Order constraints in optimisation
I have a set of many (10000+) items, from which have I have to choose exactly 20 items. I can only choose each item once. My items have profits, and costs, as well as several boolean properties (such as colour). I need to output the results in a specific order: in particular I need the first and third items to be blue, and the second and fourth items to be red.
Each item is represented as a tuple:
item = ('item name', cost, profit, is_blue, is_red)
as an example
vase = ['Ming Vase', 1000, 10000, 0, 1]
plate = ['China Plate', 10, 5, 1, 0]
and the total set of items is a list of lists:
items = [item1, item2, ..., itemN].
My profits and costs are also lists:
profits = [x[2] for x in items]
costs = [x[1] for x in items]
For each item chosen, it needs to have a minimum value, and a minimum of 5 items must have the property (is_blue) flag set to 1.
I want to choose the 20 cheapest items with the highest value, such that 5 of them have the is_blue flag set to 1, and the first and third items are blue (etc).
I'm having trouble formulating this using google OR tools.
from ortools.linear_solver import pywraplp
solver = pywraplp.Solver('SolveAssignmentProblemMIP',
pywraplp.Solver.CBC_MIXED_INTEGER_PROGRAMMING)
x = {}
for i in range(MAX_ITEMS):
x[i] = solver.BoolVar('x[%s]' % (i))
#Define the constraints
total_chosen = 20
solver.Add(solver.Sum([x[i] for i in range(MAX_ITEMS)]) == total_chosen)
blues = [x[3] for x in items]
solver.Add(solver.Sum([blues[i] * x[i] for i in .
range(MAX_ITEMS)]) >= 5)
max_cost = 5.0
for i in range(MAX_ITEMS):
solver.Add(x[i] * cost[i] <= max_cost)
solver.Maximize(solver.Sum([profits[i] * x[i] for i in range(total_chosen)]))
sol = solver.Solve()
I can get the set of items I've chosen by:
for i in range(MAX_ITEMS):
if x[i].solution_value() > 0:
print(item[i].item_name)
This works fine - it chooses the set of 20 items which maximise the profits subject to the cost constraint, but I'm stuck on how to extend this to choosing items in way that guarantees that the first is blue etc.
Any help in formulating the constraints and objective would be really helpful. Thanks!
python-3.x optimization or-tools mixed-integer-programming
add a comment |
I have a set of many (10000+) items, from which have I have to choose exactly 20 items. I can only choose each item once. My items have profits, and costs, as well as several boolean properties (such as colour). I need to output the results in a specific order: in particular I need the first and third items to be blue, and the second and fourth items to be red.
Each item is represented as a tuple:
item = ('item name', cost, profit, is_blue, is_red)
as an example
vase = ['Ming Vase', 1000, 10000, 0, 1]
plate = ['China Plate', 10, 5, 1, 0]
and the total set of items is a list of lists:
items = [item1, item2, ..., itemN].
My profits and costs are also lists:
profits = [x[2] for x in items]
costs = [x[1] for x in items]
For each item chosen, it needs to have a minimum value, and a minimum of 5 items must have the property (is_blue) flag set to 1.
I want to choose the 20 cheapest items with the highest value, such that 5 of them have the is_blue flag set to 1, and the first and third items are blue (etc).
I'm having trouble formulating this using google OR tools.
from ortools.linear_solver import pywraplp
solver = pywraplp.Solver('SolveAssignmentProblemMIP',
pywraplp.Solver.CBC_MIXED_INTEGER_PROGRAMMING)
x = {}
for i in range(MAX_ITEMS):
x[i] = solver.BoolVar('x[%s]' % (i))
#Define the constraints
total_chosen = 20
solver.Add(solver.Sum([x[i] for i in range(MAX_ITEMS)]) == total_chosen)
blues = [x[3] for x in items]
solver.Add(solver.Sum([blues[i] * x[i] for i in .
range(MAX_ITEMS)]) >= 5)
max_cost = 5.0
for i in range(MAX_ITEMS):
solver.Add(x[i] * cost[i] <= max_cost)
solver.Maximize(solver.Sum([profits[i] * x[i] for i in range(total_chosen)]))
sol = solver.Solve()
I can get the set of items I've chosen by:
for i in range(MAX_ITEMS):
if x[i].solution_value() > 0:
print(item[i].item_name)
This works fine - it chooses the set of 20 items which maximise the profits subject to the cost constraint, but I'm stuck on how to extend this to choosing items in way that guarantees that the first is blue etc.
