# Vogel's approximation method

(Redirected from VAM)
Vogel's approximation method
You are encouraged to solve this task according to the task description, using any language you may know.

Vogel's Approximation Method (VAM) is a technique for finding a good initial feasible solution to an allocation problem.

The powers that be have identified 5 tasks that need to be solved urgently. Being imaginative chaps, they have called them “A”, “B”, “C”, “D”, and “E”. They estimate that:

• A will require 30 hours of work,
• B will require 20 hours of work,
• C will require 70 hours of work,
• D will require 30 hours of work, and
• E will require 60 hours of work.

They have identified 4 contractors willing to do the work, called “W”, “X”, “Y”, and “Z”.

• W has 50 hours available to commit to working,
• X has 60 hours available,
• Y has 50 hours available, and
• Z has 50 hours available.

The cost per hour for each contractor for each task is summarized by the following table:

```   A  B  C  D  E
W 16 16 13 22 17
X 14 14 13 19 15
Y 19 19 20 23 50
Z 50 12 50 15 11
```

The task is to use VAM to allocate contractors to tasks. It scales to large problems, so ideally keep sorts out of the iterative cycle. It works as follows:

Step 1: Balance the given transportation problem if either (total supply>total demand) or (total supply<total demand)
Step 2: Determine the penalty cost for each row and column by subtracting the lowest cell cost in the row or column from the next lowest cell cost in the same row or column.
Step 3: Select the row or column with the highest penalty cost (breaking ties arbitrarily or choosing the lowest-cost cell).
Step 4: Allocate as much as possible to the feasible cell with the lowest transportation cost in the row or column with the highest penalty cost.
Step 5: Repeat steps 2, 3 and 4 until all requirements have been meet.
Step 6: Compute total transportation cost for the feasible allocations.

For this task assume that the model is balanced.

For each task and contractor (row and column above) calculating the difference between the smallest two values produces:

```        A       B       C       D       E       W       X       Y       Z
1       2       2       0       4       4       3       1       0       1   E-Z(50)
```

Determine the largest difference (D or E above). In the case of ties I shall choose the one with the lowest price (in this case E because the lowest price for D is Z=15, whereas for E it is Z=11). For your choice determine the minimum cost (chosen E above so Z=11 is chosen now). Allocate as much as possible from Z to E (50 in this case limited by Z's supply). Adjust the supply and demand accordingly. If demand or supply becomes 0 for a given task or contractor it plays no further part. In this case Z is out of it. If you choose arbitrarily, and chose D see here for the working.

Repeat until all supply and demand is met:

```2       2       2       0       3       2       3       1       0       -   C-W(50)
3       5       5       7       4      35       -       1       0       -   E-X(10)
4       5       5       7       4       -       -       1       0       -   C-X(20)
5       5       5       -       4       -       -       0       0       -   A-X(30)
6       -      19       -      23       -       -       -       4       -   D-Y(30)
-       -       -       -       -       -       -       -       -   B-Y(20)
```

Finally calculate the cost of your solution. In the example given it is £3100:

```   A  B  C  D  E
W       50
X 30    20    10
Y    20    30
Z             50
```

The optimal solution determined by GLPK is £3100:

```   A  B  C  D  E
W       50
X 10 20 20    10
Y 20       30
Z             50
```
Cf.

## 11l

Translation of: Python
```V costs = [‘W’ = [‘A’ = 16, ‘B’ = 16, ‘C’ = 13, ‘D’ = 22, ‘E’ = 17],
‘X’ = [‘A’ = 14, ‘B’ = 14, ‘C’ = 13, ‘D’ = 19, ‘E’ = 15],
‘Y’ = [‘A’ = 19, ‘B’ = 19, ‘C’ = 20, ‘D’ = 23, ‘E’ = 50],
‘Z’ = [‘A’ = 50, ‘B’ = 12, ‘C’ = 50, ‘D’ = 15, ‘E’ = 11]]
V demand = [‘A’ = 30, ‘B’ = 20, ‘C’ = 70, ‘D’ = 30, ‘E’ = 60]
V cols = sorted(demand.keys())
V supply = [‘W’ = 50, ‘X’ = 60, ‘Y’ = 50, ‘Z’ = 50]
V res = Dict(costs.keys().map(k -> (k, DefaultDict[Char, Int]())))
[Char = [Char]] g
L(x) supply.keys()
g[x] = sorted(costs[x].keys(), key' g -> :costs[@x][g])
L(x) demand.keys()
g[x] = sorted(costs.keys(), key' g -> :costs[g][@x])

L !g.empty
[Char = Int] d
L(x) demand.keys()
d[x] = I g[x].len > 1 {(costs[g[x][1]][x] - costs[g[x][0]][x])} E costs[g[x][0]][x]
[Char = Int] s
L(x) supply.keys()
s[x] = I g[x].len > 1 {(costs[x][g[x][1]] - costs[x][g[x][0]])} E costs[x][g[x][0]]
V f = max(d.keys(), key' n -> @d[n])
V t = max(s.keys(), key' n -> @s[n])
(t, f) = I d[f] > s[t] {(f, g[f][0])} E (g[t][0], t)
V v = min(supply[f], demand[t])
res[f][t] += v
demand[t] -= v
I demand[t] == 0
L(k, n) supply
I n != 0
g[k].remove(t)
g.pop(t)
demand.pop(t)
supply[f] -= v
I supply[f] == 0
L(k, n) demand
I n != 0
g[k].remove(f)
g.pop(f)
supply.pop(f)

L(n) cols
print("\t "n, end' ‘ ’)
print()
V cost = 0
L(g) sorted(costs.keys())
print(g" \t", end' ‘ ’)
L(n) cols
V y = res[g][n]
I y != 0
print(y, end' ‘ ’)
cost += y * costs[g][n]
print("\t", end' ‘ ’)
print()
print("\n\nTotal Cost =  "cost)```
Output:
```         A       B       C       D       E
W                        50
X                        20              40
Y        30      20
Z                                30      20

Total Cost =  3130
```

## C

Translation of: Kotlin
```#include <stdio.h>
#include <limits.h>

#define TRUE 1
#define FALSE 0
#define N_ROWS 4
#define N_COLS 5

typedef int bool;

int supply[N_ROWS] = { 50, 60, 50, 50 };
int demand[N_COLS] = { 30, 20, 70, 30, 60 };

int costs[N_ROWS][N_COLS] = {
{ 16, 16, 13, 22, 17 },
{ 14, 14, 13, 19, 15 },
{ 19, 19, 20, 23, 50 },
{ 50, 12, 50, 15, 11 }
};

bool row_done[N_ROWS] = { FALSE };
bool col_done[N_COLS] = { FALSE };

void diff(int j, int len, bool is_row, int res[3]) {
int i, c, min1 = INT_MAX, min2 = min1, min_p = -1;
for (i = 0; i < len; ++i) {
if((is_row) ? col_done[i] : row_done[i]) continue;
c = (is_row) ? costs[j][i] : costs[i][j];
if (c < min1) {
min2 = min1;
min1 = c;
min_p = i;
}
else if (c < min2) min2 = c;
}
res[0] = min2 - min1; res[1] = min1; res[2] = min_p;
}

void max_penalty(int len1, int len2, bool is_row, int res[4]) {
int i, pc = -1, pm = -1, mc = -1, md = INT_MIN;
int res2[3];

for (i = 0; i < len1; ++i) {
if((is_row) ? row_done[i] : col_done[i]) continue;
diff(i, len2, is_row, res2);
if (res2[0] > md) {
md = res2[0];  /* max diff */
pm = i;        /* pos of max diff */
mc = res2[1];  /* min cost */
pc = res2[2];  /* pos of min cost */
}
}

if (is_row) {
res[0] = pm; res[1] = pc;
}
else {
res[0] = pc; res[1] = pm;
}
res[2] = mc; res[3] = md;
}

void next_cell(int res[4]) {
int i, res1[4], res2[4];
max_penalty(N_ROWS, N_COLS, TRUE, res1);
max_penalty(N_COLS, N_ROWS, FALSE, res2);

if (res1[3] == res2[3]) {
if (res1[2] < res2[2])
for (i = 0; i < 4; ++i) res[i] = res1[i];
else
for (i = 0; i < 4; ++i) res[i] = res2[i];
return;
}
if (res1[3] > res2[3])
for (i = 0; i < 4; ++i) res[i] = res2[i];
else
for (i = 0; i < 4; ++i) res[i] = res1[i];
}

int main() {
int i, j, r, c, q, supply_left = 0, total_cost = 0, cell[4];
int results[N_ROWS][N_COLS] = { 0 };

for (i = 0; i < N_ROWS; ++i) supply_left += supply[i];
while (supply_left > 0) {
next_cell(cell);
r = cell[0];
c = cell[1];
q = (demand[c] <= supply[r]) ? demand[c] : supply[r];
demand[c] -= q;
if (!demand[c]) col_done[c] = TRUE;
supply[r] -= q;
if (!supply[r]) row_done[r] = TRUE;
results[r][c] = q;
supply_left -= q;
total_cost += q * costs[r][c];
}

printf("    A   B   C   D   E\n");
for (i = 0; i < N_ROWS; ++i) {
printf("%c", 'W' + i);
for (j = 0; j < N_COLS; ++j) printf("  %2d", results[i][j]);
printf("\n");
}
printf("\nTotal cost = %d\n", total_cost);
return 0;
}
```
Output:
```    A   B   C   D   E
W   0   0  50   0   0
X  30   0  20   0  10
Y   0  20   0  30   0
Z   0   0   0   0  50

Total cost = 3100
```

If the program is changed to this (to accomodate the second Ruby example):

```#include <stdio.h>
#include <limits.h>

#define TRUE 1
#define FALSE 0
#define N_ROWS 5
#define N_COLS 5

typedef int bool;

int supply[N_ROWS] = { 461, 277, 356, 488,  393 };
int demand[N_COLS] = { 278,  60, 461, 116, 1060 };

int costs[N_ROWS][N_COLS] = {
{ 46,  74,  9, 28, 99 },
{ 12,  75,  6, 36, 48 },
{ 35, 199,  4,  5, 71 },
{ 61,  81, 44, 88,  9 },
{ 85,  60, 14, 25, 79 }
};

// etc

int main() {
// etc

printf("     A    B    C    D    E\n");
for (i = 0; i < N_ROWS; ++i) {
printf("%c", 'V' + i);
for (j = 0; j < N_COLS; ++j) printf("  %3d", results[i][j]);
printf("\n");
}
printf("\nTotal cost = %d\n", total_cost);
return 0;
}
```

then the output, which agrees with the Phix output but not with the Ruby output itself is:

```     A    B    C    D    E
V    0    0  461    0    0
W  277    0    0    0    0
X    1    0    0    0  355
Y    0    0    0    0  488
Z    0   60    0  116  217

