Exercise 4.38(b): Linear matrix inequalities with one variable
randn('state',0);
n = 4;
A = randn(n); A = 0.5*(A'+A);
B = randn(n); B = B'*B;
c = -1;
cvx_begin sdp
variable t
minimize ( c*t )
A >= t * B;
cvx_end
disp('------------------------------------------------------------------------');
disp('The optimal t obtained is');
disp(t);
Calling SDPT3: 10 variables, 1 equality constraints
For improved efficiency, SDPT3 is solving the dual problem.
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num. of constraints = 1
dim. of sdp var = 4, num. of sdp blk = 1
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SDPT3: Infeasible path-following algorithms
*******************************************************************
version predcorr gam expon scale_data
HKM 1 0.000 1 0
it pstep dstep pinfeas dinfeas gap mean(obj) cputime
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0|0.000|0.000|1.5e+01|5.3e+00|8.7e+01| 6.855734e-01| 0:0:00| chol 1 1
1|0.904|0.933|1.4e+00|4.0e-01|8.6e+00|-1.316248e+00| 0:0:00| chol 1 1
2|0.940|0.381|8.5e-02|2.5e-01|1.1e+00|-5.136774e+00| 0:0:00| chol 1 1
3|0.045|0.388|8.1e-02|1.5e-01|8.6e+00|-1.785268e+01| 0:0:00| chol 1 1
4|0.164|1.000|6.8e-02|5.0e-05|4.1e+01|-5.089196e+01| 0:0:00| chol 1 1
5|1.000|0.939|1.8e-10|1.5e-03|3.5e+00|-4.734455e+01| 0:0:00| chol 1 1
6|1.000|1.000|3.2e-11|5.0e-07|3.9e-01|-4.827072e+01| 0:0:00| chol 1 1
7|0.967|0.984|1.8e-11|5.7e-08|1.1e-02|-4.835024e+01| 0:0:00| chol 1 1
8|0.982|1.000|4.2e-13|5.0e-09|6.2e-04|-4.835383e+01| 0:0:00| chol 1 1
9|0.987|1.000|1.4e-12|1.0e-12|3.4e-05|-4.835402e+01| 0:0:00| chol 1 1
10|0.980|1.000|2.3e-13|1.0e-12|1.8e-06|-4.835403e+01| 0:0:00| chol 1 1
11|1.000|1.000|7.0e-13|1.0e-12|3.5e-07|-4.835403e+01| 0:0:00|
stop: max(relative gap, infeasibilities) < 1.49e-08
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number of iterations = 11
primal objective value = -4.83540312e+01
dual objective value = -4.83540315e+01
gap := trace(XZ) = 3.51e-07
relative gap = 3.59e-09
actual relative gap = 3.59e-09
rel. primal infeas = 7.03e-13
rel. dual infeas = 9.98e-13
norm(X), norm(y), norm(Z) = 1.1e+02, 4.8e+01, 4.8e+02
norm(A), norm(b), norm(C) = 1.1e+01, 2.0e+00, 4.0e+00
Total CPU time (secs) = 0.4
CPU time per iteration = 0.0
termination code = 0
DIMACS: 7.0e-13 0.0e+00 1.7e-12 0.0e+00 3.6e-09 3.6e-09
-------------------------------------------------------------------
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Status: Solved
Optimal value (cvx_optval): +48.354
------------------------------------------------------------------------
The optimal t obtained is
-48.3540