Clarabel Solver Examples • clarabel

Introduction

The first two examples are from the original Clarabel documentation and the third is from SCS.

1. Basic Quadratic Program Example

Suppose that we want to solve the following 2-dimensional quadratic programming problem:

$\begin{array}{ll} \text{minimize} & 3x_1^2 + 2x_2^2 - x_1 - 4x_2\\ \text{subject to} & -1 \leq x \leq 1, ~ x_1 = 2x_2 \end{array}$

We will show how to solve this problem using Clarabel in R.

The first step is to put the problem data into the standard form expected by the solver.

1.1. Objective function

The Clarabel solver’s default configuration expects problem data in the form $\frac{1}{2}x^\top P x + q^\top x$ .
We therefore define the objective function data as

$P = 2 \cdot \begin{bmatrix} 3 & 0 \\ 0 & 2\end{bmatrix} \mbox{ and } q = \begin{bmatrix} -1 \\ -4\end{bmatrix}.$

1.2. Constraints

The solver’s default configuration expects constraints in the form $Ax + s = b$ , where $s \in \mathcal{K}$ for some composite cone $\mathcal{K}$ . We have 1 equality constraint and 4 inequalities, so we require the first element of $s$ to be zero (i.e. the first constraint will correspond to the equality) and all other elements $s_i \ge 0$ . Our cone constraint on $s$ is therefore

$s \in \mathcal K = \{0\}^1 \times \mathbb{R}^4_{\ge 0}.$

Define the constraint data as

$A = \begin{bmatrix} 1 & -2 \\ 1 & 0 \\ 0 & 1 \\ -1 & 0 \\ 0 & -1\end{bmatrix} \mbox{ and } b=\begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}.$

Note that Clarabel expects inputs in Compressed Sparse Column (CSC) format for both $P$ and $A$ and will try to convert them if not so.

1.3. Solution

P <- Matrix::Matrix(2 * c(3, 0, 0, 2), nrow = 2, ncol = 2, sparse = TRUE)
P <- as(P, "symmetricMatrix")  # P needs to be a symmetric matrix
q <- c(-1, -4)
A <- Matrix::Matrix(c(1, 1, 0, -1, 0, -2, 0, 1, 0, -1), ncol = 2, sparse = TRUE)
b <- c(0, 1, 1, 1, 1)
cones <- list(z = 1L, l = 4L)  ## 1 equality and 4 inequalities, in order
s <- clarabel(A = A, b = b, q = q, P = P, cones = cones)
cat(sprintf("Solution status, description: = (%d, %s)\n",
            s$status, solver_status_descriptions()[s$status]))
#> Solution status, description: = (2, Solver terminated with a solution.)
cat(sprintf("Solution: (x1, x2) = (%f, %f)\n", s$x[1], s$x[2]))
#> Solution: (x1, x2) = (0.428571, 0.214286)

2. Basic Second-order Cone Programming Example

We want to solve the following 2-dimensional optimization problem:

$\begin{array}{ll} \text{minimize} & x_2^2\\[2ex] \text{subject to} & \left\|\begin{pmatrix} 2x_1 \\ x_2 \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \end{pmatrix}\right\|_2 \le 1 \end{array}$

2.1. Objective function

The Clarabel solver’s default configuration expects problem data in the form $\frac{1}{2}x^\top P x + q^\top x$ .
We therefore define the objective function data as

$P = 2 \cdot \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix} \mbox{ and } q = \begin{bmatrix} 0 \\ 0\end{bmatrix}.$

2.2. Constraints

The solver’s default configuration expects constraints in the form $Ax + s = b$ , where $s \in \mathcal{K}$ for some composite cone $\mathcal{K}$ . We have a single constraint on the 2-norm of a vector, so we rewrite

$\left\|\begin{pmatrix} 2x_1 \\ x_2 \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \end{pmatrix}\right\|_2 \le 1 \quad \Longleftrightarrow \quad \begin{pmatrix} 1 \\ 2x_1 - 2\\ x_2 - 2 \end{pmatrix} \in \mathcal{K}_{SOC}$ which puts our constraint in the form $b - Ax \in \mathcal{K}_{SOC}$ .

