Hello Davor !
Based on your description and hints from Fouad maybe this does what you
expect (you can try it online at https://meimporta.eu/myglpk-ext/):
====
/* ASSIGN, Assignment Problem */
/* Written in GNU MathProg by Andrew Makhorin <m...@gnu.org> */
/* The assignment problem is one of the fundamental combinatorial
optimization problems.
In its most general form, the problem is as follows:
There are a number of agents and a number of tasks. Any agent can be
assigned to perform any task, incurring some cost that may vary
depending on the agent-task assignment. It is required to perform all
tasks by assigning exactly one agent to each task in such a way that
the total cost of the assignment is minimized.
(From Wikipedia, the free encyclopedia.) */
param m, integer, > 0;
/* number of agents */
param n, integer, > 0;
/* number of tasks */
set I := 1..m;
/* set of agents */
set J := 1..n;
/* set of tasks */
param c{i in I, j in J}, >= 0;
/* cost of allocating task j to agent i */
var x{i in I, j in J}, >= 0;
/* x[i,j] = 1 means task j is assigned to agent i
note that variables x[i,j] are binary, however, there is no need to
declare them so due to the totally unimodular constraint matrix */
s.t. phi{i in I}: sum{j in J} x[i,j] <= n;
/* each agent can perform at most one task */
s.t. psi{j in J}: sum{i in I} x[i,j] = 1;
/* each task must be assigned exactly to one agent */
minimize obj: sum{i in I, j in J} c[i,j] * x[i,j];
/* the objective is to find a cheapest assignment */
solve;
printf "\n";
printf "Agent Task Cost\n";
printf " ";
printf{j in J} "%5d ", j;
printf "\n";
for{i in I} {
printf "%5d ", i;
printf{j in J} "%5d ", x[i,j];
printf "%10g\n", sum{j in J} c[i,j] * x[i,j];
printf "\n";
}
printf "----------------------\n";
printf " Total: %10g\n", sum{i in I, j in J} c[i,j] * x[i,j];
printf "\n";
data;
/* These data correspond to an example from [Christofides]. */
/* Optimal solution is 76 */
param m := 4;
param n := 8;
param c : 1 2 3 4 5 6 7 8 :=
1 13 21 20 12 8 26 22 11
2 12 36 25 41 40 11 4 8
3 35 32 13 36 26 21 13 37
4 34 54 7 8 12 22 11 40 ;
end;
====
Output:
====
GLPSOL: GLPK LP/MIP Solver, v4.65-ex, glp_double size 8
Parameter(s) specified in the command line:
--math input_file
Reading model section from input_file...
80 lines were read
80 lines were read
Generating phi...
Generating psi...
Generating obj...
Model has been successfully generated
GLPK Simplex Optimizer, v4.65-ex
13 rows, 32 columns, 96 non-zeros
Preprocessing...
12 rows, 32 columns, 64 non-zeros
Scaling...
A: min|aij| = 1.000e+00 max|aij| = 1.000e+00 ratio = 1.000e+00
Problem data seem to be well scaled
Constructing initial basis...
Size of triangular part is 12
* 0: obj = 1.880000000e+02 inf = 0.000e+00 (12)
* 6: obj = 7.900000000e+01 inf = 0.000e+00 (0)
OPTIMAL LP SOLUTION FOUND
Time used: 0.0 secs
Memory used: 0.1 Mb (128934 bytes)
Agent Task Cost
1 2 3 4 5 6 7 8
1 0 1 0 0 1 0 0 0 29
2 1 0 0 0 0 1 1 1 35
3 0 0 0 0 0 0 0 0 0
4 0 0 1 1 0 0 0 0 15
----------------------
Total: 79
Model has been successfully processed
====
Cheers !
On 17/12/21 18:53, Fouad Mardini wrote:
Hi,
Maybe add multiple variables per agent and limit their sum to the max
per agent
On Fri, 17 Dec 2021 at 17:31, Andrew Makhorin <m...@gnu.org
<mailto:m...@gnu.org>> wrote:
-------- Forwarded Message --------
From: Davor Ocelic <doce...@hcoop.net <mailto:doce...@hcoop.net>>
To: m...@gnu.org <mailto:m...@gnu.org>
Subject: gmpl question
Date: Fri, 17 Dec 2021 10:07:38 +0100
Heya,
I would appreciate minimal help with gmpl if you could:
In the glpk example `assign.mod`, the constraint is that
an agent can have only one task assigned:
s.t. phi{i in I}: sum{j in J} x[i,j] <= 1;
/* each agent can perform at most one task */
I would need to change this rule so that an agent doesn't
have a limit on number of tasks, but all tasks need to be
distributed among a limited number of agents. (For example,
distribute the 8 tasks in the example to 4 agents).
Could you help me with the syntax for that?
Thank you kindly,
Davor
--
--
twitter.com/fmardini <http://twitter.com/fmardini>
