Python Mathematical Modeling PuLP Library Linear Programming Advanced Dictionary based Detailed Explanation
- 2021-12-09 09:32:57
- OfStack
1. Dictionary-based creation planning
In the previous section, we introduced the use of LpVariable pairs to define each decision variable one by one, setting the name, type and upper and lower bounds. Similarly, we need to set model parameters one by one for constraints. In large-scale planning problems, it is very cumbersome and inefficient to define variables and set model parameters one by one. The Pulp library provides a shortcut that combines the loops and containers of the Python language and uses dictionaries to create questions.
-(1) Use a shortcut to create a planning problem. You can create multiple variables with dictionary type (dict), such as:
name = ['Scrap 1', 'Scrap 2', 'Scrap 3', 'Scrap 4', 'Nickel', 'Chromium', 'Molybdenum']
# A dictionary of the costs of each of the Ingredients is created
mass = pulp. LpVariable. dicts ("feedstock", material, lowBound = 0, cat = 'Continuous')
-(2) Use the dictionary type (dict) to set the parameters of the target function and constraint, such as:
cost = {
'Scrap 1': 16,
'Scrap 2': 10,
'Scrap 3': 8,
'Scrap 4': 9,
'Nickel': 48,
'Chromium': 60,
'Molybdenum': 53}
-(3) Use the traversal loop structure to set objective functions and constraints, such as:
AlloyModel + = pulp. lpSum ([cost [item] * mass [item] for item in material]), "Total Production Cost"
AlloyModel + = pulp. lpSum ([mass [item] for item in material]) = = 1000, "Quality Constraints"
2. Cases of linear programming problems
In this paper, taking the feeding problem of alloy steel production as an example, the shortcut method of creating problems based on list and dictionary is analyzed.
Problem description:
An iron and steel plant produces alloys that meet the chemical composition requirements by smelting recycled metal scrap and adding a certain amount of new materials. It is planned to produce 1000 kg of alloys.
The main component of all metal scrap is iron, and different metal scrap also contains various trace elements.
The content ratio, available quantity and unit cost of each component of metal waste and new materials are shown in the following table. The content requirements of each component in the generated alloy are also shown in the table.
How to arrange the feeding ratio to minimize the material cost under the condition of meeting the requirements of alloy component content?
| 材料 | 碳 | 镍 | 铬 | 钼 | 可用量 | 成本 |
|---|---|---|---|---|---|---|
| 废料1 | 0.80 | 18.0 | 12.0 | 0.0 | 75 | 16 |
| 废料2 | 0.70 | 3.2 | 1.1 | 0.1 | 250 | 10 |
| 废料3 | 0.85 | 0 | 0 | 0 | 不限 | 8 |
| 废料4 | 0.40 | 0 | 0 | 0 | 不限 | 9 |
| 镍 | 0 | 100 | 0 | 0 | 不限 | 48 |
| 铬 | 0 | 0 | 100 | 0 | 不限 | 60 |
| 钼 | 0 | 0 | 0 | 100 | 不限 | 53 |
| 合金下限 | 0.65 | 3.0 | 1.0 | 1.1 | / | / |
| 合金上限 | 0.75 | 3.5 | 1.2 | 1.3 | / | / |
3. Build a model
(1) Decision variables
x1: Scrap 1 Amount (kg)
x2: Scrap 2 Amount (kg)
x3: Scrap 3 Amount (kg)
x4: Scrap 4 Amount (kg)
x5: Raw Nickel Content (kg)
x6: Chromium content (kg)
x7: Molybdenum dosage (kg)
(2) Objective function
min cost = 16*x1 + 10*x2 + 8*x3 + 9*x4 + 48*x5 + 60*x6 + 53*x7
(3) Constraints
0.8*x1 + 0.7*x2 + 0.85*x3 + 0.40*x4 > = 0.65*1000
0.8*x1 + 0.7*x2 + 0.85*x3 + 0.40*x4 < = 0.75*1000
18.0*x1 + 3.2*x2 + 100.0*x5 > = 3.0*1000
18.0*x1 + 3.2*x2 + 100.0*x5 < = 3.5*1000
12.0*x1 + 1.1*x2 + 100.0*x6 > = 1.0*1000
12.0*x1 + 1.1*x2 + 100.0*x6 > = 1.2*1000
0.1*x2 + 100.0*x7 > = 1.1*1000
0.1*x2 + 100.0*x7 > = 1.3*1000
(4) Variable value range
xi > = 0, i = 1, 2, … 7
x1 < = 75, x2 < = 250
4. PuLP Program 1: Defining variables one by one using LpVariable