2.3. Solution

P <- Matrix::Matrix(2 * c(0, 0, 0, 1), nrow = 2, ncol = 2, sparse = TRUE)
P <- as(P, "symmetricMatrix") # P needs to be a symmetric matrix
q <- c(0, 0)
A <- Matrix::Matrix(c(0, -2.0, 0, 0, 0, 1.0), nrow = 3, ncol = 2, sparse = TRUE)
b <- c(1, -2, -2)
cones <- list(q = 3L)
s <- clarabel(A = A, b = b, q = q, P = P, cones = cones)
cat(sprintf("Solution status, description: = (%d, %s)\n",
            s$status, solver_status_descriptions()[s$status]))
#> Solution status, description: = (2, Solver terminated with a solution.)
cat(sprintf("Solution (x1, x2) = (%f, %f)\n", s$x[1], s$x[2]))
#> Solution (x1, x2) = (1.000000, -1.000000)

3. Semidefinite Cone Programming

Semidefinite problems are the only ones here that use dense linear algebra, which they get from your R installation’s BLAS and LAPACK rather than from this package. That could affect both speed and results; see Section 7.

Semidefinite cones have to be specified in a particular form. We borrow from the documentation for the SCS solver which has similar calling conventions.

The symmetric positive semidefinite cone of matrices is the set

$\{S \in \mathbf{R}^{k \times k} \mid S = S^\top, x^\top S x \geq 0 \ \forall x \in \mathbb{R}^k \}$

and for short, we use $S \succeq 0$ to denote membership. Clarabel vectorizes this cone in a special way which we detail here.

Clarabel assumes that the input data corresponding to semidefinite cones have been vectorized by scaling the off-diagonal entries by $\sqrt{2}$ and stacking the upper triangular elements column-wise. (SCS uses the lower triangular elements.) For a $k \times k$ matrix variable (or data matrix) this operation would create a vector of length $k(k+1)/2$ . Scaling by $\sqrt{2}$ is required to preserve the inner-product.

This must be done for the rows of both $A$ and $b$ that correspond to semidefinite cones and must be done independently for each semidefinite cone.

More explicitly, we want to express $\text{Trace}(Y S)$ as $\text{vec}(Y)^\top \text{vec}(S)$ , where the $\text{vec}$ operation takes the (assumed to be symmetric) $k \times k$ matrix

$\begin{aligned} S = \begin{bmatrix} S_{11} & S_{12} & \ldots & S_{1k} \\ S_{21} & S_{22} & \ldots & S_{2k} \\ \vdots & \vdots & \ddots & \vdots \\ S_{k1} & S_{k2} & \ldots & S_{kk} \\ \end{bmatrix} \end{aligned}$

and produces a vector consisting of the upper triangular elements scaled and arranged as

$\text{vec}(S) = (S_{11}, \sqrt{2} S_{12}, S_{22}, \sqrt{2}S_{13}, \ldots, \sqrt{2} S_{1k}, \sqrt{2}S_{2k}, \sqrt{2}S_{3k}, \dots, \sqrt{2}S_{k-1,k}, S_{kk}) \in \mathbb{R}^{k(k+1)/2}.$

To recover the matrix solution this operation must be inverted on the components of the vectors returned by Clarabel corresponding to each semidefinite cone. That is, the off-diagonal entries must be scaled by $1/\sqrt{2}$ and the upper triangular entries are filled in by copying the values of lower triangular entries. Explicitly, the inverse operation takes vector $s \in \mathbb{R}^{k(k+1)/2}$ and produces the matrix

$\begin{aligned} \text{mat}(s) = \begin{bmatrix} s_{1} & s_{2} / \sqrt{2} & \ldots & s_{k(k-1)/2-1} / \sqrt{2} \\ s_{2} / \sqrt{2} & s_{3} & \ldots & s_{k(k-1)/2} / \sqrt{2} \\ \vdots & \vdots & \ddots & \vdots \\ s_{k(k-1)/2-1} / \sqrt{2} & s_{k(k-1)/2} / \sqrt{2} & \ldots & s_{k(k+1) / 2} \\ \end{bmatrix} \in \mathbb{R}^{k \times k}. \end{aligned}$

So the cone definition that Clarabel uses is

$\mathcal{S}_+^k = \{ \text{vec}(S) \mid S \succeq 0\} = \{s \in \mathbb{R}^{k(k+1)/2} \mid \text{mat}(s) \succeq 0 \}.$

Below are two functions to implement both $\text{vec}$ and $\text{mat}$ .

