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Solving Optimization Problems with SCIP
In scip: Interface to the SCIP Optimization Suite
knitr::opts_chunk$set(collapse = TRUE, comment = "#>")
library(scip)
Introduction
The scip package provides an R interface to
SCIP (Solving Constraint Integer Programs),
one of the fastest non-commercial solvers for mixed-integer programming
(MIP) and mixed-integer nonlinear programming (MINLP). SCIP supports
linear, quadratic, SOS, and indicator constraints with continuous,
binary, and integer variables.
This vignette demonstrates the package through examples adapted from the
SCIP documentation and example
suite (Copyright
2002--2026 Zuse Institute Berlin, Apache-2.0 license).
The package offers two interfaces:
- One-shot interface (
scip_solve): pass a constraint matrix and
get a solution back in one call.
- Model-building interface (
scip_model, scip_add_var, etc.):
build a model incrementally, useful for complex formulations.
Example 1: A Production Planning LP
A company produces two products. Each unit of Product 1 yields \$5
profit and each unit of Product 2 yields \$4 profit. Production is
limited by labour (6 hours available) and materials (8 kg available).
Product 1 requires 1 hour and 2 kg per unit; Product 2 requires 2 hours
and 1 kg per unit.
$$\max\; 5x_1 + 4x_2$$
subject to:
$$x_1 + 2x_2 \le 6 \quad\text{(labour)}$$
$$2x_1 + x_2 \le 8 \quad\text{(materials)}$$
$$x_1, x_2 \ge 0$$
A <- matrix(c(1, 2,
2, 1), nrow = 2, byrow = TRUE)
b <- c(6, 8)
## scip_solve minimizes, so negate the objective for maximization
res <- scip_solve(obj = c(-5, -4), A = A, b = b,
sense = c("<=", "<="),
control = list(verbose = FALSE))
res$status
-res$objval # negate back to get the maximized profit
res$x
Example 2: The Knapsack Problem
Adapted from SCIP's solveknapsackexactly.c unit test (Marc Pfetsch,
Gregor Hendel, Zuse Institute Berlin).
A hiker can carry at most 13 kg. Six items are available with weights
7, 2, 7, 5, 1, and 3 kg, each with equal value. Which items should the
hiker pack to carry as many as possible?
$$\max\; \sum_{i=1}^{6} x_i$$
subject to:
$$7x_1 + 2x_2 + 7x_3 + 5x_4 + x_5 + 3x_6 \le 13$$
$$x_i \in {0,1}$$
A <- matrix(c(7, 2, 7, 5, 1, 3), nrow = 1)
res <- scip_solve(obj = c(-1, -1, -1, -1, -1, -1),
A = A, b = 13, sense = "<=",
vtype = "B",
control = list(verbose = FALSE))
res$status
items_packed <- which(res$x == 1)
items_packed
total_weight <- sum(c(7, 2, 7, 5, 1, 3)[items_packed])
total_weight
Example 3: N-Queens Problem
Adapted from SCIP's Queens example (examples/Queens/src/queens.cpp,
Cornelius Schwarz, University of Bayreuth).
Place $n$ queens on an $n \times n$ chessboard so that no two queens
attack each other. This classic combinatorial problem is naturally
modelled as a binary integer program.
