ROI [1] is an R package to express optimization problems (in a rather low-level format). Its main advantage is that you can use the same model against a number of solvers. Linear models are expressed in a matrix notation. I am not a big fan of this: it restricts the applicability to small, simple models only. For larger, more complex models the matrix approach leads quickly to unreadable and unmaintainable models.
Solver interfaces are implemented as "plug-ins". A new plugin [2] allows to run solvers on NEOS [3]. This gives us access to a number of powerful solvers, without the need to install things locally.
To illustrate how we can express a simple transportation model in ROI, let's try a very simple transportation model [4]: \[\bbox[lightcyan,10px,border:3px solid darkblue]{\begin{align}\min\>&\sum_{i,j} c_{i,j} x_{i,j}\\&\sum_j x_{i,j} \le a_i\>\>\forall i\\&\sum_i x_{i,j} \ge b_j\>\>\forall j\\&x_{i,j}\ge 0\end{align}}\]
The GAMS representation is:
Note that all vectors and matrices have strings as indices. The ordering does not matter (like in a relational database table). Now let's do the same for ROI.
I just used vectors with integer indices: i.e. the position determines the meaning of a number (in other words: the ordering is important). We could have used vectors/matrices with row and column names, although I am not sure if this would make things much more readable. The results look like:
When reading the GAMS code, you can follow this is about a transportation model. The R code is much more obscure, it is close to write-only code. I don't understand how you can implement large and/or complex models in this way.
For some solvers you need to provide an e-mail address:
![]()
In this case, add email="...".
All structure is lost: we are just modeling:\[\begin{align} \min & \sum_j c_j x_j \\ & \sum_j a_{p,j} x_j \le b_{p}\\& \sum_j a_{q,j} x_j \ge b_{q}\end{align}\]
This plugin also allows quadratic models (but not general nonlinear models). In [2] a trivial non-convex QP model is presented: \[\begin{align}\max\>&F(P_F - 150) + C(P_C - 100) \\ & 2 F + C \le 500\\ & 2F + 3C \le 800 \\ & F \le 490 - P_F \\ & C \le 640 - 2P_C\\ & F, C, P_F, P_C\ge 0\end{align}\] ROI does not understand mathematics, so we have to pass this on in terms of matrices:
Unfortunately, this is not really close to the mathematical model and makes the problem setup difficult to read and understand.
In [2] a table is presented with some of the supported solvers:
![]()
For linear models, we can use the OMPR package [5,6] on top of ROI. This will give us more readable models. Here we implement the above transportation model. The model can look like:
This code at least has some resemblance to the mathematical model. The solution looks like:
We can clean up the solution a bit, so we get a more meaningful solution report:
Solver interfaces are implemented as "plug-ins". A new plugin [2] allows to run solvers on NEOS [3]. This gives us access to a number of powerful solvers, without the need to install things locally.
Linear models
To illustrate how we can express a simple transportation model in ROI, let's try a very simple transportation model [4]: \[\bbox[lightcyan,10px,border:3px solid darkblue]{\begin{align}\min\>&\sum_{i,j} c_{i,j} x_{i,j}\\&\sum_j x_{i,j} \le a_i\>\>\forall i\\&\sum_i x_{i,j} \ge b_j\>\>\forall j\\&x_{i,j}\ge 0\end{align}}\]
The GAMS representation is:
Set
i 'canning plants' / seattle, san-diego /
j 'markets' / new-york, chicago, topeka /;
Parameter
a(i) 'capacity of plant i in cases'
/ seattle 350
san-diego 600 /
b(j) 'demand at market j in cases'
/ new-york 325
chicago 300
topeka 275 /;
Table d(i,j) 'distance in thousands of miles'
new-york chicago topeka
seattle 2.51.71.8
san-diego 2.51.81.4;
Scalar f 'freight in dollars per case per thousand miles' / 90 /;
Parameter c(i,j) 'transport cost in thousands of dollars per case';
c(i,j) = f * d(i,j) / 1000;
Variable
x(i,j) 'shipment quantities in cases'
z 'total transportation costs in thousands of dollars';
Positive Variable x;
Equation
cost 'define objective function'
supply(i) 'observe supply limit at plant i'
demand(j) 'satisfy demand at market j';
cost.. z =e= sum((i,j), c(i,j)*x(i,j));
supply(i).. sum(j, x(i,j)) =l= a(i);
demand(j).. sum(i, x(i,j)) =g= b(j);
Model transport / all /;
solve transport using lp minimizing z;
display x.l, x.m;
Note that all vectors and matrices have strings as indices. The ordering does not matter (like in a relational database table). Now let's do the same for ROI.
