# Optimizing a matrix with cplexlp?

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Berkay Sen on 3 Jun 2019
Commented: Berkay Sen on 7 Jun 2019
Hi to all,
I'm working on my graduating thesis what about cross kidney transplants matching algorithms.
I've a solution with genetic algorithm but i want to know 'what is the optimum?' (of course GA is %90-99 interval, not %100-optimum-)
So;
I have a binary matrix(also symetric)
for example=
1 2 3 4 5 6 7 8 9 10
1 [0 1 0 0 0 0 0 0 0 0
2 1 0 1 0 0 0 0 0 0 0
3 0 1 0 0 0 0 0 0 0 0
4 0 0 0 0 0 0 0 0 1 0
5 0 0 0 0 0 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0 0
7 0 0 0 0 0 0 0 0 0 0
8 0 0 0 0 0 0 0 0 0 0
9 0 0 0 1 0 0 0 0 0 0
10 0 0 0 0 0 0 0 0 0 0]
I want to find optimum match number beetween column numbers. (How many matches can be made in this matrix?)
In this case; Possible matches are:
NO2 - NO1
NO2 - NO3
NO 4- NO9
But when NO2 was matched with NO1, it can no longer match with NO1. Its kinda monogamy.(If patient and donor NO2 exchanges kidneys with p and d NO1, of course NO2 can not exchange with NO3)
In this example, optimum match value is = 2.
When i solve this matrix with genetic algortihm, i had 410 as a result. ( theoretical match number is 500 ((1000/2)), of course impossible.)
HOW CAN I SOLVE THIS MATRIX USING CPLEXLP OR MILP?(or something else solver)
I will be very pleased if anybody help,
Yours truly.

Greg Heath on 6 Jun 2019
Sorry, cannot help.
Greg
Walter Roberson on 6 Jun 2019
CPLEXLP is a third party product that is outside my experience.
Berkay Sen on 6 Jun 2019
Hi Greg,
thank you anyway;
Hi Walter,
There is no restraction with solution way, its just a suggestion for solver. I can use MILP or any function, solver.
Thank you

Stephan on 6 Jun 2019
Edited: Stephan on 6 Jun 2019
Hi,
you could think about using graphs for this job:
A = [0 1 0 0 0 0 0 0 0 0;
1 0 1 0 0 0 0 0 0 0;
0 1 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 1 0;
0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0;
0 0 0 1 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0];
% create a graph object
g = graph(A);
% plot graph
plot(g);
% remove nodes who have no matches / are unconnected
g = rmnode(g,find(degree(g)==0));
% count the number of different matches
result = numel(unique(conncomp(g)));
% plot resulting graph
hold on
plot(g)
hold off

#### 1 Comment

Stephan on 6 Jun 2019
Note that the node numbers in the reduced graph have changed - you can fix this by naming the nodes. see documentation for more details.

Alan Weiss on 6 Jun 2019
I really don't know, but it is possible that the matchpairs function might be applicable.
Alan Weiss
MATLAB mathematical toolbox documentation

Berkay Sen on 6 Jun 2019
Hi Alan,
thank you for your great suggest.
But matchpairs doesn't working for me because matchpairs lookin rows(1. row, 2.row, 3.row...) and choosing best choise.
Imagine that,
No1 can match with No4,No5,No6;
But No2 can only match with No4,
And No3 can only match with No5
If matchpairs choose no4 because placement, then No2 can no longer match with anyone.
Same for no3-no5.
So algorithm must know this and choose NO6 for NO1.
I hope i could explain that :)
Christine Tobler on 6 Jun 2019
Thank you for thinking of matchpairs, Alan. Unfortunately it wouldn't work for this case: Matchpairs will match rows of A to columns, but it doesn't have an understanding that if row i of A is matched, then column i of A is also unavailable for another match.
Basically, in matchpairs the rows and columns represent two different sets of resources to be matched up, while in this problem both rows and columns represent the same set of resources.
Berkay Sen on 6 Jun 2019
Hi Christine, That's completely true, have you got any idea? Thank you for your comment

Christine Tobler on 6 Jun 2019
You can solve this using intlinprog. This is probably not the most efficient way of solving the problem, but reliable
A = [0 1 0 0 0 0 0 0 0 0;
1 0 1 0 0 0 0 0 0 0;
0 1 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 1 0;
0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0;
0 0 0 1 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0];
% Define an optimization variable: x(i,j) == 1 if (i, j) are matched, otherwise x(i, j) == 0
x = optimvar("x", size(A), "Type", "integer", "LowerBound", 0, "UpperBound", 1);
% Define an optimization problem
problem = optimproblem;
% Maximize the number of indices (i, j) for which A(i, j) == 1 and x(i, j) == 1
% (change sign because Objective is minimized in solve)
problem.Objective = -sum(sum(x.*A));
% Every column of x can only contain 1 non-zero (otherwise that column is matched to more than one row)
problem.Constraints.noDuplicates = sum(x, 1) == 1;
% Require x(i, j) == x(j, i)
problem.Constraints.symmetric = x == x';
% Solve the problem
s = solve(problem);
% Find indices i, j for which A(i, j) == 1 and s.x(i, j) == 1
[i, j] = find(triu(A & s.x))
% This returns i = [2; 4] and j = [3; 9], so two pairs (2, 3) and (4, 9)
This code is inspired by this Sudoku solver example.

Show 1 older comment
Berkay Sen on 6 Jun 2019
Christine,
I don't know how to thank you. Algorithm is running smoothly. You're superb!
Thank you to all who spent their time!
I'm gonna share the project(cross kidney transplant algorithm) when i finish it.
Christine Tobler on 6 Jun 2019
You're very welcome. I just realized you probably need an additional constraint:
% Require x(i, i) == 0
n = size(x, 1);
problem.Constraints.NonDiagonal = x(1:n+1:end) == 0;
This will prevent the solver from matching an index with itself.
Berkay Sen on 7 Jun 2019
Thanks but i've already adjust that. My matrix's diagonal is full of 0. It's working perfectly ?