Total cost = 60748
```

## C++

Translation of: Java
```#include <iostream>
#include <numeric>
#include <vector>

template <typename T>
std::ostream &operator<<(std::ostream &os, const std::vector<T> &v) {
auto it = v.cbegin();
auto end = v.cend();

os << '[';
if (it != end) {
os << *it;
it = std::next(it);
}
while (it != end) {
os << ", " << *it;
it = std::next(it);
}

return os << ']';
}

std::vector<int> demand = { 30, 20, 70, 30, 60 };
std::vector<int> supply = { 50, 60, 50, 50 };
std::vector<std::vector<int>> costs = {
{16, 16, 13, 22, 17},
{14, 14, 13, 19, 15},
{19, 19, 20, 23, 50},
{50, 12, 50, 15, 11}
};

int nRows = supply.size();
int nCols = demand.size();

std::vector<bool> rowDone(nRows, false);
std::vector<bool> colDone(nCols, false);
std::vector<std::vector<int>> result(nRows, std::vector<int>(nCols, 0));

std::vector<int> diff(int j, int len, bool isRow) {
int min1 = INT_MAX;
int min2 = INT_MAX;
int minP = -1;
for (int i = 0; i < len; i++) {
if (isRow ? colDone[i] : rowDone[i]) {
continue;
}
int c = isRow
? costs[j][i]
: costs[i][j];
if (c < min1) {
min2 = min1;
min1 = c;
minP = i;
} else if (c < min2) {
min2 = c;
}
}
return { min2 - min1, min1, minP };
}

std::vector<int> maxPenalty(int len1, int len2, bool isRow) {
int md = INT_MIN;
int pc = -1;
int pm = -1;
int mc = -1;
for (int i = 0; i < len1; i++) {
if (isRow ? rowDone[i] : colDone[i]) {
continue;
}
std::vector<int> res = diff(i, len2, isRow);
if (res[0] > md) {
md = res[0];    // max diff
pm = i;         // pos of max diff
mc = res[1];    // min cost
pc = res[2];    // pos of min cost
}
}
return isRow
? std::vector<int> { pm, pc, mc, md }
: std::vector<int>{ pc, pm, mc, md };
}

std::vector<int> nextCell() {
auto res1 = maxPenalty(nRows, nCols, true);
auto res2 = maxPenalty(nCols, nRows, false);

if (res1[3] == res2[3]) {
return res1[2] < res2[2]
? res1
: res2;
}
return res1[3] > res2[3]
? res2
: res1;
}

int main() {
int supplyLeft = std::accumulate(supply.cbegin(), supply.cend(), 0, [](int a, int b) { return a + b; });
int totalCost = 0;

while (supplyLeft > 0) {
auto cell = nextCell();
int r = cell[0];
int c = cell[1];

int quantity = std::min(demand[c], supply[r]);

demand[c] -= quantity;
if (demand[c] == 0) {
colDone[c] = true;
}

supply[r] -= quantity;
if (supply[r] == 0) {
rowDone[r] = true;
}

result[r][c] = quantity;
supplyLeft -= quantity;

totalCost += quantity * costs[r][c];
}

for (auto &a : result) {
std::cout << a << '\n';
}

std::cout << "Total cost: " << totalCost;

return 0;
}
```
Output:
```[0, 0, 50, 0, 0]
[30, 0, 20, 0, 10]
[0, 20, 0, 30, 0]
[0, 0, 0, 0, 50]
Total cost: 3100```

## D

Strongly typed version (but K is not divided in Task and Contractor types to keep code simpler).

Translation of: Python
```void main() {
import std.stdio, std.string, std.algorithm, std.range;

enum K { A, B, C, D, E,  X, Y, Z, W }
immutable int[K][K] costs = cast() //**
[K.W: [K.A: 16, K.B: 16, K.C: 13, K.D: 22, K.E: 17],
K.X: [K.A: 14, K.B: 14, K.C: 13, K.D: 19, K.E: 15],
K.Y: [K.A: 19, K.B: 19, K.C: 20, K.D: 23, K.E: 50],
K.Z: [K.A: 50, K.B: 12, K.C: 50, K.D: 15, K.E: 11]];
int[K] demand, supply;
with (K)
demand = [A: 30, B: 20, C: 70, D: 30, E: 60],
supply = [W: 50, X: 60, Y: 50, Z: 50];

auto cols = demand.keys.sort().release;
auto res = costs.byKey.zip((int[K]).init.repeat).assocArray;
K[][K] g;
foreach (immutable x; supply.byKey)
g[x] = costs[x].keys.schwartzSort!(k => cast()costs[x][k]) //**
.release;
foreach (immutable x; demand.byKey)
g[x] = costs.keys.schwartzSort!(k=> cast()costs[k][x]).release;

while (g.length) {
int[K] d, s;
foreach (immutable x; demand.byKey)
d[x] = g[x].length > 1 ?
costs[g[x][1]][x] - costs[g[x][0]][x] :
costs[g[x][0]][x];
foreach (immutable x; supply.byKey)
s[x] = g[x].length > 1 ?
costs[x][g[x][1]] - costs[x][g[x][0]] :
costs[x][g[x][0]];
auto f = d.keys.minPos!((a,b) => d[a] > d[b])[0];
auto t = s.keys.minPos!((a,b) => s[a] > s[b])[0];
if (d[f] > s[t]) {
t = f;
f = g[f][0];
} else {
f = t;
t = g[t][0];
}
immutable v = min(supply[f], demand[t]);
res[f][t] += v;
demand[t] -= v;
if (demand[t] == 0) {
foreach (immutable k, immutable n; supply)
if (n != 0)
g[k] = g[k].remove!(c => c == t);
g.remove(t);
demand.remove(t);
}
supply[f] -= v;
if (supply[f] == 0) {
foreach (immutable k, immutable n; demand)
if (n != 0)
g[k] = g[k].remove!(c => c == f);
g.remove(f);
supply.remove(f);
}
}

writefln("%-(\t%s%)", cols);
auto cost = 0;
foreach (immutable c; costs.keys.sort().release) {
write(c, '\t');
foreach (immutable n; cols) {
if (n in res[c]) {
immutable y = res[c][n];
if (y != 0) {
y.write;
cost += y * costs[c][n];
}
}
'\t'.write;
}
writeln;
}
writeln("\nTotal Cost = ", cost);
}
```
Output:
```	A	B	C	D	E
X	30		20		10
Y		20		30
Z					50
W			50

Total Cost = 3100
```

## Go

Translation of: Kotlin
```package main

import (
"fmt"
"math"
)

var supply = []int{50, 60, 50, 50}
var demand = []int{30, 20, 70, 30, 60}

var costs = make([][]int, 4)

var nRows = len(supply)
var nCols = len(demand)

var rowDone = make([]bool, nRows)
var colDone = make([]bool, nCols)
var results = make([][]int, nRows)

func init() {
costs[0] = []int{16, 16, 13, 22, 17}
costs[1] = []int{14, 14, 13, 19, 15}
costs[2] = []int{19, 19, 20, 23, 50}
costs[3] = []int{50, 12, 50, 15, 11}

for i := 0; i < len(results); i++ {
results[i] = make([]int, nCols)
}
}

func nextCell() []int {
res1 := maxPenalty(nRows, nCols, true)
res2 := maxPenalty(nCols, nRows, false)
switch {
case res1[3] == res2[3]:
if res1[2] < res2[2] {
return res1
} else {
return res2
}
case res1[3] > res2[3]:
return res2
default:
return res1
}
}

func diff(j, l int, isRow bool) []int {
min1 := math.MaxInt32
min2 := min1
minP := -1
for i := 0; i < l; i++ {
var done bool
if isRow {
done = colDone[i]
} else {
done = rowDone[i]
}
if done {
continue
}
var c int
if isRow {
c = costs[j][i]
} else {
c = costs[i][j]
}
if c < min1 {
min2, min1, minP = min1, c, i
} else if c < min2 {
min2 = c
}
}
return []int{min2 - min1, min1, minP}
}

func maxPenalty(len1, len2 int, isRow bool) []int {
md := math.MinInt32
pc, pm, mc := -1, -1, -1
for i := 0; i < len1; i++ {
var done bool
if isRow {
done = rowDone[i]
} else {
done = colDone[i]
}
if done {
continue
}
res := diff(i, len2, isRow)
if res[0] > md {
md = res[0]  // max diff
pm = i       // pos of max diff
mc = res[1]  // min cost
pc = res[2]  // pos of min cost
}
}
if isRow {
return []int{pm, pc, mc, md}
}
return []int{pc, pm, mc, md}
}

func main() {
supplyLeft := 0
for i := 0; i < len(supply); i++ {
supplyLeft += supply[i]
}
totalCost := 0
for supplyLeft > 0 {
cell := nextCell()
r, c := cell[0], cell[1]
q := demand[c]
if q > supply[r] {
q = supply[r]
}
demand[c] -= q
if demand[c] == 0 {
colDone[c] = true
}
supply[r] -= q
if supply[r] == 0 {
rowDone[r] = true
}
results[r][c] = q
supplyLeft -= q
totalCost += q * costs[r][c]
}

fmt.Println("    A   B   C   D   E")
for i, result := range results {
fmt.Printf("%c", 'W' + i)
for _, item := range result {
fmt.Printf("  %2d", item)
}
fmt.Println()
}
fmt.Println("\nTotal cost =", totalCost)
}
```
Output:
```    A   B   C   D   E
W   0   0  50   0   0
X  30   0  20   0  10
Y   0  20   0  30   0
Z   0   0   0   0  50

Total cost = 3100
```

If the program is changed as follows to accomodate the second Ruby example:

```package main

import (
"fmt"
"math"
)

var supply = []int{461, 277, 356, 488, 393}
var demand = []int{278, 60, 461, 116, 1060}

var costs = make([][]int, nRows)

var nRows = len(supply)
var nCols = len(demand)

var rowDone = make([]bool, nRows)
var colDone = make([]bool, nCols)
var results = make([][]int, nRows)

func init() {
costs[0] = []int{46, 74, 9, 28, 99}
costs[1] = []int{12, 75, 6, 36, 48}
costs[2] = []int{35, 199, 4, 5, 71}
costs[3] = []int{61, 81, 44, 88, 9}
costs[4] = []int{85, 60, 14, 25, 79}

for i := 0; i < len(results); i++ {
results[i] = make([]int, nCols)
}
}

// etc

func main() {
// etc

fmt.Println("     A    B    C    D    E")
for i, result := range results {
fmt.Printf("%c", 'V'+i)
for _, item := range result {
fmt.Printf("  %3d", item)
}
fmt.Println()
}
fmt.Println("\nTotal cost =", totalCost)
}
```

then the output, which agrees with the C and Phix output but not with the Ruby output itself, is:

```     A    B    C    D    E
V    0    0  461    0    0
W  277    0    0    0    0
X    1    0    0    0  355
Y    0    0    0    0  488
Z    0   60    0  116  217

Total cost = 60748
```

## J

Implementation:

```vam=:1 :0
:
exceeding=. 0 <. -&(+/)
D=. x,y exceeding x NB. x: demands
S=. y,x exceeding y NB. y: sources
C=. (m,.0),0        NB. m: costs
B=. 1+>./,C         NB. bigger than biggest cost
mincost=. <./@-.&0  NB. smallest non-zero cost
penalty=. |@(B * 2 -/@{. /:~ -. 0:)"1 - mincost"1
R=. C*0
while. 0 < +/D,S do.
pS=. penalty C
pD=. penalty |:C
if. pS >&(>./) pD do.
row=. (i. >./) pS
col=. (i. mincost) row { C
else.
col=. (i. >./) pD
row=. (i. mincost) col {"1 C
end.
n=. (row{S) <. col{D
S=. (n-~row{S) row} S
D=. (n-~col{D) col} D
C=. C * S *&*/ D
R=. n (<row,col)} R
end.
_1 _1 }. R
)
```

Note that for our penalty we are using the difference between the two smallest relevant costs multiplied by 1 larger than the highest represented cost and we subtract from that multiple the smallest relevant cost. This gives us the tiebreaker mechanism currently specified for this task.

```demand=: 30 20 70 30 60
src=: 50 60 50 50
cost=: 16 16 13 22 17,14 14 13 19 15,19 19 20 23 50,:50 12 50 15 11

demand cost vam src
0  0 50  0  0
30  0 20  0 10
0 20  0 30  0
0  0  0  0 50
```

