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<title>Section 5.2.5: Mixed strategies for matrix games (LP formulation)</title>
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<h1>Section 5.2.5: Mixed strategies for matrix games (LP formulation)</h1>
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<pre class="codeinput">
<span class="comment">% Boyd &amp; Vandenberghe, "Convex Optimization"</span>
<span class="comment">% Jo&euml;lle Skaf - 08/24/05</span>
<span class="comment">%</span>
<span class="comment">% Player 1 wishes to choose u to minimize his expected payoff u'Pv, while</span>
<span class="comment">% player 2 wishes to choose v to maximize u'Pv, where P is the payoff</span>
<span class="comment">% matrix, u and v are the probability distributions of the choices of each</span>
<span class="comment">% player (i.e. u&gt;=0, v&gt;=0, sum(u_i)=1, sum(v_i)=1)</span>
<span class="comment">% LP formulation: minimize t</span>
<span class="comment">% s.t. u &gt;=0 , sum(u) = 1, P'*u &lt;= t*1</span>
<span class="comment">% maximize t</span>
<span class="comment">% s.t. v &gt;=0 , sum(v) = 1, P*v &gt;= t*1</span>
<span class="comment">% Input data</span>
randn(<span class="string">'state'</span>,0);
n = 12;
m = 12;
P = randn(n,m);
<span class="comment">% Optimal strategy for Player 1</span>
fprintf(1,<span class="string">'Computing the optimal strategy for player 1 ... '</span>);
cvx_begin
variables <span class="string">u(n)</span> <span class="string">t1</span>
minimize ( t1 )
u &gt;= 0;
ones(1,n)*u == 1;
P'*u &lt;= t1*ones(m,1);
cvx_end
fprintf(1,<span class="string">'Done! \n'</span>);
<span class="comment">% Optimal strategy for Player 2</span>
fprintf(1,<span class="string">'Computing the optimal strategy for player 2 ... '</span>);
cvx_begin
variables <span class="string">v(m)</span> <span class="string">t2</span>
maximize ( t2 )
v &gt;= 0;
ones(1,m)*v == 1;
P*v &gt;= t2*ones(n,1);
cvx_end
fprintf(1,<span class="string">'Done! \n'</span>);
<span class="comment">% Displaying results</span>
disp(<span class="string">'------------------------------------------------------------------------'</span>);
disp(<span class="string">'The optimal strategies for players 1 and 2 are respectively: '</span>);
disp([u v]);
disp(<span class="string">'The expected payoffs for player 1 and player 2 respectively are: '</span>);
[t1 t2]
disp(<span class="string">'They are equal as expected!'</span>);
</pre>
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<pre class="codeoutput">
Computing the optimal strategy for player 1 ...
Calling SDPT3: 25 variables, 13 equality constraints
------------------------------------------------------------
num. of constraints = 13
dim. of linear var = 24
dim. of free var = 1 *** convert ublk to lblk
*******************************************************************
SDPT3: Infeasible path-following algorithms
*******************************************************************
version predcorr gam expon scale_data
NT 1 0.000 1 0
it pstep dstep pinfeas dinfeas gap prim-obj dual-obj cputime
-------------------------------------------------------------------
0|0.000|0.000|7.5e+01|2.0e+01|2.4e+03| 2.875957e-10 0.000000e+00| 0:0:00| chol 1 1
1|0.903|0.656|7.3e+00|7.1e+00|6.8e+02| 1.084223e+01 -8.279551e+00| 0:0:00| chol 1 1
2|1.000|0.981|2.5e-06|1.5e-01|3.2e+01| 1.144905e+01 -1.008569e+01| 0:0:00| chol 1 1
3|1.000|0.806|1.5e-05|2.9e-02|5.8e+00| 2.825809e+00 -2.488865e+00| 0:0:00| chol 1 1
4|1.000|0.067|8.2e-07|2.8e-02|3.3e+00| 6.379819e-01 -2.400851e+00| 0:0:00| chol 1 1
5|0.915|0.884|1.6e-06|3.2e-03|5.4e-01| 1.130049e-01 -4.134015e-01| 0:0:00| chol 1 1
6|1.000|0.652|9.5e-08|1.1e-03|3.5e-01| 1.157364e-01 -2.254496e-01| 0:0:00| chol 1 1
7|0.933|0.584|1.6e-07|4.7e-04|1.1e-01|-1.533932e-02 -1.260174e-01| 0:0:00| chol 1 1
8|0.942|0.375|4.8e-08|2.9e-04|6.2e-02|-3.640272e-02 -9.820508e-02| 0:0:00| chol 1 1
9|1.000|0.529|1.7e-08|1.4e-04|2.9e-02|-4.200414e-02 -7.055158e-02| 0:0:00| chol 1 1
10|1.000|0.575|2.4e-09|5.8e-05|1.2e-02|-4.410085e-02 -5.584656e-02| 0:0:00| chol 1 1
11|1.000|0.637|5.6e-10|2.1e-05|4.2e-03|-4.475713e-02 -4.888417e-02| 0:0:00| chol 1 1
12|0.989|0.963|7.6e-11|7.9e-07|1.5e-04|-4.483902e-02 -4.499178e-02| 0:0:00| chol 1 1
13|0.989|0.988|3.2e-12|1.7e-06|1.9e-06|-4.484041e-02 -4.484217e-02| 0:0:00| chol 1 1
14|1.000|0.989|3.3e-14|2.2e-08|4.1e-08|-4.484041e-02 -4.484045e-02| 0:0:00| chol 1 1
15|1.000|0.989|1.5e-15|4.6e-10|7.7e-10|-4.484042e-02 -4.484042e-02| 0:0:00|
stop: max(relative gap, infeasibilities) &lt; 1.49e-08
-------------------------------------------------------------------
number of iterations = 15
primal objective value = -4.48404220e-02
dual objective value = -4.48404227e-02
gap := trace(XZ) = 7.68e-10
relative gap = 7.05e-10
actual relative gap = 6.70e-10
rel. primal infeas = 1.54e-15
rel. dual infeas = 4.60e-10
norm(X), norm(y), norm(Z) = 9.7e-01, 4.0e-01, 7.4e-01
norm(A), norm(b), norm(C) = 1.4e+01, 2.0e+00, 2.4e+00
Total CPU time (secs) = 0.16
CPU time per iteration = 0.01
termination code = 0
DIMACS: 1.5e-15 0.0e+00 5.5e-10 0.0e+00 6.7e-10 7.1e-10
-------------------------------------------------------------------
------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): -0.0448404
Done!
