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
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};
    }
}