## Java

Works with: Java version 8
```import java.util.Arrays;
import static java.util.Arrays.stream;
import java.util.concurrent.*;

public class VogelsApproximationMethod {

final static int[] demand = {30, 20, 70, 30, 60};
final static int[] supply = {50, 60, 50, 50};
final static int[][] costs = {{16, 16, 13, 22, 17}, {14, 14, 13, 19, 15},
{19, 19, 20, 23, 50}, {50, 12, 50, 15, 11}};

final static int nRows = supply.length;
final static int nCols = demand.length;

static boolean[] rowDone = new boolean[nRows];
static boolean[] colDone = new boolean[nCols];
static int[][] result = new int[nRows][nCols];

public static void main(String[] args) throws Exception {
int supplyLeft = stream(supply).sum();
int totalCost = 0;

while (supplyLeft > 0) {
int[] cell = nextCell();
int r = cell[0];
int c = cell[1];

int quantity = Math.min(demand[c], supply[r]);
demand[c] -= quantity;
if (demand[c] == 0)
colDone[c] = true;

supply[r] -= quantity;
if (supply[r] == 0)
rowDone[r] = true;

result[r][c] = quantity;
supplyLeft -= quantity;

totalCost += quantity * costs[r][c];
}

stream(result).forEach(a -> System.out.println(Arrays.toString(a)));
System.out.println("Total cost: " + totalCost);

es.shutdown();
}

static int[] nextCell() throws Exception {
Future<int[]> f1 = es.submit(() -> maxPenalty(nRows, nCols, true));
Future<int[]> f2 = es.submit(() -> maxPenalty(nCols, nRows, false));

int[] res1 = f1.get();
int[] res2 = f2.get();

if (res1[3] == res2[3])
return res1[2] < res2[2] ? res1 : res2;

return (res1[3] > res2[3]) ? res2 : res1;
}

static int[] diff(int j, int len, boolean isRow) {
int min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;
int minP = -1;
for (int i = 0; i < len; i++) {
if (isRow ? colDone[i] : rowDone[i])
continue;
int c = isRow ? costs[j][i] : costs[i][j];
if (c < min1) {
min2 = min1;
min1 = c;
minP = i;
} else if (c < min2)
min2 = c;
}
return new int[]{min2 - min1, min1, minP};
}

static int[] maxPenalty(int len1, int len2, boolean isRow) {
int md = Integer.MIN_VALUE;
int pc = -1, pm = -1, mc = -1;
for (int i = 0; i < len1; i++) {
if (isRow ? rowDone[i] : colDone[i])
continue;
int[] res = diff(i, len2, isRow);
if (res[0] > md) {
md = res[0];  // max diff
pm = i;       // pos of max diff
mc = res[1];  // min cost
pc = res[2];  // pos of min cost
}
}
return isRow ? new int[]{pm, pc, mc, md} : new int[]{pc, pm, mc, md};
}
}
```
```[0, 0, 50, 0, 0]
[30, 0, 20, 0, 10]
[0, 20, 0, 30, 0]
[0, 0, 0, 0, 50]
Total cost: 3100```

## Julia

This solution is designed to scale well to large numbers of suppliers and customers. The opportunity cost matrix is sorted only once, and penalties are recalculated only when the relevant resources are exhausted. The solution is stored in a sparse matrix, because the number of components to a solution is less than s+c (suppliers + customers) but the size of the matrix is s*c.

This solution does not impose the requirement that the problem be balanced. `vogel` will iterate until either supply or demand is exhausted and provide a low-cost result even when the problem is unbalanced, whether this result is a good solution is left for the user to decide. The function `isbalanced` can be used to test whether a given problem is balanced.

Types

The immutable type `TProblem` stores the problem's parameters. It includes permutation matrices that allow the rows and columns of the total opportunity cost matrix to be sorted as needed.

`Resource` stores the currently available quantity of a given supply or demand as well as its penalty, cost, and some meta-data. `isavailable` indicates whether any of the given resource remains. `isless` is designed to make the currently most usable resource appear as a maximum compared to other resources.

```immutable TProblem{T<:Integer,U<:String}
sd::Array{Array{T,1},1}
toc::Array{T,2}
labels::Array{Array{U,1},1}
tsort::Array{Array{T,2}, 1}
end

function TProblem{T<:Integer,U<:String}(s::Array{T,1},
d::Array{T,1},
toc::Array{T,2},
slab::Array{U,1},
dlab::Array{U,1})
scnt = length(s)
dcnt = length(d)
size(toc) = (scnt,dcnt) || error("Supply, Demand, TOC Size Mismatch")
length(slab) == scnt || error("Supply Label Size Labels")
length(dlab) == dcnt || error("Demand Label Size Labels")
0 <= minimum(s) || error("Negative Supply Value")
0 <= minimum(d) || error("Negative Demand Value")
sd = Array{T,1}[]
push!(sd, s)
push!(sd, d)
labels = Array{U,1}[]
push!(labels, slab)
push!(labels, dlab)
tsort = Array{T,2}[]
push!(tsort, mapslices(sortperm, toc, 2))
push!(tsort, mapslices(sortperm, toc, 1))
TProblem(sd, toc, labels, tsort)
end
isbalanced(tp::TProblem) = sum(tp.sd[1]) == sum(tp.sd[2])

type Resource{T<:Integer}
dim::T
i::T
quant::T
l::T
m::T
p::T
q::T
end
function Resource{T<:Integer}(dim::T, i::T, quant::T)
zed = zero(T)
Resource(dim, i, quant, zed, zed, zed, zed)
end

isavailable(r::Resource) = 0 < r.quant
Base.isless(a::Resource, b::Resource) = a.p < b.p || (a.p == b.p && b.q < a.q)
```

Functions

`penalize!` updates the penalty, cost and some meta-data of lists of supplies and demands. It short-circuits to avoid recalculating these values when the relevant resources remain available. Sorting is provided by the permutation matrices in `TProblem`.

`vogel` implements Vogel's approximation method on a `TProblem`. It is somewhat straightforward, given the types and `penalize!`.

```function penalize!{T<:Integer,U<:String}(sd::Array{Array{Resource{T},1},1},
tp::TProblem{T,U})
avail = BitArray{1}[]
for dim in 2:-1:1
push!(avail, bitpack(map(isavailable, sd[dim])))
end
for dim in 1:2, r in sd[dim]
if r.quant == 0
r.l = r.m = r.p = r.q = 0
continue
end
r.l == 0 || !avail[dim][r.l] || !avail[dim][r.m] || continue
rsort = filter(x->avail[dim][x], vec(slicedim(tp.tsort[dim],dim,r.i)))
rcost = vec(slicedim(tp.toc, dim, r.i))[rsort]
if length(rsort) == 1
r.l = r.m = rsort[1]
r.p = r.q = rcost[1]
else
r.l, r.m = rsort[1:2]
r.p = rcost[2] - rcost[1]
r.q = rcost[1]
end
end
nothing
end

function vogel{T<:Integer,U<:String}(tp::TProblem{T,U})
sdcnt = collect(size(tp.toc))
sol = spzeros(T, sdcnt[1], sdcnt[2])
sd = Array{Resource{T},1}[]
for dim in 1:2
push!(sd, [Resource(dim, i, tp.sd[dim][i]) for i in 1:sdcnt[dim]])
end
while any(map(isavailable, sd[1])) && any(map(isavailable, sd[2]))
penalize!(sd, tp)
a = maximum([sd[1], sd[2]])
b = sd[rem1(a.dim+1,2)][a.l]
if a.dim == 2 # swap to make a supply and b demand
a, b = b, a
end
expend = min(a.quant, b.quant)
sol[a.i, b.i] = expend
a.quant -= expend
b.quant -= expend
end
return sol
end
```

Main

```using Printf

sup = [50, 60, 50, 50]
slab = ["W", "X", "Y", "Z"]
dem = [30, 20, 70, 30, 60]
dlab = ["A", "B", "C", "D", "E"]
c = [16 16 13 22 17;
14 14 13 19 15;
19 19 20 23 50;
50 12 50 15 11]

tp = TProblem(sup, dem, c, slab, dlab)
sol = vogel(tp)
cost = sum(tp.toc .* sol)

println("The solution is:")
print("        ")
for s in tp.labels[2]
print(@sprintf "%4s" s)
end
println()
for i in 1:size(tp.toc)[1]
print(@sprintf "    %4s" tp.labels[1][i])
for j in 1:size(tp.toc)[2]
print(@sprintf "%4d" sol[i,j])
end
println()
end
println("The total cost is:  ", cost)
```
Output:
```The solution is:
A   B   C   D   E
W   0   0  50   0   0
X  10  20  20   0  10
Y  20   0   0  30   0
Z   0   0   0   0  50
The total cost is:  3100
```

## Kotlin

Translation of: Java
```// version 1.1.3

val supply = intArrayOf(50, 60, 50, 50)
val demand = intArrayOf(30, 20, 70, 30, 60)

val costs = arrayOf(
intArrayOf(16, 16, 13, 22, 17),
intArrayOf(14, 14, 13, 19, 15),
intArrayOf(19, 19, 20, 23, 50),
intArrayOf(50, 12, 50, 15, 11)
)

val nRows = supply.size
val nCols = demand.size

val rowDone = BooleanArray(nRows)
val colDone = BooleanArray(nCols)
val results = Array(nRows) { IntArray(nCols) }

fun nextCell(): IntArray {
val res1 = maxPenalty(nRows, nCols, true)
val res2 = maxPenalty(nCols, nRows, false)
if (res1[3] == res2[3])
return if (res1[2] < res2[2]) res1 else res2
return if (res1[3] > res2[3]) res2 else res1
}

fun diff(j: Int, len: Int, isRow: Boolean): IntArray {
var min1 = Int.MAX_VALUE
var min2 = min1
var minP = -1
for (i in 0 until len) {
val done = if (isRow) colDone[i] else rowDone[i]
if (done) continue
val c = if (isRow) costs[j][i] else costs[i][j]
if (c < min1) {
min2 = min1
min1 = c
minP = i
}
else if (c < min2) min2 = c
}
return intArrayOf(min2 - min1, min1, minP)
}

fun maxPenalty(len1: Int, len2: Int, isRow: Boolean): IntArray {
var md = Int.MIN_VALUE
var pc = -1
var pm = -1
var mc = -1
for (i in 0 until len1) {
val done = if (isRow) rowDone[i] else colDone[i]
if (done) continue
val res = diff(i, len2, isRow)
if (res[0] > md) {
md = res[0]  // max diff
pm = i       // pos of max diff
mc = res[1]  // min cost
pc = res[2]  // pos of min cost
}
}
return if (isRow) intArrayOf(pm, pc, mc, md) else
intArrayOf(pc, pm, mc, md)
}

fun main(args: Array<String>) {
var supplyLeft = supply.sum()
var totalCost = 0
while (supplyLeft > 0) {
val cell = nextCell()
val r = cell[0]
val c = cell[1]
val q = minOf(demand[c], supply[r])
demand[c] -= q
if (demand[c] == 0) colDone[c] = true
supply[r] -= q
if (supply[r] == 0) rowDone[r] = true
results[r][c] = q
supplyLeft -= q
totalCost += q * costs[r][c]
}

println("    A   B   C   D   E")
for ((i, result) in results.withIndex()) {
print(('W'.toInt() + i).toChar())
for (item in result) print("  %2d".format(item))
println()
}
println("\nTotal Cost = \$totalCost")
}
```
Output:
```    A   B   C   D   E
W   0   0  50   0   0
X  30   0  20   0  10
Y   0  20   0  30   0
Z   0   0   0   0  50

Total Cost = 3100
```

## Lua

Translation of: Kotlin
```function initArray(n,v)
local tbl = {}
for i=1,n do
table.insert(tbl,v)
end
return tbl
end

function initArray2(m,n,v)
local tbl = {}
for i=1,m do
table.insert(tbl,initArray(n,v))
end
return tbl
end

supply = {50, 60, 50, 50}
demand = {30, 20, 70, 30, 60}
costs = {
{16, 16, 13, 22, 17},
{14, 14, 13, 19, 15},
{19, 19, 20, 23, 50},
{50, 12, 50, 15, 11}
}

nRows = table.getn(supply)
nCols = table.getn(demand)

rowDone = initArray(nRows, false)
colDone = initArray(nCols, false)
results = initArray2(nRows, nCols, 0)

function diff(j,le,isRow)
local min1 = 100000000
local min2 = min1
local minP = -1
for i=1,le do
local done = false
if isRow then
done = colDone[i]
else
done = rowDone[i]
end
if not done then
local c = 0
if isRow then
c = costs[j][i]
else
c = costs[i][j]
end
if c < min1 then
min2 = min1
min1 = c
minP = i
elseif c < min2 then
min2 = c
end
end
end
return {min2 - min1, min1, minP}
end

function maxPenalty(len1,len2,isRow)
local md = -100000000
local pc = -1
local pm = -1
local mc = -1

for i=1,len1 do
local done = false
if isRow then
done = rowDone[i]
else
done = colDone[i]
end
if not done then
local res = diff(i, len2, isRow)
if res[1] > md then
md = res[1] -- max diff
pm = i      -- pos of max diff
mc = res[2] -- min cost
pc = res[3] -- pos of min cost
end
end
end

if isRow then
return {pm, pc, mc, md}
else
return {pc, pm, mc, md}
end
end

function nextCell()
local res1 = maxPenalty(nRows, nCols, true)
local res2 = maxPenalty(nCols, nRows, false)
if res1[4] == res2[4] then
if res1[3] < res2[3] then
return res1
else
return res2
end
else
if res1[4] > res2[4] then
return res2
else
return res1
end
end
end

function main()
local supplyLeft = 0
for i,v in pairs(supply) do
supplyLeft = supplyLeft + v
end
local totalCost = 0
while supplyLeft > 0 do
local cell = nextCell()
local r = cell[1]
local c = cell[2]
local q = math.min(demand[c], supply[r])
demand[c] = demand[c] - q
if demand[c] == 0 then
colDone[c] = true
end
supply[r] = supply[r] - q
if supply[r] == 0 then
rowDone[r] = true
end
results[r][c] = q
supplyLeft = supplyLeft - q
totalCost = totalCost + q * costs[r][c]
end

print("    A   B   C   D   E")
local labels = {'W','X','Y','Z'}
for i,r in pairs(results) do
io.write(labels[i])
for j,c in pairs(r) do
io.write(string.format("  %2d", c))
end
print()
end
print("Total Cost = " .. totalCost)
end

main()
```
Output:
```    A   B   C   D   E
W   0   0  50   0   0
X  30   0  20   0  10
Y   0  20   0  30   0
Z   0   0   0   0  50
Total Cost = 3100```

