Solve linear assignment problem



M = matchpairs(Cost,costUnmatched) solves the linear assignment problem for the rows and columns of the matrix Cost. Each row is assigned to a column in such a way that the total cost is minimized. costUnmatched specifies the cost per row of not assigning each row, and also the cost per column of not having a row assigned to each column.

[M,uR,uC] = matchpairs(Cost,costUnmatched) additionally returns indices for unmatched rows in uR and indices for unmatched columns in uC.


[___] = matchpairs(Cost,costUnmatched,goal) specifies the goal of the optimization using any of the output argument combinations in previous syntaxes. goal can be 'min' or 'max' to produce matches that either minimize or maximize the total cost.


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Assign salespeople to flights such that the total cost of transportation is minimized.

A company has four salespeople who need to travel to key cities around the country. The company must book their flights, and wants to spend as little money as possible. These salespeople are based in different parts of the country, so the cost for them to fly to each city varies.

This table shows the cost for each salesperson to fly to each key city.

DallasChicagoNew York CitySt. LouisFred$600$670$960$560Beth$900$280$970$540Sue$310$350$950$820Greg$325$290$600$540

Each city represents a sales opportunity. If a city is missed, then the company loses out on an average revenue gain of $2,000.

Create a cost matrix to represent the cost of each salesperson flying to each city.

C = [600 670 960 560
     900 280 970 540
     310 350 950 820
     325 290 600 540];

Use matchpairs to assign the salespeople to the cities with minimal cost. Specify the cost of unassignment as 1000, since the cost of unassignment is counted twice if a row and a column remain unmatched.

M = matchpairs(C,1000)
M = 4×2

     3     1
     2     2
     4     3
     1     4

matchpairs calculates the least expensive way to get a salesperson to each city.

DallasChicagoNew York CitySt. LouisFred$600$670$960$560Beth$900$280$970$540Sue$310$350$950$820Greg$325$290$600$540

Match rows to columns when you have many more columns than rows in the cost matrix.

Create a 3-by-8 cost matrix. Since you have only three rows, matchpairs can produce at most three matches with the eight columns.

rng default % for reproducibility
C = randi([10 100], 3, 8)
C = 3×8

    84    93    35    97    97    22    82    13
    92    67    59    24    54    48    97    87
    21    18    97    98    82    93    69    94

Use matchpairs to match the rows and columns of the cost matrix. To get the maximum number of matches, use a large cost of unassignment (relative to the magnitude of the entries in the cost matrix). Specify three outputs to return the indices of unmatched rows and columns.

[M,uR,uC] = matchpairs(C,1e4)
M = 3×2

     3     2
     2     4
     1     8

uR =

  0x1 empty double column vector
uC = 5×1


Five of the columns in C are not matched with any rows.

Assign taxis to routes such that the profit is maximized.

A taxi company has several ride requests from across the city. The company wants to dispatch its limited number of taxis in a way that makes the most money.

This table shows the estimated taxi fare for each of five ride requests. Only three of the five ride requests can be filled.

Ride 1Ride 2Ride 3Ride 4Ride 5Cab A$5.70$6.30$3.10$4.80$3.50Cab B$5.80$6.40$3.30$4.70$3.20Cab C$5.70$6.30$3.20$4.90$3.40

Create a profits matrix to represent the profits of each taxi ride.

P = [5.7 6.3 3.1 4.8 3.5
     5.8 6.4 3.3 4.7 3.2
     5.7 6.3 3.2 4.9 3.4];

Use matchpairs to match the taxis to the most profitable rides. Specify three outputs to return any unmatched rows and columns, and the 'max' option to maximize the profits. Specify the cost of unassignment as zero, since the company makes no money from unfilled taxis or ride requests.

costUnmatched = 0;
[M,uR,uC] = matchpairs(P,costUnmatched,'max')
M = 3×2

     1     1
     2     2
     3     4

uR =

  0x1 empty double column vector
uC = 2×1


matchpairs calculates the most profitable rides to fill. The solution leaves ride requests 3 and 5 unfilled.

Ride 1Ride 2Ride 3Ride 4Ride 5Cab A$5.70$6.30$3.10$4.80$3.50Cab B$5.80$6.40$3.30$4.70$3.20Cab C$5.70$6.30$3.20$4.90$3.40

Calculate the total profits for the calculated solution. Since costUnmatched is zero, you only need to add together the profits from each match.

TotalProfits = sum(P(sub2ind(size(P), M(:,1), M(:,2))))
TotalProfits = 17

Use matchpairs to track the movement of several points by minimizing the total changes in distance.

Plot a grid of points at time t=0 in green. At time t=1, some of the points move a small amount in a random direction.

[x,y] = meshgrid(4:6:16);
x0 = x(:)';
y0 = y(:)';
hold on
rng default % for reproducibility
x1 = x0 + randn(size(x0));
y1 = y0 + randn(size(y0));

Use matchpairs to match the points at t=0 with the points at t=1. To do this, first calculate a cost matrix where C(i,j) is the Euclidean distance from point i to point j.