Any help in formulating the constraints and objective would be really helpful. Thanks!
python-3.x optimization or-tools mixed-integer-programming
Why do you care that the first is blue? You will have at least 5 blue items, so just reorder your solution with 1st and 3rd items being blue
– juvian
Nov 6 '18 at 16:18
add a comment |
I have a set of many (10000+) items, from which have I have to choose exactly 20 items. I can only choose each item once. My items have profits, and costs, as well as several boolean properties (such as colour). I need to output the results in a specific order: in particular I need the first and third items to be blue, and the second and fourth items to be red.
Each item is represented as a tuple:
item = ('item name', cost, profit, is_blue, is_red)
as an example
vase = ['Ming Vase', 1000, 10000, 0, 1]
plate = ['China Plate', 10, 5, 1, 0]
and the total set of items is a list of lists:
items = [item1, item2, ..., itemN].
My profits and costs are also lists:
profits = [x[2] for x in items]
costs = [x[1] for x in items]
For each item chosen, it needs to have a minimum value, and a minimum of 5 items must have the property (is_blue) flag set to 1.
I want to choose the 20 cheapest items with the highest value, such that 5 of them have the is_blue flag set to 1, and the first and third items are blue (etc).
I'm having trouble formulating this using google OR tools.
from ortools.linear_solver import pywraplp
solver = pywraplp.Solver('SolveAssignmentProblemMIP',
pywraplp.Solver.CBC_MIXED_INTEGER_PROGRAMMING)
x = {}
for i in range(MAX_ITEMS):
x[i] = solver.BoolVar('x[%s]' % (i))
#Define the constraints
total_chosen = 20
solver.Add(solver.Sum([x[i] for i in range(MAX_ITEMS)]) == total_chosen)
blues = [x[3] for x in items]
solver.Add(solver.Sum([blues[i] * x[i] for i in .
range(MAX_ITEMS)]) >= 5)
max_cost = 5.0
for i in range(MAX_ITEMS):
solver.Add(x[i] * cost[i] <= max_cost)
solver.Maximize(solver.Sum([profits[i] * x[i] for i in range(total_chosen)]))
sol = solver.Solve()
I can get the set of items I've chosen by:
for i in range(MAX_ITEMS):
if x[i].solution_value() > 0:
print(item[i].item_name)
This works fine - it chooses the set of 20 items which maximise the profits subject to the cost constraint, but I'm stuck on how to extend this to choosing items in way that guarantees that the first is blue etc.
Any help in formulating the constraints and objective would be really helpful. Thanks!
python-3.x optimization or-tools mixed-integer-programming
I have a set of many (10000+) items, from which have I have to choose exactly 20 items. I can only choose each item once. My items have profits, and costs, as well as several boolean properties (such as colour). I need to output the results in a specific order: in particular I need the first and third items to be blue, and the second and fourth items to be red.
Each item is represented as a tuple:
item = ('item name', cost, profit, is_blue, is_red)
as an example
vase = ['Ming Vase', 1000, 10000, 0, 1]
plate = ['China Plate', 10, 5, 1, 0]
and the total set of items is a list of lists:
items = [item1, item2, ..., itemN].
My profits and costs are also lists:
profits = [x[2] for x in items]
costs = [x[1] for x in items]
For each item chosen, it needs to have a minimum value, and a minimum of 5 items must have the property (is_blue) flag set to 1.
I want to choose the 20 cheapest items with the highest value, such that 5 of them have the is_blue flag set to 1, and the first and third items are blue (etc).
I'm having trouble formulating this using google OR tools.
from ortools.linear_solver import pywraplp
solver = pywraplp.Solver('SolveAssignmentProblemMIP',
pywraplp.Solver.CBC_MIXED_INTEGER_PROGRAMMING)
x = {}
for i in range(MAX_ITEMS):
x[i] = solver.BoolVar('x[%s]' % (i))
#Define the constraints
total_chosen = 20
solver.Add(solver.Sum([x[i] for i in range(MAX_ITEMS)]) == total_chosen)
blues = [x[3] for x in items]
solver.Add(solver.Sum([blues[i] * x[i] for i in .
range(MAX_ITEMS)]) >= 5)
max_cost = 5.0
for i in range(MAX_ITEMS):
solver.Add(x[i] * cost[i] <= max_cost)
solver.Maximize(solver.Sum([profits[i] * x[i] for i in range(total_chosen)]))
sol = solver.Solve()
I can get the set of items I've chosen by:
for i in range(MAX_ITEMS):
if x[i].solution_value() > 0:
print(item[i].item_name)
This works fine - it chooses the set of 20 items which maximise the profits subject to the cost constraint, but I'm stuck on how to extend this to choosing items in way that guarantees that the first is blue etc.