## Nim

Translation of: Kotlin
```import math, sequtils, strutils

var
supply = [50, 60, 50, 50]
demand = [30, 20, 70, 30, 60]

let
costs = [[16, 16, 13, 22, 17],
[14, 14, 13, 19, 15],
[19, 19, 20, 23, 50],
[50, 12, 50, 15, 11]]

nRows = supply.len
nCols = demand.len

var
rowDone = newSeq[bool](nRows)
colDone = newSeq[bool](nCols)
results = newSeqWith(nRows, newSeq[int](nCols))

proc diff(j, len: int; isRow: bool): array[3, int] =
var min1, min2 = int.high
var minP = -1
for i in 0..<len:
let done = if isRow: colDone[i] else: rowDone[i]
if done: continue
let c = if isRow: costs[j][i] else: costs[i][j]
if c < min1:
min2 = min1
min1 = c
minP = i
elif c < min2:
min2 = c
result = [min2 - min1, min1, minP]

proc maxPenalty(len1, len2: int; isRow: bool): array[4, int] =
var md = int.low
var pc, pm, mc = -1
for i in 0..<len1:
let done = if isRow: rowDone[i] else: colDone[i]
if done: continue
let res = diff(i, len2, isRow)
if res[0] > md:
md = res[0]  # max diff
pm = i       # pos of max diff
mc = res[1]  # min cost
pc = res[2]  # pos of min cost
result = if isRow: [pm, pc, mc, md] else: [pc, pm, mc, md]

proc nextCell(): array[4, int] =
let res1 = maxPenalty(nRows, nCols, true)
let res2 = maxPenalty(nCols, nRows, false)
if res1[3] == res2[3]:
return if res1[2] < res2[2]: res1 else: res2
result = if res1[3] > res2[3]: res2 else: res1

when isMainModule:

var supplyLeft = sum(supply)
var totalCost = 0

while supplyLeft > 0:
let cell = nextCell()
let r = cell[0]
let c = cell[1]
let q = min(demand[c], supply[r])
dec demand[c], q
if demand[c] == 0: colDone[c] = true
dec supply[r], q
if supply[r] == 0: rowDone[r] = true
results[r][c] = q
dec supplyLeft, q
inc totalCost, q * costs[r][c]

echo "    A   B   C   D   E"
for i, result in results:
stdout.write chr(i + ord('W'))
for item in result:
stdout.write "  ", (\$item).align(2)
echo()
echo "\nTotal cost = ", totalCost
```
Output:
```    A   B   C   D   E
W   0   0  50   0   0
X  30   0  20   0  10
Y   0  20   0  30   0
Z   0   0   0   0  50

Total cost = 3100```

## Perl

```#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Vogel%27s_approximation_method
use warnings;
use List::AllUtils qw( max_by nsort_by min );

my \$data = <<END;
A=30 B=20 C=70 D=30 E=60
W=50 X=60 Y=50 Z=50
AW=16 BW=16 CW=13 DW=22 EW=17
AX=14 BX=14 CX=13 DX=19 EX=15
AY=19 BY=19 CY=20 DY=23 EY=50
AZ=50 BZ=12 CZ=50 DZ=15 EZ=11
END
my \$table = sprintf +('%4s' x 6 . "\n") x 5,
map {my \$t = \$_; map "\$_\$t", '', 'A' .. 'E' } '' , 'W' .. 'Z';

my (\$cost, %assign) = (0);
while( \$data =~ /\b\w=\d/ )
{
my @penalty;
for ( \$data =~ /\b(\w)=\d/g )
{
my @all = map /(\d+)/, nsort_by { /\d+/ && \$& }
grep { my (\$t, \$c) = /(.)(.)=/; \$data =~ /\b\$c=\d/ and \$data =~ /\b\$t=\d/ }
\$data =~ /\$_\w=\d+|\w\$_=\d+/g;
push @penalty, [ \$_, (\$all[1] // 0) - \$all[0] ];
}
my \$rc = (max_by { \$_->[1] } nsort_by
{ my \$x = \$_->[0]; \$data =~ /(?:\$x\w|\w\$x)=(\d+)/ && \$1 } @penalty)->[0];
my @lowest = nsort_by { /\d+/ && \$& }
grep { my (\$t, \$c) = /(.)(.)=/; \$data =~ /\b\$c=\d/ and \$data =~ /\b\$t=\d/ }
\$data =~ /\$rc\w=\d+|\w\$rc=\d+/g;
my (\$t, \$c) = \$lowest[0] =~ /(.)(.)/;
my \$allocate = min \$data =~ /\b[\$t\$c]=(\d+)/g;
\$table =~ s/\$t\$c/ sprintf "%2d", \$allocate/e;
\$cost += \$data =~ /\$t\$c=(\d+)/ && \$1 * \$allocate;
\$data =~ s/\b\$_=\K\d+/ \$& - \$allocate || '' /e for \$t, \$c;
}
print "cost \$cost\n\n", \$table =~ s/[A-Z]{2}/--/gr;
```
Output:
```cost 3100

A   B   C   D   E
W  --  --  50  --  --
X  30  --  20  --  10
Y  --  20  --  30  --
Z  --  --  --  --  50
```

## Phix

See Transportation_problem#Phix for optimal results.

Translation of: YaBasic
Translation of: Go
```with javascript_semantics
sequence supply = {50,60,50,50},
demand = {30,20,70,30,60},
costs = {{16,16,13,22,17},
{14,14,13,19,15},
{19,19,20,23,50},
{50,12,50,15,11}}

sequence row_done = repeat(false,length(supply)),
col_done = repeat(false,length(demand))

function diff(integer j, leng, bool is_row)
integer min1 = #3FFFFFFF, min2 = min1, min_p = -1
for i=1 to leng do
if not iff(is_row?col_done:row_done)[i] then
integer c = iff(is_row?costs[j,i]:costs[i,j])
if c<min1 then
min2 = min1
min1 = c
min_p = i
elsif c<min2 then
min2 = c
end if
end if
end for
return {min2-min1,min1,min_p,j}
end function

function max_penalty(integer len1, len2, bool is_row)
integer pc = -1, pm = -1, mc = -1, md = -#3FFFFFFF
for i=1 to len1 do
if not iff(is_row?row_done:col_done)[i] then
sequence res2 = diff(i, len2, is_row)
if res2[1]>md then
{md,mc,pc,pm} = res2
end if
end if
end for
return {md,mc}&iff(is_row?{pm,pc}:{pc,pm})
end function

integer supply_left = sum(supply),
total_cost = 0

sequence results = repeat(repeat(0,length(demand)),length(supply))

while supply_left>0 do
sequence cell = min(max_penalty(length(supply), length(demand), true),
max_penalty(length(demand), length(supply), false))
integer {{},{},r,c} = cell,
q = min(demand[c], supply[r])
demand[c] -= q
col_done[c] = (demand[c]==0)
supply[r] -= q
row_done[r] = (supply[r]==0)
results[r, c] = q
supply_left -= q
total_cost += q * costs[r, c]
end while

printf(1,"     A   B   C   D   E\n")
for i=1 to length(supply) do
printf(1,"%c ",'Z'-length(supply)+i)
for j=1 to length(demand) do
printf(1,"%4d",results[i,j])
end for
printf(1,"\n")
end for
printf(1,"\nTotal cost = %d\n", total_cost)
```
Output:
```     A   B   C   D   E
W    0   0  50   0   0
X   30   0  20   0  10
Y    0  20   0  30   0
Z    0   0   0   0  50

Total cost = 3100
```

Using the sample from Ruby:

```sequence supply = {461, 277, 356, 488,   393},
demand = {278, 60, 461, 116, 1060},
costs  = {{46, 74,  9, 28, 99},
{12, 75,  6, 36, 48},
{35, 199, 4,  5, 71},
{61, 81, 44, 88,  9},
{85, 60, 14, 25, 79}}
```
Output:

Note this agrees with C and Go but not Ruby

```     A   B   C   D   E
V    0   0 461   0   0
W  277   0   0   0   0
X    1   0   0   0 355
Y    0   0   0   0 488
Z    0  60   0 116 217

Total cost = 60748
```

## Python

Translation of: Ruby
```from collections import defaultdict

costs  = {'W': {'A': 16, 'B': 16, 'C': 13, 'D': 22, 'E': 17},
'X': {'A': 14, 'B': 14, 'C': 13, 'D': 19, 'E': 15},
'Y': {'A': 19, 'B': 19, 'C': 20, 'D': 23, 'E': 50},
'Z': {'A': 50, 'B': 12, 'C': 50, 'D': 15, 'E': 11}}
demand = {'A': 30, 'B': 20, 'C': 70, 'D': 30, 'E': 60}
cols = sorted(demand.iterkeys())
supply = {'W': 50, 'X': 60, 'Y': 50, 'Z': 50}
res = dict((k, defaultdict(int)) for k in costs)
g = {}
for x in supply:
g[x] = sorted(costs[x].iterkeys(), key=lambda g: costs[x][g])
for x in demand:
g[x] = sorted(costs.iterkeys(), key=lambda g: costs[g][x])

while g:
d = {}
for x in demand:
d[x] = (costs[g[x][1]][x] - costs[g[x][0]][x]) if len(g[x]) > 1 else costs[g[x][0]][x]
s = {}
for x in supply:
s[x] = (costs[x][g[x][1]] - costs[x][g[x][0]]) if len(g[x]) > 1 else costs[x][g[x][0]]
f = max(d, key=lambda n: d[n])
t = max(s, key=lambda n: s[n])
t, f = (f, g[f][0]) if d[f] > s[t] else (g[t][0], t)
v = min(supply[f], demand[t])
res[f][t] += v
demand[t] -= v
if demand[t] == 0:
for k, n in supply.iteritems():
if n != 0:
g[k].remove(t)
del g[t]
del demand[t]
supply[f] -= v
if supply[f] == 0:
for k, n in demand.iteritems():
if n != 0:
g[k].remove(f)
del g[f]
del supply[f]

for n in cols:
print "\t", n,
print
cost = 0
for g in sorted(costs):
print g, "\t",
for n in cols:
y = res[g][n]
if y != 0:
print y,
cost += y * costs[g][n]
print "\t",
print
print "\n\nTotal Cost = ", cost
```
Output:
```    A   B   C   D   E
W           50
X   30      20      10
Y       20      30
Z                   50

Total Cost =  3100```

## Racket

Losley:

