randn('state',0);
n = 10;
W = randn(n); W = 0.5*(W + W');
fprintf(1,'Solving the dual of the two-way partitioning problem...');
cvx_begin sdp
variable nu(n)
maximize ( -sum(nu) )
W + diag(nu) >= 0;
cvx_end
fprintf(1,'Done! \n');
opt1 = cvx_optval;
fprintf(1,'Solving the SDP relaxation of the two-way partitioning problem...');
cvx_begin sdp
variable X(n,n) symmetric
minimize ( trace(W*X) )
diag(X) == 1;
X >= 0;
cvx_end
fprintf(1,'Done! \n');
opt2 = cvx_optval;
disp('------------------------------------------------------------------------');
disp('The optimal value of the Lagrange dual and the SDP relaxation fo the ');
disp('two-way partitioning problem are, respectively, ');
disp([opt1 opt2])
disp('They are equal as expected!');
Solving the dual of the two-way partitioning problem...
Calling SDPT3 4.0: 55 variables, 10 equality constraints
For improved efficiency, SDPT3 is solving the dual problem.
------------------------------------------------------------
num. of constraints = 10
dim. of sdp var = 10, num. of sdp blk = 1
*******************************************************************
SDPT3: Infeasible path-following algorithms
*******************************************************************
version predcorr gam expon scale_data
HKM 1 0.000 1 0
it pstep dstep pinfeas dinfeas gap prim-obj dual-obj cputime
-------------------------------------------------------------------
0|0.000|0.000|6.8e+00|4.3e+00|1.0e+03| 0.000000e+00 0.000000e+00| 0:0:00| chol 1 1
1|1.000|1.000|2.6e-06|4.2e-02|8.6e+01|-7.238328e+00 -9.272163e+01| 0:0:00| chol 1 1
2|1.000|0.880|7.1e-08|8.7e-03|1.6e+01|-1.728912e+01 -3.312548e+01| 0:0:00| chol 1 1
3|0.884|1.000|1.6e-08|4.2e-04|4.7e+00|-2.624691e+01 -3.095297e+01| 0:0:00| chol 1 1
4|0.962|0.948|4.7e-09|6.1e-05|2.5e-01|-2.863288e+01 -2.888462e+01| 0:0:00| chol 1 1
5|0.979|0.989|4.8e-10|4.8e-06|1.9e-02|-2.881025e+01 -2.882921e+01| 0:0:00| chol 1 1
6|0.946|0.982|2.8e-10|5.0e-07|9.2e-04|-2.882492e+01 -2.882583e+01| 0:0:00| chol 1 1
7|1.000|1.000|1.9e-09|5.6e-11|8.9e-05|-2.882561e+01 -2.882569e+01| 0:0:00| chol 1 1
8|0.974|0.984|1.6e-10|8.6e-11|2.2e-06|-2.882567e+01 -2.882568e+01| 0:0:00| chol 1 1
9|1.000|1.000|5.4e-11|3.1e-11|1.4e-07|-2.882567e+01 -2.882568e+01| 0:0:00|
stop: max(relative gap, infeasibilities) < 1.49e-08
-------------------------------------------------------------------
number of iterations = 9
primal objective value = -2.88256749e+01
dual objective value = -2.88256750e+01
gap := trace(XZ) = 1.41e-07
relative gap = 2.40e-09
actual relative gap = 2.35e-09
rel. primal infeas (scaled problem) = 5.44e-11
rel. dual " " " = 3.10e-11
rel. primal infeas (unscaled problem) = 0.00e+00
rel. dual " " " = 0.00e+00
norm(X), norm(y), norm(Z) = 8.6e+00, 1.0e+01, 1.2e+01
norm(A), norm(b), norm(C) = 4.2e+00, 4.2e+00, 7.6e+00
Total CPU time (secs) = 0.11
CPU time per iteration = 0.01
termination code = 0
DIMACS: 1.1e-10 0.0e+00 9.7e-11 0.0e+00 2.4e-09 2.4e-09
-------------------------------------------------------------------
------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): -26.6924
Done!
Solving the SDP relaxation of the two-way partitioning problem...
Calling SDPT3 4.0: 55 variables, 10 equality constraints
------------------------------------------------------------
num. of constraints = 10
dim. of sdp var = 10, num. of sdp blk = 1
*******************************************************************
SDPT3: Infeasible path-following algorithms
*******************************************************************
version predcorr gam expon scale_data
HKM 1 0.000 1 0
it pstep dstep pinfeas dinfeas gap prim-obj dual-obj cputime
-------------------------------------------------------------------
0|0.000|0.000|6.8e+00|4.3e+00|1.0e+03| 0.000000e+00 0.000000e+00| 0:0:00| chol 1 1
1|1.000|1.000|2.6e-06|4.2e-02|8.6e+01|-7.238328e+00 -9.272163e+01| 0:0:00| chol 1 1
2|1.000|0.880|7.1e-08|8.7e-03|1.6e+01|-1.728912e+01 -3.312548e+01| 0:0:00| chol 1 1
3|0.884|1.000|1.6e-08|4.2e-04|4.7e+00|-2.624691e+01 -3.095297e+01| 0:0:00| chol 1 1
4|0.962|0.948|4.7e-09|6.1e-05|2.5e-01|-2.863288e+01 -2.888462e+01| 0:0:00| chol 1 1
5|0.979|0.989|4.8e-10|4.8e-06|1.9e-02|-2.881025e+01 -2.882921e+01| 0:0:00| chol 1 1
6|0.946|0.982|2.8e-10|5.0e-07|9.2e-04|-2.882492e+01 -2.882583e+01| 0:0:00| chol 1 1
7|1.000|1.000|1.9e-09|5.6e-11|8.9e-05|-2.882561e+01 -2.882569e+01| 0:0:00| chol 1 1
8|0.974|0.984|1.6e-10|8.6e-11|2.2e-06|-2.882567e+01 -2.882568e+01| 0:0:00| chol 1 1
9|1.000|1.000|5.4e-11|3.1e-11|1.4e-07|-2.882567e+01 -2.882568e+01| 0:0:00|
stop: max(relative gap, infeasibilities) < 1.49e-08
-------------------------------------------------------------------
number of iterations = 9
primal objective value = -2.88256749e+01
dual objective value = -2.88256750e+01
gap := trace(XZ) = 1.41e-07
relative gap = 2.40e-09
actual relative gap = 2.35e-09
rel. primal infeas (scaled problem) = 5.44e-11
rel. dual " " " = 3.10e-11
rel. primal infeas (unscaled problem) = 0.00e+00
rel. dual " " " = 0.00e+00
norm(X), norm(y), norm(Z) = 8.6e+00, 1.0e+01, 1.2e+01
norm(A), norm(b), norm(C) = 4.2e+00, 4.2e+00, 7.6e+00
Total CPU time (secs) = 0.09
CPU time per iteration = 0.01
termination code = 0
DIMACS: 1.1e-10 0.0e+00 9.7e-11 0.0e+00 2.4e-09 2.4e-09
-------------------------------------------------------------------
------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): -26.6924
Done!
------------------------------------------------------------------------
The optimal value of the Lagrange dual and the SDP relaxation fo the
two-way partitioning problem are, respectively,
-26.6924 -26.6924
They are equal as expected!