Tabu search in R

Question

Good evening,

As part of a data analysis course we have been thrown into the Metaheuristics realm.....and I am really struggling to understand how to implement a Tabu search in R since my background in programming is rather limited.

I haven't found any R or Python example on Google or youtube either so I'm really praying I'll find something here.

The problem I have is similar to the "location problem" in optimisation. I need to find the best combination of Hubs that minimizes the total distance between Hubs and nodes.

I need to find 5 hubs, and the total capacity for each one is 120

nodes <- structure(list(node_number = 1:50, 
               x = c(2L, 80L, 36L, 57L, 33L, 76L, 77L, 94L, 
                     89L, 59L, 39L, 87L, 44L, 2L, 19L, 5L, 
                     58L, 14L, 43L, 87L, 11L, 31L, 51L, 55L, 
                     84L, 12L, 53L, 53L, 33L, 69L, 43L, 10L, 
                     8L, 3L, 96L, 6L, 59L, 66L, 22L, 75L, 4L, 
                     41L, 92L, 12L, 60L, 35L, 38L, 9L, 54L, 1L), 
               y = c(62L, 25L, 88L, 23L, 17L, 43L, 85L, 6L, 11L, 
                     72L, 82L, 24L, 76L, 83L, 43L, 27L, 72L, 50L, 
                     18L, 7L, 56L, 16L, 94L, 13L, 57L, 2L, 33L, 10L, 
                     32L, 67L, 5L, 75L, 26L, 1L, 22L, 48L, 22L, 69L,
                     50L, 21L, 81L, 97L, 34L, 64L, 84L, 100L, 2L, 9L, 59L, 58L), 
               node_demand = c(3L, 14L, 1L, 14L, 19L, 2L, 14L, 6L, 
                               7L, 6L, 10L, 18L, 3L, 6L, 20L, 4L, 
                               14L, 11L, 19L,  15L, 15L, 4L, 13L, 
                               13L, 5L, 16L, 3L, 7L, 14L, 17L, 
                               3L, 3L, 12L, 14L, 20L, 13L, 10L, 
                               9L, 6L, 18L, 7L, 20L, 9L, 1L, 8L, 
                               5L, 1L, 7L, 9L, 2L)), 
          .Names = c("node_number", "x", "y", "node_demand"), 
          class = "data.frame", row.names = c(NA, -50L))


hubs_required = 5
total_capacity = 120

My strategy was to create a distance matrix, then I will create another 50 x 50 matrix to represent wether a node becomes a hub or not, and finally I will multiply both and add all the distances to get the total distance.

I created the dataframe:

nodes_df <- as.data.frame(nodes)
colnames(nodes_df) <- c("x", "y", "node_demand")
rownames(nodes_df) <- paste('Node',1:50)

I created the distance matrix

distance_df <-as.data.frame(as.matrix(round(dist(nodes_df,method = "euclidean",diag = TRUE,upper = TRUE))))
colnames(distance_df) <- paste("Node",1:50)

I created the node demand matrix:

demand <- as.vector(rep(c(nodes_df[,'node_demand']),50)) 
demand_matrix <- matrix(demand,nrow=50,ncol=50,byrow = TRUE)
diag(demand_matrix) <- 0
demand_matrix <- as.data.frame(demand_matrix)

I created an empty matrix to show whether a node becomes a hub "1" or not "0"

hubs_matrix <- matrix(0,nrow = 50,ncol = 50,byrow = TRUE)
colnames(hubs_matrix) <- paste("Hub",1:50)
rownames(hubs_matrix) <- paste("Node",1:50)

Then to create the initial solution I randomly assign Hubs and calculate the distance and demand.

set.seed(37)
hubs_matrix <- do.call("cbind", lapply(1:50, function(x) sample(c(1, rep(0, 49)), 50)))
sum_distances <- (hubs_matrix * distance_df)
sum(rowSums(sum_distances))

The idea is to try different combinations of '1'' and '0' as to minimise the total distance but I am having the following issues:

• I got no idea how to do the local search and do the permutations from the initial solution.

• I got no idea how to prevent R to use the best solution for a certain period of time, i.e the Tabu list

• I got no idea how to deal with the supply restriction for each node ( total demand from each node < 120), I could do it with a loop but since in this case I'm multiplying matrices I'm pretty lost.

Anybody could give me a hand???

Many thanks!