Translation of: Ruby

Strangely, due to the sub-deterministic nature of the hash tables, resources were allocated differently to the #Ruby version; but somehow at the same total cost!

```#lang racket
(define-values (1st 2nd 3rd) (values first second third))

(define-syntax-rule (?: x t f) (if (zero? x) f t))

(define (hash-ref2
hsh# key-1 key-2
#:fail-2 (fail-2 (λ () (error 'hash-ref2 "key-2:~a is not found in hash" key-2)))
#:fail-1 (fail-1 (λ () (error 'hash-ref2 "key-1:~a is not found in hash" key-1))))
(hash-ref (hash-ref hsh# key-1 fail-1) key-2 fail-2))

(define (VAM costs all-supply all-demand)
(define (reduce-g/x g/x x#-- x x-v y y-v)
(for/fold ((rv (?: x-v g/x (hash-remove g/x x))))
(#:when (zero? y-v) ((k n) (in-hash x#--)) #:unless (zero? n))
(hash-update rv k (curry remove y))))

(define (cheapest-candidate/tie-break candidates)
(define cand-max3 (3rd (argmax 3rd candidates)))
(argmin 2nd (for/list ((cand candidates) #:when (= (3rd cand) cand-max3)) cand)))

(let vam-loop
((res (hash))
(supply all-supply)
(g/supply
(for/hash ((x (in-hash-keys all-supply)))
(define costs#x (hash-ref costs x))
(define key-fn (λ (g) (hash-ref costs#x g)))
(values x (sort (hash-keys costs#x) < #:key key-fn #:cache-keys? #t))))
(demand all-demand)
(g/demand
(for/hash ((x (in-hash-keys all-demand)))
(define key-fn (λ (g) (hash-ref2 costs g x)))
(values x (sort (hash-keys costs) < #:key key-fn #:cache-keys? #t)))))
(cond
[(and (hash-empty? supply) (hash-empty? demand)) res]
[(or (hash-empty? supply) (hash-empty? demand)) (error 'VAM "Unbalanced supply / demand")]
[else
(define D
(let ((candidates
(for/list ((x (in-hash-keys demand)))
(match-define (hash-table ((== x) (and g#x (list g#x.0 _ ...))) _ ...) g/demand)
(define z (hash-ref2 costs g#x.0 x))
(match g#x
[(list _ g#x.1 _ ...) (list x z (- (hash-ref2 costs g#x.1 x) z))]
[(list _) (list x z z)]))))
(cheapest-candidate/tie-break candidates)))

(define S
(let ((candidates
(for/list ((x (in-hash-keys supply)))
(match-define (hash-table ((== x) (and g#x (list g#x.0 _ ...))) _ ...) g/supply)
(define z (hash-ref2 costs x g#x.0))
(match g#x
[(list _ g#x.1 _ ...) (list x z (- (hash-ref2 costs x g#x.1) z))]
[(list _) (list x z z)]))))
(cheapest-candidate/tie-break candidates)))

(define-values (d s)
(let ((t>f? (if (= (3rd D) (3rd S)) (> (2nd S) (2nd D)) (> (3rd D) (3rd S)))))
(if t>f? (values (1st D) (1st (hash-ref g/demand (1st D))))
(values (1st (hash-ref g/supply (1st S))) (1st S)))))

(define v (min (hash-ref supply s) (hash-ref demand d)))

(define d-v (- (hash-ref demand d) v))
(define s-v (- (hash-ref supply s) v))

(define demand-- (?: d-v (hash-set demand d d-v) (hash-remove demand d)))
(define supply-- (?: s-v (hash-set supply s s-v) (hash-remove supply s)))

(vam-loop
(hash-update res s (λ (h) (hash-update h d (λ (x) (+ v x)) 0)) hash)
supply-- (reduce-g/x g/supply supply-- s s-v d d-v)
demand-- (reduce-g/x g/demand demand-- d d-v s s-v))])))

(define (vam-solution-cost costs demand?cols solution)
(match demand?cols
[(? list? demand-cols)
(for*/sum ((g (in-hash-keys costs)) (n (in-list demand-cols)))
(* (hash-ref2 solution g n #:fail-2 0) (hash-ref2 costs g n)))]
[(hash-table (ks _) ...) (vam-solution-cost costs (sort ks symbol<? solution))]))

(define (describe-VAM-solution costs demand sltn)
(define demand-cols (sort (hash-keys demand) symbol<?))
(string-join
(map
(curryr string-join "\t")
`(,(map ~a (cons "" demand-cols))
,@(for/list ((g (in-hash-keys costs)))
(cons (~a g) (for/list ((c demand-cols)) (~a (hash-ref2 sltn g c #:fail-2 "-")))))
()
("Total Cost:" ,(~a (vam-solution-cost costs demand-cols sltn)))))
"\n"))

;; --------------------------------------------------------------------------------------------------
(let ((COSTS (hash 'W (hash 'A 16 'B 16 'C 13 'D 22 'E 17)
'X (hash 'A 14 'B 14 'C 13 'D 19 'E 15)
'Y (hash 'A 19 'B 19 'C 20 'D 23 'E 50)
'Z (hash 'A 50 'B 12 'C 50 'D 15 'E 11)))
(DEMAND (hash 'A 30 'B 20 'C 70 'D 30 'E 60))
(SUPPLY (hash 'W 50 'X 60 'Y 50 'Z 50)))
(displayln (describe-VAM-solution COSTS DEMAND (VAM COSTS SUPPLY DEMAND))))
```
Output:
```	A	B	C	D	E
W	-	-	50	-	-
X	10	20	20	-	10
Y	20	-	-	30	-
Z	-	-	-	-	50

Total Cost:	3100```

## Raku

(formerly Perl 6)

Works with: Rakudo version 2019.03.1
Translation of: Sidef
```my  %costs =
:W{:16A, :16B, :13C, :22D, :17E},
:X{:14A, :14B, :13C, :19D, :15E},
:Y{:19A, :19B, :20C, :23D, :50E},
:Z{:50A, :12B, :50C, :15D, :11E};

my %demand = :30A, :20B, :70C, :30D, :60E;
my %supply = :50W, :60X, :50Y, :50Z;

my @cols = %demand.keys.sort;

my %res;
my %g = (|%supply.keys.map: -> \$x { \$x => [%costs{\$x}.sort(*.value)».key]}),
(|%demand.keys.map: -> \$x { \$x => [%costs.keys.sort({%costs{\$_}{\$x}})]});

while (+%g) {
my @d = %demand.keys.map: -> \$x
{[\$x, my \$z = %costs{%g{\$x}[0]}{\$x},%g{\$x}[1] ?? %costs{%g{\$x}[1]}{\$x} - \$z !! \$z]}

my @s = %supply.keys.map: -> \$x
{[\$x, my \$z = %costs{\$x}{%g{\$x}[0]},%g{\$x}[1] ?? %costs{\$x}{%g{\$x}[1]} - \$z !! \$z]}

@d = |@d.grep({ (.[2] == max @d».[2]) }).&min: :by(*.[1]);
@s = |@s.grep({ (.[2] == max @s».[2]) }).&min: :by(*.[1]);

my (\$t, \$f) = @d[2] == @s[2] ?? (@s[1],@d[1]) !! (@d[2],@s[2]);
my (\$d, \$s) = \$t > \$f ?? (@d[0],%g{@d[0]}[0]) !! (%g{@s[0]}[0], @s[0]);

my \$v = %supply{\$s} min %demand{\$d};

%res{\$s}{\$d} += \$v;
%demand{\$d} -= \$v;

if (%demand{\$d} == 0) {
%supply.grep( *.value != 0 )».key.map: -> \$v
{ %g{\$v}.splice((%g{\$v}.first: * eq \$d, :k),1) };
%g{\$d}:delete;
%demand{\$d}:delete;
}

%supply{\$s} -= \$v;

if (%supply{\$s} == 0) {
%demand.grep( *.value != 0 )».key.map: -> \$v
{ %g{\$v}.splice((%g{\$v}.first: * eq \$s, :k),1) };
%g{\$s}:delete;
%supply{\$s}:delete;
}
}

say join "\t", flat '', @cols;
my \$total;
for %costs.keys.sort -> \$g {
print "\$g\t";
for @cols -> \$col {
print %res{\$g}{\$col} // '-', "\t";
\$total += (%res{\$g}{\$col} // 0) * %costs{\$g}{\$col};
}
print "\n";
}
say "\nTotal cost: \$total";
```
Output:
```	A	B	C	D	E
W	-	-	50	-	-
X	30	-	20	-	10
Y	-	20	-	30	-
Z	-	-	-	-	50

