how to optimize variables combination in a dictionary using Pulp python?

Question

Please forgive me if this is a poor question. But how do I optimize a set of variables in a dictionary?

I posted an excel img of my problem. Where the max cost is 169 and I have to select 4/6 five cost to get the max profit? So my goal is to have a combination of 4/6 (max profits) with a cost limit of $169.

I was able to learn to get the max from a basic equation but could not extend to help solve my example.

 x1 = pulp.LpVariable("x1", 0, 40)   # 0<= x1 <= 40
 x2 = pulp.LpVariable("x2", 0, 1000) # 0<= x2 <= 1000

 prob = pulp.LpProblem("problem", pulp.LpMaximize)

 prob += 2*x1+x2 <= 100 
 prob += x1+x2 <= 80


 prob += 3*x1+2*x2

 status = prob.solve()
 pulp.LpStatus[status]

# print the results x1 = 20, x2 = 60
  pulp.value(x1)
  pulp.value(x2)

Source:https://thomas-cokelaer.info/blog/2012/11/solving-a-linear-programming-problem-with-python-pulp/

I can only pick 4 locations out of the 5. Sorry, should have been clearer. — TroyMcClure8998
– TroyMcClure8998, Commented Feb 10, 2020 at 17:18
This is a trivial optimization model. Just use binary variables x[i] indicating if project i is selected. — Erwin Kalvelagen
– Erwin Kalvelagen, Commented Feb 10, 2020 at 18:22

kabdulla · Accepted Answer · 2020-02-11 20:18:12Z

This is an example of the "knapsack" problem - where you want to pick the most valuable items, subject to a limit on the number/cost that can the selected.

The following:

from pulp import *

# PROBLEM DATA:
costs = [15, 25, 35, 40, 45, 55]
profits = [1.7, 2, 2.4, 3.2, 5.6, 6.2]
max_cost = 169
max_to_pick = 4

# DECLARE PROBLEM OBJECT:
prob = LpProblem("Mixed Problem", LpMaximize)

# VARIABLES
# x_i - whether to include item i (1), or not (0)
n = len(costs)
N = range(n)
x = LpVariable.dicts('x', N, cat="Binary")

# OBJECTIVE
prob += lpSum([profits[i]*x[i] for i in N])

# CONSTRAINTS
prob += lpSum([x[i] for i in N]) <= max_to_pick        # Limit number to include
prob += lpSum([x[i]*costs[i] for i in N]) <= max_cost  # Limit max. cost

# SOLVE & PRINT RESULTS
prob.solve()
print(LpStatus[prob.status])
print('Profit = ' + str(value(prob.objective)))
print('Cost = ' + str(sum([x[i].varValue*costs[i] for i in N])))

for v in prob.variables ():
    print (v.name, "=", v.varValue)

Returns:

Optimal
Profit = 17.0
Cost = 165.0
('x_0', '=', 0.0)
('x_1', '=', 1.0)
('x_2', '=', 0.0)
('x_3', '=', 1.0)
('x_4', '=', 1.0)
('x_5', '=', 1.0)

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