Let $x_{i,j} \in {0,1}$ indicate whether a queen is placed at row
$i$, column $j$. The constraints are:
- Rows: exactly one queen per row, $\sum_j x_{i,j} = 1$
- Columns: exactly one queen per column, $\sum_i x_{i,j} = 1$
- Diagonals: at most one queen per diagonal, $\sum_{(i,j)\in D} x_{i,j} \le 1$
solve_nqueens <- function(n) {
m <- scip_model(paste0(n, "-queens"))
scip_set_objective_sense(m, "maximize")
## Create n*n binary variables x[i,j] with objective 1
## Variable (i,j) is stored at index (i-1)*n + j
idx <- function(i, j) (i - 1L) * n + j
scip_add_vars(m, obj = rep(1, n * n), lb = 0, ub = 1, vtype = "B",
names = sprintf("x%d_%d", rep(1:n, each = n), rep(1:n, n)))
## Row constraints: exactly one queen per row
for (i in 1:n) {
scip_add_linear_cons(m, vars = idx(i, 1:n), coefs = rep(1, n),
lhs = 1, rhs = 1, name = sprintf("row_%d", i))
}
## Column constraints: exactly one queen per column
for (j in 1:n) {
scip_add_linear_cons(m, vars = idx(1:n, j), coefs = rep(1, n),
lhs = 1, rhs = 1, name = sprintf("col_%d", j))
}
## Diagonal constraints: at most one queen per diagonal
## Down-right diagonals
for (d in (-(n - 2)):(n - 2)) {
rows <- max(1, 1 + d):min(n, n + d)
cols <- rows - d
if (length(rows) >= 2) {
scip_add_linear_cons(m, vars = idx(rows, cols),
coefs = rep(1, length(rows)),
rhs = 1, name = sprintf("diag_down_%d", d))
}
}
## Down-left diagonals (anti-diagonals)
for (s in 3:(2 * n - 1)) {
rows <- max(1, s - n):min(n, s - 1)
cols <- s - rows
if (length(rows) >= 2) {
scip_add_linear_cons(m, vars = idx(rows, cols),
coefs = rep(1, length(rows)),
rhs = 1, name = sprintf("diag_up_%d", s))
}
}
scip_optimize(m)
sol <- scip_get_solution(m)
## Extract queen positions
x_mat <- matrix(round(sol$x), nrow = n, byrow = TRUE)
positions <- which(x_mat == 1, arr.ind = TRUE)
colnames(positions) <- c("row", "col")
list(status = scip_get_status(m),
n_queens = as.integer(sol$objval),
positions = positions[order(positions[, "row"]), ],
board = x_mat)
}
result <- solve_nqueens(8)
result$status
result$n_queens
result$positions
Visualize the board (. = empty, Q = queen):
board_str <- apply(result$board, 1, function(row) {
paste(ifelse(row == 1, "Q", "."), collapse = " ")
})
cat(board_str, sep = "\n")
Example 4: Circle Packing (Quadratic Constraints)
Adapted from SCIP's CallableLibrary example
(examples/CallableLibrary/src/circlepacking.c, Jose Salmeron and
Stefan Vigerske, Zuse Institute Berlin).
Pack $n$ circles with given radii into a rectangle of minimum area. For
each circle $i$ with radius $r_i$, find center coordinates $(x_i,
y_i)$. The rectangle has width $W$ and height $H$.
Constraints:
- Circles stay within the rectangle: $r_i \le x_i \le W - r_i$ and
$r_i \le y_i \le H - r_i$
- Circles do not overlap:
$(x_i - x_j)^2 + (y_i - y_j)^2 \ge (r_i + r_j)^2$ for all $i < j$
Objective: Minimize $W \times H$ (area).
Since bilinear objectives are harder to handle, we use a simpler
approach: minimize $W + H$ as a proxy and fix one dimension. Here we
minimize the width given a fixed height, packing 3 circles.
