Divide Items based on their available stock to other ID - r

Question

Hi Everyone i am facing a unique problem . I want to find out Transfer of Inventory based on condition if Quantity required for a particular Item id is more than Stock on hand. we should transfer the inventory from other ID.
For Example. item I60 is available for 7 IDs. For E1, E6 available stock is less than quantity so what i want to do is Transfer the excess inventory from E3 (i.e 6-2 =4) to E1 and E6. So transfer for E1 will be 1 and E6 will be 2 and remaining SOH from E3 will be 3. I hope everyone can understand it.
structure(list(ID = structure(c(1L, 6L, 7L, 3L, 5L, 2L, 4L, 8L,
1L, 7L, 3L, 5L, 2L, 9L, 8L, 1L, 7L, 3L, 5L, 2L, 9L, 8L, 1L, 7L,
3L, 5L, 2L, 9L, 8L, 1L, 7L, 3L, 5L, 2L, 9L, 8L, 1L, 7L, 3L, 5L,
2L, 9L, 8L, 1L, 7L, 3L, 5L, 2L, 9L), .Label = c("E1", "E2", "E3",
"E4", "E5", "E6", "E7", "E8", "E9"), class = "factor"), Item.Code = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L,
3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L,
5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 7L
), .Label = c("I60", "I67", "I68", "I69", "I70", "I71", "I72"
), class = "factor"), Quantity = c(2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 4L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), Stock_on_hand = c(1L,
0L, 2L, 6L, 2L, 2L, 2L, 0L, 6L, 3L, -1L, 1L, 2L, 9L, 1L, 5L,
-1L, 9L, 3L, 38L, 5L, 10L, 2L, 3L, 2L, 2L, 1L, 8L, 0L, 2L, 2L,
4L, 2L, 1L, 5L, 1L, -1L, 4L, 3L, 1L, 2L, 11L, 1L, 2L, 0L, 3L,
1L, 4L, 1L), Transfer = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 4L, 0L,
0L, 0L, 0L, 0L, 0L, 3L, 0L, 7L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 6L, 0L, 0L, 0L, 0L, 0L, 0L, 3L, 0L, 0L, 2L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 1L, 0L, 2L, 0L, 0L)), class = "data.frame", row.names = c(NA,
-49L))
highlighted columns should be generated with R code