#' Return an vectorization of symmetric matrix using the upper triangular part,
#' still in column order.
#' @param S a symmetric matrix
#' @return vector of values
vec <- function(S) {
  n <- nrow(S)
  sqrt2 <- sqrt(2.0)
  upper_tri <- upper.tri(S, diag = FALSE)
  S[upper_tri] <- S[upper_tri] * sqrt2
  S[upper.tri(S, diag = TRUE)]
}

#' Return the symmetric matrix from the [vec] vectorization
#' @param v a vector
#' @return a symmetric matrix
mat <- function(v) {
  n <- (sqrt(8 * length(v) + 1) - 1) / 2
  sqrt2 <- sqrt(2.0)
  S <- matrix(0, n, n)
  upper_tri <- upper.tri(S, diag = TRUE)
  S[upper_tri] <- v / sqrt2
  S <- S + t(S)
  diag(S) <- diag(S) / sqrt(2)
  S
}

3.1. Example

Consider the problem:

$\begin{array}{ll} \text{minimize} & x_1 - x_2 + x_3\\ \text{subject to} & B_1 - A_{11}x_1 - A_{12}x_2 - A_{13}x_3 \succeq 0\\ & B_2 - A_{21}x_1 - A_{22}x_2 - A_{23}x_3 \succeq 0 \end{array}$ where $A_{11}=\begin{bmatrix} -7 & -11 \\ -11 & 3\end{bmatrix}\mbox{, } A_{12}=\begin{bmatrix} 7 & -18 \\ -18 & 8\end{bmatrix}\mbox{, } A_{13}=\begin{bmatrix} -2 & -8 \\ -8 & 1\end{bmatrix},$

$A_{21}=\begin{bmatrix} -21 & -11 & 0\\ -11 & 10 & 8\\ 0 & 8 & 5\end{bmatrix}\mbox{, } A_{22}=\begin{bmatrix} 0 & 10 & 16\\ 10 & -10 & -10\\ 16 & -10 & 3\end{bmatrix}\mbox{, } A_{23}=\begin{bmatrix} -5 & 2 & -17\\ 2 & -6 & 8\\ -17 & 8 & 6\end{bmatrix},$

and

$B_1=\begin{bmatrix} 33 & -9 \\ -9 & 26\end{bmatrix}\mbox{, } B_2=\begin{bmatrix} 14 & 9 & 40\\ 9 & 91 & 10\\ 40 & 10 & 15\end{bmatrix}.$

The constraints involve symmetric positive semidefinite cones over variables $x \in \mathbb{R}^n$ and $S \in \mathbb{R}^{k \times k}$

$B - \sum_{i=1}^n \mathcal{A}_i x_i = S \succeq 0$

where data $B, \mathcal{A}_1, \ldots, \mathcal{A}_n \in \mathbb{R}^{k \times k}$ are symmetric. We can write this in the canonical form over a new variable $s \in \mathcal{S}_+^k$ :

$\begin{aligned} \begin{align} s &= \text{vec}(S)\\ &= \text{vec}(B - \sum_{i=1}^n \mathcal{A}_i x_i) \\ &= \text{vec}(B) - \sum_{i=1}^n \text{vec}(\mathcal{A}_i) x_i \\ &= b - Ax \end{align} \end{aligned}$

using the fact that $\text{vec}$ is linear, where $b = \text{vec}(B)$ and

$A = \begin{bmatrix} \text{vec}(\mathcal{A}_1) & \text{vec}(\mathcal{A}_2) & \cdots & \text{vec}(\mathcal{A}_n) \end{bmatrix}$

i.e., the vectors $\text{vec}(\mathcal{A}_i)$ stacked columnwise. This is in a form that we can input into Clarabel. To recover the matrix solution from the optimal solution returned by Clarabel, we simply use $S^\star = \text{mat}(s^\star)$ .

We have two such constraints and will therefore construct the vectors for both cones in order and specify the appropriate dimensions, 2 and 3, respectively.

q <- c(1, -1, 1) # objective: x_1 - x2 + x_3
A11 <- matrix(c(-7, -11, -11, 3), nrow = 2)
A12 <- matrix(c(7, -18, -18, 8), nrow = 2)
A13 <- matrix(c(-2, -8, -8, 1), nrow = 2)

A21 <- matrix(c(-21, -11, 0, -11, 10, 8, 0, 8, 5), nrow = 3)
A22 <- matrix(c(0, 10, 16, 10, -10, -10, 16, -10, 3), nrow = 3)
A23 <- matrix(c(-5, 2, -17, 2, -6, 8, -17, 8, 6), nrow = 3)

B1 <- matrix(c(33, -9, -9, 26), nrow = 2)
B2 <- matrix(c(14, 9, 40, 9, 91, 10, 40, 10, 15), nrow = 3)