C = zeros(size(x).^2);
for k = 1:length(y1)
  C(k,:) = vecnorm([x1(k)-x0; y1(k)-y0],2,1)';
C = 9×9

    2.8211    3.2750    9.2462    6.1243    6.3461   10.7257   11.7922   11.9089   14.7169
    4.9987    2.2771    7.5752    6.2434    4.3794    8.4485   11.1792   10.2553   12.5447
   15.2037    9.3130    3.7833   17.1539   12.2408    8.7988   20.7211   16.8803   14.5783
    6.9004    8.6551   13.1987    1.1267    5.3446   11.3075    5.1888    7.3633   12.3901
    8.6703    6.3191    8.7571    5.9455    0.3249    6.0714    8.2173    5.6816    8.3089
   13.5530    8.1918    4.7464   12.7818    6.8409    1.4903   14.6652    9.9242    7.3426
   11.5682   13.1257   16.8150    5.5702    8.3359   13.4144    0.4796    6.2201   12.2127
   13.6699   12.3432   13.7784    8.6461    6.3438    8.8167    5.8858    0.3644    6.1337
   20.6072   17.2853   15.6495   16.5444   12.1590    9.6935   13.9562    8.3006    3.8761

Next, use matchpairs to match the rows and columns in the cost matrix. Specify the cost of unassignment as 1. With such a low cost of unassignment relative to the entries in the cost matrix, it is likely matchpairs will leave some points unmatched.

M = matchpairs(C,1)
M = 5×2

     4     4
     5     5
     6     6
     7     7
     8     8

The values M(:,2) correspond to the original points (x0,y0), while the values M(:,1) correspond to the moved points (x1,y1).

Plot the matched pairs of points. The points that moved farther than 2*costUnmatched away from the original point remain unmatched.

xc = [x0(M(:,2)); x1(M(:,1))];
yc = [y0(M(:,2)); y1(M(:,1))];

Input Arguments

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Cost matrix. Each entry Cost(i,j) specifies the cost of assigning row i to column j.

Data Types: single | double

Cost of not matching, specified as a scalar. matchpairs compares the value of 2*costUnmatched to the entries in Cost to determine whether it is more beneficial for a row or column to remain unmatched. Use this parameter to make matches more or less likely in the algorithm. For more information, see linear assignment problem.

Example: M = matchpairs(C,10) specifies a cost of 10 for not matching a row or column of C.

Data Types: single | double

Optimization goal, specified as either 'min' or 'max'. The optimization goal specifies whether the total cost should be minimized or maximized.

Example: M = matchpairs(Cost,costUnmatched,'max') specifies that the rows and columns of Cost should be matched together to maximize the total cost.

Output Arguments

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Matches, returned as a matrix. M is a p-by-2 matrix, where M(i,1) and M(i,2) are the row and column indices of a matched pair in the cost matrix. The rows of M are sorted with the second column in ascending order.

  • Each row and column can be matched a single time only, so each M(i,1) value and each M(i,2) value is unique.

  • M contains p matches, and p is less than or equal to the maximum number of matches min(size(Cost)).

  • The cost of the matches in M is sum([Cost(M(1,1),M(1,2)), Cost(M(2,1),M(2,2)), ..., Cost(M(p,1),M(p,2))]).

Unassigned rows, returned as a column vector of indices. The entries in uR indicate which rows in Cost are unassigned. Each entry in uR and uC contributes to the total cost of the solution according to costUnassigned.

Unassigned columns, returned as a column vector of indices. The entries in uC indicate which columns in Cost are unassigned. Each entry in uR and uC contributes to the total cost of the solution according to costUnassigned.

More About

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Linear Assignment Problem

The linear assignment problem is a way of assigning rows to columns such that each row is assigned to a column and the total cost of the assignments is minimized (or maximized). The cost of assigning each row to each column is captured in a cost matrix. The entry Cost(i,j) is the cost of assigning row i to column j.

The cost of unassignment assigns a cost to any row or column that is not matched. This practice allows for minimum-cost solutions that do not assign all rows or columns. If a row and column are not matched, this increases the total cost by 2*costUnmatched.

The total cost of a solution M is the sum of the cost of all matched pairs added to the cost of all unmatched pairs:


In code the total cost is

CostAssigned = sum(Cost(sub2ind(size(Cost), M(:,1), M(:,2))));
CostUnassigned = costUnmatched*(sum(size(Cost))-2*size(M,1));
TotalCost = CostAssigned + CostUnassigned;

  • Cost is an m-by-n matrix.

  • M is a p-by-2 matrix, where M(i,1) and M(i,2) are the row and column of a matched pair.

  • (m+n-2*p) is the total number of unmatched rows and columns.


[1] Duff, I.S. and J. Koster. "On Algorithms For Permuting Large Entries to the Diagonal of a Sparse Matrix." SIAM J. Matrix Anal. & Appl. 22(4), 2001. pp 973–996.

Extended Capabilities

Introduced in R2019a