#
# data
#
n <- 2# number of plants (sources)
m <- 3# number of markets (distinations)
si <- c('seattle', 'san-diego')
sj <- c('new-york', 'chicago', 'topeka')
a <- c(350, 600) # capacity
b <- c(325, 300, 275) # demand
# distances
d <- matrix(c(2.5, 1.7, 1.8, 2.5, 1.8, 1.4), nrow=n, ncol=m, byrow=TRUE)
f <- 90# unit freight cost
c <- d*f/1000# unit transportation cost
# flatten c row wise
cflat <- as.vector(t(c))
# create constraint matrix .... somewhat torturous code
nvar <- n * m;
A <- numeric(0)
varnames <- character(0);
for (i in1:n) {
row <- rep(0,nvar)
for (j in1:m) {
row[(i-1)*m+j] <- 1
varnames <- c(varnames,paste(si[i],sj[j],sep=":"))
}
A<-rbind(A,row)
}
for (j in1:m) {
row <- rep(0,nvar)
for (i in1:n)
row[(i-1)*m+j] <- 1
A<-rbind(A,row)
}
rhs <- c(a,b)
dir <- c(rep('<=',n),rep('>=',m))
LP <- OP(L_objective(cflat,names=varnames),
L_constraint(A,dir,rhs),
max=FALSE)
(sol <- ROI_solve(LP, solver = "glpk"))
I just used vectors with integer indices: i.e. the position determines the meaning of a number (in other words: the ordering is important). We could have used vectors/matrices with row and column names, although I am not sure if this would make things much more readable. The results look like:
> (sol <- ROI_solve(LP, solver = "glpk"))
Optimal solution found.
The objective value is: 1.536750e+02
> solution(sol)
seattle:new-york seattle:chicago seattle:topeka san-diego:new-york san-diego:chicago san-diego:topeka
5030002750275
When reading the GAMS code, you can follow this is about a transportation model. The R code is much more obscure, it is close to write-only code. I don't understand how you can implement large and/or complex models in this way.
NEOS
To use a NEOS solver we just need to change the last line:
> (sol <- ROI_solve(LP, solver = "neos", method = "mosek"))
Optimal solution found.
The objective value is: 1.536750e+02
For some solvers you need to provide an e-mail address:
In this case, add email="...".
Behind the scenes
Interestingly, under the hood, a GAMS model is generated and sent to NEOS. It looks like:
> model_call <- ROI_solve(LP, solver = "neos", method = "mosek",
+ dry_run = TRUE)
> cat(as.list(model_call)$xmlstring)
<?xml version="1.0" encoding="UTF-8"?>
<document>
<category>milp</category>
<solver>MOSEK</solver>
<inputMethod>GAMS</inputMethod>
<model><![CDATA[Option IntVarUp = 0;
Set i / R1*R5 / ;
Set ileq(i) / R1, R2 / ;
Set igeq(i) / R3, R4, R5 / ;
Set j / C1*C6 / ;
Parameter objL(j)
/C1 0.225
C2 0.153
C3 0.162
C4 0.225
C5 0.162
C6 0.126/ ;
Parameter rhs(i)
/R1 350
R2 600
R3 325
R4 300
R5 275/ ;
Parameter A
/R1.C1 1
R3.C1 1
R1.C2 1
R4.C2 1
R1.C3 1
R5.C3 1
R2.C4 1
R3.C4 1
R2.C5 1
R4.C5 1
R2.C6 1
R5.C6 1/;
Variables obj;
Positive Variables x(j);
Equations
ObjSum
LinLeq(ileq)
LinGeq(igeq);
ObjSum .. obj =e= sum(j, x(j) * objL(j)) ;
LinLeq(ileq) .. sum(j, A(ileq, j) * x(j)) =l= rhs(ileq) ;
LinGeq(igeq) .. sum(j, A(igeq, j) * x(j)) =g= rhs(igeq) ;
Model LinearProblem /all/ ;
Solve LinearProblem using LP minimizing obj ;
option decimals = 8;
display '---BEGIN.SOLUTION---', x.l, '---END.SOLUTION---';
file results /results.txt/;
results.nw = 0;
results.nd = 15;
results.nr = 2;
results.nz = 0;
put results;
put 'solution:'/;
loop(j, put, x.l(j)/);
put 'objval:'/;
put LinearProblem.objval/;
put 'solver_status:'/;
put LinearProblem.solvestat/;
put 'model_status:'/;
put LinearProblem.modelstat/;]]></model>
<options><![CDATA[]]></options>
<gdx><![CDATA[]]></gdx>
<wantgdx><![CDATA[]]></wantgdx>
<wantlog><![CDATA[]]></wantlog>
<comments><![CDATA[]]></comments>