radii <- c(1.0, 1.5, 0.8)
n <- length(radii)
H <- 5.0 # fixed height
m <- scip_model("circlepacking")
## Variables: x_i, y_i for each circle, plus W
x_idx <- integer(n)
y_idx <- integer(n)
for (i in seq_len(n)) {
x_idx[i] <- scip_add_var(m, obj = 0, lb = radii[i], ub = 100,
name = sprintf("x%d", i))
y_idx[i] <- scip_add_var(m, obj = 0, lb = radii[i], ub = H - radii[i],
name = sprintf("y%d", i))
}
w_idx <- scip_add_var(m, obj = 1, lb = 0, ub = 100, name = "W")
## x_i <= W - r_i => x_i - W <= -r_i
for (i in seq_len(n)) {
scip_add_linear_cons(m, vars = c(x_idx[i], w_idx), coefs = c(1, -1),
rhs = -radii[i],
name = sprintf("fit_x_%d", i))
}
## Non-overlap: (x_i - x_j)^2 + (y_i - y_j)^2 >= (r_i + r_j)^2
## Expand: x_i^2 - 2*x_i*x_j + x_j^2 + y_i^2 - 2*y_i*y_j + y_j^2 >= (r_i+r_j)^2
for (i in 1:(n - 1)) {
for (j in (i + 1):n) {
min_dist_sq <- (radii[i] + radii[j])^2
scip_add_quadratic_cons(m,
quadvars1 = c(x_idx[i], x_idx[i], x_idx[j], y_idx[i], y_idx[i], y_idx[j]),
quadvars2 = c(x_idx[i], x_idx[j], x_idx[j], y_idx[i], y_idx[j], y_idx[j]),
quadcoefs = c(1, -2, 1, 1, -2, 1),
lhs = min_dist_sq,
name = sprintf("nooverlap_%d_%d", i, j))
}
}
scip_optimize(m)
cat("Status:", scip_get_status(m), "\n")
sol <- scip_get_solution(m)
W_opt <- sol$x[w_idx]
cat(sprintf("Minimum width: %.3f (height fixed at %.1f)\n", W_opt, H))
for (i in seq_len(n)) {
cat(sprintf(" Circle %d (r=%.1f): center = (%.3f, %.3f)\n",
i, radii[i], sol$x[x_idx[i]], sol$x[y_idx[i]]))
}
Example 5: Facility Location (Mixed-Integer)
Adapted from SCIP's SCFLP example
(examples/SCFLP/doc/xternal_scflp.c, Zuse Institute Berlin).
A company must decide which of $p$ potential warehouse locations to
open. Each warehouse $i$ has a fixed opening cost $f_i$ and a capacity
$k_i$. Each of $q$ customers $j$ has a demand $d_j$ and a
per-unit shipping cost $c_{ij}$ from warehouse $i$. Minimize total cost.
$$\min\; \sum_i f_i y_i + \sum_{i,j} c_{ij} x_{ij}$$
subject to:
$$\sum_i x_{ij} \ge d_j \quad\forall j \quad\text{(demand)}$$
$$\sum_j x_{ij} \le k_i\, y_i \quad\forall i \quad\text{(capacity)}$$
$$y_i \in {0,1},\; x_{ij} \ge 0$$
## Problem data
p <- 3 # warehouses
q <- 4 # customers
fixed_cost <- c(100, 150, 120)
capacity <- c(50, 60, 40)
demand <- c(20, 25, 15, 30)
## Shipping cost matrix (warehouses x customers)
ship_cost <- matrix(c(
4, 8, 5, 6,
6, 3, 7, 4,
5, 5, 4, 8
), nrow = p, byrow = TRUE)
m <- scip_model("facility_location")
## Binary variables y_i: open warehouse i?
y_idx <- integer(p)
for (i in 1:p) {
y_idx[i] <- scip_add_var(m, obj = fixed_cost[i], vtype = "B",
name = sprintf("y%d", i))
}
## Continuous variables x_ij: amount shipped from i to j
x_idx <- matrix(0L, nrow = p, ncol = q)
for (i in 1:p) {
for (j in 1:q) {
x_idx[i, j] <- scip_add_var(m, obj = ship_cost[i, j],
lb = 0, ub = max(capacity),
name = sprintf("x%d_%d", i, j))
}
}
## Demand constraints: sum_i x_ij >= d_j
for (j in 1:q) {
scip_add_linear_cons(m, vars = x_idx[, j], coefs = rep(1, p),
lhs = demand[j],
name = sprintf("demand_%d", j))
}
## Capacity constraints: sum_j x_ij - k_i * y_i <= 0
for (i in 1:p) {
scip_add_linear_cons(m,
vars = c(x_idx[i, ], y_idx[i]),
coefs = c(rep(1, q), -capacity[i]),
rhs = 0,
name = sprintf("capacity_%d", i))
}
scip_optimize(m)
sol <- scip_get_solution(m)
cat("Status:", scip_get_status(m), "\n")
cat("Total cost:", sol$objval, "\n")
open_warehouses <- which(round(sol$x[y_idx]) == 1)
cat("Open warehouses:", open_warehouses, "\n")
shipments <- matrix(sol$x[x_idx], nrow = p)
cat("\nShipment plan (rows=warehouses, cols=customers):\n")
rownames(shipments) <- paste0("W", 1:p)
colnames(shipments) <- paste0("C", 1:q)
print(round(shipments, 1))
Example 6: Indicator Constraints
Adapted from SCIP's indicator test instances
(check/instances/Indicator/, Zuse Institute Berlin).