Computing the optimal strategy for player 2 ...
Calling SDPT3: 25 variables, 13 equality constraints
------------------------------------------------------------
num. of constraints = 13
dim. of linear var = 24
dim. of free var = 1 *** convert ublk to lblk
*******************************************************************
SDPT3: Infeasible path-following algorithms
*******************************************************************
version predcorr gam expon scale_data
NT 1 0.000 1 0
it pstep dstep pinfeas dinfeas gap prim-obj dual-obj cputime
-------------------------------------------------------------------
0|0.000|0.000|7.9e+01|2.0e+01|2.4e+03|-2.875957e-10 0.000000e+00| 0:0:00| chol 1 1
1|0.904|0.649|7.5e+00|7.2e+00|6.8e+02| 1.098635e+01 -8.128345e+00| 0:0:00| chol 1 1
2|1.000|0.981|2.7e-06|1.4e-01|3.1e+01| 1.147210e+01 -9.969407e+00| 0:0:00| chol 1 1
3|1.000|0.815|1.6e-05|2.7e-02|5.5e+00| 2.771718e+00 -2.299255e+00| 0:0:00| chol 1 1
4|1.000|0.064|8.1e-07|2.6e-02|3.2e+00| 6.734830e-01 -2.219711e+00| 0:0:00| chol 1 1
5|1.000|0.856|1.6e-06|3.7e-03|9.5e-01| 4.523807e-01 -4.802449e-01| 0:0:00| chol 1 1
6|0.972|0.412|2.0e-07|2.2e-03|5.5e-01| 1.986795e-01 -3.427010e-01| 0:0:00| chol 1 1
7|1.000|0.376|3.1e-07|1.4e-03|3.7e-01| 1.401321e-01 -2.248989e-01| 0:0:00| chol 1 1
8|1.000|0.388|6.3e-08|8.4e-04|2.3e-01| 9.168130e-02 -1.354947e-01| 0:0:00| chol 1 1
9|1.000|0.404|2.4e-08|5.0e-04|1.4e-01| 7.060727e-02 -6.977975e-02| 0:0:00| chol 1 1
10|1.000|0.611|7.1e-09|1.9e-04|5.2e-02| 4.979776e-02 -1.869084e-03| 0:0:00| chol 1 1
11|0.924|0.758|1.2e-09|4.7e-05|1.2e-02| 4.559065e-02 3.325888e-02| 0:0:00| chol 1 1
12|0.996|0.859|5.6e-11|6.6e-06|1.7e-03| 4.486235e-02 4.319745e-02| 0:0:00| chol 1 1
13|0.988|0.983|8.2e-12|1.9e-05|3.0e-05| 4.484071e-02 4.481317e-02| 0:0:00| chol 1 1
14|0.989|0.989|1.1e-13|3.4e-07|3.3e-07| 4.484043e-02 4.484012e-02| 0:0:00| chol 1 1
15|1.000|0.989|5.0e-15|3.8e-09|7.0e-09| 4.484042e-02 4.484042e-02| 0:0:00|
stop: max(relative gap, infeasibilities) &lt; 1.49e-08
-------------------------------------------------------------------
number of iterations = 15
primal objective value = 4.48404239e-02
dual objective value = 4.48404172e-02
gap := trace(XZ) = 7.00e-09
relative gap = 6.43e-09
actual relative gap = 6.14e-09
rel. primal infeas = 5.04e-15
rel. dual infeas = 3.77e-09
norm(X), norm(y), norm(Z) = 7.4e-01, 4.4e-01, 9.7e-01
norm(A), norm(b), norm(C) = 1.4e+01, 2.0e+00, 2.4e+00
Total CPU time (secs) = 0.16
CPU time per iteration = 0.01
termination code = 0
DIMACS: 5.0e-15 0.0e+00 4.6e-09 0.0e+00 6.1e-09 6.4e-09
-------------------------------------------------------------------
------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): -0.0448404
Done!
------------------------------------------------------------------------
The optimal strategies for players 1 and 2 are respectively:
0.2695 0.0686
0.0000 0.1619
0.0973 0.0000
0.1573 0.2000
0.1145 0.0000
0.0434 0.1545
0.0000 0.1146
0.0000 0.0000
0.2511 0.1030
0.0670 0.0000
0.0000 0.0000
0.0000 0.1974
The expected payoffs for player 1 and player 2 respectively are:
ans =
-0.0448 -0.0448
They are equal as expected!
</pre>
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