Introduction

The rcbc package provides an interface to the CBC (COIN-OR branch and cut) solver (Forrest & Lougee-Heimer 2005). Specifically, CBC is an open-source mixed integer programming solver that is developed as part of the Computational Infrastructure for Operations Research (COIN-OR) project. By interfacing with the CBC solver, the rcbc package can be used to generate optimal solutions to optimization problems. To showcase this package, we will define an optimization problem and then find an optimal solution for it.

Formulating the problem

Our optimization problem is an integer programming problem (Wolsey 2020). This means that is is an optimization problem with certain characteristics that, in turn, means it can be solved by the CBC solver. It is formulated following:

\begin{align} \text{maximize: } & x + 2 \times y + 0.5 \times z & \tag{1a} \\ \text{subject to: } & x + y \leq 1 & \tag{1b} \\ & 3 \times x + 4 \times z \geq 5 & \tag{1c} \\ & x + z = 8 & \tag{1d} \\ & x \in \{ 0, 1, 2, \dots, 10 \} & \tag{1e} \\ & y \in \{ 0, 1, 2, \dots, 11 \} & \tag{1f} \\ & z \in \{ 0, 1, 2, \dots, 13 \} & \tag{1g} \\ \end{align}

Here, the $$x$$, $$y$$, and $$z$$ variables are the decision variables. In other words, we don’t know what values these variables should have in an optimal solution, and so we can use mathematical optimization techniques to find this out. Our objective function is to maximize a weighted sum of the $$x$$, $$y$$, and $$z$$ variables (eqn 1a). We also have constraints that limit what values the decision variables can have (eqns 1b – 1g). A couple of our constraints specify inequalities (eqns 1b – 1c), meaning that certain combinations of variables cannot exceed (or be smaller than) certain thresholds (e.g. $$x$$ plus $$y$$ cannot exceed a value of 1). We also have an equality constraint (eqn 1d), meaning that certain combinations of variables must equal a certain value. Finally, we also have constraints (eqns 1e – 1g) that specify that the decision variables must have integer values that range between certain thresholds (e.g. $$x$$ must be an integer value between 0 and 10).

Encoding the problem

After defining the problem formulation mathematically, we need to encode objects within our R session to describe this problem. We can achieve this by creating a series of R objects that – in tandem – work together to describe the optimization problem:

# load rcbc package
library(rcbc)

# create vector to describe the objective function (eqn 1a)
## each element corresponds to a decision variable,
## and the values are the coefficients
obj <- c("x" = 1, "y" = 2, "z" = 0.5)

## print object
print(obj)
##   x   y   z
## 1.0 2.0 0.5
# create vector to indicate which variables are integer (eqns 1e -- 1g)
## TRUE means that the variable is an integer
## FALSE means that the variable is not an integer (i.e. continuous)
is_int <- c("x" = TRUE, "y" = TRUE, "z" = TRUE)

## print object
print(is_int)
##    x    y    z
## TRUE TRUE TRUE
# create vector to indicate the maximum value for each variable (eqns 1e -- 1g)
## create object
col_ub <- c("x" = 10, "y" = 11, "z" = 13)

## print object
print(col_ub)
##  x  y  z
## 10 11 13
# create vector to indicate the minimum value for each variable (eqns 1e -- 1g)
## create object
col_lb <- c("x" = 0, "y" = 0, "z" = 0)

## print object
print(col_lb)
## x y z
## 0 0 0
# create matrix to describe the constraint coefficients (eqns 1b -- 1d)
## this matrix will describe the inequality and equality constraints,
## and for these constraints it will contain the coefficients that
## are on the left hand side of the ">=", "<=", and "=" symbols
A <- matrix(c(1, 1, 0, 3, 0, 4, 1, 0, 1), byrow = TRUE, ncol = 3)
rownames(A) <- c("(eqn 1b)", "(eqn 1c)", "(eqn 1d)")
colnames(A) <- c("x", "y", "z")

