Logical constraints
Supply Chain Analytics in Python
Aaren Stubberfield
Supply Chain Analytics Mgr.
Example problem
Maximum Weight 20,000 lbs
| Product | Weight (lbs) | Profitability ($US) |
|---|---|---|
| A | 12,800 | 77,878 |
| B | 10,900 | 82,713 |
| C | 11,400 | 82,728 |
| D | 2,100 | 68,423 |
| E | 11,300 | 84,119 |
| F | 2,300 | 77,765 |
- Select most profitable product to ship without exceeding weight limit
- Decision Variables:
- X$_{i}$ = 1 if product _i_ is selected else 0
- Objective:
- Maximize z = $\sum$ Profitability$_{i}$X$_{i}$
- Constraint:
- $\sum$ Weight$_{i}$X$_{i}$ < 20,0000
prod = ['A', 'B', 'C', 'D', 'E', 'F'] weight = {'A':12800, 'B':10900, 'C':11400, 'D':2100, 'E':11300, 'F':2300} prof = {'A':77878, 'B':82713, 'C':82728, 'D':68423, 'E':84119, 'F':77765}# Initialize Class model = LpProblem("Loading Truck Problem", LpMaximize) # Define Decision Variables x = LpVariable.dicts('ship_', prod, cat='Binary')# Define Objective model += lpSum([prof[i]*x[i] for i in prod]) # Define Constraint model += lpSum([weight[i]*x[i] for i in prod]) <= 20000# Solve Model model.solve() for i in prod: print("{} status {}".format(i, x[i].varValue))
Example result
Maximum Weight 20,000 lbs
| Product | Ship or Not |
|---|---|
| A | No |
| B | No |
| C | No |
| D | Yes |
| E | Yes |
| F | Yes |
Result
- Profitability: $230,307
- Weight of Products: 15,700 lbs
Logical constraint example 1
Either product E is selected or product D is selected, but not both.
- X$_{E}$ = 1 if product _i_ is selected else 0
- X$_{D}$ = 1 if product _i_ is selected else 0
- Constraint
- X$_{E}$ + X$_{D}$ ≤ 1
Code example - logical constraint example 1
model += x['E'] + x['D'] <= 1
prod = ['A', 'B', 'C', 'D', 'E', 'F']
weight = {'A':12800, 'B':10900, 'C':11400,
'D':2100, 'E':11300, 'F':2300}
prof = {'A':77878, 'B':82713, 'C':82728,
'D':68423, 'E':84119, 'F':77765}
# Initialize Class
model = LpProblem("Loading Truck Problem",
LpMaximize)
# Define Decision Variables
x = LpVariable.dicts('ship_', prod,
cat='Binary')
# Define Objective
model += lpSum([prof[i]*x[i] for i in prod])
# Define Constraint
model +=
lpSum([weight[i]*x[i] for i in prod]) <= 20000
model += x['E'] + x['D'] <= 1
# Solve Model
model.solve()
for i in prod:
print("{} status {}".format(i, x[i].varValue))
Logical constraint 1 example result
Maximum Weight 20,000 lbs
| Product | Ship or Not |
|---|---|
| A | No |
| B | No |
| C | Yes |
| D | Yes |
| E | No |
| F | Yes |
Result
- Profitability: $228,916
- Weight of Products: 15,800 lbs
Logical constraint example 2
If product D is selected then product B must also be selected.
- X$_{D}$ = 1 if product _i_ is selected else 0
- X$_{B}$ = 1 if product _i_ is selected else 0
- Constraint
- X$_{D}$ ≤ X$_{B}$
Code example - logical constraint example 2
model += x['D'] <= x['B']prod = ['A', 'B', 'C', 'D', 'E', 'F'] weight = {'A':12800, 'B':10900, 'C':11400, 'D':2100, 'E':11300, 'F':2300} prof = {'A':77878, 'B':82713, 'C':82728, 'D':68423, 'E':84119, 'F':77765} # Initialize Class model = LpProblem("Loading Truck Problem", LpMaximize) # Define Decision Variables x = LpVariable.dicts('ship_', prod, cat='Binary')
# Define Objective
model += lpSum([prof[i]*x[i] for i in prod])
# Define Constraint
model +=
lpSum([weight[i]*x[i] for i in prod]) <= 20000
model += x['D'] <= x['B']
# Solve Model
model.solve()
for i in prod:
print("{} status {}".format(i, x[i].varValue))
Logical constraint 2 example result
Maximum Weight 20,000 lbs
| Product | Ship or Not |
|---|---|
| A | No |
| B | Yes |
| C | No |
| D | Yes |
| E | No |
| F | Yes |
Result
- Profitability: $228,901
- Weight of Products: 15,300 lbs
Other logical constraints
| Logical Constraint | Constraint |
|---|---|
| If item _i_ is selected, then item _j_ is also selected. | x$_{i}$ - x$_{j}$ ≤ 0 |
| Either item _i_ is selected or item _j_ is selected, but not both. | x$_{i}$ + x$_{j}$ = 1 |
| If item _i_ is selected, then item _j_ is not selected. | x$_{i}$ - x$_{j}$ ≤ 1 |
| If item _i_ is not selected, then item _j_ is not selected. | -x$_{i}$ + x$_{j}$ ≤ 0 |
| At most one of items _i_, _j_, and _k_ are selected. | x$_{i}$ + x$_{j}$ + x$_{k}$ ≤ 1 |
1 James Orlin, and Ebrahim Nasrabadi. 15.053 Optimization Methods in Management Science. Spring 2013. Massachusetts Institute of Technology: MIT OpenCourseWare. License: Creative Commons BY-NC-SA.
Summary
- Reviewed examples of logical constraints
- Listed a table of other logical constraints
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Supply Chain Analytics in Python
