问题描述

制造商需要提炼几种原料油，并将它们混合在一起，才能生产出一种可出售的食品。所需生油分为两类:

Category	Oil
Vegetable oils:	VEG 1 VEG 2
Non-vegetable oils:	OIL 1 OIL 2 OIL 3

制造商可以选择购买当月的原料油和/或在期货市场上购买并在下一个月交付。即期交割和期货市场的价格以美元/吨计算如下:

Month	VEG 1	VEG 2	OIL 1	OIL 2	OIL 3
January	110	120	130	110	115
February	130	130	110	90	115
March	110	140	130	100	95
April	120	110	120	120	125
May	100	120	150	110	105
June	90	100	140	80	135

还有一些其他的因素必须加以考虑。这些包括:

1. 最终的食品售价为每吨150美元。
2. 每种油(蔬菜油和非蔬菜油)都需要在不同的生产线上精炼。
3. 精炼能力有限，在任何一个月最多可精炼200吨植物油和250吨非植物油。
4. 此外，在精炼过程中没有浪费，所以精炼的原料油的总和将等于可用的精炼油的数量。
5. 提炼这些油的成本可以忽略不计。

除了上面提到的炼油限制之外，还有储存以供未来使用的原料油数量的限制，每储存一吨油都有成本。每种原料油限量1000吨，储存费为每月每吨$5$美元。生产企业不得储存生产出来的食品或成品油。

最终的食品产品的硬度必须在3到6之间的一个给定的硬度等级。在模型中，硬度线性混合，各原料油的硬度为:

Oils	Hardness
VEG 1	8.8
VEG 2	6.1
OIL 1	2.0
OIL2	4.2
OIL 3	5.0

在1月初，每种原料油有500吨储存。同样地，原油在6月底每种原料油也需要有500吨存储。

根据上述信息，每月应该做出什么样的购买和制造决策，以实现利润最大化?

数学模型

集合及索引

• $t \in \text{Months}=\{\text{Jan},\text{Feb},\text{Mar},\text{Apr},\text{May},\text{Jun}\}$: 月份集合
• $V=\{\text{VEG1},\text{VEG2}\}$: 植物油集合
• $N=\{\text{OIL1},\text{OIL2},\text{OIL3}\}$: 非植物油集合
• $o \in \text{Oils} = V \cup N$: 所有原料油种类的集合

参数

• $\text{price} \in \mathbb{R}^+$: 每吨售价
• $\text{init_store} \in \mathbb{R}^+$: 初始存储量(吨)
• $\text{target_store} \in \mathbb{R}^+$: 目标存储量(吨)
• $\text{holding_cost} \in \mathbb{R}^+$: 每吨油每月的存储费用
• $\text{veg_cap} \in \mathbb{R}^+$: 植物油每月精炼能力
• $\text{oil_cap} \in \mathbb{R}^+$: 非植物油每月精炼能力
• $\text{min_hardness} \in \mathbb{R}^+$: 成品油硬度下限
• $\text{max_hardness} \in \mathbb{R}^+$: 成品油硬度上限
• $\text{hardness}_o \in \mathbb{R}^+$: 油$o$的硬度
• $\text{cost}_{t,o} \in \mathbb{R}^+$:油 $o$ 在月份 $t$的预计费用

决策变量

• $\text{produce}_t \in \mathbb{R}^+$:月份$t$的生产量.
• $\text{buy}_{t,o} \in \mathbb{R}^+$: 月份$t$购买油 $o$的量
• $\text{consume}_{t,o} \in \mathbb{R}^+$: 月份$t$油 $o$的消耗量
• $\text{store}_{t,o} \in \mathbb{R}^+$: 月份$t$油 $o$的存储量

目标函数

• 利润: 研究时间范围内最大化利润 $$ \begin{equation} \text{Maximize} \quad Z = \sum_{t \in \text{Months}}\text{price}*\text{produce}_t - \sum_{t \in \text{Months}}\sum_{o \in \text{Oils}}(\text{cost}_{t,o}*\text{consume}_{t,o} + \text{holding_cost}*\text{store}_{t,o}) \tag{0} \end{equation} $$

