PIQP Solver Interface • piqp

1. Introduction

PIQP solves quadratic programs of the form

$\begin{aligned} \min_{x} \quad & \frac{1}{2} x^\top P x + c^\top x \\ \text {s.t.}\quad & Ax=b, \\ & h_l \leq Gx \leq h_u, \\ & x_l \leq x \leq x_u \end{aligned}$

with primal decision variables $x \in \mathbb{R}^n$ , matrices $P\in \mathbb{S}_+^n$ , $A \in \mathbb{R}^{p \times n}$ , $G \in \mathbb{R}^{m \times n}$ , and vectors $c \in \mathbb{R}^n$ , $b \in \mathbb{R}^p$ , $h_l \in \mathbb{R}^m$ , $h_u \in \mathbb{R}^m$ , $x_l \in \mathbb{R}^n$ , and $x_u \in \mathbb{R}^n$ .

2. The Problem Solver Interface

Consider:

$\begin{aligned} \min_{x} \quad & \frac{1}{2} x^\top \begin{bmatrix} 6 & 0 \\ 0 & 4 \end{bmatrix} x + \begin{bmatrix} -1 \\ -4 \end{bmatrix}^\top x \\ \text {s.t.}\quad & \begin{bmatrix} 1 & -2 \end{bmatrix} x = 1, \\ & \begin{bmatrix} 1 & -1 \\ 2 & 0 \end{bmatrix} x \leq \begin{bmatrix} 0.2 \\ -1 \end{bmatrix}, \\ & -1 \leq x_1 \leq 1. \end{aligned}$

The data for this problem can be specified as below.

P <- matrix(c(6, 0, 0, 4), nrow = 2)
c <- c(-1, -4)
A <- matrix(c(1, -2), nrow = 1)
b <- 1
G <- matrix(c(1, 2, -1, 0), nrow = 2)
h_u <- c(0.2, -1)
x_l <- c(-1, -Inf)  ## 2 variables
x_u <- c(1, Inf)    ## 2 variables

The problem can now be solved via a call to solve_piqp().

sol <- solve_piqp(P, c, A, b, G, h_u = h_u, x_l = x_l, x_u = x_u, backend = "auto")
cat(sprintf("(Solution status, description): = (%d, %s)\n",
            sol$status, sol$info$status_desc))
#> (Solution status, description): = (1, solved)
cat(sprintf("Objective: %f, solution: (x1, x2) = (%f, %f)\n", sol$info$primal_obj, sol$x[1], sol$x[2]))
#> Objective: 6.160000, solution: (x1, x2) = (-0.600000, -0.800000)

sol contains many components as str(sol) will display but the most important ones are:

status : 1 if all goes well (more below),
x : solution vector
y : dual solution for the equality constraints
z_l : dual solution for the lower inequality constraints
z_u : dual solution for the upper inequality constraints
z_bl : dual solution of lower bound box constraints
z_bu : dual solution of upper bound box constraints
info$status_desc: a descriptive string of the status
info$primal_obj : primal objective value
info$run_time : total runtime, if asked for in settings (see below).

One can always construct the descriptive string for the status using:

status_description(sol$status)
#> [1] "Solver solved problem up to given tolerance."

Note that PIQP can handle infinite box constraints well, i.e. when elements of $x_l$ or $x_u$ are $-\infty$ or $\infty$ , respectively. The inequality constraints now support double-sided bounds $h_l \leq Gx \leq h_u$ . For one-sided inequalities $Gx \leq h$ , simply pass h_u = h (the default h_l is -Inf).

3. The Solver Model Object

Users who wish to solve QP problems will mostly use the solve_piqp() function. Behind the scenes, solve_piqp() creates a solver object and calls methods on the object to obtain the solution. The solver object can be created explicitly using piqp() and provides more elaborate facilities for updating problem data and warm-starting subsequent solves, which can be very efficient when one is solving a sequence of related problems.