## print object
print(A)
##          x y z
## (eqn 1b) 1 1 0
## (eqn 1c) 3 0 4
## (eqn 1d) 1 0 1
# create vector to describe the constraint upper limits (eqns 1b -- 1d)
## specifically, these describe the maximum possible value when
## applying the equations on the left hand side of the ">=", "<=", and "="
## symbols. note that eqn 1c has a maximum possible value of Inf
## because it has a >= inequality
row_ub <- c("(eqn 1b)" = 1, "(eqn 1c)" = Inf, "(eqn 1d)" = 8)

## print object
print(row_ub)
## (eqn 1b) (eqn 1c) (eqn 1d)
##        1      Inf        8
# create vector to describe the constraint lower limits (eqns 1b -- 1d)
## specifically, these describe the minimum possible value when
## applying the equations on the left hand side of the ">=", "<=", and "="
## symbols. note that eqn 1b has a minimum possible value of -Inf
## because it has a <= inequality
row_lb <- c("(eqn 1b)" = -Inf, "(eqn 1c)" = 5, "(eqn 1d)" = 8)

## print object
print(row_lb)
## (eqn 1b) (eqn 1c) (eqn 1d)
##     -Inf        5        8

After running this code, we will have generated seven R objects (i.e. obj, is_int, A, col_ub, col_lb, row_ub, and row_lb). If you have previous experience with mixed integer programming and are confused about the row_ub and row_lb objects (and expect to see objects for the constraint senses and right-hand-side values), please see below for details on alternate constraint specifications. Looking at this specification, we can see that each of these objects contains elements that correspond to decision variables or constraints in the problem formulation. We can also see that multiple objects are sometimes used to encode a single equation (e.g. A, row_ub, and row_lb encode eqn 1c). Although we have assigned names (e.g. to values, rows or columns) within these objects to help describe their purpose, these names are actually unnecessary. Indeed, it is the order of these values that is important. For instance, the $$x$$ decision variable is always in the first position among the R objects that relate to the decision variables (e.g. first index in obj and first column in A).

Solving the problem

After formulating the optimization problem, we can now solve it to optimality. This process is much simpler than encoding the optimization problem:

# solve problem to optimality
## note that we specify max = TRUE because our problem is to maximize
## the objective value (per eqn 1a)
result <- cbc_solve(
obj = obj, mat = A,
col_ub = col_ub, col_lb = col_lb,
row_ub = row_ub, row_lb = row_lb,
is_integer = is_int, max = TRUE)

We can now extract information from the result object. Specifically, we can extract optimal values for the $$x$$, $$y$$, and $$z$$ decision variables in the solution. We can also extract the optimal objective value (i.e. the value of eqn 1a given optimal values for the $$x$$, $$y$$, and $$z$$ variables). Finally, we can extract information on the status of the solver. This status information is important to verify that the CBC solver did indeed find an optimal solution, and did not end the optimization process for other reasons. For example, if we made a mistake when encoding the problem formulation, the solver might end the optimization process once it realizes that there is no possible solution (i.e. the input problem specification is “infeasible”).

# solution values
## extract values
sol <- column_solution(result)

## set variable names
names(sol) <- c("x", "y", "z")

## print object
print(sol)
## x y z
## 0 1 8
# extract objective value
## extract value
objv <- objective_value(result)

## print object
print(objv)
## [1] 6
# status
## extract information
status <- solution_status(result)

## print value
print(status)
## [1] "optimal"

The solution status (i.e. optimal) confirms that the optimization process solved the problem to optimality. Thus we can be confident that we now have optimal values for the $$x$$, $$y$$, and $$z$$ variables. Although this example completed quickly, the CBC solver will take more time to solve larger and more complex optimization problems (see Mittelmann 2021a and 2021b for benchmarks). In such cases, it is often helpful to customize the optimization process (e.g. specify that you only need near-optimal solutions). For more information on such customizations, please see documentation for the cbc_solve() function. You might also find it useful to try other solvers too (e.g. Gurobi or IBM CPLEX).