约束条件

• 初始存储量: 一月份油$o$的购买量+初始存储量等于这个月油的消费量+存储量 $$ \begin{equation} \text{init_store} + \text{buy}_{Jan, o} = \text{consume}_{Jan,o} + \text{store}_{Jan,o} \quad \forall o \in \text{Oils} \tag{1} \end{equation} $$

• 平衡: 上个月的存储量+购买量=这个月的消耗量+存储量 $$ \begin{equation} \text{store}_{t-1,o} + \text{buy}_{t,o} = \text{consume}_{t,o} + \text{store}_{t,o} \quad \forall (t,o) \in \text{Months} \setminus \{\text{Jan}\} \times \text{Oils} \tag{2} \end{equation} $$

• 存储目标: 最后一个月的存储量要达到目标 $$ \begin{equation} \text{store}_{Jun,o} = \text{target_store} \quad \forall o \in \text{Oils} \tag{3} \end{equation} $$

• 精炼能力: 每个月的精炼量不能超出精炼能力 $$ \begin{equation} \sum_{o \in V}\text{consume}_{t,o} \leq \text{veg_cap} \quad \forall t \in \text{Months} \tag{4.1} \end{equation} $$

$$ \begin{equation} \sum_{o \in N}\text{consume}_{t,o} \leq \text{oil_cap} \quad \forall t \in \text{Months} \tag{4.2} \end{equation} $$

• 硬度: 成品油硬度值应满足硬度要求 $$ \begin{equation} \text{min_hardness}*\text{produce}_t \leq \sum_{o \in \text{Oils}} \text{hardness}_o*\text{consume}_{t,o} \leq \text{max_hardness}*\text{produce}_t \quad \forall t \in \text{Months} \tag{5} \end{equation} $$

• 质量守恒:消耗的原料油的量等于产出的成品油的量 $$ \begin{equation} \sum_{o \in \text{Oils}}\text{consume}_{t,o} = \text{produce}_t \quad \forall t \in \text{Months} \tag{6} \end{equation} $$

参考资料：

• H. Paul Williams, Model Building in Mathematical Programming, fifth edition (Page 253-254, 296-298, 349-350)

---

import numpy as np
import pandas as pd

import pulp as lp

# Parameters

months = ["Jan", "Feb", "Mar", "Apr", "May", "Jun"]

oils = ["VEG1", "VEG2", "OIL1", "OIL2", "OIL3"]

cost = {
    ('Jan', 'VEG1'): 110,
    ('Jan', 'VEG2'): 120,
    ('Jan', 'OIL1'): 130,
    ('Jan', 'OIL2'): 110,
    ('Jan', 'OIL3'): 115,
    ('Feb', 'VEG1'): 130,
    ('Feb', 'VEG2'): 130,
    ('Feb', 'OIL1'): 110,
    ('Feb', 'OIL2'): 90,
    ('Feb', 'OIL3'): 115,
    ('Mar', 'VEG1'): 110,
    ('Mar', 'VEG2'): 140,
    ('Mar', 'OIL1'): 130,
    ('Mar', 'OIL2'): 100,
    ('Mar', 'OIL3'): 95,
    ('Apr', 'VEG1'): 120,
    ('Apr', 'VEG2'): 110,
    ('Apr', 'OIL1'): 120,
    ('Apr', 'OIL2'): 120,
    ('Apr', 'OIL3'): 125,
    ('May', 'VEG1'): 100,
    ('May', 'VEG2'): 120,
    ('May', 'OIL1'): 150,
    ('May', 'OIL2'): 110,
    ('May', 'OIL3'): 105,
    ('Jun', 'VEG1'): 90,
    ('Jun', 'VEG2'): 100,
    ('Jun', 'OIL1'): 140,
    ('Jun', 'OIL2'): 80,
    ('Jun', 'OIL3'): 135
}


hardness = {"VEG1": 8.8, "VEG2": 6.1, "OIL1": 2.0, "OIL2": 4.2, "OIL3": 5.0}

price = 150
init_store = 500
target_store = 500
veg_cap = 200
oil_cap = 250

min_hardness = 3
max_hardness = 6
holding_cost = 5

food = lp.LpProblem(name='Food_Manufacture_I',sense=lp.LpMaximize)
# 每个月生产的油量
produce = lp.LpVariable.dicts(name="Produce",indexs=months,lowBound=0,cat='Continuous')
# 每个月购买油的量
buy = lp.LpVariable.dicts(name = "Buy",
                          indexs=[(m,o) for m in months for o in oils],
                          lowBound=0,
                          cat='Continuous')
# 每个月用掉的油的量
consume = lp.LpVariable.dicts( name = "Use", 
                              indexs=[(m,o) for m in months for o in oils],
                              lowBound=0,
                              cat='Continuous')
# 每个月的存储量
store = lp.LpVariable.dicts( name = "Store", 
                              indexs=[(m,o) for m in months for o in oils],
                            lowBound=0,
                              cat='Continuous')