Indicator constraints model logical implications: "if a binary variable
is 1, then a linear constraint must hold." They are widely used in
scheduling, network design, and disjunctive programming.
Here we model a simple production problem with setup costs: a machine
must be "turned on" ($z_i = 1$) before it can produce. Turning on a
machine costs \$50. Each unit produced on machine $i$ earns revenue
$r_i$. Machine $i$ can produce at most $u_i$ units, but only if turned
on.
$$\max\; \sum_i (r_i x_i - 50 z_i)$$
subject to:
$$z_i = 1 \implies x_i \le u_i \quad \text{(capacity when on)}$$
$$z_i = 0 \implies x_i = 0 \quad \text{(nothing when off)}$$
$$x_i \ge 0,\; z_i \in {0,1}$$
We model $z_i = 0 \implies x_i \le 0$ as an indicator constraint on the
negated variable. Equivalently, we add the big-M constraint
$x_i \le u_i z_i$.
revenue <- c(12, 8, 15)
max_prod <- c(10, 20, 8)
setup_cost <- 50
m <- scip_model("setup_production")
scip_set_objective_sense(m, "maximize")
z_idx <- integer(3)
x_idx <- integer(3)
for (i in 1:3) {
z_idx[i] <- scip_add_var(m, obj = -setup_cost, vtype = "B",
name = sprintf("z%d", i))
x_idx[i] <- scip_add_var(m, obj = revenue[i], lb = 0, ub = max_prod[i],
name = sprintf("x%d", i))
## x_i <= max_prod[i] * z_i => x_i - max_prod[i] * z_i <= 0
scip_add_linear_cons(m, vars = c(x_idx[i], z_idx[i]),
coefs = c(1, -max_prod[i]),
rhs = 0,
name = sprintf("link_%d", i))
}
scip_optimize(m)
sol <- scip_get_solution(m)
cat("Status:", scip_get_status(m), "\n")
cat("Profit:", sol$objval, "\n")
for (i in 1:3) {
on <- round(sol$x[z_idx[i]])
prod <- sol$x[x_idx[i]]
cat(sprintf(" Machine %d: %s, produce %.0f units\n",
i, if (on) "ON" else "OFF", prod))
}
Solver Controls
Use scip_control() with the one-shot interface, or scip_set_param()
with the model-building interface, to tune solver behavior.
## One-shot: time limit and gap tolerance
ctrl <- scip_control(verbose = FALSE, time_limit = 60, gap_limit = 0.01)
## Model-building: set SCIP parameters directly
m <- scip_model("tuning_example")
scip_set_param(m, "display/verblevel", 0L) # suppress output
scip_set_param(m, "limits/time", 60.0) # 60-second time limit
scip_set_param(m, "limits/gap", 0.01) # 1% optimality gap
Common SCIP parameters:
| Parameter | Type | Description |
|-----------|------|-------------|
| display/verblevel | int | Verbosity (0=silent, 5=max) |
| limits/time | real | Time limit in seconds |
| limits/gap | real | Relative MIP gap tolerance |
| limits/nodes | longint | Maximum B&B nodes |
| limits/solutions | int | Stop after this many solutions |
See the SCIP parameter documentation
for the complete list.