Any help in formulating the constraints and objective would be really helpful. Thanks!
python-3.x optimization or-tools mixed-integer-programming
python-3.x optimization or-tools mixed-integer-programming
asked Nov 6 '18 at 16:07
Tom KealyTom Kealy
78211031
78211031
Why do you care that the first is blue? You will have at least 5 blue items, so just reorder your solution with 1st and 3rd items being blue
– juvian
Nov 6 '18 at 16:18
add a comment |
Why do you care that the first is blue? You will have at least 5 blue items, so just reorder your solution with 1st and 3rd items being blue
– juvian
Nov 6 '18 at 16:18
Why do you care that the first is blue? You will have at least 5 blue items, so just reorder your solution with 1st and 3rd items being blue
– juvian
Nov 6 '18 at 16:18
Why do you care that the first is blue? You will have at least 5 blue items, so just reorder your solution with 1st and 3rd items being blue
– juvian
Nov 6 '18 at 16:18
add a comment |
1 Answer
1
active
oldest
votes
Instead of expressing chosen items with BoolVar, consider making a list of 20 IntVar with domain of 0..MAX_ITEMS. From there it should be fairly easy to do something like this:
solver.Add(chosens[0].IndexOf(all_items)[3] == 1)
solver.Add(chosens[2].IndexOf(all_items)[3] == 1)
chosens[i].IndexOf(all_items) simply means all_items[IndexOfChosen], I.E: whichever item is chosen for the Ith place. If you go with this approach, do not forget to MakeAllDifferent!
@Tom Kealy Also consider using cp_sat as it is shinier, superior and in most cases, faster.
– Burak
Nov 19 '18 at 8:48
add a comment |
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1 Answer
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Instead of expressing chosen items with BoolVar, consider making a list of 20 IntVar with domain of 0..MAX_ITEMS. From there it should be fairly easy to do something like this:
solver.Add(chosens[0].IndexOf(all_items)[3] == 1)
solver.Add(chosens[2].IndexOf(all_items)[3] == 1)
chosens[i].IndexOf(all_items) simply means all_items[IndexOfChosen], I.E: whichever item is chosen for the Ith place. If you go with this approach, do not forget to MakeAllDifferent!
@Tom Kealy Also consider using cp_sat as it is shinier, superior and in most cases, faster.
– Burak
Nov 19 '18 at 8:48
add a comment |
Instead of expressing chosen items with BoolVar, consider making a list of 20 IntVar with domain of 0..MAX_ITEMS. From there it should be fairly easy to do something like this:
solver.Add(chosens[0].IndexOf(all_items)[3] == 1)
solver.Add(chosens[2].IndexOf(all_items)[3] == 1)
chosens[i].IndexOf(all_items) simply means all_items[IndexOfChosen], I.E: whichever item is chosen for the Ith place. If you go with this approach, do not forget to MakeAllDifferent!
@Tom Kealy Also consider using cp_sat as it is shinier, superior and in most cases, faster.
– Burak
Nov 19 '18 at 8:48
add a comment |
Instead of expressing chosen items with BoolVar, consider making a list of 20 IntVar with domain of 0..MAX_ITEMS. From there it should be fairly easy to do something like this:
solver.Add(chosens[0].IndexOf(all_items)[3] == 1)
solver.Add(chosens[2].IndexOf(all_items)[3] == 1)
chosens[i].IndexOf(all_items) simply means all_items[IndexOfChosen], I.E: whichever item is chosen for the Ith place. If you go with this approach, do not forget to MakeAllDifferent!
Instead of expressing chosen items with BoolVar, consider making a list of 20 IntVar with domain of 0..MAX_ITEMS. From there it should be fairly easy to do something like this:
solver.Add(chosens[0].IndexOf(all_items)[3] == 1)
solver.Add(chosens[2].IndexOf(all_items)[3] == 1)
chosens[i].IndexOf(all_items) simply means all_items[IndexOfChosen], I.E: whichever item is chosen for the Ith place. If you go with this approach, do not forget to MakeAllDifferent!
edited Nov 19 '18 at 8:40
answered Nov 19 '18 at 8:34
BurakBurak
736
736
@Tom Kealy Also consider using cp_sat as it is shinier, superior and in most cases, faster.
– Burak
Nov 19 '18 at 8:48
add a comment |
@Tom Kealy Also consider using cp_sat as it is shinier, superior and in most cases, faster.
– Burak
Nov 19 '18 at 8:48
@Tom Kealy Also consider using cp_sat as it is shinier, superior and in most cases, faster.
– Burak
Nov 19 '18 at 8:48
@Tom Kealy Also consider using cp_sat as it is shinier, superior and in most cases, faster.
– Burak
Nov 19 '18 at 8:48
add a comment |
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Why do you care that the first is blue? You will have at least 5 blue items, so just reorder your solution with 1st and 3rd items being blue
– juvian
Nov 6 '18 at 16:18