uk + Xk – Nk) + Fk+1(Uk+ Xk – Nk) } Uk≠0
min{ 0.5*(Xk – Nk) + Fk+1(Xk – Nk) } Uk=0
F5(X5)=0
四个月内的最优成本为F1(X1)=F1(0)
详细计算步骤如下:
(1)k=4时
0≤X4≤4,max{0,4 – X4}≤U4≤min{6,4-X4}
X4 | U4 | X5 | V4 | F5(X5) | V4 + F5(X5) |
0 | 4 | 0 | 7 | 0 | 7=F4(0) |
1 | 3 | 0 | 6 | 0 | 6=F4(1) |
2 | 2 | 0 | 5 | 0 | 5=F4(2) |
3 | 1 | 0 | 4 | 0 | 4=F4(3) |
4 | 0 | 0 | 0 | 0 | 0=F4(4) |
即对于状态X4的每个取值,都有唯一确定的决策变量U4使得F4(X4)最优
(2)k=3时
0≤X3≤6,max{0,2 – X3}≤U3≤min{6,6-X3}
X3 | U3 | X4 | V3 | F4(X4) | V3 + F4(X4) |
0 | 2 | 0 | 5 | 7 | 12 |
3 | 1 | 6.5 | 6 | 12.5 | |
4 | 2 | 8 | 5 | 13 | |
5 | 3 | 9.5 | 4 | 13.5 | |
6 | 4 | 11 | 0 | 11=F3(0) | |
1 | 1 | 0 | 4 | 7 | 11 |
2 | 1 | 5.5 | 6 | 11.5 | |
3 | 2 | 7 | 5 | 12 | |
4 | 3 | 8.5 | 4 | 12.5 | |
5 | 4 | 10 | 0 | 10=F3(1) | |
2 | 0 | 0 | 0 | 7 | 7=F3(2) |
1 | 1 | 4.5 | 6 | 10.5 | |
2 | 2 | 6 | 5 | 11 | |
3 | 3 | 7.5 | 4 | 11.5 | |
4 | 4 | 9 | 0 | 9 | |
3 | 0 | 1 | 0.5 | 6 | 6.5=F3(3) |
1 | 2 | 5 | 5 | 10 | |
2 | 3 | 6.5 | 4 | 10.5 | |
3 | 4 | 8 | 0 | 8 | |
4 | 0 | 2 | 1 | 5 | 6=F3(4) |
1 | 3 | 5.5 | 4 | 9.5 | |
2 | 4 | 7 | 0 | 7 | |
5 | 0 | 3 | 1.5 | 4 | 5.5=F3(5) |
1 | 4 | 6 | 0 | 6 | |
6 | 0 | 4 | 3 | 0 | 2=F3(6) |
(3)k=2时
0≤X2≤9,max{0,3 – X2}≤U2≤min{6,9-X2}
X2 | U2 | X3 | V2 | F3(X3) | V2 + F3(X3) |
0 | 3 | 0 | 6 | 11 | 17 |
4 | 1 | 7.5 | 10 | 17.5 | |
5 | 2 | 9 | 7 | 16=F2(0) | |
6 | 3 | 10.5 | 6.5 | 17 | |
1 | 2 | 0 | 5 | 11 | 16 |
3 | 1 | 6.5 | 10 | 16.5 | |
4 | 2 | 8 | 7 | 15=F2(1) | |
5 | 3 | 9.5 | 6.5 | 16 | |
6 | 4 | 11 | 6 | 17 | |
2 | 1 | 0 | 4 | 11 | 15 |
2 | 1 | 5.5 | 10 | 15.5 | |
3 | 2 | 7 | 7 | 14=F2(2) | |
4 | 3 | 8.5 | 6.5 | 15 | |
5 | 4 | 10 | 6 | 16 | |
6 | 5 | 11.5 | 5.5 | 17 | |
3 | 0 | 0 | 0 | 11 | 11=F2(3) |
1 | 1 | 4.5 | 10 | 14.5 | |
2 | 2 | 6 | 7 | 13 | |
3 | 3 | 7.5 | 6.5 | 14 | |
4 | 4 | 9 | 6 | 15 | |
5 | 5 | 10.5 | 5.5 | 16 | |
6 | 6 | 12 | 2 | 14 | |
4 | 0 | 1 | 0.5 | 10 | 10.5=F2(4) |
1 | 2 | 5 | 7 | 13 | |
2 | 3 | 6.5 | 6.5 | 13 | |
3 | 4 | 8 | 6 | 14 | |
4 | 5 | 9.5 | 5.5 | 15 | |
5 | 6 | 11 | 2 | 13 | |
5 | 0 | 2 | 1 | 7 | 8=F2(5) |
1 | 3 | 5.5 | 6.5 | 12 | |
2 | 4 | 7 | 6 | 13 | |
3 | 5 | 8.5 | 5.5 | 14 | |
4 | 6 | 10 | 2 | 12 | |
6 | 0 | 3 | 1.5 | 6.5 | 8=F2(6) |
1 | 4 | 6 | 6 | 12 | |
2 | 5 | 7.5 | 5.5 | 13 | |
3 | 6 | 9 | 2 | 11 | |
7 | 0 | 4 | 2 | 6 | 8=F2(7) |
1 | 5 | 6.5 | 5.5 | 12 | |
2 | 6 | 8 | 2 | 10 | |
8 | 0 | 5 | 2.5 | 5.5 | 8=F2(8) |
1 | 6 | 7 | 2 | 9 | |
9 | 0 | 6 | 3 | 2 | 5=F2(9) |
(4)k=1时
X1=0,max{0,2}≤U1≤min{6,11}
X1 | U1 | X2 | V1 | F2(X2) | V1 + F2(X2) |
0 | 2 | 0 | 5 | 16 | 21 |
3 | 1 | 6.5 | 15 | 21.5 | |
4 | 2 | 8 | 14 | 22 | |
5 | 3 | 9.5 | 11 | 20.5=F1(0) | |
6 | 4 | 11 | 10.5 | 21.5 |
由以上计算可得,4个月的总最优成本为F1(0) = 20.5(千元)
从k=1回溯,可得最优结果中各阶段的状态变量Xk和决策变量Uk如下表:
月份k | 产量Uk | 月初库存量Xk | 需求量Nk | 每月成本Vk |
1 | 5 | 0 | 2 | 9.5 |
2 | 0 | 3 | 3 | 0 |
3 | 6 | 0 | 2 | 11 |
4 | 0 | 4 | 4 | 0 |