special cases of lp - technological university of panama

Special cases of LP

Minimum flow problems

Description

The minimum cost flow problem holds a central position among network optimization models.

It encompasses such a broad class of applications and because it can be solved extremely efficiently.

It considers: flow through a network with limited arc capacities.

flow through an arc.

Multiple sources (supply nodes) and multiple destinations (demand nodes) for the flow, with associated costs

H. R. Alvarez A., Ph. D.


Problem Statement

They are LP problems with special structures

They allow special algorithms

They take advantages of their structure to use a network approach.

This structure allows the solution of large problems by the network approach.


Elements of a minimum flow problem

Given a number of sources and sinks

Every source and sink has a maximum capacity to absorb

They might have intermediate nodes

There are arches that: With a maximum capacity

They have an associated cost to a flow unit.


Elements

Sources

Arches

Sinks

Nodes

Formulation:

Consider a directed and connected network where the n nodes include at least one supply node and at least one demand node.

The decision variables are


General Formulation:

Includes the following information:

The value of b depends on the nature of the node:

The objective is to minimize the total cost of sending the available supply through the network to satisfy the given demand.

In a feasible solution, the total flow being generated at the supply nodes equals the total flow being absorbed at the demand nodes.


L. P. formulation


Example:


The Assignment Problem


The Assignment Problem

Suppose that there are n working centers and n possible workers to be assigned, each of one can be assigned to only one of the n centers.

Suppose that there is a cost ci,j of assigning a worker i to a center j.

The objective is to minimize the total assignment cost.


otherwise 0

j center to assigned is i workerif X

n ..., 2, 1, i X

n ..., 2, 1,j X

:.t.s

XCZ.min

j,i

j

ji,

i

j,i

i j

j,ij,i

1

1

1

General formulation


Solution

SIMPLEX Method or binary integer programming

Row Column Method or the Hungarian Method.


The Hungarian Method

The Hungarian method is a combinatorial optimization algorithm which solves the assignment problem in polynomial time.

It was developed and published by Harold Kuhn in 1955.

"Hungarian method" because the algorithm was largely based on the earlier works of two Hungarian mathematicians: Dénes Kőnig and Jenő Egerváry.

The algorithm is known also as Kuhn-Munkres algorithm or Munkres assignment algorithm.


The matrix form of the problem

Given the cost coefficients of an assignment problem such that


Algorithm

1. Row reduction: the lowest of all cij (from j = 1 to n) is taken and is subtracted from each element in that row. This procedure is repeated for all rows.

2. Column reduction: the lowest of all cij (from 1= 1 to m) is taken and is subtracted from each element in that column. This procedure is repeated for all column. The resulting matrix is called the cost reduced matrix.

3. Covering zeroes: Cover all the zeroes in the cost reduced matrix with minimum number of horizontal and vertical lines. If the number of lines is equal to n, the optimal solution was obtained. Otherwise go to the next step.

4. Creating new zeroes: From the covered matrix generated in three, find the smallest element not covered by lines. Subtract this number to all the not covered numbers and add it to the numbers in the intersections of the lines. All the other elements do not change. Go back to step 3.


Example

A restaurant manager wants to serve different customers in different service areas. The manager knows that different client/server combinations may vary service costs due to customer characteristics and servers capabilities. Following is a cost matrix for customers and servers:


Cost for server

Server cost

Customer 1 2 3

1 12.90 11.90 12.10

2 15.30 15.50 14.30

3 11.90 13.90 13.00


Final solution: network representation

One formulation in MPL


Solution


Non zero Matrix - MPL


The transportation problem


The Transportation Problem

Is concerned with distributing any commodity from any group of supply centers, called sources, to any group of receiving centers, called destinations, in such a way as to minimize the total distribution cost.

Each source has a certain supply of units to distribute to the destinations, and each destination has a certain demand for units to be received from the sources.

Assumptions The requirements assumption: means that it needs to be a

balance between the total supply s from all sources and the total demand d at all destinations.

The feasible solutions property: A transportation problem will have feasible solutions if and only if s = d

The cost assumption: The cost of distributing units from any particular source to any particular destination is directly proportional to the number of units distributed.

The model: Any problem (whether involving transportation or not) fits the model for a transportation problem if it can be described completely in terms of a parameter table and it satisfies both the requirements assumption and the cost assumption.


Description A set of m supply points from which a good is

shipped. Supply point i can supply at most si

units.

A set of n demand points to which the good is shipped. Demand point j must receive at least dj units of the shipped good.

Each unit produced at supply point i and shipped to demand point j incurs a variable cost of cij


The parameter table


General formulation


If it is a balanced problem


The general formulation of the balanced problem

A requirement of this problem is that the flow from the sources must be equal to the capacity of the destinations. Otherwise, it is necessary to add either dummy sources or dummy destinations before solving the problem.

The network representation



Solution

Simplex method

Transportation algorithm

Initial Tableau

Initial solution

Optimality test

Redistribution of deliveries


Example

A business delivers products from four areas in the country to four distribution areas including two international . Data shows sources and destination capacities in tons, and the corresponding cost to the different destinations.