A <- rbind(
  cbind(vec(A11), vec(A12), vec(A13)), # first psd constraint
  cbind(vec(A21), vec(A22), vec(A23))  # second psd constraint
)
b <- c(vec(B1), vec(B2)) # stack both psd constraints
cones <- list(s = c(2, 3)) # cone dimensions
s <- clarabel(A = A, b = b, q = q, cones = cones)
cat(sprintf("Solution status, description: = (%d, %s)\n",
            s$status, solver_status_descriptions()[s$status]))
#> Solution status, description: = (2, Solver terminated with a solution.)
cat(sprintf("Solution (x1, x2, x3) = (%f, %f, %f)\n", s$x[1], s$x[2], s$x[3]))
#> Solution (x1, x2, x3) = (-0.367746, 1.898333, -0.887466)

3.2. Chordal Decomposition of Sparse Semidefinite Programs

The cost of a semidefinite program grows very quickly with the side length of its semidefinite cones. When the aggregate sparsity pattern of those constraints is chordal, however, a large cone can be replaced by several small ones that overlap only on their shared entries, and the solver works with the small ones instead. This is enabled by default through chordal_decomposition_enable.

The following builds a linear matrix inequality $F_0 + \sum_i x_i F_i \succeq 0$ in which every data matrix is banded with half-bandwidth 2. A banded pattern is chordal, with cliques of size 3, so an $n \times n$ cone decomposes into roughly $n$ tiny ones.

banded_sdp <- function(n, w = 2, m = 5, seed = 7) {
  band_sym <- function(s) {
    set.seed(s)
    M <- matrix(0, n, n)
    for (i in seq_len(n))
      for (j in i:min(i + w, n)) M[i, j] <- M[j, i] <- rnorm(1)
    M
  }
  Fs <- lapply(seq_len(m), function(k) band_sym(1000 * seed + k))
  F0 <- band_sym(1000 * seed + m + 1)
  ## shift F0 so that x = 0 is strictly feasible
  F0 <- F0 + diag(n) * (abs(min(eigen(F0, only.values = TRUE)$values)) + 1)
  list(A = -do.call(cbind, lapply(Fs, vec)),
       b = vec(F0),
       q = vapply(Fs, function(Fi) sum(diag(Fi)), numeric(1)),
       cones = list(s = as.integer(n)))
}

solve_timed <- function(p, chordal) {
  ctrl <- clarabel_control(verbose = FALSE,
                           chordal_decomposition_enable = chordal)
  elapsed <- system.time(
    s <- clarabel(A = p$A, b = p$b, q = p$q, cones = p$cones, control = ctrl)
  )[["elapsed"]]
  data.frame(chordal = chordal,
             seconds = unname(elapsed),
             status = names(solver_status_descriptions())[s$status],
             objective = s$obj_val)
}

p40 <- banded_sdp(n = 40)
rbind(solve_timed(p40, FALSE), solve_timed(p40, TRUE))
#>   chordal seconds       status objective
#> 1   FALSE   0.473 AlmostSolved -6.603129
#> 2    TRUE   0.003       Solved -6.603129

Two things are worth reading off that comparison. The decomposed solve is faster by more than two orders of magnitude, and the two solves do not terminate at the same accuracy: on this instance the undecomposed solve stops at AlmostSolved while the decomposed one reaches Solved. That is not a general rule — which side stops early depends on the instance, and it goes both ways — but it is a reminder that decomposition produces a genuinely different numerical problem, and that the status is worth checking rather than assuming.

The gap widens sharply with the size of the cone. Measured separately, since the larger rows are too slow to run while this vignette is built, on R’s reference BLAS (Section 7):

$n$	cone length	decomposition off	on	speedup	status (off / on)
20	210	0.013s	0.001s	13x	Solved / Solved
40	820	0.420s	0.002s	210x	AlmostSolved / Solved
60	1830	4.393s	0.004s	1098x	Solved / Solved
80	3240	29.855s	0.004s	7464x	Solved / Solved
120	7260	424.978s	0.008s	53122x	Solved / AlmostSolved

The undecomposed column grows roughly like $n^6$ ; the decomposed one is close to linear in $n$ . Seven minutes becomes eight milliseconds. Note also that the reduced-accuracy termination lands on the decomposed side at $n = 120$ and on the undecomposed side at $n = 40$ , which is what is meant above by it going both ways.

