Programming absolute deviation as linear program

Question

I am attempting to convert a sum of absolute deviations to a linear programming problem so that I can utilize CPLEX (or other solver). I am stuck on how the matrices are to be set up. The problem is as follows:

minimize abs(x1 - 5) + abs(x2 - 3)
s.t. x1 + x2 = 10

I have the following constraints set up to transform the problem into a linear form:

  x1 - 5  <= t1
-(x1 - 5) <= t1 and 

  x2 - 3  <= t2
-(x2 - 3) <= t2

I've set up the objective function as

c = [0,0,1,1]

but I am lost on how to set up

Ax <= b

in matrix form. What I have so far is:

A = [[ 1, -1, 0, 0],
     [-1, -1, 0, 0],
     [ 0,  0, 1,-1],
     [ 0,  0,-1,-1]]
b =  [ 5, -5, 3,-3] 

I have set up the other constraint in matrix for as:

B =  [1, 1, 0, 0]
b2 = [10] 

When I run the following:

linprog(c,A_ub=A,b_ub=b,A_eq=B,b_eq=b2,bounds=[(0,None),(0,None)])

I get the following error message back:

ValueError: Invalid input for linprog: A_eq must have exactly two dimensions, and the number of columns in A_eq must be equal to the size of c

I know there is a solution because when I use scipy.optimize.minimize it solves to [6,4]. I'm sure the issue is I am not formulating the input matrices correctly but I am not sure how to set them up so that it runs.

Edit - here is the code that does not run:

import numpy as np
from scipy.optimize import linprog, minimize

c = np.block([np.zeros(2),np.ones(2)])
print("c =>",c)

A = [[ 1, -1, 0, 0],
     [-1, -1, 0, 0],
     [ 0,  0, 1,-1],
     [ 0,  0,-1,-1]]

b =  [[ 5, -5, 3,-3]]
print(A)
print(np.multiply(A,b))

B = [ 1, 1, 0, 0]
b2 = [10]
print(np.multiply(B,b2))

linprog(c,A_ub=A,b_ub=b,A_eq=B,b_eq=b2,bounds=[(0,None),(0,None)],
        options={'disp':True})

Answer 1

I think the message is quite good. B should be 2-dimensional matrix instead of a 1-dimensional vector. So:

B =  [[1, 1, 0, 0]]

Secondly, the bounds array is too short. Thirdly, your ordering of variables is inconsistent. The columns in A are x1,t1,x2,t2 while the columns in B (and c) seem to be x1,x2,t1,t2 . They need to follow the same scheme.