Alternate constraint specifications

If you have previous experience with mixed integer programming solvers, you might expect to see R objects that encode sense and right-hand-side (“rhs”) values for the constraints (instead of row_ub and row_lb). For example, the Rsymphony package provides an interface to the SYMPHONY solver (Ralphs and Güzelsoy 2005). Within the Rsymphony package, the Rsymphony_solve_LP function – an equivalent to the cbc_solve() function – has dir and rhs parameters for encoding constraints (instead of row_ub and row_lb). For example, following the alternate format (per Rsymphony_solve_LP), we would create the following R objects (instead of row_lb and row_ub):

# create dir object following Rsymphony_solve_LP format (per eqn 1)
## create object
dir <- c("(eqn 1b)" = "<=", "(eqn 1c)" = ">=", "(eqn 1d)" = "=")

## print object
print(dir)
## (eqn 1b) (eqn 1c) (eqn 1d)
##     "<="     ">="      "="
# create rhs object following Rsymphony_solve_LP format (per eqn 1)
## create object
rhs <- c("(eqn 1b)" = 1, "(eqn 1c)" = 5, "(eqn 1d)" = 8)

## print object
print(rhs)
## (eqn 1b) (eqn 1c) (eqn 1d)
##        1        5        8

If you already have your optimization problem encoded in such a format, you can use the following code to convert it into the correct format for rcbc (i.e. create row_ub and row_lb objects based on dir and rhs objects):

# initialize objects for rcbc
row_ub2 <- numeric(length(dir))
row_lb2 <- numeric(length(dir))

# convert >= constraints
idx <- dir == ">="
row_lb2[idx] <- rhs[idx]
row_ub2[idx] <- Inf

# convert <= constraints
idx <- dir == "<="
row_lb2[idx] <- -Inf
row_ub2[idx] <- rhs[idx]

# convert = constraints
idx <- dir == "="
row_lb2[idx] <- rhs[idx]
row_ub2[idx] <- rhs[idx]

We can compare these new R objects – based on converting the rhs and dir R objects – with the R objects we created earlier to verify that the conversion worked correctly.

# print constraint upper bound objects (note that names are unnecessary)
## print row_ub object we created earlier
print(row_ub)
## (eqn 1b) (eqn 1c) (eqn 1d)
##        1      Inf        8
## print row_ub2 object converted from rhs and dir objects
print(row_ub2)
## [1]   1 Inf   8
# print constraint lower bound objects (note that names are unnecessary)
## print row_lb object we created earlier
print(row_lb)
## (eqn 1b) (eqn 1c) (eqn 1d)
##     -Inf        5        8
## print row_lb2 object converted from rhs and dir objects
print(row_lb2)
## [1] -Inf    5    8

Since these new objects (i.e. row_lb2 and row_ub2) contain the same values as those we created earlier (i.e. row_lb and row_ub), we can be confident that the conversion code was successful.

Conclusion

Hopefully, this vignette has provided an informative tutorial on using the rcbc package to solve optimization problems. Although we only covered integer programming problems here, the CBC solver can also generate solutions for linear programming problems and mixed integer linear programming problems. If you need to solve other types of optimization problems (e.g. nonlinear problems), you will need to use other software (see the Optimization and Mathematical Programming CRAN Task View for suggestions). If you encounter any issues or have any questions about using the rcbc package, please create an issue on the package’s online code repository. Additionally, if you have questions about the CBC solver specifically (e.g. concerning the effect of certain parameters), please create an issue on the CBC online code repository.

Citation

Please cite the rcbc R package and the CBC solver in publications.


To cite the rcbc package in publications, use:

Schumacher D, Ooms J, and Hanson JO (2021) rcbc: COIN CBC MILP Solver
Bindings. R package version 0.1.0.9001.
https://github.com/dirkschumacher/rcbc

Forrest J and Lougee-Heimer R (2005) CBC User Guide. In Emerging
theory, Methods, and Applications (pp. 257--277). INFORMS,
Catonsville, MD.

Please cite both COIN-OR CBC and this package.
To see these entries in BibTeX format, use 'print(<citation>,
bibtex=TRUE)', 'toBibtex(.)', or set
'options(citation.bibtex.max=999)'.