Total cost: 3100```

## REXX

Translation of: java

### Vogel's Approximation

```/* REXX ***************************************************************
* Solve the Transportation Problem using Vogel's Approximation
Default Input
2 3        # of sources / # of demands
25 35      sources
20 30 10   demands
3 5 7      cost matrix    <
3 2 5
* 20201210 support no input file -courtesy GS
*          Note: correctness of input is not checked
* 20210102 restored Vogel's Approximation and added Optimization
* 20210103 eliminated debug code
**********************************************************************/
Signal On Halt
Signal On Novalue
Signal On Syntax

Parse Arg fid
If fid='' Then
fid='input1.txt'
Call init
m.=0
Do Forever
dmax.=0
dmax=0
Do r=1 To rr
dr.r=''
Do c=1 To cc
If cost.r.c<>'*' Then
dr.r=dr.r cost.r.c
End
dr.r=words(dr.r) dr.r
dr.r=diff(dr.r)
If dr.r>dmax Then Do; dmax=dr.r; dmax.0='R'; dmax.1=r; dmax.2=dr.r; End
End
Do c=1 To cc
dc.c=''
Do r=1 To rr
If cost.r.c<>'*' Then
dc.c=dc.c cost.r.c
End
dc.c=words(dc.c) dc.c
dc.c=diff(dc.c)
If dc.c>dmax Then Do; dmax=dc.c; dmax.0='C'; dmax.1=c; dmax.2=dc.c; End
End
cmin=999
Select
When dmax.0='R' Then Do
r=dmax.1
Do c=1 To cc
If cost.r.c<>'*' &,
cost.r.c<cmin Then Do
cmin=cost.r.c
cx=c
End
End
Call allocate r cx
End
When dmax.0='C' Then Do
c=dmax.1
Do r=1 To rr
If cost.r.c<>'*' &,
cost.r.c<cmin Then Do
cmin=cost.r.c
rx=r
End
End
Call allocate rx c
End
Otherwise
Leave
End
End

Do r=1 To rr
Do c=1 To cc
If cost.r.c<>'*' Then Do
Call allocate r c
cost.r.c='*'
End
End
End

Call show_alloc 'Vogel''s Approximation'

Do r=1 To rr
Do c=1 To cc
cost.r.c=word(matrix.r.c,3)   /* restore cost.*.* */
End
End

Call steppingstone
Exit

/**********************************************************************
* Subroutines for Vogel's Approximation
**********************************************************************/

init:
If lines(fid)=0 Then Do
fid='Default input'
in.1=sourceline(4)
Parse Var in.1 numSources .
Do i=2 To numSources+3
in.i=sourceline(i+3)
End
End
Else Do
Do i=1 By 1 while lines(fid)>0
in.i=linein(fid)
End
End
Parse Var in.1 numSources numDestinations . 1 rr cc .
source.=0
demand.=0
source_sum=0
Do i=1 To numSources
Parse Var in.2 source.i in.2
ss.i=source.i
source_in.i=source.i
source_sum=source_sum+source.i
End
l=linein(fid)
demand_sum=0
Do i=1 To numDestinations
Parse Var in.3 demand.i in.3
dd.i=demand.i
demand_in.i=demand.i
demand_sum=demand_sum+demand.i
End
Do i=1 To numSources
j=i+3
l=in.j
Do j=1 To numDestinations
Parse Var l cost.i.j l
End
End
Do i=1 To numSources
ol=format(source.i,3)
Do j=1 To numDestinations
ol=ol format(cost.i.j,4)
End
End
ol='   '
Do j=1 To numDestinations
ol=ol format(demand.j,4)
End

Select
When source_sum=demand_sum Then Nop  /* balanced */
When source_sum>demand_sum Then Do   /* unbalanced - add dummy demand */
Say 'This is an unbalanced case (sources exceed demands). We add a dummy consumer.'
cc=cc+1
demand.cc=source_sum-demand_sum
demand_in.cc=demand.cc
dd.cc=demand.cc
Do r=1 To rr
cost.r.cc=0
End
End
Otherwise /* demand_sum>source_sum */ Do /* unbalanced - add dummy source */
Say 'This is an unbalanced case (demands exceed sources). We add a dummy source.'
rr=rr+1
source.rr=demand_sum-source_sum
source_in.rr=source.rr
ss.rr=source.rr
Do c=1 To cc
cost.rr.c=0
End
End
End

Say 'Sources / Demands / Cost'
ol='    '
Do c=1 To cc
ol=ol format(demand.c,3)
End
Say ol

Do r=1 To rr
ol=format(source.r,4)
Do c=1 To cc
ol=ol format(cost.r.c,3)
matrix.r.c=r c cost.r.c 0
End
Say ol
End
Return

allocate: Procedure Expose m. source. demand. cost. rr cc matrix.
Parse Arg r c
sh=min(source.r,demand.c)
source.r=source.r-sh
demand.c=demand.c-sh
m.r.c=sh
matrix.r.c=subword(matrix.r.c,1,3) sh
If source.r=0 Then Do
Do c=1 To cc
cost.r.c='*'
End
End
If demand.c=0 Then Do
Do r=1 To rr
cost.r.c='*'
End
End
Return

diff: Procedure
Parse Value arg(1) With n list
If n<2 Then Return 0
list=wordsort(list)
Return word(list,2)-word(list,1)

wordsort: Procedure
/**********************************************************************
* Sort the list of words supplied as argument. Return the sorted list
**********************************************************************/
Parse Arg wl
wa.=''
wa.0=0
Do While wl<>''
Parse Var wl w wl
Do i=1 To wa.0
If wa.i>w Then Leave
End
If i<=wa.0 Then Do
Do j=wa.0 To i By -1
ii=j+1
wa.ii=wa.j
End
End
wa.i=w
wa.0=wa.0+1
End
swl=''
Do i=1 To wa.0
swl=swl wa.i
End
/* Say swl */
Return strip(swl)

show_alloc: Procedure Expose matrix. rr cc demand_in. source_in.
Return
Say ''
total=0
ol='    '
Do c=1 to cc
ol=ol format(demand_in.c,3)
End
Say ol
as=''
Do r=1 to rr
ol=format(source_in.r,4)
a=word(matrix.r.1,4)
If a=0.0000000001 Then a=0
If a>0 Then
ol=ol format(a,3)
Else
ol=ol ' - '
total=total+word(matrix.r.1,4)*word(matrix.r.1,3)
Do c=2 To cc
a=word(matrix.r.c,4)
If a=0.0000000001 Then a=0
If a>0 Then
ol=ol format(a,3)
Else
ol=ol ' - '
total=total+word(matrix.r.c,4)*word(matrix.r.c,3)
as=as a
End
Say ol
End
Say 'Total costs:' format(total,4,1)
Return

/**********************************************************************
* Subroutines for Optimization
**********************************************************************/

steppingstone: Procedure Expose matrix. cost. rr cc matrix. demand_in.,
source_in. ms fid move cnt.
maxReduction=0
move=''
Call fixDegenerateCase
Do r=1 To rr
Do c=1 To cc
Parse Var matrix.r.c r c cost qrc
If qrc=0 Then Do
path=getclosedpath(r,c)
If pelems(path)<4 Then Do
Iterate
End
reduction = 0
lowestQuantity = 1e10
leavingCandidate = ''
plus=1
pathx=path
Do While pathx<>''
Parse Var pathx s '|' pathx
If plus Then
reduction=reduction+word(s,3)
Else Do
reduction=reduction-word(s,3)
If word(s,4)<lowestQuantity Then Do
leavingCandidate = s
lowestQuantity = word(s,4)
End
End
plus=\plus
End
If reduction < maxreduction Then Do
move=path
leaving=leavingCandidate
maxReduction = reduction
End
End
End
End
if move<>'' Then Do
quant=word(leaving,4)
If quant=0 Then Do
Call show_alloc 'Optimum'
Exit
End
plus=1
Do While move<>''
Parse Var move m '|' move
Parse Var m r c cpu qrc
Parse Var matrix.r.c vr vc vcost vquant
If plus Then
nquant=vquant+quant
Else
nquant=vquant-quant
matrix.r.c = vr vc vcost nquant
plus=\plus
End
move=''
Call steppingStone
End
Else
Call show_alloc 'Optimal Solution' fid
Return

getclosedpath: Procedure Expose matrix. cost. rr cc matrix.
Parse Arg rd,cd
path=rd cd cost.rd.cd word(matrix.rd.cd,4)
do r=1 To rr
Do c=1 To cc
If word(matrix.r.c,4)>0 Then Do
path=path'|'r c cost.r.c word(matrix.r.c,4)
End
End
End
path=magic(path)
Return stones(path)

magic: Procedure
Parse Arg list
Do Forever
list_1=remove_1(list)
If list_1=list Then Leave
list=list_1
End
Return list_1

remove_1: Procedure
Parse Arg list
cntr.=0
cntc.=0
Do i=1 By 1 While list<>''
parse Var list e.i '|' list
Parse Var e.i r c .
cntr.r=cntr.r+1
cntc.c=cntc.c+1
End
n=i-1
keep.=1
Do i=1 To n
Parse Var e.i r c .
If cntr.r<2 |,
cntc.c<2 Then Do
keep.i=0
End
End
list=e.1
Do i=2 To n
If keep.i Then
list=list'|'e.i
End
Return list

stones: Procedure
Parse Arg lst
tstc=lst
Do i=1 By 1 While tstc<>''
Parse Var tstc o.i '|' tstc
end
stones=lst
o.0=i-1
prev=o.1
Do i=1 To o.0
st.i=prev
k=i//2
nbrs=getNeighbors(prev,lst)
Parse Var nbrs n.1 '|' n.2
If k=0 Then
prev=n.2
Else
prev=n.1
End
stones=st.1
Do i=2 To o.0
stones=stones'|'st.i
End
Return stones

getNeighbors: Procedure Expose o.
parse Arg s, lst
Do i=1 To 4
Parse Var lst o.i '|' lst
End
nbrs.=''
sr=word(s,1)
sc=word(s,2)
Do i=1 To o.0
If o.i<>s Then Do
or=word(o.i,1)
oc=word(o.i,2)
If or=sr & nbrs.0='' Then
nbrs.0 = o.i
else if oc=sc & nbrs.1='' Then
nbrs.1 = o.i
If nbrs.0<>'' & nbrs.1<>'' Then
Leave
End
End
return nbrs.0'|'nbrs.1

m1: Procedure
Parse Arg z
Return z-1

pelems: Procedure
Call Trace 'O'
Parse Arg p
n=0
Do While p<>''
Parse Var p x '|' p
If x<>'' Then n=n+1
End
Return n

fixDegenerateCase: Procedure Expose matrix. rr cc ms
Call matrixtolist
If (rr+cc-1)<>ms Then Do
Do r=1 To rr
Do c=1 To cc
If word(matrix.r.c,4)=0 Then Do
matrix.r.c=subword(matrix.r.c,1,3) 1.e-10
Return
End
End
End
End
Return

matrixtolist: Procedure Expose matrix. rr cc ms
ms=0
list=''
Do r=1 To rr
Do c=1 To cc
If word(matrix.r.c,4)>0 Then Do
list=list'|'matrix.r.c
ms=ms+1
End
End
End
Return strip(list,,'|')

Novalue:
Say 'Novalue raised in line' sigl
Say sourceline(sigl)
Say 'Variable' condition('D')
Signal lookaround

Syntax:
Say 'Syntax raised in line' sigl
Say sourceline(sigl)
Say 'rc='rc '('errortext(rc)')'

halt:
lookaround:
If fore() Then Do
Say 'You can look around now.'
Trace ?R
Nop
End
Exit 12
```
Output:
```F:\>regina tpv vv.txt
Sources / Demands / Cost
30  20  70  30  60
50  16  16  13  22  17
60  14  14  13  19  15
50  19  19  20  23  50
50  50  12  50  15  11

Vogel's Approximation
30  20  70  30  60
50  -   -   50  -   -
60  -   -   20  -   40
50  30  20  -   -   -
50  -   -   -   30  20
Total costs: 3130.0

Optimum
30  20  70  30  60
50  -   -   50  -   -
60  30  -   20  -   10
50  -   20  -   30  -
50  -   -   -   -   50
Total costs: 3100.0```

### Low Cost Algorithm

```/* REXX ***************************************************************
* Solve the Transportation Problem using the Least Cost Method
Default Input
2 3        # of sources / # of demands
25 35      sources
20 30 10   demands
3 5 7      cost matrix
3 2 5
* 20201228 corresponds to NWC above
*          Note: correctness of input is not checked
* 20210103 remove debug code
**********************************************************************/
Signal On Halt
Signal On Novalue
Signal On Syntax

Parse Arg fid
If fid='' Then
fid='input1.txt'
Call init
Do r=1 To rr
Do c=1 To cc
matrix.r.c=r c cost.r.c 0
End
End
Do Until source_sum=0
mincost=1e10
Do r=1 To rr
If source.r>0 Then Do
Do c=1 To cc
If demand.c>0 Then Do
cost=word(matrix.r.c,3)
If cost>0 & cost<mincost |,
source_sum=source.r |,
demand_sum=demand.c Then Do
tgt=r c cost
mincost=cost
End
End
End
End
End
Parse Var tgt tr tc .
a=min(source.tr,demand.tc)
matrix.tr.tc=subword(matrix.tr.tc,1,3) word(matrix.tr.tc,4)+a
source.tr=source.tr-a
demand.tc=demand.tc-a
source_sum=source_sum-a
demand_sum=demand_sum-a

End
Call show_alloc 'Low Cost Algorithm'
Call steppingstone
Exit

/**********************************************************************
* Subroutines for Low Cost Algorithm
**********************************************************************/

init:
If lines(fid)=0 Then Do
fid='Default input'
in.1=sourceline(4)
Parse Var in.1 numSources .
Do i=2 To numSources+3
in.i=sourceline(i+3)
End
End
Else Do
Do i=1 By 1 while lines(fid)>0
in.i=linein(fid)
End
End
Parse Var in.1 numSources numDestinations . 1 rr cc .
source_sum=0
Do i=1 To numSources
Parse Var in.2 source.i in.2
ss.i=source.i
source_sum=source_sum+source.i
source_in.i=source.i
End
demand_sum=0
Do i=1 To numDestinations
Parse Var in.3 demand.i in.3
dd.i=demand.i
demand_in.i=demand.i
demand_sum=demand_sum+demand.i
End
Do i=1 To numSources
j=i+3
l=in.j
Do j=1 To numDestinations
Parse Var l cost.i.j l
End
End
Do i=1 To numSources
ol=format(source.i,3)
Do j=1 To numDestinations
ol=ol format(cost.i.j,4)
End
End
Select
When source_sum=demand_sum Then Nop  /* balanced */
When source_sum>demand_sum Then Do   /* unbalanced - add dummy demand */
Say 'This is an unbalanced case (sources exceed demands). We add a dummy consumer.'
cc=cc+1
demand.cc=source_sum-demand_sum
demand_in.cc=demand.cc
dd.cc=demand.cc
Do r=1 To rr
cost.r.cc=0
End
End
Otherwise /* demand_sum>source_sum */ Do /* unbalanced - add dummy source */
Say 'This is an unbalanced case (demands exceed sources). We add a dummy source.'
rr=rr+1
source.rr=demand_sum-source_sum
ss.rr=source.rr
source_in.rr=source.rr
Do c=1 To cc
cost.rr.c=0
End
End
End

Say 'Sources / Demands / Cost'
ol='    '
Do c=1 To cc
ol=ol format(demand.c,3)
End
Say ol
Do r=1 To rr
ol=format(source.r,4)
Do c=1 To cc
ol=ol format(cost.r.c,3)
End
Say ol
End
Return

show_alloc: Procedure Expose matrix. rr cc demand_in. source_in.
Return
Say ''
total=0
ol='    '
Do c=1 to cc
ol=ol format(demand_in.c,3)
End
Say ol
as=''
Do r=1 to rr
ol=format(source_in.r,4)
a=word(matrix.r.1,4)
If a=0.0000000001 Then a=0
If a>0 Then
ol=ol format(a,3)
Else
ol=ol ' - '
total=total+word(matrix.r.1,4)*word(matrix.r.1,3)
Do c=2 To cc
a=word(matrix.r.c,4)
If a=0.0000000001 Then a=0
If a>0 Then
ol=ol format(a,3)
Else
ol=ol ' - '
total=total+word(matrix.r.c,4)*word(matrix.r.c,3)
as=as a
End
Say ol
End
Say 'Total costs:' format(total,4,1)
Return