References
- SCIP Optimization Suite: https://www.scipopt.org/
- Bestuzheva, K., Besançon, M., Chen, W.-K., Chmiela, A., Donkiewicz,
T., van Doornmalen, J., et al. (2021). The SCIP Optimization Suite
8.0. ZIB-Report 21-41, Zuse Institute Berlin.
- Examples in this vignette are adapted from the SCIP example suite
(Copyright 2002--2026 Zuse Institute Berlin, Apache-2.0 license),
available at
examples/ in the
SCIP source.
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knitr::opts_chunk$set(collapse = TRUE, comment = "#>") library(scip)
Introduction
The scip package provides an R interface to
SCIP (Solving Constraint Integer Programs),
one of the fastest non-commercial solvers for mixed-integer programming
(MIP) and mixed-integer nonlinear programming (MINLP). SCIP supports
linear, quadratic, SOS, and indicator constraints with continuous,
binary, and integer variables.
This vignette demonstrates the package through examples adapted from the SCIP documentation and example suite (Copyright 2002--2026 Zuse Institute Berlin, Apache-2.0 license).
The package offers two interfaces:
- One-shot interface (
scip_solve): pass a constraint matrix and get a solution back in one call. - Model-building interface (
scip_model,scip_add_var, etc.): build a model incrementally, useful for complex formulations.
Example 1: A Production Planning LP
A company produces two products. Each unit of Product 1 yields \$5 profit and each unit of Product 2 yields \$4 profit. Production is limited by labour (6 hours available) and materials (8 kg available). Product 1 requires 1 hour and 2 kg per unit; Product 2 requires 2 hours and 1 kg per unit.
$$\max\; 5x_1 + 4x_2$$ subject to: $$x_1 + 2x_2 \le 6 \quad\text{(labour)}$$ $$2x_1 + x_2 \le 8 \quad\text{(materials)}$$ $$x_1, x_2 \ge 0$$
A <- matrix(c(1, 2, 2, 1), nrow = 2, byrow = TRUE) b <- c(6, 8) ## scip_solve minimizes, so negate the objective for maximization res <- scip_solve(obj = c(-5, -4), A = A, b = b, sense = c("<=", "<="), control = list(verbose = FALSE)) res$status -res$objval # negate back to get the maximized profit res$x
Example 2: The Knapsack Problem
Adapted from SCIP's solveknapsackexactly.c unit test (Marc Pfetsch,
Gregor Hendel, Zuse Institute Berlin).
A hiker can carry at most 13 kg. Six items are available with weights 7, 2, 7, 5, 1, and 3 kg, each with equal value. Which items should the hiker pack to carry as many as possible?
$$\max\; \sum_{i=1}^{6} x_i$$ subject to: $$7x_1 + 2x_2 + 7x_3 + 5x_4 + x_5 + 3x_6 \le 13$$ $$x_i \in {0,1}$$
A <- matrix(c(7, 2, 7, 5, 1, 3), nrow = 1) res <- scip_solve(obj = c(-1, -1, -1, -1, -1, -1), A = A, b = 13, sense = "<=", vtype = "B", control = list(verbose = FALSE)) res$status items_packed <- which(res$x == 1) items_packed total_weight <- sum(c(7, 2, 7, 5, 1, 3)[items_packed]) total_weight
Example 3: N-Queens Problem
Adapted from SCIP's Queens example (examples/Queens/src/queens.cpp,
Cornelius Schwarz, University of Bayreuth).
Place $n$ queens on an $n \times n$ chessboard so that no two queens attack each other. This classic combinatorial problem is naturally modelled as a binary integer program.
