Cpu time per iteration and number of iterations

1

I am using cvx 2.1 Matlab interface. I want to find only solution time of my optimization problem. The report after the solution of the optimization problem showing the following:

number of iterations = 13
. … .
Total CPU time (secs) = 0.14
CPU time per iteration = 0.01

What is ‘Total CPU time’ in CVX 2.1 version ? Is it problem conversion + solution ?

Does ‘number of iterations’ include only iterations of the solution algorithm ?

Is the measure for ‘CPU time per iteration’ in ‘sec’ ? Does it mean average of problem conversion + solution or Does it indicate solution time only ?

Below the entire output is given:

Calling SDPT3 4.0: 35 variables, 12 equality constraints
For improved efficiency, SDPT3 is solving the dual problem.

num. of constraints = 12
dim. of socp var = 13, num. of socp blk = 1
dim. of linear var = 22

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|1.3e+000|3.8e+000|1.7e+007| 2.170741e+006 0.000000e+000| 0:0:00| chol 1 1
1|0.764|0.768|3.0e-001|8.9e-001|4.8e+006| 1.743033e+006 2.631242e+004| 0:0:00| chol 1 1
2|0.596|0.600|1.2e-001|3.6e-001|2.8e+006| 1.505080e+006 3.484280e+004| 0:0:00| chol 1 1
3|0.587|0.583|5.0e-002|1.5e-001|1.7e+006| 1.164930e+006 3.563589e+004| 0:0:00| chol 1 1
4|0.615|0.583|1.9e-002|6.3e-002|1.0e+006| 8.294539e+005 3.346315e+004| 0:0:00| chol 1 1
5|0.711|0.590|5.6e-003|3.0e-002|6.3e+005| 5.433060e+005 3.221201e+004| 0:0:00| chol 1 1
6|1.000|0.545|2.8e-008|1.5e-002|2.8e+005| 2.781313e+005 4.417208e+004| 0:0:00| chol 1 1
7|0.856|1.000|6.4e-009|1.5e-004|1.0e+005| 1.071360e+005 6.847232e+003| 0:0:00| chol 1 1
8|0.990|0.877|1.0e-009|8.4e-005|4.3e+003| 1.551193e+004 1.169171e+004| 0:0:00| chol 1 1
9|0.929|0.961|2.3e-009|3.9e-005|3.3e+002| 1.178394e+004 1.168164e+004| 0:0:01| chol 1 1
10|0.982|1.000|2.0e-010|1.9e-005|1.8e+001| 1.147925e+004 1.156999e+004| 0:0:01| chol 1 1
11|0.985|0.991|9.0e-011|9.4e-006|2.7e-001| 1.146170e+004 1.151650e+004| 0:0:01| chol 2 2
12|0.957|1.000|7.8e-009|1.8e-011|1.2e-002| 1.146145e+004 1.146143e+004| 0:0:01| chol
linsysolve: Schur complement matrix not positive definite
switch to LU factor. lu 28 5
13|0.943|1.000|2.4e-009|1.4e-011|8.9e-004| 1.146144e+004 1.146144e+004| 0:0:01| lu 11 ^ 6
14|0.858|0.875|1.7e-008|3.7e-012|1.3e-004| 1.146144e+004 1.146144e+004| 0:0:01| lu 13 ^18
15|0.943|0.817|9.3e-009|9.8e-013|2.0e-005| 1.146143e+004 1.146144e+004| 0:0:01|
stop: max(relative gap, infeasibilities) < 1.49e-008

number of iterations = 15
primal objective value = 1.14614348e+004
dual objective value = 1.14614350e+004
gap := trace(XZ) = 2.00e-005
relative gap = 8.72e-010
actual relative gap = -8.41e-009
rel. primal infeas = 9.32e-009
rel. dual infeas = 9.84e-013
norm(X), norm(y), norm(Z) = 1.6e+005, 8.9e+003, 1.3e+004
norm(A), norm(b), norm© = 6.9e+000, 3.3e+003, 2.5e+002
Total CPU time (secs) = 0.59
CPU time per iteration = 0.04
termination code = 0
DIMACS: 1.8e-008 0.0e+000 2.0e-012 0.0e+000 -8.4e-009 8.7e-010

Status: Solved
Optimal value (cvx_optval): -111763

2

Where are you seeing this? CVX itself doesn’t provide that kind of information. Perhaps the solver is offering that?

3

It was in matlab. Using ‘cvx_begin’ and ‘cvx_end’.

4

Yes, I know you were using CVX. Perhaps you should edit your post with the entire output you’re seeing. Like I said, my suspicion is that it is the solver providing this information.

5

I have edited the post with complete output.

6

Great. As I suspected, it is the output of the SDPT3 solver where you are seeing that information. So it is only the solution time, not the conversion time.

7

Then is it possible to extract the solution time through cvx command or some matlab command? Or I can only get this by reading the output of the SDPT3? I wish to do some comparison according to the solution time.

8

You could use the MATLAB diary command to write the output to a file, then use something like PERL or sed to extract the time from the file.

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