Example

Sources Tons

Chiriquí 2,500

Azuero 1,250

Darién 850

Coclé 1,000

Destinations Tons

Panamá 1980

Colón 750

Puerto Balboa 1000

Puerto Cristóbal

1870


Example - Costs

From/To: Panamá Colón Balboa Cristóbal

Chiriquí 50 55 50 55

Azuero 40 48 39 42

Darién 15 25 18 26

Coclé 22 28 25 29


General LP formulation Minimize 50x1,1 + 55x1,2 + … + 25x4,3 + 29x4,4 s. t. : x1,1 + x1,2 + … + x1,4 ≤ 2500 . . . . . . . . . . . . . . . x4,1 + x4,2 + … + x4,4 ≤ 1000 x1,1 + x2,1 + … + x4,1 ≥ 1980 . . . . . . . . . . . . . . . x1,4 + x2,4 + … + x4,4 ≥ 1870

Sources

Destinations


Balanced LP formulation Minimize 50x1,1 + 55x1,2 + … + 25x4,3 + 29x4,4 s. t. : x1,1 + x1,2 + … + x1,4 = 2500 . . . . . . . . . . . . . . . x4,1 + x4,2 + … + x4,4 = 1000 x1,1 + x2,1 + … + x4,1 = 1980 . . . . . . . . . . . . . . . x1,4 + x2,4 + … + x4,4 = 1870

Sources

Destinations


The parameter table:


The network representation


Initial solution– Norwest corner


Final solution


Final solution: network representation

One code in MPL


The optimal solution



Solver

MPL

The transhipment problem


Description

Normally goods are not directly deliverd to their final destination.

The use intermediate point or distribution centers.

The problem can be described as a minimal flow problem, but with intermediate nodes.


Example

Source Tons

Chiriquí 2,500

Azuero 1,250

Darién 850

Coclé 1,000

Destination Tons

Panamá 1,980

Colón 750

Puerto Balboa 1,000

Puerto Cristóbal

1,870

Distribution Tons

In 3,500

Out 3,500


Costs table

De/A: Panamá Colón Balboa Cristóbal Entrada Salida

Chiriquí 50 55 50 55 20

Azuero 40 48 39 42 18

Darién 15 25 18 26 30

Coclé 22 28 25 29 10

Entrada 5

Salida 15 20 15 20


Chiriquí

Azuero

Darién

Coclé

Entrada Salida

Panamá

Colón

Balboa

Cristóbal


Standard formulation Minimizar 50x1,1 + 55x1,2 + … + 25x4,3 + 29x4,4+ 20x1,5+ 18x2,5+… +10x4,5 + 5x5,6 + 15x6,1 + 20x6,2 + … + 20x6,5

Sujeto a: x1,1 + x1,2 + … + x1,4 + x1, 5 = 2500 . . . . . . . . . . . . x4,1 + x4,2 + … + x4,4 + + x4, 5 = 1000 x5,,6 ≤ 3500 X1,5 + x2,5 + x3,5 + x4,5 = X6,1 + x6,2 + x6,3 + x6,4 = x5,6 x1,1 + x2,1 + … + x4,1 + x6,1 = 1980 . . . . . . . . . . . . x1,4 + x2,4 + … + x4,4 + x6,4 = 1870

Sources

Destination

Flow constraint

Violates the tipical formulation of the transportation problem…


Formulation as a minimal flow problem


Final solution

Formulation and solution in Solver


Comparing solutions



Comparing solutions

Direct Transhipment

Costo 219,100 194,100


Integer linear programming

Only accepts integer solutions

Although the formulation is similar to the LP problem, it has the integrality constraint:

xi {I ≥ 0}


Different types of solutions:

Integer solutions

Binary solutions (0, 1)

Mixed solutions


Solution:

Relaxation: the model does not have the integrality constraint.

The general solution will have all the possible integer solutions.

It is an upper bound in the solution

Rounding up the solution might violate the feasibilty of the problem.


Branch and bound method

Introduced by Land and Doig in 1960

It is a sequential search algorithm

Implicitly enumerates most of the possible solutions of the problem.

It divides the set of possible solution in different subsets.

For every subset, both the upper limits and the feasibility criterium will be used to limit the solution.


General algorithm

1. Find the upper boundary given by the solution of the relaxed model.

2. Define two solution subsets such that d + 1 ≤ xk ≤ d where d is a constant defined by the smallest integer

for the solution for xk

3. For each subset define a new optimal solution: A subset will be fathomed if

- The solution is not feasible - There is a better solution

4. The process stops when an optimal solution with integer variables is found.


Example

Given the following problem

Max.: x = 4x1 + 11x2

s. t.

2x1 – x2 ≤ 4

2x1 + 5x2 ≤ 16

- x1 + 2x2 ≤ 4

x1 y x2 ≥ 0 and integers


x1 ≤ 1 X1 ≥ 2

X2 ≤ 2 X2 ≥ 3

Fathomed

Upper bound solution


Optimal Integer solution in MPL


A slight different formulation


special cases of lp - technological university of panama

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