[Help-glpk] Problem using Rglpk with Ruby

[Lukas Meyer]

Hello

I would like to use a GLPK linear programming model in Ruby using the Rglpk wrapper. 

Unfortunately, I do not get the same solution when comparing the glpk and Rglpk outputs. Despite I am setting the column kind to Binary (cols.kind = Rglpk::GLP_BV), the solution shows integer values instead of boolean.

I want to model the following linear programming capital budgeting problem:

Maximize NPV of projects (x1…x5) by taking into account Budget and Resource constraints (c1, c2). Solution can only be 0 or 1

Maximize: 928 x1 + 908 x2 + 801 x3 + 543 x4 + 944 x5

Constraints:

c1: 398 x1 + 151 x2 + 129 x3 + 275 x4 + 291 x5 <= 800

c2: 111 x1 + 139 x2 + 56 x3 + 54 x4 + 123 x5 <= 200

Result is binary (0 or 1): x1 x2 x3 x4 x5

The model is rather simple, but somehow I am missing something in the ruby code.  

Do you have any hint what could be wrong?

Many thanks in advance and best regards,

Lukas

Please find below my code (1) ruby using Rglpk module, 2) GLPK in mathprog)

1) Ruby with Rglpk (returns WRONG solution):

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require 'rglpk'

p = Rglpk::Problem.new

p.name = "Project selection"

p.obj.dir = Rglpk::GLP_MAX

rows = p.add_rows(2)

rows[0].name = "p"

rows[0].set_bounds(Rglpk::GLP_UP, 0, 800)

rows[1].name = "q"

rows[1].set_bounds(Rglpk::GLP_UP, 0, 200)

cols = p.add_cols(5)

cols[0].name = "x1"

cols[0].kind = Rglpk::GLP_BV

cols[0].set_bounds(Rglpk::GLP_BV, 0, 1)

cols[1].name = "x2"

cols[1].kind = Rglpk::GLP_BV

cols[1].set_bounds(Rglpk::GLP_BV, 0, 1)

cols[2].name = "x3"

cols[2].kind = Rglpk::GLP_BV

cols[2].set_bounds(Rglpk::GLP_BV, 0, 1)

cols[3].name = "x4"

cols[3].kind = Rglpk::GLP_BV

cols[3].set_bounds(Rglpk::GLP_BV, 0, 1)

cols[4].name = "x5"

cols[4].kind = Rglpk::GLP_BV

cols[4].set_bounds(Rglpk::GLP_BV, 0, 1)

p.obj.coefs = [928, 908, 801, 543, 944]

p.set_matrix([

 398, 151, 129, 275, 291,

111, 139, 56, 54, 123

])

p.simplex

z = p.obj.get

x1 = cols[0].get_prim

x2 = cols[1].get_prim

x3 = cols[2].get_prim

x4 = cols[3].get_prim

x5 = cols[4].get_prim

printf("z = %g; x1 = %g; x2 = %g; x3 = %g; x4 = %g; x5 = %g\n", z, x1, x2, x3, x4, x5)

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2) glpk in mathprog format (returns CORRECT solution):

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set PROJECT;

param Budget {i in PROJECT};

param Resources {i in PROJECT};

param NPV     {i in PROJECT};

var x {i in PROJECT} >=0, binary;

maximize z: sum{i in PROJECT} NPV[i]*x[i];

s.t. Bud : sum{i in PROJECT} Budget[i]*x[i] <= 800;

s.t. Res : sum{i in PROJECT} Resources[i]*x[i] <= 200;

data;

set PROJECT := p1 p2 p3 p4 p5;

param Budget:=

p1  398

p2 151

p3 129

p4 275

p5 291;

param Resources:=

p1 111

p2 139

p3 56

p4 54

p5 123;

param NPV:=

p1 928

p2 908

p3 801

p4 543

p5 944;

end;

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