This procedure is the same as the previous method, using LpVariable to define variables one by one. The complete program code is as follows:
import pulp # Import pulp Library
# 1. Establish optimization problems AlloyLP: Find the minimum (LpMinimize)
AlloyLP = pulp.LpProblem(" Optimization of alloy production materials ", sense=pulp.LpMinimize) # Define the problem and find the minimum value
# 2. Define decision variables x1~x7
x1 = pulp.LpVariable(' Waste 1#', lowBound=0, upBound=75.0, cat='Continuous') # Definition x1
x2 = pulp.LpVariable(' Waste 2#', lowBound=0, upBound=250., cat='Continuous') # Definition x2
x3 = pulp.LpVariable(' Waste 3#', lowBound=0, cat='Continuous') # Definition x3
x4 = pulp.LpVariable(' Waste 4#', lowBound=0, cat='Continuous') # Definition x4
x5 = pulp.LpVariable(' Raw nickel ', lowBound=0, cat='Continuous') # Definition x5
x6 = pulp.LpVariable(' Raw material chromium ', lowBound=0, cat='Continuous') # Definition x6
x7 = pulp.LpVariable(' Molybdenum ', lowBound=0, cat='Continuous') # Definition x7
# 3. Define the objective function cost
AlloyLP += (16*x1 + 10*x2 + 8*x3 + 9*x4 + 48*x5 + 60*x6 + 53*x7) # Feeding cost
# 4. Set constraints
AlloyLP += (x1 + x2 + x3 + x4 + x5 + x6 + x7 == 1000) # Equality constraint
AlloyLP += (0.8*x1 + 0.7*x2 + 0.85*x3 + 0.4*x4 >= 0.65*1000) # Inequality constraint
AlloyLP += (0.8*x1 + 0.7*x2 + 0.85*x3 + 0.4*x4 <= 0.75*1000) # Inequality constraint
AlloyLP += (18.0*x1 + 3.2*x2 + 100.0*x5 >= 3.0*1000) # Inequality constraint
AlloyLP += (18.0*x1 + 3.2*x2 + 100.0*x5 <= 3.5*1000) # Inequality constraint
AlloyLP += (12.0*x1 + 1.1*x2 + 100.0*x6 >= 1.0*1000) # Inequality constraint
AlloyLP += (12.0*x1 + 1.1*x2 + 100.0*x6 <= 1.2*1000) # Inequality constraint
AlloyLP += (0.1*x2 + 100.0*x7 >= 1.1*1000) # Inequality constraint
AlloyLP += (0.1*x2 + 100.0*x7 <= 1.3*1000) # Inequality constraint
AlloyLP += (x1 + x2 + x3 + x4 + x5 + x6 + x7 == 1000) # Equality constraint
# 5. Solving linear programming problems
AlloyLP.solve()
# 6. Output optimization results
print(AlloyLP) # Output problem setting parameters and conditions
# print(" Solving state :", pulp.LpStatus[AlloyLP.status]) # Output solution status
for v in AlloyLP.variables():
print(v.name, " = ", v.varValue) # Output the optimal value of each variable
print(" Minimum material cost = ", pulp.value(AlloyLP.objective)) # Output the objective function value of the optimal solution
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5. PuLP Program 2: Defining Decision Variables and Constraints Using dict
This program uses dict to define variables, objective functions and constraint parameters, which is convenient for parameter setting of complex problems.
import pulp # Import pulp Library
# 1. Establish optimization problems AlloyLP: Find the minimum (LpMinimize)
AlloyLP = pulp.LpProblem(" Optimization of alloy production materials ", sense=pulp.LpMinimize) # Define the problem and find the minimum value
# 2. Define decision variables x1~x7
x1 = pulp.LpVariable(' Waste 1#', lowBound=0, upBound=75.0, cat='Continuous') # Definition x1
x2 = pulp.LpVariable(' Waste 2#', lowBound=0, upBound=250., cat='Continuous') # Definition x2
x3 = pulp.LpVariable(' Waste 3#', lowBound=0, cat='Continuous') # Definition x3
x4 = pulp.LpVariable(' Waste 4#', lowBound=0, cat='Continuous') # Definition x4
x5 = pulp.LpVariable(' Raw nickel ', lowBound=0, cat='Continuous') # Definition x5
x6 = pulp.LpVariable(' Raw material chromium ', lowBound=0, cat='Continuous') # Definition x6
x7 = pulp.LpVariable(' Molybdenum ', lowBound=0, cat='Continuous') # Definition x7
# 3. Define the objective function cost
AlloyLP += (16*x1 + 10*x2 + 8*x3 + 9*x4 + 48*x5 + 60*x6 + 53*x7) # Feeding cost