/**********************************************************************
* Subroutines for Optimization
**********************************************************************/

steppingstone: Procedure Expose matrix. cost. rr cc matrix. demand_in.,
source_in. fid move cnt.
maxReduction=0
move=''
Call fixDegenerateCase
Do r=1 To rr
Do c=1 To cc
Parse Var matrix.r.c r c cost qrc
If qrc=0 Then Do
path=getclosedpath(r,c)
If pelems(path)<4 then
Iterate
reduction = 0
lowestQuantity = 1e10
leavingCandidate = ''
plus=1
pathx=path
Do While pathx<>''
Parse Var pathx s '|' pathx
If plus Then
reduction=reduction+word(s,3)
Else Do
reduction=reduction-word(s,3)
If word(s,4)<lowestQuantity Then Do
leavingCandidate = s
lowestQuantity = word(s,4)
End
End
plus=\plus
End
If reduction < maxreduction Then Do
move=path
leaving=leavingCandidate
maxReduction = reduction
End
End
End
End
if move<>'' Then Do
quant=word(leaving,4)
If quant=0 Then Do
Call show_alloc 'Optimum'
Exit
End
plus=1
Do While move<>''
Parse Var move m '|' move
Parse Var m r c cpu qrc
Parse Var matrix.r.c vr vc vcost vquant
If plus Then
nquant=vquant+quant
Else
nquant=vquant-quant
matrix.r.c = vr vc vcost nquant
plus=\plus
End
move=''
Call steppingStone
End
Else
Call show_alloc 'Optimal Solution' fid
Return

getclosedpath: Procedure Expose matrix. cost. rr cc
Parse Arg rd,cd
path=rd cd cost.rd.cd word(matrix.rd.cd,4)
do r=1 To rr
Do c=1 To cc
If word(matrix.r.c,4)>0 Then Do
path=path'|'r c cost.r.c word(matrix.r.c,4)
End
End
End
path=magic(path)
Return stones(path)

magic: Procedure
Parse Arg list
Do Forever
list_1=remove_1(list)
If list_1=list Then Leave
list=list_1
End
Return list_1

remove_1: Procedure
Parse Arg list
cntr.=0
cntc.=0
Do i=1 By 1 While list<>''
parse Var list e.i '|' list
Parse Var e.i r c .
cntr.r=cntr.r+1
cntc.c=cntc.c+1
End
n=i-1
keep.=1
Do i=1 To n
Parse Var e.i r c .
If cntr.r<2 |,
cntc.c<2 Then Do
keep.i=0
End
End
list=e.1
Do i=2 To n
If keep.i Then
list=list'|'e.i
End
Return list

stones: Procedure
Parse Arg lst
stones=lst
tstc=lst
Do i=1 By 1 While tstc<>''
Parse Var tstc o.i '|' tstc
End
o.0=i-1
prev=o.1
Do i=1 To o.0
st.i=prev
k=i//2
nbrs=getNeighbors(prev, lst)
Parse Var nbrs n.1 '|' n.2
If k=0 Then
prev=n.2
Else
prev=n.1
End
stones=st.1
Do i=2 To o.0
stones=stones'|'st.i
End
Return stones

getNeighbors: Procedure
parse Arg s, lst
Do i=1 By 1 While lst<>''
Parse Var lst o.i '|' lst
End
o.0=i-1
nbrs.=''
sr=word(s,1)
sc=word(s,2)
Do i=1 To o.0
If o.i<>s Then Do
or=word(o.i,1)
oc=word(o.i,2)
If or=sr & nbrs.0='' Then
nbrs.0 = o.i
else if oc=sc & nbrs.1='' Then
nbrs.1 = o.i
If nbrs.0<>'' & nbrs.1<>'' Then
Leave
End
End
return nbrs.0'|'nbrs.1

m1: Procedure
Parse Arg z
Return z-1

pelems: Procedure
Call Trace 'O'
Parse Arg p
n=0
Do While p<>''
Parse Var p x '|' p
If x<>'' Then n=n+1
End
Return n

fixDegenerateCase: Procedure Expose matrix. rr cc ms ms demand_in. source_in. move cnt.
Call matrixtolist
If (rr+cc-1)<>ms Then Do
Do r=1 To rr
Do c=1 To cc
If word(matrix.r.c,4)=0 Then Do
matrix.r.c=subword(matrix.r.c,1,3) 1.e-10
Return
End
End
End
End
Return

matrixtolist: Procedure Expose matrix. rr cc ms
ms=0
list=''
Do r=1 To rr
Do c=1 To cc
If word(matrix.r.c,4)>0 Then Do
list=list'|'matrix.r.c
ms=ms+1
End
End
End
Return strip(list,,'|')

Novalue:
Say 'Novalue raised in line' sigl
Say sourceline(sigl)
Say 'Variable' condition('D')
Signal lookaround

Syntax:
Say 'Syntax raised in line' sigl
Say sourceline(sigl)
Say 'rc='rc '('errortext(rc)')'

halt:
lookaround:
If fore() Then Do
Say 'You can look around now.'
Trace ?R
Nop
End
Exit 12
```
Output:
```F:\>rexx tpl vv.txt
Sources / Demands / Cost
30  20  70  30  60
50  16  16  13  22  17
60  14  14  13  19  15
50  19  19  20  23  50
50  50  12  50  15  11

Low Cost Algorithm
30  20  70  30  60
50  -   -   50  -   -
60  30  10  20  -   -
50  -   10  -   30  10
50  -   -   -   -   50
Total costs: 3400.0

Optimum
30  20  70  30  60
50  -   -   50  -   -
60  30  -   20  -   10
50  -   20  -   30  -
50  -   -   -   -   50
Total costs: 3100.0```

## Ruby

Breaks ties using lowest cost cell.

```# VAM
#
#  Nigel_Galloway
#  September 1st., 2013
COSTS  = {W: {A: 16, B: 16, C: 13, D: 22, E: 17},
X: {A: 14, B: 14, C: 13, D: 19, E: 15},
Y: {A: 19, B: 19, C: 20, D: 23, E: 50},
Z: {A: 50, B: 12, C: 50, D: 15, E: 11}}
demand = {A: 30, B: 20, C: 70, D: 30, E: 60}
supply = {W: 50, X: 60, Y: 50, Z: 50}
COLS = demand.keys
res = {}; COSTS.each_key{|k| res[k] = Hash.new(0)}
g = {}; supply.each_key{|x| g[x] = COSTS[x].keys.sort_by{|g| COSTS[x][g]}}
demand.each_key{|x| g[x] = COSTS.keys.sort_by{|g| COSTS[g][x]}}

until g.empty?
d = demand.collect{|x,y| [x, z = COSTS[g[x][0]][x], g[x][1] ? COSTS[g[x][1]][x] - z : z]}
dmax = d.max_by{|n| n[2]}
d = d.select{|x| x[2] == dmax[2]}.min_by{|n| n[1]}
s = supply.collect{|x,y| [x, z = COSTS[x][g[x][0]], g[x][1] ? COSTS[x][g[x][1]] - z : z]}
dmax = s.max_by{|n| n[2]}
s = s.select{|x| x[2] == dmax[2]}.min_by{|n| n[1]}
t,f = d[2]==s[2] ? [s[1], d[1]] : [d[2],s[2]]
d,s = t > f ? [d[0],g[d[0]][0]] : [g[s[0]][0],s[0]]
v = [supply[s], demand[d]].min
res[s][d] += v
demand[d] -= v
if demand[d] == 0 then
supply.reject{|k, n| n == 0}.each_key{|x| g[x].delete(d)}
g.delete(d)
demand.delete(d)
end
supply[s] -= v
if supply[s] == 0 then
demand.reject{|k, n| n == 0}.each_key{|x| g[x].delete(s)}
g.delete(s)
supply.delete(s)
end
end

COLS.each{|n| print "\t", n}
puts
cost = 0
COSTS.each_key do |g|
print g, "\t"
COLS.each do |n|
y = res[g][n]
print y if y != 0
cost += y * COSTS[g][n]
print "\t"
end
puts
end
print "\n\nTotal Cost = ", cost
```
Output:
```        A       B       C       D       E
W                       50
X       30              20              10
Y               20              30
Z                                       50

Total Cost = 3100
```

### Reference Example

Replacing the data in the Task Example with:

```COSTS  = {S1: {D1: 46, D2:  74, D3:  9, D4: 28, D5: 99},
S2: {D1: 12, D2:  75, D3:  6, D4: 36, D5: 48},
S3: {D1: 35, D2: 199, D3:  4, D4:  5, D5: 71},
S4: {D1: 61, D2:  81, D3: 44, D4: 88, D5:  9},
S5: {D1: 85, D2:  60, D3: 14, D4: 25, D5: 79}}
demand = {D1: 278, D2: 60, D3: 461, D4: 116, D5: 1060}
supply = {S1: 461, S2: 277, S3: 356, S4: 488, S5: 393}
```

Produces:

```        D1      D2      D3      D4      D5
S1      1       60      68              332
S2      277
S3                              116     240
S4                                      488
S5                      393

Total Cost = 68804
```

## Sidef

Translation of: Ruby
```var costs = :(
W => :(A => 16, B => 16, C => 13, D => 22, E => 17),
X => :(A => 14, B => 14, C => 13, D => 19, E => 15),
Y => :(A => 19, B => 19, C => 20, D => 23, E => 50),
Z => :(A => 50, B => 12, C => 50, D => 15, E => 11)
)

var demand = :(A => 30, B => 20, C => 70, D => 30, E => 60)
var supply = :(W => 50, X => 60, Y => 50, Z => 50)

var cols = demand.keys.sort

var (:res, :g)
supply.each {|x| g{x} = costs{x}.keys.sort_by{|g| costs{x}{g} }}
demand.each {|x| g{x} = costs   .keys.sort_by{|g| costs{g}{x} }}

while (g) {
var d = demand.collect {|x|
[x, var z = costs{g{x}[0]}{x}, g{x}[1] ? costs{g{x}[1]}{x}-z : z]
}

var s = supply.collect {|x|
[x, var z = costs{x}{g{x}[0]}, g{x}[1] ? costs{x}{g{x}[1]}-z : z]
}

d.grep! { .[2] == d.max_by{ .[2] }[2] }.min_by! { .[1] }
s.grep! { .[2] == s.max_by{ .[2] }[2] }.min_by! { .[1] }

var (t,f) = (d[2] == s[2] ? ((s[1], d[1])) : ((d[2], s[2])))
(d,s) = (t > f ? ((d[0], g{d[0]}[0])) : ((g{s[0]}[0],s[0])))

var v = (supply{s} `min` demand{d})

res{s}{d} := 0 += v
demand{d} -= v

if (demand{d} == 0) {
supply.grep {|_,n| n != 0 }.each {|x| g{x}.delete(d) }
g.delete(d)
demand.delete(d)
}

supply{s} -= v

if (supply{s} == 0) {
demand.grep {|_,n| n != 0 }.each {|x| g{x}.delete(s) }
g.delete(s)
supply.delete(s)
}
}

say("\t", cols.join("\t"))

var cost = 0
costs.keys.sort.each { |g|
print(g, "\t")
cols.each { |n|
if (defined(var y = res{g}{n})) {
print(y)
cost += (y * costs{g}{n})
}
print("\t")
}
print("\n")
}

say "\n\nTotal Cost = #{cost}"
```
Output:
```	A	B	C	D	E
W			50
X	30		20		10
Y		20		30
Z					50

Total Cost = 3100
```