Let $x_{i,j} \in {0,1}$ indicate whether a queen is placed at row $i$, column $j$. The constraints are:
- Rows: exactly one queen per row, $\sum_j x_{i,j} = 1$
- Columns: exactly one queen per column, $\sum_i x_{i,j} = 1$
- Diagonals: at most one queen per diagonal, $\sum_{(i,j)\in D} x_{i,j} \le 1$
solve_nqueens <- function(n) { m <- scip_model(paste0(n, "-queens")) scip_set_objective_sense(m, "maximize") ## Create n*n binary variables x[i,j] with objective 1 ## Variable (i,j) is stored at index (i-1)*n + j idx <- function(i, j) (i - 1L) * n + j scip_add_vars(m, obj = rep(1, n * n), lb = 0, ub = 1, vtype = "B", names = sprintf("x%d_%d", rep(1:n, each = n), rep(1:n, n))) ## Row constraints: exactly one queen per row for (i in 1:n) { scip_add_linear_cons(m, vars = idx(i, 1:n), coefs = rep(1, n), lhs = 1, rhs = 1, name = sprintf("row_%d", i)) } ## Column constraints: exactly one queen per column for (j in 1:n) { scip_add_linear_cons(m, vars = idx(1:n, j), coefs = rep(1, n), lhs = 1, rhs = 1, name = sprintf("col_%d", j)) } ## Diagonal constraints: at most one queen per diagonal ## Down-right diagonals for (d in (-(n - 2)):(n - 2)) { rows <- max(1, 1 + d):min(n, n + d) cols <- rows - d if (length(rows) >= 2) { scip_add_linear_cons(m, vars = idx(rows, cols), coefs = rep(1, length(rows)), rhs = 1, name = sprintf("diag_down_%d", d)) } } ## Down-left diagonals (anti-diagonals) for (s in 3:(2 * n - 1)) { rows <- max(1, s - n):min(n, s - 1) cols <- s - rows if (length(rows) >= 2) { scip_add_linear_cons(m, vars = idx(rows, cols), coefs = rep(1, length(rows)), rhs = 1, name = sprintf("diag_up_%d", s)) } } scip_optimize(m) sol <- scip_get_solution(m) ## Extract queen positions x_mat <- matrix(round(sol$x), nrow = n, byrow = TRUE) positions <- which(x_mat == 1, arr.ind = TRUE) colnames(positions) <- c("row", "col") list(status = scip_get_status(m), n_queens = as.integer(sol$objval), positions = positions[order(positions[, "row"]), ], board = x_mat) }
result <- solve_nqueens(8) result$status result$n_queens result$positions
Visualize the board (. = empty, Q = queen):
board_str <- apply(result$board, 1, function(row) { paste(ifelse(row == 1, "Q", "."), collapse = " ") }) cat(board_str, sep = "\n")
Example 4: Circle Packing (Quadratic Constraints)
Adapted from SCIP's CallableLibrary example
(examples/CallableLibrary/src/circlepacking.c, Jose Salmeron and
Stefan Vigerske, Zuse Institute Berlin).
Pack $n$ circles with given radii into a rectangle of minimum area. For each circle $i$ with radius $r_i$, find center coordinates $(x_i, y_i)$. The rectangle has width $W$ and height $H$.
Constraints:
- Circles stay within the rectangle: $r_i \le x_i \le W - r_i$ and $r_i \le y_i \le H - r_i$
- Circles do not overlap: $(x_i - x_j)^2 + (y_i - y_j)^2 \ge (r_i + r_j)^2$ for all $i < j$
Objective: Minimize $W \times H$ (area).
Since bilinear objectives are harder to handle, we use a simpler approach: minimize $W + H$ as a proxy and fix one dimension. Here we minimize the width given a fixed height, packing 3 circles.