Two consequences are worth knowing. Dual variables reported for semidefinite constraints are not in general the same as those obtained without decomposition, because the decomposition is reversed and the dual completed afterwards, and a positive semidefinite completion is not unique. And a problem that was actually decomposed cannot have its data updated in place, which matters for the warm starts described next. Pass chordal_decomposition_enable = FALSE to recover the previous behavior in either case.

4. Updating Problem Data (Warm Starts)

When solving a sequence of related problems that share the same sparsity structure, it is more efficient to create a persistent solver and update only the data that changes between solves. This avoids rebuilding the solver’s internal data structures each time.

Updates are refused when the solver has altered the structure of the problem, since the stored factorization no longer corresponds to the data you would be updating. Three settings can do that: presolve_enable, input_sparse_dropzeros, and chordal_decomposition_enable.

What matters is whether the transformation actually happened, not whether it was permitted. Presolve only rewrites the problem if it finds something to remove, and chordal decomposition only engages for semidefinite constraints whose sparsity pattern is decomposable. Linear, quadratic and second-order cone problems are never chordally decomposed, so leaving chordal_decomposition_enable at its default costs them nothing. The example below is a semidefinite program, but its two blocks are small and dense, hence not decomposable, so its updates are allowed even with chordal decomposition enabled.

Rather than reasoning about which transformation might apply, ask the solver: solver_is_update_allowed() reports whether the instance you actually built can be updated.

4.1. Example: SDP with a changed constraint

We revisit the semidefinite program from Section 3, but now suppose the constraint matrix $B_1$ changes from $\begin{bmatrix} 33 & -9 \\ -9 & 26 \end{bmatrix}$ to $\begin{bmatrix} 40 & -12 \\ -12 & 30 \end{bmatrix}$ .

First, we set up and solve the original problem using a persistent solver.

q_sdp <- c(1, -1, 1)
A11 <- matrix(c(-7, -11, -11, 3), nrow = 2)
A12 <- matrix(c(7, -18, -18, 8), nrow = 2)
A13 <- matrix(c(-2, -8, -8, 1), nrow = 2)

A21 <- matrix(c(-21, -11, 0, -11, 10, 8, 0, 8, 5), nrow = 3)
A22 <- matrix(c(0, 10, 16, 10, -10, -10, 16, -10, 3), nrow = 3)
A23 <- matrix(c(-5, 2, -17, 2, -6, 8, -17, 8, 6), nrow = 3)

B1 <- matrix(c(33, -9, -9, 26), nrow = 2)
B2 <- matrix(c(14, 9, 40, 9, 91, 10, 40, 10, 15), nrow = 3)

A_sdp <- rbind(
  cbind(vec(A11), vec(A12), vec(A13)),
  cbind(vec(A21), vec(A22), vec(A23))
)
b_sdp <- c(vec(B1), vec(B2))
cones_sdp <- list(s = c(2, 3))

## Create a persistent solver with updates enabled
ctrl <- clarabel_control(presolve_enable = FALSE, verbose = FALSE)
solver <- clarabel_solver(A = A_sdp, b = b_sdp, q = q_sdp, cones = cones_sdp, control = ctrl)
solver_is_update_allowed(solver)  # should be TRUE
#> [1] TRUE
sol1 <- solver_solve(solver)
cat(sprintf("Original solution (x1, x2, x3) = (%f, %f, %f)\n",
            sol1$x[1], sol1$x[2], sol1$x[3]))
#> Original solution (x1, x2, x3) = (-0.367746, 1.898333, -0.887466)

Now update the constraint with the new $B_1$ and re-solve.

B1_new <- matrix(c(40, -12, -12, 30), nrow = 2)
b_sdp_new <- c(vec(B1_new), vec(B2))  # only B1 changes
solver_update(solver, b = b_sdp_new)
sol2 <- solver_solve(solver)
cat(sprintf("Updated solution  (x1, x2, x3) = (%f, %f, %f)\n",
            sol2$x[1], sol2$x[2], sol2$x[3]))
#> Updated solution  (x1, x2, x3) = (-0.404810, 2.166303, -0.649288)

The solver reuses its internal factorization, making the second solve faster than constructing a new solver from scratch.

5. Cone Specifications

The following cones can be specified in Clarabel.

