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我想找到一个优化问题的解决方案。其目的是通过以低价购买并以更高的价格出售来最大化利润。存在诸如最大库存水平和最大买卖单位数量等限制。此外，买卖限制取决于库存水平。我问了一个类似的问题，尽管这里没有最后一个条件R optimization buy sell。

这是一个例子：

price = c(12, 11, 12, 13, 16, 17, 18, 17, 18, 16, 17, 13)
capacity = 25
max_units_buy_30 = 4 # when inventory level is lower then 30% it is possible to buy 0 to 4 units
max_units_buy_65 = 3 # when inventory level is between 30% and 65% it is possible to buy 0 to 3 units
max_units_buy_100 = 2 # when inventory level is between 65% and 100% it is possible to buy 0 to 2 units
max_units_sell_30 = 4 # when inventory level is lower then 30% it is possible to sell 0 to 4 units
max_units_sell_70 = 6 # when inventory level is between 30% and 70% it is possible to sell 0 to 6 units
max_units_sell_100 = 8 # when inventory level is between 70% and 100% it is possible to sell 0 to 8 units
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1 回答 1

1

这里发生了很多事情。

  1. 描述

描述好像有问题。“最高卖出/价格取决于库存水平。 ”这似乎是错误的。从数据来看，价格看起来是恒定的，但买卖限制取决于库存水平。

  1. 时间

把握好时机很重要。通常，我们将buy和视为在tsell期间发生的事情（我们称它们为流变量）。是一个存量变量，在t期末测量。这么说和依赖有点奇怪（我们在时间上倒退了）。当然，我们可以对其建模并求解（我们求解为联立方程，所以我们可以做这些事情）。但是，这在现实世界中可能没有意义。或许我们应该看看以改变和。invsell[t]buy[t]inv[t]inv[t-1]buy[t]sell[t]

  1. 细分库存水平。

我们需要将库存水平分成多个部分。我们有以下部分：

0%-30%
30%-65%
65%-70%
70%-100%

我们将二进制变量与每个段相关联：

inventory in [0%-30%]  <=> δ[1,t] = 1, all other zero
             [30%-65%]     δ[2,t] = 1 
             [65%-70%]     δ[3,t] = 1 
             [70%-100%]    δ[4,t] = 1

因为我们需要在所有时间段都这样做，所以我们添加了一个额外的索引 t。警告：我们将关联δ[k,t]t 期开始时的库存，即inv[t-1]。我们可以通过根据我们所在的细分更改下限和上限δ[k,t]来链接。inv[t-1]

  1. 买入/卖出界限

类似于库存的界限，我们有以下买卖上限：

     segment     buy   sell
     0%-30%       4     4 
     30%-65%      3     6
     65%-70%      2     6
     70%-100%     2     8

第一步是建立一个数学模型。这里发生了太多事情，我们可以立即编写代码。数学模型是我们的“设计”。所以我们开始：

在此处输入图像描述

有了这个，我们可以开发一些 R 代码。在这里，我们使用 CVXR 作为建模工具，使用 GLPK 作为 MIP 求解器。

> library(CVXR)
> 
> # data
> price = c(12, 11, 12, 13, 16, 17, 18, 17, 18, 16, 17, 13)
> capacity = 25
> max_units_buy = 4
> max_units_sell = 8
> 
> # capacity segments
> s <- c(0,0.3,0.65,0.7,1)
> 
> # corresponding lower and upper bounds
> invlb <- s[1:(length(s)-1)] * capacity
> invlb
[1]  0.00  7.50 16.25 17.50
> invub <- s[2:length(s)] * capacity
> invub
[1]  7.50 16.25 17.50 25.00
> 
> buyub <- c(4,3,2,2)
> sellub <- c(4,6,6,8)
> 
> # number of time periods
> NT <- length(price)
> NT
[1] 12
> 
> # number of capacity segments
> NS <- length(s)-1
> NS
[1] 4
> 
> # Decision variables
> inv = Variable(NT,integer=T)
> buy = Variable(NT,integer=T)
> sell = Variable(NT,integer=T)
> delta = Variable(NS,NT,boolean=T)
> 
> # Lag operator
> L = cbind(rbind(0,diag(NT-1)),0)
> 
> # optimization model
> problem <- Problem(Maximize(sum(price*(sell-buy))),
+                    list(inv == L %*% inv + buy - sell,
+                         sum_entries(delta,axis=2)==1, 
+                         L %*% inv >= t(delta) %*% invlb,
+                         L %*% inv <= t(delta) %*% invub,
+                         buy <= t(delta) %*% buyub,
+                         sell <= t(delta) %*% sellub,
+                         inv >= 0, inv <= capacity,
+                         buy >= 0, sell >= 0))
> result <- solve(problem,verbose=T)
GLPK Simplex Optimizer, v4.47
120 rows, 84 columns, 369 non-zeros
      0: obj =  0.000000000e+000  infeas = 1.200e+001 (24)
*    23: obj =  0.000000000e+000  infeas = 0.000e+000 (24)
*    85: obj = -9.875986758e+001  infeas = 0.000e+000 (2)
OPTIMAL SOLUTION FOUND
GLPK Integer Optimizer, v4.47
120 rows, 84 columns, 369 non-zeros
84 integer variables, 48 of which are binary
Integer optimization begins...
+    85: mip =     not found yet >=              -inf        (1; 0)
+   123: >>>>> -8.800000000e+001 >= -9.100000000e+001   3.4% (17; 0)
+   126: >>>>> -9.000000000e+001 >= -9.100000000e+001   1.1% (9; 11)
+   142: mip = -9.000000000e+001 >=     tree is empty   0.0% (0; 35)
INTEGER OPTIMAL SOLUTION FOUND
> cat("status:",result$status)
status: optimal
> cat("objective:",result$value)
objective: 90
> print(result$getValue(buy))
      [,1]
 [1,]    3
 [2,]    4
 [3,]    4
 [4,]    3
 [5,]    3
 [6,]    1
 [7,]    0
 [8,]    0
 [9,]    0
[10,]    4
[11,]    0
[12,]    0
> print(result$getValue(sell))
      [,1]
 [1,]    0
 [2,]    0
 [3,]    0
 [4,]    0
 [5,]    0
 [6,]    0
 [7,]    8
 [8,]    6
 [9,]    4
[10,]    0
[11,]    4
[12,]    0
> print(result$getValue(inv))
      [,1]
 [1,]    3
 [2,]    7
 [3,]   11
 [4,]   14
 [5,]   17
 [6,]   18
 [7,]   10
 [8,]    4
 [9,]    0
[10,]    4
[11,]    0
[12,]    0
> print(result$getValue(delta))
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,]    1    1    1    0    0    0    0    0    1     1     1     1
[2,]    0    0    0    1    1    0    0    1    0     0     0     0
[3,]    0    0    0    0    0    1    0    0    0     0     0     0
[4,]    0    0    0    0    0    0    1    0    0     0     0     0
>

所以，我想有人为此欠我一瓶好干邑。

于 2020-05-20T10:44:10.077 回答