radii <- c(1.0, 1.5, 0.8) n <- length(radii) H <- 5.0 # fixed height m <- scip_model("circlepacking") ## Variables: x_i, y_i for each circle, plus W x_idx <- integer(n) y_idx <- integer(n) for (i in seq_len(n)) { x_idx[i] <- scip_add_var(m, obj = 0, lb = radii[i], ub = 100, name = sprintf("x%d", i)) y_idx[i] <- scip_add_var(m, obj = 0, lb = radii[i], ub = H - radii[i], name = sprintf("y%d", i)) } w_idx <- scip_add_var(m, obj = 1, lb = 0, ub = 100, name = "W") ## x_i <= W - r_i => x_i - W <= -r_i for (i in seq_len(n)) { scip_add_linear_cons(m, vars = c(x_idx[i], w_idx), coefs = c(1, -1), rhs = -radii[i], name = sprintf("fit_x_%d", i)) } ## Non-overlap: (x_i - x_j)^2 + (y_i - y_j)^2 >= (r_i + r_j)^2 ## Expand: x_i^2 - 2*x_i*x_j + x_j^2 + y_i^2 - 2*y_i*y_j + y_j^2 >= (r_i+r_j)^2 for (i in 1:(n - 1)) { for (j in (i + 1):n) { min_dist_sq <- (radii[i] + radii[j])^2 scip_add_quadratic_cons(m, quadvars1 = c(x_idx[i], x_idx[i], x_idx[j], y_idx[i], y_idx[i], y_idx[j]), quadvars2 = c(x_idx[i], x_idx[j], x_idx[j], y_idx[i], y_idx[j], y_idx[j]), quadcoefs = c(1, -2, 1, 1, -2, 1), lhs = min_dist_sq, name = sprintf("nooverlap_%d_%d", i, j)) } } scip_optimize(m) cat("Status:", scip_get_status(m), "\n") sol <- scip_get_solution(m) W_opt <- sol$x[w_idx] cat(sprintf("Minimum width: %.3f (height fixed at %.1f)\n", W_opt, H)) for (i in seq_len(n)) { cat(sprintf(" Circle %d (r=%.1f): center = (%.3f, %.3f)\n", i, radii[i], sol$x[x_idx[i]], sol$x[y_idx[i]])) }
Example 5: Facility Location (Mixed-Integer)
Adapted from SCIP's SCFLP example
(examples/SCFLP/doc/xternal_scflp.c, Zuse Institute Berlin).
A company must decide which of $p$ potential warehouse locations to open. Each warehouse $i$ has a fixed opening cost $f_i$ and a capacity $k_i$. Each of $q$ customers $j$ has a demand $d_j$ and a per-unit shipping cost $c_{ij}$ from warehouse $i$. Minimize total cost.
$$\min\; \sum_i f_i y_i + \sum_{i,j} c_{ij} x_{ij}$$ subject to: $$\sum_i x_{ij} \ge d_j \quad\forall j \quad\text{(demand)}$$ $$\sum_j x_{ij} \le k_i\, y_i \quad\forall i \quad\text{(capacity)}$$ $$y_i \in {0,1},\; x_{ij} \ge 0$$
## Problem data p <- 3 # warehouses q <- 4 # customers fixed_cost <- c(100, 150, 120) capacity <- c(50, 60, 40) demand <- c(20, 25, 15, 30) ## Shipping cost matrix (warehouses x customers) ship_cost <- matrix(c( 4, 8, 5, 6, 6, 3, 7, 4, 5, 5, 4, 8 ), nrow = p, byrow = TRUE) m <- scip_model("facility_location") ## Binary variables y_i: open warehouse i? y_idx <- integer(p) for (i in 1:p) { y_idx[i] <- scip_add_var(m, obj = fixed_cost[i], vtype = "B", name = sprintf("y%d", i)) } ## Continuous variables x_ij: amount shipped from i to j x_idx <- matrix(0L, nrow = p, ncol = q) for (i in 1:p) { for (j in 1:q) { x_idx[i, j] <- scip_add_var(m, obj = ship_cost[i, j], lb = 0, ub = max(capacity), name = sprintf("x%d_%d", i, j)) } } ## Demand constraints: sum_i x_ij >= d_j for (j in 1:q) { scip_add_linear_cons(m, vars = x_idx[, j], coefs = rep(1, p), lhs = demand[j], name = sprintf("demand_%d", j)) } ## Capacity constraints: sum_j x_ij - k_i * y_i <= 0 for (i in 1:p) { scip_add_linear_cons(m, vars = c(x_idx[i, ], y_idx[i]), coefs = c(rep(1, q), -capacity[i]), rhs = 0, name = sprintf("capacity_%d", i)) } scip_optimize(m) sol <- scip_get_solution(m) cat("Status:", scip_get_status(m), "\n") cat("Total cost:", sol$objval, "\n") open_warehouses <- which(round(sol$x[y_idx]) == 1) cat("Open warehouses:", open_warehouses, "\n") shipments <- matrix(sol$x[x_idx], nrow = p) cat("\nShipment plan (rows=warehouses, cols=customers):\n") rownames(shipments) <- paste0("W", 1:p) colnames(shipments) <- paste0("C", 1:q) print(round(shipments, 1))
Example 6: Indicator Constraints
Adapted from SCIP's indicator test instances
(check/instances/Indicator/, Zuse Institute Berlin).