</document>
>
All structure is lost: we are just modeling:\[\begin{align} \min & \sum_j c_j x_j \\ & \sum_j a_{p,j} x_j \le b_{p}\\& \sum_j a_{q,j} x_j \ge b_{q}\end{align}\]
Quadratic models
This plugin also allows quadratic models (but not general nonlinear models). In [2] a trivial non-convex QP model is presented: \[\begin{align}\max\>&F(P_F - 150) + C(P_C - 100) \\ & 2 F + C \le 500\\ & 2F + 3C \le 800 \\ & F \le 490 - P_F \\ & C \le 640 - 2P_C\\ & F, C, P_F, P_C\ge 0\end{align}\] ROI does not understand mathematics, so we have to pass this on in terms of matrices:
Q <- simple_triplet_matrix(i = 1:4, j = c(2, 1, 4, 3), rep(1, 4))
as.matrix(Q)
var_names <- c("full", "price_full", "compact", "price_compact")
o <- OP(
Q_objective(Q = Q, L = c(-150, 0, -100, 0), names = var_names),
L_constraint(rbind(c(2, 0, 1, 0), c(2, 0, 3, 0),
c(1, 1, 0, 0), c(0, 0, 1, 2)),
dir = leq(4), rhs = c(500, 800, 490, 640)),
maximum = TRUE)
(sol <- ROI_solve(o, solver = "neos", method = "BARON"))
Unfortunately, this is not really close to the mathematical model and makes the problem setup difficult to read and understand.
In [2] a table is presented with some of the supported solvers:
Linear models: OMPR
For linear models, we can use the OMPR package [5,6] on top of ROI. This will give us more readable models. Here we implement the above transportation model. The model can look like:
library(ompr)
library(ompr.roi)
library(dplyr)
#
# data
#
n <- 2# number of plants (sources)
m <- 3# number of markets (distinations)
si <- c('seattle', 'san-diego')
sj <- c('new-york', 'chicago', 'topeka')
a <- c(350, 600) # capacity
b <- c(325, 300, 275) # demand
# distances
d <- matrix(c(2.5, 1.7, 1.8, 2.5, 1.8, 1.4), nrow=n, ncol=m, byrow=TRUE)
f <- 90# unit freight cost
c <- d*f/1000# unit transportation cost
#
# model + solve
#
(sol <- MIPModel() %>%
add_variable(x[i,j],i=1:n,j=1:m,type = "continuous", lb = 0) %>%
set_objective(sum_expr(c[i,j]*x[i,j],i=1:n,j=1:m), "min") %>%
add_constraint(sum_expr(x[i,j],j=1:m)<=a[i], i=1:n) %>%
add_constraint(sum_expr(x[i,j],i=1:n)>=b[j], j=1:m) %>%
solve_model(with_ROI(solver = "neos", method="mosek")))
#
# solution
#
sol %>%
get_solution(x[i, j]) %>%
filter(value > 0)
This code at least has some resemblance to the mathematical model. The solution looks like:
Status: optimal
Objective value: 153.675
variable i j value
1 x 1150
2 x 21275
3 x 12300
4 x 23275
We can clean up the solution a bit, so we get a more meaningful solution report:
> sol %>%
+ get_solution(x[i, j]) %>%
+ filter(value > 0) %>%
+ mutate(Plant = si[i],Market = sj[j]) %>%
+ select(Plant,Market,Shipment=value)
Plant Market Shipment
1 seattle new-york 50
2 san-diego new-york 275
3 seattle chicago 300
4 san-diego topeka 275
References
- Many Solvers, One Interface, ROI, the R Optimization Infrastructure Package, https://www.r-project.org/conferences/useR-2010/slides/Theussl+Hornik+Meyer.pdf
- Ronald Hochreiter and Florian Schwendinger, ROI Plug-in NEOS, https://cran.r-project.org/web/packages/ROI.plugin.neos/vignettes/ROI.plugin.neos_Introduction.pdf
- NEOS Server: State-of-the-Art Solvers for Numerical Optimization, https://neos-server.org/neos/
- Matlab vs GAMS: linear programming, http://yetanothermathprogrammingconsultant.blogspot.com/2016/09/matlab-vs-gams-linear-programming.html
- Mixed Integer Linear Programming in R, https://dirkschumacher.github.io/ompr/
- MIP model in R using the OMPR package, http://yetanothermathprogrammingconsultant.blogspot.com/2017/04/mip-model-in-r-using-ompr-package.html