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以R-多个数据集作为输入的LPSolve

lpsolve linear-programming mathematical-optimization r

Amit Madan · 技术社区 · 7 年前

我正在使用R进行LPSolve,我的输入数据是多个CSV文件的形式,每个文件都有一个表。其中2个表如下所述:

生产厂房的总流出量=源自该厂房的路线的总和(路线体积)
生产厂房总流出量<=生产能力

`Production Total Outflow = â(Route Volume) where (Production House ID from table_1)==(Originating from Prod House ID from table_2)`

Production Total Outflow <= Production Capacity

实际上,我有数千行。我试图为上述两个约束编写以下代码。将有2个约束:

#Reading Data from files
routeData = read.csv("Route.csv", header = TRUE)
ProductionData = read.csv("Production.csv", header = TRUE)

#Fetching variable columns
routeID = routeData$RouteID
productionID = ProductionData$ProductionID
productionCapacity = ProductionData$Supply.Capacity

numberOfColumns = length(routeID) + length(productionID) #4+2 decision variables
model <- make.lp(nrow=0, ncol=numberOfColumns, verbose="important")

for(i in 1:length(productionID)){
  add.constraint(model, 1, "<=", productionCapacity[i]) #Something wrong here
}
#I haven't attempted to write the other constraint

我无法进一步编写约束。请帮助大家。我没有分享这个目标,因为它还有很多其他限制。

1 回复 | 直到 7 年前

Karsten W. 7 年前

下面是一个示例,它试图在生产车间上均匀分布路线体积

library(lpSolveAPI)

prodcap <- setNames(c(50,100), c(1,2))
route <- data.frame(rid=1:4, pid_from=rep(1:2, each=2))
route_volume <- 125 # example

nvars <- nrow(route)+1 # example: evenly distribute production house output relative to capacity
lprec <- make.lp(0, nvars)

set.objfn(lprec, obj=1, indices=nvars)

# capacity constraints
for (i in seq(1, length(prodcap))) {
    route_ids <- which(route[,"pid_from"]==i)
    add.constraint(lprec, xt=rep(1, length(route_ids)), type="<=", rhs=prodcap[i], indices=route_ids)
}

# total outflow constraint
add.constraint(lprec, xt=rep(1, nrow(route)), type="=", rhs=route_volume, indices=seq(1, nvars-1))

# example: define the last decision variable as maximum flow over each production house
for (i in seq(1, length(prodcap))) {
    route_ids <- which(route[,"pid_from"]==i)
    add.constraint(lprec, xt=c(rep(1/prodcap[i], length(route_ids)), -1), type="<=", rhs=0, indices=c(route_ids, nvars))
}

# solve
status <- solve(lprec)
if(status!=0) stop("no solution found, error code=", status)
get.variables(lprec)[seq(1, nrow(route))]
#[1] 41.66667  0.00000 83.33333  0.00000

请注意,如果您有数千条路线/生产房屋,那么在中预分配约束可能更有效 make.lp 和使用 set.row 而不是 add.constraint .以下是一个示例,根据注释中的要求,将route_volume作为附加决策变量:

library(lpSolveAPI)

prodcap <- setNames(c(50,100), c(1,2))
route <- data.frame(rid=1:4, pid_from=rep(1:2, each=2))
route_volume <- 125 # example

# the first nrow(route) vars are the outflows, 
# then 1 variable for maximum flow (relative to capacity) over all production house
# then 1 last variable for the route volume
nvars <- nrow(route)+2 
ncons <- 2*length(prodcap)+3

# pre-allocate the constraints
lprec <- make.lp(ncons, nvars)

# set objective: minimize maximum flow relative to capacity (example)
set.objfn(lprec, obj=1, indices=nvars-1)

# capacity constraints
rownum <- 1
for (i in seq(1, length(prodcap))) {
    route_ids <- which(route[,"pid_from"]==i)
    set.row(lprec, row=rownum, xt=rep(1, length(route_ids)), indices=route_ids)
    set.rhs(lprec, prodcap[i], constraints=rownum)
    rownum <- rownum + 1
}

# total outflow constraint ("=" resolves to two constraints)
set.row(lprec, row=rownum, xt=c(rep(1, nrow(route)), -1), indices=c(seq(1, nvars-2), nvars))
set.rhs(lprec, 0, constraints=rownum)
rownum <- rownum + 1
set.row(lprec, row=rownum, xt=c(rep(-1, nrow(route)), 1), indices=c(seq(1, nvars-2), nvars))
set.rhs(lprec, 0, constraints=rownum)
rownum <- rownum + 1

# additional constraint for route volume
set.row(lprec, row=rownum, xt=-1, indices=nvars)
set.rhs(lprec, -125, constraints=rownum) #example: route_volume >= 125
rownum <- rownum + 1

# example: define the second last decision variable as maximum flow (relative to capacity) over all production houses
# rhs is 0, which is preset
for (i in seq(1, length(prodcap))) {
    route_ids <- which(route[,"pid_from"]==i)
    set.row(lprec, row=rownum, xt=c(rep(1/prodcap[i], length(route_ids)), -1), indices=c(route_ids, nvars-1))
    set.rhs(lprec, 0, constraints=rownum)
    rownum <- rownum + 1
}

# solve
status <- solve(lprec)
if(status!=0) stop("no solution found, error code=", status)
get.variables(lprec)[seq(1, nrow(route))]