## Tcl

Works with: Tcl version 8.6
```package require Tcl 8.6

# A sort that works by sorting by an auxiliary key computed by a lambda term
proc sortByFunction {list lambda} {
lmap k [lsort -index 1 [lmap k \$list {
list \$k [uplevel 1 [list apply \$lambda \$k]]
}]] {lindex \$k 0}
}

# A simple way to pick a “best” item from a list
proc minimax {list maxidx minidx} {
set max -Inf; set min Inf
foreach t \$list {
if {[set m [lindex \$t \$maxidx]] > \$max} {
set best \$t
set max \$m
set min Inf
} elseif {\$m == \$max && [set m [lindex \$t \$minidx]] < \$min} {
set best \$t
set min \$m
}
}
return \$best
}

# The approximation engine. Note that this does not change the provided
# arguments at all since they are copied on write.
proc VAM {costs demand supply} {
# Initialise the sorted sequence of pairs and the result dictionary
foreach x [dict keys \$demand] {
dict set g \$x [sortByFunction [dict keys \$supply] {g {
upvar 1 costs costs x x; dict get \$costs \$g \$x
}}]
dict set row \$x 0
}
foreach x [dict keys \$supply] {
dict set g \$x [sortByFunction [dict keys \$demand] {g {
upvar 1 costs costs x x; dict get \$costs \$x \$g
}}]
dict set res \$x \$row
}

# While there's work to do...
while {[dict size \$g]} {
# Select "best" demand
lassign [minimax [lmap x [dict keys \$demand] {
if {![llength [set gx [dict get \$g \$x]]]} continue
set z [dict get \$costs [lindex \$gx 0] \$x]
if {[llength \$gx] > 1} {
list \$x \$z [expr {[dict get \$costs [lindex \$gx 1] \$x] - \$z}]
} else {
list \$x \$z \$z
}
}] 2 1] d dVal dCost

# Select "best" supply
lassign [minimax [lmap x [dict keys \$supply] {
if {![llength [set gx [dict get \$g \$x]]]} continue
set z [dict get \$costs \$x [lindex \$gx 0]]
if {[llength \$gx] > 1} {
list \$x \$z [expr {[dict get \$costs \$x [lindex \$gx 1]] - \$z}]
} else {
list \$x \$z \$z
}
}] 2 1] s sVal sCost

# Compute how much to transfer, and with which "best"
if {\$sCost == \$dCost ? \$sVal > \$dVal : \$sCost < \$dCost} {
set s [lindex [dict get \$g \$d] 0]
} else {
set d [lindex [dict get \$g \$s] 0]
}
set v [expr {min([dict get \$supply \$s], [dict get \$demand \$d])}]

# Transfer some supply to demand
dict update res \$s inner {dict incr inner \$d \$v}
dict incr demand \$d -\$v
if {[dict get \$demand \$d] == 0} {
dict for {k n} \$supply {
if {\$n != 0} {
# Filter list in dictionary to remove element
dict set g \$k [lmap x [dict get \$g \$k] {
if {\$x eq \$d} continue; set x
}]
}
}
dict unset g \$d
dict unset demand \$d
}
dict incr supply \$s -\$v
if {[dict get \$supply \$s] == 0} {
dict for {k n} \$demand {
if {\$n != 0} {
dict set g \$k [lmap x [dict get \$g \$k] {
if {\$x eq \$s} continue; set x
}]
}
}
dict unset g \$s
dict unset supply \$s
}
}
return \$res
}
```

Demonstration:

```set COSTS {
W {A 16 B 16 C 13 D 22 E 17}
X {A 14 B 14 C 13 D 19 E 15}
Y {A 19 B 19 C 20 D 23 E 50}
Z {A 50 B 12 C 50 D 15 E 11}
}
set DEMAND {A 30 B 20 C 70 D 30 E 60}
set SUPPLY {W 50 X 60 Y 50 Z 50}

set RES [VAM \$COSTS \$DEMAND \$SUPPLY]

puts \t[join [dict keys \$DEMAND] \t]
set cost 0
foreach g [dict keys \$SUPPLY] {
puts \$g\t[join [lmap n [dict keys \$DEMAND] {
set c [dict get \$RES \$g \$n]
incr cost [expr {\$c * [dict get \$COSTS \$g \$n]}]
expr {\$c ? \$c : ""}
}] \t]
}
puts "\nTotal Cost = \$cost"
```
Output:
```        A       B       C       D       E
W                       50
X       10      20      20              10
Y       20                      30
Z                                       50

Total Cost = 3100
```

## Wren

Translation of: Kotlin
Library: Wren-math
Library: Wren-fmt
```import "./math" for Nums
import "./fmt" for Fmt

var supply = [50, 60, 50, 50]
var demand = [30, 20, 70, 30, 60]

var costs = [
[16, 16, 13, 22, 17],
[14, 14, 13, 19, 15],
[19, 19, 20, 23, 50],
[50, 12, 50, 15, 11]
]

var nRows = supply.count
var nCols = demand.count

var rowDone = List.filled(nRows, false)
var colDone = List.filled(nCols, false)
var results = List.filled(nRows, null)
for (i in 0...nRows) results[i] = List.filled(nCols, 0)

var diff = Fn.new { |j, len, isRow|
var min1 = Num.maxSafeInteger
var min2 = min1
var minP = -1
for (i in 0...len) {
var done = isRow ? colDone[i] : rowDone[i]
if (!done) {
var c = isRow ? costs[j][i] : costs[i][j]
if (c < min1) {
min2 = min1
min1 = c
minP = i
} else if (c < min2) min2 = c
}
}
return [min2 - min1, min1, minP]
}

var maxPenalty = Fn.new { |len1, len2, isRow|
var md = Num.minSafeInteger
var pc = -1
var pm = -1
var mc = -1
for (i in 0...len1) {
var done = isRow ? rowDone[i] : colDone[i]
if (!done) {
var res = diff.call(i, len2, isRow)
if (res[0] > md) {
md = res[0]  // max diff
pm = i       // pos of max diff
mc = res[1]  // min cost
pc = res[2]  // pos of min cost
}
}
}
return isRow ? [pm, pc, mc, md] : [pc, pm, mc, md]
}

var nextCell = Fn.new {
var res1 = maxPenalty.call(nRows, nCols, true)
var res2 = maxPenalty.call(nCols, nRows, false)
if (res1[3] == res2[3]) return (res1[2] < res2[2]) ? res1 : res2
return (res1[3] > res2[3]) ? res2 : res1
}

var supplyLeft = Nums.sum(supply)
var totalCost = 0
while (supplyLeft > 0) {
var cell = nextCell.call()
var r = cell[0]
var c = cell[1]
var q = demand[c].min(supply[r])
demand[c] = demand[c] - q
if (demand[c] == 0) colDone[c] = true
supply[r] = supply[r] - q
if (supply[r] == 0) rowDone[r] = true
results[r][c] = q
supplyLeft = supplyLeft - q
totalCost = totalCost + q*costs[r][c]
}

System.print("    A   B   C   D   E")
var i = 0
for (result in results) {
Fmt.write("\$c", "W".bytes[0] + i)
for (item in result) Fmt.write("  \$2d", item)
System.print()
i = i + 1
}
System.print("\nTotal Cost = %(totalCost)")
```
Output:
```    A   B   C   D   E
W   0   0  50   0   0
X  30   0  20   0  10
Y   0  20   0  30   0
Z   0   0   0   0  50

Total Cost = 3100
```

## Yabasic

Translation of: C
```N_ROWS = 4 : N_COLS = 5

dim supply(N_ROWS)
dim demand(N_COLS)

restore sup
for n = 0 to N_ROWS - 1
next n

restore dem
for n = 0 to N_COLS - 1
next n

label sup
data 50, 60, 50, 50

label dem
data 30, 20, 70, 30, 60

dim costs(N_ROWS, N_COLS)

label cost
data 16, 16, 13, 22, 17
data 14, 14, 13, 19, 15
data 19, 19, 20, 23, 50
data 50, 12, 50, 15, 11

restore cost
for i = 0 to N_ROWS - 1
for j = 0 to N_COLS - 1
next j
next i

dim row_done(N_ROWS)
dim col_done(N_COLS)

sub diff(j, leng, is_row, res())
local i, c, min1, min2, min_p, test

min1 = 10e300 : min2 = min1 : min_p = -1

for i = 0 to leng - 1
if is_row then
test = col_done(i)
else
test = row_done(i)
end if
if test continue
if is_row then
c = costs(j, i)
else
c = costs(i, j)
end if
if c < min1 then
min2 = min1
min1 = c
min_p = i
elseif c < min2 then
min2 = c
end if
next i
res(0) = min2 - min1
res(1) = min1
res(2) = min_p
end sub

sub max_penalty(len1, len2, is_row, res())
local i, pc, pm, mc, md, res2(3), test

pc = -1 : pm = -1 : mc = -1 : md = -10e300

for i = 0 to len1 - 1
if is_row then
test = row_done(i)
else
test = col_done(i)
end if
if test continue
diff(i, len2, is_row, res2())
if res2(0) > md then
md = res2(0)  //* max diff */
pm = i        //* pos of max diff */
mc = res2(1)  //* min cost */
pc = res2(2)  //* pos of min cost */
end if
next i

if is_row then
res(0) = pm : res(1) = pc
else
res(0) = pc : res(1) = pm
end if
res(2) = mc : res(3) = md
end sub

sub next_cell(res())
local i, res1(4), res2(4)

max_penalty(N_ROWS, N_COLS, TRUE, res1())
max_penalty(N_COLS, N_ROWS, FALSE, res2())

if res1(3) = res2(3) then
if res1(2) < res2(2) then
for i = 0 to 3 : res(i) = res1(i) : next i
else
for i = 0 to 3 : res(i) = res2(i) : next i
end if
return
end if
if res1(3) > res2(3) then
for i = 0 to 3 : res(i) = res2(i) : next i
else
for i = 0 to 3 : res(i) = res1(i) : next i
end if
end sub

supply_left = 0 : total_cost = 0 : dim cell(4)

dim results(N_ROWS, N_COLS)

for i = 0 to N_ROWS - 1 : supply_left = supply_left + supply(i) : next i

while(supply_left > 0)
next_cell(cell())
r = cell(0)
c = cell(1)
q = min(demand(c), supply(r))
demand(c) = demand(c) - q
if not demand(c) col_done(c) = TRUE
supply(r) = supply(r) - q
if not supply(r) row_done(r) = TRUE
results(r, c) = q
supply_left = supply_left - q
total_cost = total_cost + q * costs(r, c)
wend

print "    A   B   C   D   E\n"
for i = 0 to N_ROWS - 1
print chr\$(asc("W") + i), " ";
for j = 0 to N_COLS - 1
print results(i, j) using "###";
next j
print
next i
print "\nTotal cost = ", total_cost```

## zkl

Translation of: Python
Translation of: Ruby
```costs:=Dictionary(
"W",Dictionary("A",16, "B",16, "C",13, "D",22, "E",17),
"X",Dictionary("A",14, "B",14, "C",13, "D",19, "E",15),
"Y",Dictionary("A",19, "B",19, "C",20, "D",23, "E",50),
demand:=Dictionary("A",30, "B",20, "C",70, "D",30, "E",60);  // gonna be modified
supply:=Dictionary("W",50, "X",60, "Y",50, "Z",50);	    // gonna be modified```
```cols:=demand.keys.sort();
res :=vogel(costs,supply,demand);
cost:=0;
println("\t",cols.concat("\t"));
foreach g in (costs.keys.sort()){
print(g,"\t");
foreach n in (cols){
y:=res[g].find(n);
if(y){ y=y[0]; print(y); cost+=y*costs[g][n]; }
print("\t");
}
println();
}
println("\nTotal Cost = ",cost);```
```fcn vogel(costs,supply,demand){
// a Dictionary can be created via a list of (k,v) pairs
res:= Dictionary(costs.pump(List,fcn([(k,_)]){ return(k,D()) }));
g  := Dictionary(); // cross index costs and make writable
supply.pump(Void,'wrap([(k,_)]){ g[k] =
costs[k].keys.sort('wrap(a,b){ costs[k][a]<costs[k][b] }).copy() });
demand.pump(Void,'wrap([(k,_)]){ g[k] =
costs.keys.sort('wrap(a,b){ costs[a][k]<costs[b][k] }).copy() });

while(g){
d:=Dictionary(demand.pump(List,'wrap([(k,_)]){ return(k,
g[k][0,2].apply('wrap(gk){ costs[gk][k] }).reverse().reduce('-)) }));
s:=Dictionary(supply.pump(List,'wrap([(k,_)]){ return(k,
g[k][0,2].apply('wrap(gk){ costs[k][gk] }).reverse().reduce('-)) }));
f:=(0).max(d.values); f=d.filter('wrap([(_,v)]){ v==f })[-1][0];
t:=(0).max(s.values); t=s.filter('wrap([(_,v)]){ v==t })[-1][0];
t,f=(if(d[f]>s[t]) T(f,g[f][0]) else T(g[t][0],t));
v:=supply[f].min(demand[t]);
res[f].appendV(t,v);  // create t:(v) or append v to t:(...)
if(0 == (demand[t]-=v)){
supply.pump(Void,'wrap([(k,n)]){ if(n!=0) g[k].remove(t) });
g.del(t); demand.del(t);
}
if(0 == (supply[f]-=v)){
demand.pump(Void,'wrap([(k,n)]){ if(n!=0) g[k].remove(f) });
g.del(f); supply.del(f);
}
}//while
res
}```
Output:
```	A	B	C	D	E
W			50
X	10	20	20		10
Y	20			30
Z					50

Total Cost = 3100
```