# 4. Set constraints
AlloyLP += (x1 + x2 + x3 + x4 + x5 + x6 + x7 == 1000) # Equality constraint
AlloyLP += (0.8*x1 + 0.7*x2 + 0.85*x3 + 0.4*x4 >= 0.65*1000) # Inequality constraint
AlloyLP += (0.8*x1 + 0.7*x2 + 0.85*x3 + 0.4*x4 <= 0.75*1000) # Inequality constraint
AlloyLP += (18.0*x1 + 3.2*x2 + 100.0*x5 >= 3.0*1000) # Inequality constraint
AlloyLP += (18.0*x1 + 3.2*x2 + 100.0*x5 <= 3.5*1000) # Inequality constraint
AlloyLP += (12.0*x1 + 1.1*x2 + 100.0*x6 >= 1.0*1000) # Inequality constraint
AlloyLP += (12.0*x1 + 1.1*x2 + 100.0*x6 <= 1.2*1000) # Inequality constraint
AlloyLP += (0.1*x2 + 100.0*x7 >= 1.1*1000) # Inequality constraint
AlloyLP += (0.1*x2 + 100.0*x7 <= 1.3*1000) # Inequality constraint
AlloyLP += (x1 + x2 + x3 + x4 + x5 + x6 + x7 == 1000) # Equality constraint
# 5. Solving linear programming problems
AlloyLP.solve()
# 6. Output optimization results
print(AlloyLP) # Output problem setting parameters and conditions
# print(" Solving state :", pulp.LpStatus[AlloyLP.status]) # Output solution status
for v in AlloyLP.variables():
print(v.name, " = ", v.varValue) # Output the optimal value of each variable
print(" Minimum material cost = ", pulp.value(AlloyLP.objective)) # Output the objective function value of the optimal solution
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6. Python program and running results
Program 1 and Program 2 run the same result as follows:
Welcome to the CBC MILP Solver
Version: 2.9.0
Build Date: Feb 12 2015
Steel production problems :
MINIMIZE
16* Raw materials _ Waste 1 + 10* Raw materials _ Waste 2 + 8* Raw materials _ Waste 3 + 9* Raw materials _ Waste 4 + 53* Raw materials _ Molybdenum + 60* Raw materials _ Chromium + 48* Raw materials _ Nickel + 0
SUBJECT TO
Quality constraint : Raw materials _ Waste 1 + Raw materials _ Waste 2 + Raw materials _ Waste 3 + Raw materials _ Waste 4 + Raw materials _ Molybdenum + Raw materials _ Chromium + Raw materials _ Nickel = 1000
Minimum proportion of carbon : 0.8 Raw materials _ Waste 1 + 0.7 Raw materials _ Waste 2 + 0.85 Raw materials _ Waste 3 + 0.4 Raw materials _ Waste 4 >= 650
Maximum proportion of carbon : 0.8 Raw materials _ Waste 1 + 0.7 Raw materials _ Waste 2 + 0.85 Raw materials _ Waste 3 + 0.4 Raw materials _ Waste 4 <= 750
Minimum proportion of nickel : 18 Raw materials _ Waste 1 + 3.2 Raw materials _ Waste 2 + 100 Raw materials _ Nickel >= 3000
Maximum proportion of nickel : 18 Raw materials _ Waste 1 + 3.2 Raw materials _ Waste 2 + 100 Raw materials _ Nickel <= 3500
Minimum proportion of chromium : 12 Raw materials _ Waste 1 + 1.1 Raw materials _ Waste 2 + 100 Raw materials _ Chromium >= 1000
Maximum proportion of chromium : 12 Raw materials _ Waste 1 + 1.1 Raw materials _ Waste 2 + 100 Raw materials _ Chromium <= 1200
Minimum proportion of molybdenum : 0.1 Raw materials _ Waste 2 + 100 Raw materials _ Molybdenum >= 1100
Maximum proportion of molybdenum : 0.1 Raw materials _ Waste 2 + 100 Raw materials _ Molybdenum <= 1300
Waste 1 Available dosage : Raw materials _ Waste 1 <= 75
Waste 2 Available dosage : Raw materials _ Waste 2 <= 250
VARIABLES
Raw materials _ Waste 1 Continuous
Raw materials _ Waste 2 Continuous
Raw materials _ Waste 3 Continuous
Raw materials _ Waste 4 Continuous
Raw materials _ Molybdenum Continuous
Raw materials _ Chromium Continuous
Raw materials _ Nickel Continuous
Optimized state : Optimal
Raw materials _ Waste 1 = 75.0
Raw materials _ Waste 2 = 90.909091
Raw materials _ Waste 3 = 672.28283
Raw materials _ Waste 4 = 137.30808
Raw materials _ Molybdenum = 10.909091
Raw materials _ Chromium = 0.0
Raw materials _ Nickel = 13.590909
Optimal total cost = 9953.671725000002
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