The above problem could be solved using the solver model object thus:

model <- piqp(P, c, A, b, G, h_u = h_u, x_l = x_l, x_u = x_u)
sol2 <- solve(model)
identical(sol, sol2)
#> [1] FALSE

Indeed, this is exactly what solve_piqp() does. The generic functions solve(), update(), get_settings(), get_dims(), and update_settings() are available on the model object.

get_dims(model)
#> $n
#> [1] 2
#> 
#> $p
#> [1] 1
#> 
#> $m
#> [1] 2

Updating problem data and re-solving

The real advantage of the model object is the ability to update problem data and re-solve. When update() is called, only the changed data is passed to the solver, and the solver can warm-start from the previous solution. This is much more efficient than creating a new model from scratch each time.

Suppose we want to tighten the inequality bounds and change the linear cost. We update the model in place and re-solve:

update(model, c = c(-2, -3), h_u = c(0.1, -1.5))
sol3 <- solve(model)
cat(sprintf("Status: %s\n", sol3$info$status_desc))
#> Status: solved
cat(sprintf("Objective: %f, solution: (x1, x2) = (%f, %f)\n",
            sol3$info$primal_obj, sol3$x[1], sol3$x[2]))
#> Objective: 7.840000, solution: (x1, x2) = (-0.800000, -0.900000)

We can continue updating. Here we relax the variable bounds and change the equality constraint:

update(model, A = matrix(c(1, -1), nrow = 1), b = 0,
       x_l = c(-2, -2), x_u = c(2, 2))
sol4 <- solve(model)
cat(sprintf("Status: %s\n", sol4$info$status_desc))
#> Status: solved
cat(sprintf("Objective: %f, solution: (x1, x2) = (%f, %f)\n",
            sol4$info$primal_obj, sol4$x[1], sol4$x[2]))
#> Objective: 6.562500, solution: (x1, x2) = (-0.750000, -0.750000)

Dimension mismatches are caught:

update(model, b = c(5, 2))
#> Error in `update.piqp::piqp_model`:
#> ! Update parameters do not match original problem dimensions

Settings can also be updated between solves:

update_settings(model, new_settings = list(verbose = FALSE, max_iter = 100L))

4. Dense and Sparse Interfaces

PIQP supports dense and sparse problem formulations. For small and dense problems the dense interface is preferred since vectorized instructions and cache locality can be exploited more efficiently, but for sparse problems the sparse interface and result in significant speedups.

Either interface can be requested explicitly via the backend parameter which can take on any value among "dense", "sparse", or "auto", the default. The last value will automatically switch to a sparse interface if any of the supplied inputs ( $A$ , $P$ , or $G$ ) is a sparse matrix; otherwise it uses the dense interface.

sparse_sol <- solve_piqp(P, c, A, b, G, h_u = h_u, x_l = x_l, x_u = x_u, backend = "sparse")
str(sparse_sol)
#> List of 12
#>  $ status: int 1
#>  $ x     : num [1:2] -0.6 -0.8
#>  $ y     : num -11.8
#>  $ z_l   : num [1:2] 0 0
#>  $ z_u   : num [1:2] 1.64e+01 8.96e-11
#>  $ z_bl  : num [1:2] 2.18e-10 0.00
#>  $ z_bu  : num [1:2] 1.55e-12 0.00
#>  $ s_l   : num [1:2] 1e+30 1e+30
#>  $ s_u   : num [1:2] 5.67e-12 2.00e-01
#>  $ s_bl  : num [1:2] 4e-01 1e+30
#>  $ s_bu  : num [1:2] 1.6 1.0e+30
#>  $ info  :List of 32
#>   ..$ status_desc       : chr "solved"
#>   ..$ iter              : num 7
#>   ..$ rho               : num 1e-10
#>   ..$ delta             : num 1e-10
#>   ..$ mu                : num 5.02e-11
#>   ..$ sigma             : num 1e-06
#>   ..$ primal_step       : num 0.99
#>   ..$ dual_step         : num 0.99
#>   ..$ primal_res        : num 3.13e-13
#>   ..$ primal_res_rel    : num 1.96e-13
#>   ..$ dual_res          : num 1.5e-10
#>   ..$ dual_res_rel      : num 2.08e-11
#>   ..$ primal_res_reg    : num 3.13e-11
#>   ..$ primal_res_reg_rel: num 1.96e-11
#>   ..$ dual_res_reg      : num 1.5e-08
#>   ..$ dual_res_reg_rel  : num 2.08e-09
#>   ..$ primal_prox_inf   : num 0
#>   ..$ dual_prox_inf     : num 0
#>   ..$ primal_obj        : num 6.16
#>   ..$ dual_obj          : num 6.16
#>   ..$ duality_gap       : num 3.27e-10
#>   ..$ duality_gap_rel   : num 2.77e-11
#>   ..$ factor_retires    : num 0
#>   ..$ reg_limit         : num 1e-10
#>   ..$ no_primal_update  : num 1
#>   ..$ no_dual_update    : num 0
#>   ..$ setup_time        : num 0
#>   ..$ update_time       : num 0
#>   ..$ solve_time        : num 0
#>   ..$ kkt_factor_time   : num 0
#>   ..$ kkt_solve_time    : num 3442235
#>   ..$ run_time          : num 0

