Transportation problem: Difference between revisions

 

=={{header|1C}}==

 

перем m,n; // Table size

Процедура КомандаРассчитать(Команда)

РешениеТранспортнойЗадачи();

 

=={{header|C sharp|C#}}==

{{trans|Java}}

using System.Collections.Generic;

using System.IO;

}

}

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<pre>Optimal solution input1.txt

=={{header|C++}}==

{{trans|Kotlin}}

#include <iomanip>

#include <iostream>

#include <string>

#include <vector>

 

using namespace std;

 

return 0;

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<pre>input1.txt

=={{header|D}}==

{{trans|Java}}

 

final class Shipment {

printResult(fileName, demand, supply, costs, matrix);

}

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<pre>Optimal solution transportation_problem1.txt

 

=={{header|Glagol}}==

ИСПОЛЬЗУЕТ

Вывод ИЗ "...\Отделы\Обмен\",

ВывестиПлан

 

 

=== Input ===

=={{header|Go}}==

{{trans|Java}}

 

import (

t.printResult()

}

 

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In other words:

 

ndxasgn=: conjunction define

:

r=. n (<ndxs)} r

end.

 

Task data:

 

supply=: 25 35

 

Task example:

 

20 0 5

 

=={{header|Java}}==

{{works with|Java|8}}

import java.util.*;

import static java.util.Arrays.stream;

}

}

 

<pre>input1.txt

=={{header|Julia}}==

Code taken from [https://github.com/dylanomics/transportation_problem here] using [https://jump.dev/JuMP.jl/stable/ JuMP].

 

# cost vector

λ = [JuMP.dual(C1[1]),JuMP.dual(C1[2])]

μ = [JuMP.dual(C2[1]),JuMP.dual(C2[2]),JuMP.dual(C2[3])]

 

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=={{header|Kotlin}}==

{{trans|Java}}

 

import java.io.File

}

}

 

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=={{header|Nim}}==

{{trans|Go}}

 

type

tr.northWestCornerRule()

tr.steppingStone()

 

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=={{header|Pascal}}==

{ based on the program of <Svetlana Belashova> }

 

GotoXY(40,1);

l1: d:=ReadKey;

 

=={{header|Phix}}==

The simplest solution I could think of.<br>

Assumes 0 cost is not allowed, but using say -1 as the "done" cost instead should be fine.

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<pre>

Obviously I did not really quite understand the problem when I rattled out the above... this does much better.

{{trans|Go}}

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(Obviously the other eight tests all work fine and produce similar output.)

Using <code>lpSolve::lp.transport</code>.

 

 

# cost matrix

rownames(sol) <- c("Supplier 1", "Supplier 2")

colnames(sol) <- c("Customer 1", "Customer 2", "Customer 3")

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<pre>Success: the objective function is 180

Using <code>typed/racket</code>, to keep track of Vectors of Vectors of data.

 

;; {{trans|Java}}

(define-type (V2 A) (Vectorof (Vectorof A)))

30 40 35 45

$

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Output of: <code>raco test Transportation-problem.rkt</code>:

{{works with|Rakudo|2019.03.1}}

Using [[Vogel's_approximation_method#Raku|Vogel's approximation method]]:

my %demand = :20C1, :30C2, :10C3;

my %supply = :25S1, :35S2;

print "\n";

}

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<pre> C1 C2 C3

=={{header|REXX}}==

{{trans|Java}}

* Solve the Transportation Problem using the Northwest Corner Method

Default Input

Nop

End

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<pre>F:\>rexx tpx2 input1.txt

Use network solver in SAS/OR:

 

data cost_data;

input from $ to $ cost;

print _OROPTMODEL_NUM_['OBJECTIVE'];

print flow;

 

Output:

=={{header|Visual Basic .NET}}==

{{trans|C#}}

 

Class Shipment

End Sub

 

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<pre>Optimal solution input1.txt

{{libheader|Wren-seq}}

{{libheader|Wren-fmt}}

 

var Shipment = Struct.create("Shipment", ["quantity", "costPerUnit", "r", "c"])

t.steppingStone()

t.printResult()

 

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