# Storage complexity of SDP solver SCS

This is a follow up question to this question.

Consider the following SDP in standard form:

\begin{align} &\min_{X\in S^n, X>0} \operatorname{tr}(AX)\\ &\mbox{subject to}\; \operatorname{tr}(B_iX)\leq b_i, i=1,\dots,m\\ \end{align}

Here, $A\in S^n,B_i \in S^n$, $b_i\in \mathbb{R}$ and $m=O(n)$.

Can anyone please explain the expected storage complexity of this problem in the general case?

My understanding is that first order solver like SCS will form a solve a $n^2$ sized linear system using gradient information only, and hence the storage complexity will be $O(n^4)$ which reflects the size of gradient matrix which will be $n^2\times n^2$.

Is this understanding correct? My confusion is regarding the possibility of dual problem being smaller size. Is that taken care of by CVX automatically? For example, if $n=1000$ and $m=1000$, it seems like primal form as mentioned above would form the smaller problem.

Edit After Johan's Answer: I am currently using CVX. I tried an example similar to what you gave. Using notation in my post above, I am using $n=100,m=10$. $X\in S^n$. If I submit my problem in format above, I get following output:

Calling SCS 1.1.7: 5050 variables, 10 equality constraints

Lin-sys: sparse-indirect, nnz in A = 50500, CG tol ~ 1/iter^(2.00) eps = 1.35e-06, alpha = 1.50, max_iters = 10000, normalize = 1, scale = 1.00 Variables n = 10, constraints m = 5050 Cones: sd vars: 5050, sd blks: 1 Setup time: 3.02e-03s

Now if I submit my problem by forming a dual of this, I get the message that CVX will solve the dual of my dual (i.e. the original problem), and I get exactly same number of constraints and variables as above.

So my question is whether CVX is really solving the way you have described ? What is the expected space complexity given the above variables ?

• If you want to get the definitive word on what CVX is doing, post your question at ask.cvxr.com .If you don't get it there, you won't get it anywhere. Oct 20 '15 at 22:47
• The notation used by SCS when displaying dimensions is a bit confusing. 10 variables in the dual, ok, but to call the vectorized size of the dual LMI constraints is odd. The efficient primal form has 5050 "variables" (vectorization of $X$) and 10 constraints. Note though that there is no chance that it is working in the "wrong" form, as that would require working with a matrix of size $125250 \times 125250$ for my example below, and that just isn't reasonable considering how fast the problem is solved. BTW, I presume you are reading the SCS implementation... Oct 21 '15 at 6:26
• After browsing the paper, I am a bit confused notation-wise, as they appear to have switched the notation ($y$ is the cone etc). Hence, take everything I've said with a large dose of salt. However, by reading the paper, you will clearly see space required as the algorithm is outlined in detail. Oct 21 '15 at 6:46
• ...the general conclusion is sort of correct any way. For the case of the indirect method, if the problem is dense, the solver will possibly work with the $m\times m$ matrix. For other cases, it works in an sparse format with various compositions of $A$ and $A^T$ (sparse LDL etc), where it is hard to talk about space complexity as it depends on the sparsity of all the matrices that arise. This is all discussed in Section 4 of the paper. Oct 21 '15 at 7:03
• @JohanLöfberg Thanks for the clarifications. Actually, from reading the paper and looking at SCS output above: In notation given in output, the matrix $A$(of the paper) is $m\times n$, where $m=5050$ and $n=10$. The linear system is of same order. Hence if we instead use the notation I mentioned in question (n=size of SDP cone=100, m=constraints in primal=10), the linear system being solved is size $O(n^2\times m)$. Oct 22 '15 at 3:01

No, your understanding is not correct. By introducing a slack $s\in R^{m}_+$ you can write the constraints in the form $\operatorname{tr}(B_iX) + s_i = b_i$. You now have a standard semidefinite program, in primal form, over the product of a semidefinite cone of size $n$, and an LP cone of size $m$. The linear system will be a linear system of size $m \times m$ as there are $m$ equalities in the primal ($m$ variables in the dual). A very bad model would be to do what you implicitly hint, i.e., to introduce the elements of $X$ as variables, leading to a linear system of size $O(n^2)$. When solving a primal-dual pair in semidefinite programming, you never explicitly solve for the elements of the primal cone, the matrix as a whole is reconstructed from the solution of the line-search for the duals.

The stuff here might be a relevant read (and the linked paper there)

EDIT: Trivial example in YALMIP (disclaimer, developed by me)

You don't want YALMIP to interpret this as a problem defined by the individual variables in $X$ and fit data into a dual description, which is default (leads to 125250 variables in dual /125250 equalities in primal)

X = sdpvar(500);
optimize([X>=0,trace(X)==1],trace(randn(500)*X),sdpsettings('solver','scs'))


Instead, you want YALMIP to interpret it from a primal point of view (1 equality in primal / 1 variable in dual)

optimize([X>=0,trace(X)==1],trace(randn(500)*X),sdpsettings('solver','scs','dualize',1))

• I presume you mean X = sdpvar(500,500); Oct 20 '15 at 14:13
• sdpvar(500) and sdpvar(500,500) are equivalent Oct 21 '15 at 5:54