Transforming a logic constraint into python pulp code - linear-programming

Question

I started working on a problem in the past several days...
A company plans its business in a three month period. It can produce
110 units at a cost of 600 each. The minimum amount it must produce
per month is 15 units if active (but of course, it can choose to be closed
during the month, and produce 0 units). Each month it can subcotract the
prodution of 60 units, at a cost of 660 each. Storing a unit for one month
costs 20$ per unit per month. The marketing department has forcasted
sales of 100, 130 and 150 units for the next three months, respectively.
The goal is to meet the demand each month while minimizing the total
cost.
I deduced that we need to have an objective function of form min[Sum(i=0..3) 600*x1+660*x2+20*x3].
We need to add some constrains on x1>=15, and on x2 0<=x2<=60
Also we will also need another constraint for each month...
For the first one i=1 => x1+x2 = 100 - x3last (x3last is an extra variable that should hold the amount existing in deposit from the previous month), and for i=2 and i=3 same constraints.
I don't have any idea how to write this in pulp, and i would appreciate some help. Thx ^_^

Answer 1

I'd tend to agree with #Erwin that you should focus on formulating the problem as a Linear Program. It is then easy to translate this into code in PULP or one of many other PULP libraries/tools/languages.
As an example of this - lets work through this process for the example problem you have written out in your question.
Decision Variables
The first thing to decide is what you can/should decide. This set of information is called the decision variables. Picking the best/easiest decision variables for your problem comes with practice - the important thing is that once you know the values of the variables you have a unique solution to the problem.
Here I would suggest the following. These assume that the forecasts for demand are perfect. For each month i:
Whether the production line should be open - o[i]
How much to produce in that month - p[i]
How much to hold in storage for next month - s[i]
How much to get made externally - e[i]
Objective Function
The objective in your case is obvious - minimise the total cost. So we can just write this down: sum(i=0...2)[p[i]*600 + s[i]*20 + e[i]*660]
Constraints
Let's lift these directly our of your problem description:
"It can produce 110 units at a cost of 600 each. The minimum amount it must produce per month is 15 units if active (but of course, it can choose to be closed during the month, and produce 0 units)."
p[i] >= o[i]*15
p[i] <= o[i]*110
The first constraint forces the minimum production about to be 15 if the production is open that month (o[i] == 1), if the production is not open this constraint has not effect. The second constraint sets a maximum value on p[i] of 110 if the production is open and a maximum production of 0 if the production is closed that month (o[i] == 0).
"Each month it can subcotract the prodution of 60 units, at a cost of 660 each"
e[i] <= 60
"The marketing department has forcasted sales of 100, 130 and 150 units for the next three months, respectively. The goal is to meet the demand each month while minimizing the total cost." If we declare the sales in each mongth to be sales[i], we can define our "flow constraint" as:
p[i] + e[i] + s[i-1] == s[i] + sales[i]
The way to think of this constraint is inputs on the left, and outputs on the right. Inputs of units are production, external production, and stuff taken out of storage from last month. Outputs are units left/put in storage for next month and sales.
Finally in code:
from pulp import *
all_i = [1,2,3]
all_i_with_0 = [0,1,2,3]
sales = {1:100, 2:130, 3:150}
o = LpVariable.dicts('open', all_i, cat='Binary')
p =LpVariable.dicts('production', all_i, cat='Linear')
s =LpVariable.dicts('stored', all_i_with_0, lowBound=0, cat='Linear')
e =LpVariable.dicts('external', all_i, lowBound=0, cat='Linear')
prob = LpProblem("MinCost", LpMinimize)
prob += lpSum([p[i]*600 + s[i]*20 + e[i]*660 for i in all_i]) # Objective
for i in all_i:
prob += p[i] >= o[i]*15
prob += p[i] <= o[i]*110
prob += e[i] <= 60
prob += p[i] + e[i] + s[i-1] == sales[i] + s[i]
prob += s[0] == 0 # No stock inherited from previous monts
prob.solve()
# The status of the solution
print ("Status:", LpStatus [prob.status])
# Dislay the optimums of each var
for v in prob.variables ():
print (v.name, "=", v.varValue)
# Objective fcn
print ("Obj. Fcn: ", value(prob.objective))
Which returns:
Status: Optimal
external_1 = 0.0
external_2 = 10.0
external_3 = 40.0
open_1 = 1.0
open_2 = 1.0
open_3 = 1.0
production_1 = 110.0
production_2 = 110.0
production_3 = 110.0
stored_0 = 0.0
stored_1 = 10.0
stored_2 = 0.0
stored_3 = 0.0
Obj. Fcn: 231200.0