[R] linear programming in R | limits to what it can do, or my mistake?

[Martin Becker]

Apart from the fact that the statement "such that t1+t2+t3+t4=2970 (as 
it must)" is not correct, the LP can be implemented as follows:

library(lpSolve)
LHS <- rbind(
c(0,0,0,0, 1, 0, 0,0),
c(1,0,0,0,-1, 1, 0,0),
c(0,1,0,0, 0,-1, 1,0),
c(0,0,1,0, 0, 0,-1,1),
cbind(-diag(4),diag(4)),
c(0,0,0,0,0,1,0,0),
c(0,0,0,0,0,0,1,0),
c(0,0,0,0,0,0,0,1)
)

RHS <- c(640,825,580,925,0,0,0,0,1000,1000,1000)

DIR <- c(rep("==",4),rep(">=",3),"=",rep("<=",3))

OBJ <- c(35,55,50,65,0,0,0,0)

lp("min",OBJ,LHS,DIR,RHS)

Best,
Martin


Am 29.01.24 um 22:28 schrieb Evan Cooch:
> Question for 'experts' in LP using R (using the lpSolve package, say) --
> which does not apply to me for the sort of problem I describe below.
> I've run any number of LP's using lpSolve in R, but all of them to date
> have objective and constraint functions that both contain the same
> variables. This lets you set up a LHS and RHS matrix/vector that are
> symmetrical.
> 
> But, for a problem a student posed in class, I'm stuck with how to do it
> in R, if its even possible (its trivial in Maxima, Maple...even using
> Solver in Excel, but I haven't been remotely successful in getting
> anything to work in R).
> 
> Suppose you have a production system that at 4 sequential time steps
> generate 640, 825, 580, and 925 units. At each time step, you need to
> decide how many of those units need to be 'quality control' (QC) checked
> in some fashion, subject to some constraints.
> 
>    --> at no point in time can the number of units in the system be >1000
>    --> at the end of the production cycle, there can be no units left
>    --> 'QC checking' costs money, varying as a function of the time step
> -- 35, 55, 50 and 65 for each unit, for each time step in turn.
> 
> Objective is to minimize total cost. The total cost objective function
> is trivial. Let p1 = number sent out time step 1, p2 number sent out at
> time step 3, and so on. So, total cost function we want to minimize is
> simply
> 
>     cost=(35*p1)+(55*p2)+(50*p3)+(65*p4)
> 
> where p1+p2+p3+p4=(640+825+580+925)=2970 (i.e., all the products get
> checked). The question is, what number do you send out at each time step
> to minimize cost?
> 
> Where I get hung up in R is the fact that if I let t(i) be the number of
> products at each time step, then
> 
>       t1=640,
>       t2=t1-p1+825
>       t3=t2-p2+580
>       t4=t3-p3+925
> 
> such that t1+t2+t3+t4=2970 (as it must), with additional constraints being
> 
>     p1<=t1, p2<=t2, p3<=t3, p4<=t4, {t1..t4}<=1000, and t4-p4=0.
> 
> There may be algebraic ways to reduce the number of functions needed to
> describe the constraints, but I can't for the life of me see how I can
> create a coefficient matrix (typically, the LHS) since each line of said
> matrix, which corresponds to the constraints, needs to be a function of
> the unknowns in the objective function -- being, p1, p2, p3 and p4.
> 
> In Maple (for example), this is trivial:
> 
>        cost:=35*p10+55*p12+50*p14+65*p16;
> cnsts:={t10=640,t12=t10-p10+825,t14=t12-p12+580,t16=t14-p14+925,t16-p16=0,p10<=t10,p12<=t12,p14<=t14,p16<=t16,t10<=1000,t12<=1000,t14<=1000,t16<=1000};
>        Minimize(cost,cnsts,assume={nonnegative});
> 
> which yields (correctly):
> 
> p1=640, p2=405, p3=1000, p4=925
> 
> for minimized cost of 154800.
> 
> Took only a minute to also set this up in Maxima, and using Solver in
> Excel. But danged if I can suss out any way to do this in R.
> 
> Pointers to the obvious welcome.
> 	[[alternative HTML version deleted]]
> 
> 

-- 
apl. Prof. PD Dr. Martin Becker, Akad. Oberrat
Lehrstab Statistik
Quantitative Methoden
Fakultät für Empirische Humanwissenschaften und Wirtschaftswissenschaft
Universität des Saarlandes
Campus C3 1, Raum 2.17
66123 Saarbrücken
Deutschland