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I think I'm facing a Milp problem but I'm not sure.

The problem in a simplified form is:
There are 3 suppliers of materials (trucks) for 3 cities. The real problem is 30 Suppliers and 100 cities...

Supliers capacity: a:1; b:2; c:3.
Cities Demand: a:2; b:3; c:1

Distances Supplier(Cities):

  • a(a:2;b:4;c:6)
  • b(a:4;b:2;c:4)
  • c(a:6;b:4;c:2)

like this with each Capacity and Demand

Sa1 - Ca2
Sb2 - Cb3
Sc3 - Cc1

The goal its optimize the suply but there is one (devil) condition:

  • Just one supplier per city.

Whitout the contidion the problem is a simple problem to solve with basic Linear Programming.
With the condition I think that could be solved with Mixed Integer Linear Programming - MILP. But not figure it out how to solve this with MILP Method and Pulp (python module).

If someone can help me
Thanks!

My first try

from scipy.optimize import linprog

c = [2,4,6,4,2,4,6,4,2]
Ae = [[1,1,1,0,0,0,0,0,0],
      [0,0,0,1,1,1,0,0,0],
      [0,0,0,0,0,0,1,1,1],
      [1,0,0,1,0,0,1,0,0],
      [0,1,0,0,1,0,0,1,0],
      [0,0,1,0,0,1,0,0,1],
     ]

be = [1,2,3,2,3,1]


x0_bounds = (0,None)
x1_bounds = (0,None)
x2_bounds = (0,None)
x3_bounds = (0,None)
x4_bounds = (0,None)
x5_bounds = (0,None)
x6_bounds = (0,None)
x7_bounds = (0,None)
x8_bounds = (0,None)

sol = linprog(c, A_eq= Ae, b_eq = be, bounds = ((x0_bounds,x1_bounds,x2_bounds,x3_bounds,x4_bounds,x5_bounds,x6_bounds,x7_bounds,x8_bounds)) )

print(sol)

      fun: 18.0
 message: 'Optimization terminated successfully.'
     nit: 10
   slack: array([], dtype=float64)
  status: 0
 success: True
       x: array([1., 0., 0., 0., 2., 0., 1., 1., 1.])

Process finished with exit code 0
luiz felipe Sousa
  • 1
  • 1

1 Answers1

0

I've done! Videos of Caylie Cincera from youtube help me a lot. Illustration of the problem. Each location could receive at most of one supplier. https://imgur.com/O2CNa9M

from pulp import *

#Pulp way to start a LP problem
prob = LpProblem("testpulp",LpMinimize)

#The 9 Arcs Origin and destiny
x1 = LpVariable("x1_11",0,None,LpInteger)
x2 = LpVariable("x2_12",0,None,LpInteger)
x3 = LpVariable("x3_13",0,None,LpInteger)
x4 = LpVariable("x4_21",0,None,LpInteger)
x5 = LpVariable("x5_22",0,None,LpInteger)
x6 = LpVariable("x6_23",0,None,LpInteger)
x7 = LpVariable("x7_31",0,None,LpInteger)
x8 = LpVariable("x8_32",0,None,LpInteger)
x9 = LpVariable("x9_33",0,None,LpInteger)

#Auxiliar Variables z y and k
z1 = LpVariable("z1",0,1,LpBinary)
z2 = LpVariable("z2",0,1,LpBinary)
z3 = LpVariable("z3",0,1,LpBinary)

y1 = LpVariable("y1",0,1,LpBinary)
y2 = LpVariable("y2",0,1,LpBinary)
y3 = LpVariable("y3",0,1,LpBinary)

k1 = LpVariable("k1",0,1,LpBinary)
k2 = LpVariable("k2",0,1,LpBinary)
k3 = LpVariable("k3",0,1,LpBinary)

#Objective Function
prob += 2*x1 + 4*x2+ 6*x3 + 4*x4 + 2*x5 + 4*x6 + 6*x7 + 4*x8 + 2*x9, "fobj"

#Constraints
#Supply constraints
prob += x1 + x2 + x3 == 1, "m1"
prob += x4 + x5 + x6 == 2, "m2"
prob += x7 + x8 + x9 == 3, "m3"

#Demand constraints
prob += x1 + x4 + x7 == 2, "d1"
prob += x2 + x5 + x8 == 3, "r2"
prob += x3 + x6 + x9 == 1, "r4"

#Trick to force unique supplier for each location
prob += x1 <= 2*y1, "yx1"
prob += x4 <= 2*y2, "yx2"
prob += x7 <= 2*y3, "yx3"

prob += x2 <= 3*z1, "zx1"
prob += x5 <= 3*z2, "zx2"
prob += x8 <= 3*z3, "zx3"

prob += x3 <= 1*k1, "kx1"
prob += x6 <= 1*k2, "kx2"
prob += x9 <= 1*k3, "kx3"


prob += y1 + y2 + y3 == 1, "yk"
prob += z1 + z2 + z3 == 1, "zk"
prob += k1 + k2 + k3 == 1, "kk"


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

print("Total Profit: ",value(prob.objective))

#The "optimal" solution of this problem is the unique solution
#The hard part is to force unique supplier for each location

The output:

k1  =  1.0
k2  =  0.0
k3  =  0.0
x1_11  =  0.0
x2_12  =  0.0
x3_13  =  1.0
x4_21  =  2.0
x5_22  =  0.0
x6_23  =  0.0
x7_31  =  0.0
x8_32  =  3.0
x9_33  =  0.0
y1  =  0.0
y2  =  1.0
y3  =  0.0
z1  =  0.0
z2  =  0.0
z3  =  1.0
Total Profit:  26.0

Process finished with exit code 0
luiz felipe Sousa
  • 1
  • 1