Indicator constraints model logical implications: "if a binary variable is 1, then a linear constraint must hold." They are widely used in scheduling, network design, and disjunctive programming.
Here we model a simple production problem with setup costs: a machine must be "turned on" ($z_i = 1$) before it can produce. Turning on a machine costs \$50. Each unit produced on machine $i$ earns revenue $r_i$. Machine $i$ can produce at most $u_i$ units, but only if turned on.
$$\max\; \sum_i (r_i x_i - 50 z_i)$$ subject to: $$z_i = 1 \implies x_i \le u_i \quad \text{(capacity when on)}$$ $$z_i = 0 \implies x_i = 0 \quad \text{(nothing when off)}$$ $$x_i \ge 0,\; z_i \in {0,1}$$
We model $z_i = 0 \implies x_i \le 0$ as an indicator constraint on the negated variable. Equivalently, we add the big-M constraint $x_i \le u_i z_i$.
revenue <- c(12, 8, 15) max_prod <- c(10, 20, 8) setup_cost <- 50 m <- scip_model("setup_production") scip_set_objective_sense(m, "maximize") z_idx <- integer(3) x_idx <- integer(3) for (i in 1:3) { z_idx[i] <- scip_add_var(m, obj = -setup_cost, vtype = "B", name = sprintf("z%d", i)) x_idx[i] <- scip_add_var(m, obj = revenue[i], lb = 0, ub = max_prod[i], name = sprintf("x%d", i)) ## x_i <= max_prod[i] * z_i => x_i - max_prod[i] * z_i <= 0 scip_add_linear_cons(m, vars = c(x_idx[i], z_idx[i]), coefs = c(1, -max_prod[i]), rhs = 0, name = sprintf("link_%d", i)) } scip_optimize(m) sol <- scip_get_solution(m) cat("Status:", scip_get_status(m), "\n") cat("Profit:", sol$objval, "\n") for (i in 1:3) { on <- round(sol$x[z_idx[i]]) prod <- sol$x[x_idx[i]] cat(sprintf(" Machine %d: %s, produce %.0f units\n", i, if (on) "ON" else "OFF", prod)) }
Solver Controls
Use scip_control() with the one-shot interface, or scip_set_param()
with the model-building interface, to tune solver behavior.
## One-shot: time limit and gap tolerance ctrl <- scip_control(verbose = FALSE, time_limit = 60, gap_limit = 0.01) ## Model-building: set SCIP parameters directly m <- scip_model("tuning_example") scip_set_param(m, "display/verblevel", 0L) # suppress output scip_set_param(m, "limits/time", 60.0) # 60-second time limit scip_set_param(m, "limits/gap", 0.01) # 1% optimality gap
Common SCIP parameters:
| Parameter | Type | Description |
|-----------|------|-------------|
| display/verblevel | int | Verbosity (0=silent, 5=max) |
| limits/time | real | Time limit in seconds |
| limits/gap | real | Relative MIP gap tolerance |
| limits/nodes | longint | Maximum B&B nodes |
| limits/solutions | int | Stop after this many solutions |
See the SCIP parameter documentation for the complete list.
References
- SCIP Optimization Suite: https://www.scipopt.org/
- Bestuzheva, K., Besançon, M., Chen, W.-K., Chmiela, A., Donkiewicz, T., van Doornmalen, J., et al. (2021). The SCIP Optimization Suite 8.0. ZIB-Report 21-41, Zuse Institute Berlin.
- Examples in this vignette are adapted from the SCIP example suite
(Copyright 2002--2026 Zuse Institute Berlin, Apache-2.0 license),
available at
examples/in the SCIP source.
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