#0. 目标函数
obj = lp.lpSum(price*produce[m] for m in months)-lp.lpSum(buy[m,o]*cost[m,o]+holding_cost*store[m,o] for m in months for o in oils)
food += obj

# 约束条件
#1. Initial Balance
for oil in oils:
    food += init_store+buy[months[0],oil]==consume[months[0], oil] + store[months[0], oil]

#2. Balance
for oil in oils:
    for month in months:
        if month != month[0]:
            food += store[months[months.index(month)-1], oil] + buy[month, oil]== consume[month, oil] + store[month, oil]

#3. Inventory Target
for oil in oils:
    food += store[months[-1], oil] == target_store

#4.1 Vegetable Oil Capacity
for month in months:
    food += lp.lpSum(consume[month, oil] for oil in oils if "VEG" in oil)<= veg_cap

#4.2 Non-vegetable Oil Capacity
for month in months:
    food += lp.lpSum(consume[month, oil] for oil in oils if "OIL" in oil)<= oil_cap

#5. Hardness
for month in months:
    food += lp.lpSum(hardness[oil]*consume[month, oil] for oil in oils)>= min_hardness*produce[month]

for month in months:
    food += lp.lpSum(hardness[oil]*consume[month, oil] for oil in oils) <= max_hardness*produce[month]

#6. Mass Conservation
for month in months:
    food += lp.lpSum(consume[month,oil] for oil in oils) == produce[month]

food.solve()

1

lp.value(food.objective)

107842.59259499997

rows = months.copy()
columns = oils.copy()
purchase_plan = pd.DataFrame(columns=columns, index=rows, data=0.0)

for month, oil in buy.keys():
    if (abs(lp.value(buy[month, oil])) > 1e-6):
        purchase_plan.loc[month, oil] = np.round(lp.value(buy[month, oil]), 1)
purchase_plan

	VEG1	VEG2	OIL1	OIL2	OIL3
Jan	0.0	0.0	0.0	0.0	0.0
Feb	0.0	0.0	0.0	0.0	0.0
Mar	0.0	0.0	0.0	0.0	0.0
Apr	0.0	0.0	0.0	0.0	0.0
May	0.0	0.0	0.0	0.0	750.0
Jun	659.3	540.7	0.0	750.0	0.0

rows = months.copy()
columns = oils.copy()
reqs = pd.DataFrame(columns=columns, index=rows, data=0.0)

for month, oil in consume.keys():
    if (abs(lp.value(consume[month, oil])) > 1e-6):
        reqs.loc[month, oil] = np.round(lp.value(consume[month, oil]), 1)
reqs

	VEG1	VEG2	OIL1	OIL2	OIL3
Jan	159.3	40.7	0.0	250.0	0.0
Feb	85.2	114.8	0.0	0.0	250.0
Mar	85.2	114.8	0.0	0.0	250.0
Apr	159.3	40.7	0.0	250.0	0.0
May	11.1	188.9	0.0	0.0	250.0
Jun	159.3	40.7	0.0	250.0	0.0

rows = months.copy()
columns = oils.copy()
store_plan = pd.DataFrame(columns=columns, index=rows, data=0.0)

for month, oil in store.keys():
    if (abs(lp.value(store[month, oil])) > 1e-6):
        store_plan.loc[month, oil] = np.round(lp.value(store[month, oil]), 1)
store_plan

	VEG1	VEG2	OIL1	OIL2	OIL3
Jan	340.7	459.3	500.0	250.0	500.0
Feb	255.6	344.4	500.0	250.0	250.0
Mar	170.4	229.6	500.0	250.0	0.0
Apr	11.1	188.9	500.0	0.0	0.0
May	0.0	0.0	500.0	0.0	500.0
Jun	500.0	500.0	500.0	500.0	500.0

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