# PIQP Solver Interface

## 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, \\ & Gx \leq h, \\ & x_{lb} \leq x \leq x_{ub} \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 $$c \in \mathbb{R}^n$$, $$b \in \mathbb{R}^p$$, $$h \in \mathbb{R}^m$$, $$x_{lb} \in \mathbb{R}^n$$, and $$x_{ub} \in \mathbb{R}^n$$.

## 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 <- c(0.2, -1)
x_lb <- c(-1, -Inf)  ## 2 variables
x_ub <- 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, x_lb = x_lb, x_ub = x_ub, 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$x[2])) #> Objective: 6.160000, solution: (x1, x2) = (-0.600000, -0.800000) sol contains many components as str(sol) will display but the most important ones are: • status : 1 if all goes well (more below), • x : solution vector • y : dual solution for the equality constraints • z : dual solution for the inequality constraints • z_lb : dual solution of lower bound box constraints • z_ub : dual solution of upper bound box constraints • info$status_desc: a descriptive string of the status
• info$primal_pobj : primal objective value • info$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 $$x_{lb}$$ or $$x_{ub}$$ are $$-\infty$$ or $$\infty$$, respectively. On the contrary, infinite values in the general inequalities $$Gx \leq h = \pm \infty$$ can cause problems, i.e., they are converted internally to -1e30 and 1e30, respectively. ## 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 explitly using piqp() and provides more elaborate facilities for updating problem data. This can be very efficient when one is solving the same kind of problem over and over. The above problem could be solved using the solver model object thus: model <- piqp(P, c, A, b, G, h, x_lb = x_lb, x_ub = x_ub) sol2 <- model$solve()
identical(sol, sol2)
#> [1] TRUE

Indeed, this is exactly what solve_piqp() does. But this interface allows us to update the settings, the bounds, etc. and resolve the same problem more efficiently.

model$update(x_lb = c(0, 0)) sol3 <- model$solve()
cat(sprintf("(Solution status, description): = (%d, %s)\n",
sol3$status, status_description(sol3$status)))
#> (Solution status, description): = (-2, The problem is primal infeasible.)

Setting the lower bounds made the problem infeasible.

## Try to give an inappropriate b value
model$update(b = c(5, 2)) #> Error in model$update(b = c(5, 2)): Update parameters not match original problem dimensions

The error message correctly indicates that b has wrong dimensions.

The methods exposed by the model object can be seen in the documentation for the object piqp_model.

For example, we could query the problem dimensions.

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

## 3. 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, x_lb, x_ub, backend = "sparse")
str(sparse_sol)
#> List of 15
#>  $status: int 1 #>$ x     : num [1:2] -0.6 -0.8
#>  $y : num -11.8 #>$ z     : num [1:2] 1.64e+01 1.01e-08
#>  $z_lb : num [1:2] 5.61e-10 0.00 #>$ z_ub  : num [1:2] 3.2e-10 0.0
#>  $s : num [1:2] 5.19e-11 2.00e-01 #>$ s_lb  : num [1:2] 0.4 Inf
#>  $s_ub : num [1:2] 1.6 Inf #>$ zeta  : num [1:2] -0.6 -0.8
#>  $lambda: num -11.8 #>$ nu    : num [1:2] 1.64e+01 1.01e-08
#>  $nu_lb : num [1:2] 5.61e-10 0.00 #>$ nu_ub : num [1:2] 3.2e-10 0.0
#>  $info :List of 24 #> ..$ status_desc     : chr "solved"
#>   ..$iter : num 6 #> ..$ rho             : num 1e-10
#>   ..$delta : num 1e-10 #> ..$ mu              : num 9.01e-10
#>   ..$sigma : num 1e-06 #> ..$ primal_step     : num 0.99
#>   ..$dual_step : num 0.99 #> ..$ primal_inf      : num 4.44e-11
#>   ..$primal_rel_inf : num 1.6 #> ..$ dual_inf        : num 8.11e-10
#>   ..$dual_rel_inf : num 7.2 #> ..$ primal_obj      : num 6.16
#>   ..$dual_obj : num 6.16 #> ..$ duality_gap     : num 3.88e-09
#>   ..$duality_gap_rel : num 11.8 #> ..$ factor_retires  : num 0
#>   ..$reg_limit : num 1e-10 #> ..$ no_primal_update: num 0
#>   ..$no_dual_update : num 0 #> ..$ setup_time      : num 0
#>   ..$update_time : num 0 #> ..$ solve_time      : num 0
#>   ..$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_lb <- rep(-1.0, 2) x_ub <- rep(1.0, 2) sol <- solve_piqp(P = P, c = c, A = A, b = b, x_lb = x_lb, x_ub = x_ub) 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 <- c(1, 1)
sol <- solve_piqp(P = P, c = c, A = A, b = b, G = G, h = h,
x_lb = c(-1, -1), x_ub = 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$x[2])) #> Objective: -0.642857, solution: (x1, x2) = (0.428571, 0.214286) Or we can move both of them into the inequalities. G <- Matrix::Matrix(c(1, 0, -1, 0, 0, 1, 0, -1), byrow = TRUE, nrow = 4, sparse = TRUE) h <- rep(1, 4) sol <- solve_piqp(A = A, b = b, c = c, P = P, G = G, h = h) cat(sprintf("(Solution status, description): = (%d, %s)\n", sol$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$x[2])) #> Objective: -0.642857, solution: (x1, x2) = (0.428571, 0.214286) All of them will yield the same result. ## 5. Solver parameters PIQP has a number of parameters that control its behavior, including verbosity, tolerances, etc.; see help on piqp_settings(). As an example, in the last problem, we can reduce the number of iterations. s <- solve_piqp(P = P, c = c, A = A, b = b, G = G, h = h, settings = list(max_iter = 3)) ## Reduced number of iterations cat(sprintf("(Solution status, description): = (%d, %s)\n", s$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) = (0.428294, 0.214147)

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