how to do assignment of numbers in r (one machine n jobs) - r

Question

I am working on assignment problem in R. I have following dataframe in r
cycle_time TAT ready_for_next ITV_no
2 10 12 0
4 12 16 0
6 13 19 0
8 11 19 0
10 15 25 0
12 17 29 0
14 13 27 0
16 13 29 0
18 12 30 0
20 16 36 0
22 13 35 0
24 12 36 0
26 15 41 0
28 14 42 0
30 17 47 0
My desired dataframe would be
cycle_time TAT ready_for_next ITV_no wait_time
2 10 12 1 0
4 12 16 2 0
6 13 19 3 0
8 11 19 4 0
10 15 25 5 0
12 17 29 1 0
14 13 27 6 0
16 13 29 2 0
18 12 30 3 1
20 16 36 4 1
22 13 35 5 3
24 12 36 6 3
26 15 41 2 3
28 14 42 3 2
30 17 47 5 5
cycle_time = crane cycle time
TAT(in mins) = turn around time of truck
ready_for_next(in mins) = ready to take next container
ITV_no = ITV no to be assigned for that job
***There are only 6 unique trucks available***
Idea here is to assign trucks such that waiting time is minimum.
In first five observations all 5 trucks are assigned.
For the next container i.e row number 6 (on 12th min) ITV_no 1 is coming back from its job so that will get assigned to this job.
7th observation(i.e 14th min) there are no trucks available,so we will have to assign new truck (i.e ITV_no 6)
8th observation(16 min) ITV_no 2 is coming back from its job,so that will get assigned to this job and so on.
If there are no trucks available then it has to wait till the nearest truck comes back from job.
How can I implement this in R?
I have build some logic
cycle_time <- c(2,4,6,8,10,12,14,16,18,20,22,24,26,28,30)
ITV_no <- c(1,2,3,4,5,6,7)
temp <- c()
TAT <- c(10,12,13,11,15,17,13,13,12,16,13,12,15,14,17)
ready_for_next <- cycle_time + TAT
assignment <- data.frame(cycle_time,TAT,ready_for_next)
assignment$ITV_no <- 0
for(i in 1:nrow(assignment)) {
for(j in 1:length(ITV_no)){
assignment$ITV_no[i] <- ifelse(assignment$cycle_time <= assignment$ready_for_next,ITV_no[j],
ifelse())
## I am not able to update the count of trucks which are already assigned
# and which are free to be assigned
}
}
Logic
1. first row increment ITV_no by 1. directly assign truck to that job
2. check if cycle_time <= previous all ready_for_next(i.e 12), if yes then increment ITV_no by 1,if no then assign previous ITV_no for that job(i.e 1)
e.g
for row 6, cycle time will get compared to all previous ready_for_next column values (25,19,19,16,12) it finds the match at first row then that ITV_no(i.e 2) is assigned to 6th row
for row 7, cycle time will get compared to all previous ready_for_next column values (25,19,19,16) **12 should be removed from comparison because the truck is already assigned to the job** match at first row then that ITV_no(i.e 2) is assigned to 6th row. No match,so new truck is assigned to that job

Answer 1

I have come up with some solution...
It is working with sample data
rm(list=ls())
df <- data.frame(qc_time = seq(2,40,2),itv_tat=c(10,15,12,18,25,19,18,16,14,10,12,15,17,19,13,12,8,15,9,14))
itv_number_vec <- vector()
itv_number_vec <- 0
itvno_time <- list()
for (i in 1:nrow(df))
{
#### Initialisation ####
if (i==1)
{
df$itv_available_time[i] <- sum(df$qc_time[i] + df$itv_tat[i])
itvno_time[[i]] <- df$itv_available_time[i]
df$delay[i] <- 0
df$itv_number[i] <- 1
itv_number_vec <- 1
}
if(i!=1)
{
if (df$qc_time[i] >= min(unlist(itvno_time)))
{
for (j in 1:length(itvno_time))
{
if (itvno_time[[j]] <= df$qc_time[i])
{
df$itv_number[i] <- j
df$itv_available_time[i] <- sum(df$qc_time[i] + df$itv_tat[i])
itvno_time[[j]] <- df$itv_available_time[i]
break
}
}
}else{
if (max(itv_number_vec)<7)
{
df$itv_number[i] <- max(itv_number_vec) + 1
itv_number_vec <- c(itv_number_vec,(max(itv_number_vec) + 1))
df$delay[i] <- 0
df$itv_available_time[i] <- sum(df$qc_time[i] + df$itv_tat[i])
itvno_time[[max(itv_number_vec)]] <- df$itv_available_time[i]
}else{
df$delay[i] <- (min(unlist(itvno_time)) - df$qc_time[i])
df$itv_number[i] <- which.min(itvno_time)
df$itv_available_time[i] <- sum(df$qc_time[i], df$itv_tat[i] ,df$delay[i])
itvno_time[[which.min(itvno_time)]] <- df$itv_available_time[i]
}
}
}
}