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.
C
<lang c>#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;
}</lang>
- 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): <lang go>#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;
}</lang>
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
D
Strongly typed version (but K is not divided in Task and Contractor types to keep code simpler).
<lang d>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];
immutable 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);
}</lang>
- Output:
A B C D E W 50 X 20 40 Y 30 20 Z 30 20 Total Cost = 3130
Go
<lang go>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)
}</lang>
- 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: <lang go>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)
}</lang>
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:
<lang J>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
)</lang>
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:
<lang J>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</lang>
Java
<lang java>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};
}
}</lang>
[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.
<lang Julia>
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) </lang>
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!.
<lang Julia>
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 </lang>
Main <lang Julia>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) </lang>
- 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
<lang scala>// 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")
}</lang>
- 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 6
<lang perl6>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";</lang>
- Output:
A B C D E W - - 50 - - X 30 - 20 - 10 Y - 20 - 30 - Z - - - - 50 Total cost: 3100
Phix
See Transportation_problem#Phix for optimal results.
<lang Phix>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)</lang>
- 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: <lang Phix>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}}</lang>
- Output:
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
<lang python>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</lang>
- Output:
A B C D E W 50 X 30 20 10 Y 20 30 Z 50 Total Cost = 3100
Racket
Losley:
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>#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))))</lang>
- Output:
A B C D E W - - 50 - - X 10 20 20 - 10 Y 20 - - 30 - Z - - - - 50 Total Cost: 3100
Ruby
Breaks ties using lowest cost cell.
Task Example
<lang ruby># 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</lang>
- 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: <lang ruby>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}</lang> Produces:
D1 D2 D3 D4 D5 S1 1 60 68 332 S2 277 S3 116 240 S4 488 S5 393 Total Cost = 68804
Sidef
<lang 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}"</lang>
- Output:
A B C D E W 50 X 30 20 10 Y 20 30 Z 50 Total Cost = 3100
Tcl
<lang tcl>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
}</lang> Demonstration: <lang tcl>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"</lang>
- Output:
A B C D E W 50 X 10 20 20 10 Y 20 30 Z 50 Total Cost = 3100
Yabasic
<lang Yabasic> 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</lang>
zkl
<lang zkl>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</lang> <lang zkl>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);</lang> <lang zkl>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
}</lang>
- Output:
A B C D E W 50 X 10 20 20 10 Y 20 30 Z 50 Total Cost = 3100