5. Another Example

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

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

Since the solver expects the objective in the form $\frac{1}{2}x^\top P x + c^\top x$ , we define

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

We have one equality constraint and box constraints. This leads to the following straight-forward formulation.

P <- matrix(2 * c(3, 0, 0, 2), nrow = 2, ncol = 2)
c <- c(-1, -4)
A <- matrix(c(1, -2), ncol = 2)
b <- 0
x_l <- rep(-1.0, 2)
x_u <- rep(1.0, 2)
sol <- solve_piqp(P = P, c = c, A = A, b = b, x_l = x_l, x_u = x_u)
cat(sprintf("(Solution status, description): = (%d, %s)\n",
            sol$status, sol$info$status_desc))
#> (Solution status, description): = (1, solved)
cat(sprintf("Objective: %f, solution: (x1, x2) = (%f, %f)\n", sol$info$primal_obj, sol$x[1], sol$x[2]))
#> Objective: -0.642857, solution: (x1, x2) = (0.428571, 0.214286)

But we can also choose to move the upper box constraints into the inequalities.

G <- diag(2)
h_u <- c(1, 1)
sol <- solve_piqp(P = P, c = c, A = A, b = b, G = G, h_u = h_u,
                  x_l = c(-1, -1), x_u = c(Inf, Inf))
cat(sprintf("(Solution status, description): = (%d, %s)\n",
            sol$status, sol$info$status_desc))
#> (Solution status, description): = (1, solved)
cat(sprintf("Objective: %f, solution: (x1, x2) = (%f, %f)\n", sol$info$primal_obj, sol$x[1], sol$x[2]))
#> Objective: -0.642857, solution: (x1, x2) = (0.428571, 0.214286)

Or we can move both of them into the inequalities.

G <- Matrix::Matrix(c(1, 0, -1, 0, 0, 1, 0, -1), byrow = TRUE,
                    nrow = 4, sparse = TRUE)
h_u <- rep(1, 4)

sol <- solve_piqp(A = A, b = b, c = c, P = P, G = G, h_u = h_u)
cat(sprintf("(Solution status, description): = (%d, %s)\n",
            sol$status, status_description(sol$status)))
#> (Solution status, description): = (1, Solver solved problem up to given tolerance.)
cat(sprintf("Objective: %f, solution: (x1, x2) = (%f, %f)\n", sol$info$primal_obj, sol$x[1], sol$x[2]))
#> Objective: -0.642857, solution: (x1, x2) = (0.428571, 0.214286)

All of them will yield the same result.

6. Solver parameters

PIQP has a number of parameters that control its behavior, including verbosity, tolerances, etc.; see help on piqp_settings(). As an example, in the last problem, we can reduce the number of iterations.

s <- solve_piqp(P = P, c = c, A = A, b = b, G = G, h_u = h_u,
          settings = list(max_iter = 3)) ## Reduced number of iterations
cat(sprintf("(Solution status, description): = (%d, %s)\n",
            s$status, s$info$status_desc))
#> (Solution status, description): = (-1, max iterations reached)
cat(sprintf("Objective: %f, solution: (x1, x2) = (%f, %f)\n", s$info$primal_obj, s$x[1], s$x[2]))
#> Objective: -0.642857, solution: (x1, x2) = (0.428564, 0.214282)

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