Vogel's approximation method

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.

• Transportation problem

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.

Task example:

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];

    static ExecutorService es = Executors.newFixedThreadPool(2);

    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
  Say 'Input file not specified or not found. Using default input instead.'
  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.
Parse Arg header
If header='' Then
  Return
Say ''
Say header
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
* 20210102 add optimization
* 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
  Say 'Input file not specified or not found. Using default input instead.'
  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.
Parse Arg header
If header='' Then
  Return
Say ''
Say header
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.

Task Example

# 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
	read supply(n)
next n

restore dem
for n = 0 to N_COLS - 1
	read demand(n)
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
		read costs(i, j)
	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),
   "Z",Dictionary("A",50, "B",12, "C",50, "D",15, "E",11)).makeReadOnly();
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