Answer 1

Looks like an integer programming problem with Updated_SOH as the variable and objective function minimizing absolute difference between Quantity and SOH subject to constraint that total sum of SOH stays constant.
Here is a heuristic approach to solve this optimization problem:
1) Calculate the difference which is to be used to sort the dataset.
2) In a similar approach as here but shifting positive values and different aggregation, we use these excess SOH to net off SOH deficit in previous rows.
3) The final output is the sum of i) existing Quantity, ii) any unfulfilled Quantity and iii) excess SOH.
setDT(df)
df[, Diff := Stock_on_hand - Quantity]
setorder(df, Item.Code, Diff)
df[, Updated_SOH := {
posVal <- replace(Diff, Diff<0, 0)
negVal <- replace(Diff, Diff>0, 0)
n <- 1L
while (any(negVal < 0) && n < .N) {
negVal <- replace(negVal, negVal>0, 0) +
shift(posVal, 1L, type="lead", fill=0) +
c(posVal[1L], rep(0, .N-1L)) #for case where there are more Quantity than SOH
posVal <- replace(negVal, negVal<0, 0)
n <- n + 1L
}
excess <- negVal[negVal > 0]
Quantity + #existing Quantity
replace(negVal, negVal>0, 0) + #unfulfilled Quantity
c(rep(0, .N - length(excess)), excess) #shifting back down excess SOH
}, by=.(Item.Code)]
output:
ID Item.Code Quantity Stock_on_hand Transfer Diff Updated_SOH
1: E6 I60 2 0 0 -2 2
2: E1 I60 2 1 0 -1 2
3: E7 I60 2 2 0 0 2
4: E5 I60 2 2 0 0 2
5: E2 I60 2 2 0 0 2
6: E4 I60 2 2 0 0 2
7: E3 I60 2 6 0 4 3
8: E3 I67 2 -1 0 -3 2
9: E8 I67 2 0 4 -2 2
10: E5 I67 2 1 0 -1 2
11: E2 I67 2 2 0 0 2
12: E7 I67 2 3 0 1 2
13: E1 I67 2 6 0 4 2
14: E9 I67 2 9 0 7 8
15: E7 I68 2 -1 7 -3 2
16: E8 I68 2 1 3 -1 2
17: E5 I68 2 3 0 1 2
18: E1 I68 2 5 0 3 2
19: E9 I68 2 5 0 3 5
20: E3 I68 2 9 0 7 9
21: E2 I68 4 38 0 34 38
22: E2 I69 2 1 6 -1 2
23: E1 I69 2 2 0 0 2
24: E3 I69 2 2 0 0 2
25: E5 I69 2 2 0 0 2
26: E7 I69 2 3 0 1 2
27: E9 I69 2 8 0 6 8
28: E8 I69 2 10 0 8 10
29: E8 I70 2 0 0 -2 2
30: E2 I70 2 1 3 -1 2
31: E1 I70 2 2 0 0 2
32: E7 I70 2 2 0 0 2
33: E5 I70 2 2 0 0 2
34: E3 I70 2 4 0 2 2
35: E9 I70 2 5 0 3 4
36: E1 I71 2 -1 2 -3 2
37: E8 I71 2 1 0 -1 2
38: E5 I71 2 1 0 -1 2
39: E2 I71 2 2 0 0 2
40: E3 I71 2 3 0 1 2
41: E7 I71 2 4 0 2 2
42: E9 I71 2 11 0 9 9
43: E7 I72 2 0 1 -2 0
44: E8 I72 2 1 0 -1 2
45: E5 I72 2 1 2 -1 2
46: E9 I72 2 1 0 -1 2
47: E1 I72 2 2 0 0 2
48: E3 I72 2 3 0 1 2
49: E2 I72 2 4 0 2 2
ID Item.Code Quantity Stock_on_hand Transfer Diff Updated_SOH
data:
library(data.table)
df <- structure(list(ID = structure(c(1L, 6L, 7L, 3L, 5L, 2L, 4L, 8L,
1L, 7L, 3L, 5L, 2L, 9L, 8L, 1L, 7L, 3L, 5L, 2L, 9L, 8L, 1L, 7L,
3L, 5L, 2L, 9L, 8L, 1L, 7L, 3L, 5L, 2L, 9L, 8L, 1L, 7L, 3L, 5L,
2L, 9L, 8L, 1L, 7L, 3L, 5L, 2L, 9L), .Label = c("E1", "E2", "E3",
"E4", "E5", "E6", "E7", "E8", "E9"), class = "factor"), Item.Code = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L,
3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L,
5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 7L
), .Label = c("I60", "I67", "I68", "I69", "I70", "I71", "I72"
), class = "factor"), Quantity = c(2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 4L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), Stock_on_hand = c(1L,
0L, 2L, 6L, 2L, 2L, 2L, 0L, 6L, 3L, -1L, 1L, 2L, 9L, 1L, 5L,
-1L, 9L, 3L, 38L, 5L, 10L, 2L, 3L, 2L, 2L, 1L, 8L, 0L, 2L, 2L,
4L, 2L, 1L, 5L, 1L, -1L, 4L, 3L, 1L, 2L, 11L, 1L, 2L, 0L, 3L,
1L, 4L, 1L), Transfer = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 4L, 0L,
0L, 0L, 0L, 0L, 0L, 3L, 0L, 7L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 6L, 0L, 0L, 0L, 0L, 0L, 0L, 3L, 0L, 0L, 2L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 1L, 0L, 2L, 0L, 0L)), class = "data.frame", row.names = c(NA,
-49L))