动态规划解生产存储问题
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 |