Parameter	Type	Length	Description	Definition (per parameter element)
z	integer	1	Number of primal zero cones (dual free cones), which corresponds to the primal equality constraints	$\{ 0 \}^{z}$
l	integer	1	Number of linear cones (non-negative cones)	$\{ x \in \mathbb{R}^{l} : x_i \ge 0, \forall i=1,\dots,l \}$
q	integer	>= 1	Vector of second-order cone sizes	$\{ (t,x) \in \mathbb{R}^{q} : \lVert x\rVert_2 \leq t \}$
s	integer	>= 1	Vector of positive semidefinite cone sizes	Upper triangular part of the positive semidefinite cone $S^s_+$ . The elements $x$ of this cone represent the columnwise stacking of the upper triangular part of a positive semidefinite matrix $X \in S^s_+$ , so that $x \in R^d$ with $d = s(s+1)/2$
ep	integer	1	Number of primal exponential cones	$\{(x, y, z) : y > 0,~~ ye^{x/y} \le z \}$
p	numeric	>= 1	Vector of primal power cone parameters	$\{(x, y, z) : x^p y^{(1-p)} \ge \lVert z\rVert,~ (x,y) \ge 0 \}$ with $p \in (0,1)$
gp	list	>= 1	List of named lists of two items, `a` : the numeric vector of at least 2 exponent terms, and `n` : an integer dimension of generalized power cone parameters	$\{(x, y) \in R^{len(a)} \times R^n : \prod\limits_{a_i \in a} x_i^{a_i} \ge \lVert y\rVert_2,~ x \ge 0 \}$ with $a_i \in (0,1)$ and $\sum a_i = 1$

Generalized power cone parameters are specified as list of two-item lists, with component named $a$ denoting the exponents and the named component $n$ denoting the dimension.

One can specify cones in any order if strict_cone_order is set to FALSE in the call to clarabel() but one has to ensure that parameter types are strictly specified for the values, e.g. 5L for integers, 0. for reals etc.

6. Control parameters

Clarabel has a number of parameters that control its behavior, including verbosity, time limits, and tolerances; see help on clarabel_control(). As an example, in the last problem, we can reduce the number of iterations.

P <- Matrix::Matrix(2 * c(0, 0, 0, 1), nrow = 2, ncol = 2, sparse = TRUE)
P <- as(P, "symmetricMatrix") # P needs to be a symmetric matrix
q <- c(0, 0)
A <- Matrix::Matrix(c(0, -2.0, 0, 0, 0, 1.0), nrow = 3, ncol = 2, sparse = TRUE)
b <- c(1, -2, -2)
cones <- list(q = 3L)
s <- clarabel(A = A, b = b, q = q, P = P, cones = cones,
              control = list(max_iter = 3)) ## Reduced number of iterations
cat(sprintf("Solution status, description: = (%d, %s)\n",
            s$status, solver_status_descriptions()[s$status]))
#> Solution status, description: = (5, Solver terminated with a solution (reduced accuracy))
cat(sprintf("Solution (x1, x2) = (%f, %f)\n", s$x[1], s$x[2]))
#> Solution (x1, x2) = (1.000000, -0.999998)

Note the different status, which should always be checked in code.

7. A Note on BLAS and LAPACK

This package links whatever BLAS and LAPACK your R installation uses; sessionInfo() reports which one is active.

On macOS the CRAN binary ships both R’s reference BLAS and a vecLib (Accelerate) build, and defaults to the reference one. Switching is a symlink:

cd /Library/Frameworks/R.framework/Resources/lib
ln -sf libRblas.vecLib.dylib libRblas.dylib   # vecLib
ln -sf libRblas.0.dylib      libRblas.dylib   # reference: the default

Repeating the Section 3.2 sweep under both libraries on a ARM Mac, moved solve times by under 1% at every size, in both directions:

$n$	reference, off	vecLib, off	reference, on	vecLib, on
20	0.015s	0.016s	0.001s	0.002s
40	0.429s	0.444s	0.003s	0.003s
60	4.475s	4.576s	0.003s	0.004s
80	30.078s	30.535s	0.005s	0.005s
120	426.752s	429.727s	0.008s	0.009s

For these problems not much time is not spent in the BLAS and LAPACK (dgemm, dsyrk, dpotrf, dsyevr) routines: the dense operations act on the $n \times n$ cone, while the cost is dominated by factorizing a KKT system whose size follows the vectorized cone length $n(n+1)/2$ , which is 7260 rows at $n = 120$ .

The results do differ between the two libraries. At $n = 40$ the decomposed dual differed by $1.3 \times 10^{-1}$ and the termination status moved from Solved to AlmostSolved. (See the R for macOS FAQ and R Installation and Administration on how Accelerate differs from the reference BLAS.)

The Section 3.2 timings were measured on the reference BLAS.