Hi Sven, May be this helps: If you look into ?simplex.object " solved This indicates whether the problem was solved.? A value of ????????? ?-1? indicates that no feasible solution could be found.? A ????????? value of ?0? that the maximum number of iterations was ????????? reached without termination of the second stage.? This may ????????? indicate an unbounded function or simply that more iterations ????????? are needed. A value of ?1? indicates that an optimal solution ????????? has been found. " ?s1 <- simplex(a=price * -1, A1=A, b1=b, A2=A, b2=b2, maxi=T) # This cant find a solution s1$solved #[1] -1 A.K. Dear All, I do have a problem with a linear optimisation I am trying to achieve: library(boot) price <- c(26.93, 26.23, 25.64, 25.97, 26.12, 26.18, 26.49, 27, 27.32, 27.93, 27.72, 27.23) A <- matrix(0, nrow=length(price)+1, ncol=length(price)) A[1,] <- 1 for(i in 1:length(price)) A[i+1,i] <- 1 b <- c(864000.01, 288000, 288000, 288000, 288000, 288000, 288000, 288000, 288000, 288000, 288000, 288000, 288000) b2 <- c(864000, 216000, 216000, 216000, 216000, 216000, 216000, 216000, 216000, 216000, 216000, 216000, 216000) simplex(a=price, A1=A, b1=b) # This works simplex(a=price, A1=A, b1=b, maxi=T) # This is the maxing function and works as well simplex(a=price * -1, A1=A, b1=b, A2=A, b2=b2, maxi=T) # This cant find a solution The result I would expect is that it picks the 4 highest(note I multiplied "price" with -1) numbers and multiplies them with constraint in b2. For example this works: library(boot) price <- c(1, 2, 3 , 4) A <- matrix(0, nrow=length(price)+1, ncol=length(price)) A[1,] <- 1 for(i in 1:length(price)) A[i+1,i] <- 1 b <- c(8.01, 2.5, 2.5, 2.5, 2.5) b2 <- c(8, 2, 2, 2, 2) simplex(a=price, A1=A, b1=b) simplex(a=price, A1=A, b1=b, maxi=T) simplex(a=price * -1, A1=A, b1=b, A2=A, b2=b2, maxi=T) I cant see a logical difference between the two, why would it not find a solution for the first problem? Thank you. Sven