LPSolve with R - 多个数据集作为输入

Question

我正在使用R的LPSolve工作，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个表格如下：

有关约束的说明 -

• 有来自每个制作公司的路线
• 来自生产厂房的总流出量=源自它的路线的总和（路线容积）
• 生产厂总流出量＆lt; =生产能力
• 路由卷本身是一个决策变量，取决于此帖中未提及的其他变量

约束的数学表示如下：

`Production Total Outflow = ∑(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。我尝试为以上2个约束编写以下代码。将有2个限制：

#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

我无法继续编写约束。请帮帮我们我还没有分享这个目标，因为它还有许多其他限制因素。

Answer 1

以下是一个尝试在生产厂房均匀分配路线量的示例

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/制作公司，则在make.lp中预先分配约束并使用set.row代替add.constraint可能更有效。以下是此示例，并将route_volume作为附加决策变量，如评论中所请求的那样：

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))]