DISTRIBUTION-MODEL (1).pptx

DISTRIBUTION MODEL
INTRODUCTION
A special type of an Linear Programming Problem in the area of physical
distribution of goods and services coming from several supply locations
that are to be delivered to demand centers.

TWO TYPES OF DISTRIBUTION
MODELS
● TRANSPORTATION MODEL
It involves shipments from a number of sources to a number of
destinations
● ASSIGNMENT PROBLEM
It deals with situations wherein a given number of candidates are
assigned to a number of positions.

FRANK LAUREN HITCHCOCK
1941
● An American mathematician and physicist, presented the simplest form
of transportation model in his study “The distribution of a product
from several sources to numerous localities”

TJALLING CHARLES KOOPMANS
1947
• A Dutch-American mathematician and economist. He
wrote “Optimum Utilization of the Transportation
System”

ABRAHAM CHARNES &
WILLIAM WAGER COOPER 1953
● Two American Operation researchers, developed the Stepping stone
method.
● The modified distribution method was developed in 1955.

7.1
Characteristics
and
Assumptions of
the
Transportation
problem

The main objective of the transportation model is to determine the cheapest
routes from the suppliers to the destinations. Figures 7.1 shows the
relationships between the suppliers and destinations. A transportation
problem exhibits the following characteristics and assumptions.
1. SUPPLY OR SOURCES – certain sources have limited available quantity of one
commodity.
2. DEMAND – the demand comes from several destinations, such as warehouses,
distributions centers, shops, etc.
3. QUANTITIES – the quantities available at each source and the demands of each
destinations are constant.
4. SHIPPING COST – is based on per unit cost of the commodity from the source to
each destination. It is usually based on the distance between the two points.
5. It is assumed that there will be no shipments between sources or between
destinations as they will require special adjustments to the transportation model
6. Demand and supply quantities are presented in whole numbers.

Sample Problem:
Mr. Danny Atienza is the owner of Cement Exchange, a cement
compositions manufacturer with three plant locations. Cement-bonded
composites are important constructions materials that are made of
hydrated cement paste that binds wood or fibers to make precast for
building components. Cement Exchange is currently a top manufacturer of
cement composites and has been an exclusive supplier to four major
distributors in Metro Manila. The total monthly purchase of each of the four
distributors are 1500, 2000, 2400, and 3500 boxes. The four distributors
may receive from any of the three manufacturing plants. However, there
are limitations to the production capacity of each plant:

PLANT 1 – 3000
PLANT 2 – 2700
PLANT 3 – 3700
Mr. Atienza would like to reduce the logistics expenses by proper
scheduling of shipments to the four distributors. The company’s chief
accountant submitted an estimate of the per box/unit shipping cost from
each plant to each distributors.

PLANTS DISTRIBUTORS
1 2 3 4
1 15 18 22 26
2 21 21 16 23
3 14 19 20 24
The shipping costs are in Philippines currency:
Based on the available data, the owner decided to use the transportation method to compute for
the cheapest delivery schedule.
Cement exchange’s distribution problem when converted to LP will show:

DECISIONS VARIABLES:
X11 = number of boxes shipped from plant 1 to distributor 1
The goal of Mr. Atienza is to minimize the total shipping cost. The transportations cost is
computed by multiplying the amount shipped over each route by the per unit shipping cost for
that route, and these costs are summed up to get the total transportation cost.

OBJECTIVE FUNCTION:
Minimum C= 15x11+18x12+22x13+26x14+21x21+25x22+16x23+23x24+14x31+19x32+20x33+24x34
The corresponding constraints are:
X11+X21+X31=1500 (requirements of distributor 1)
Take note that the demand of the distributors is equal to the total capacity of Cement
Exchange plants. Therefore, the supply constraints must indicate that the goods shipped will
not exceed the plant capacity. To mathematically formulate the constraints, we have:
X11+X12+X13+X14=3000 (capacity of plant 1)

1. Each row in the transportation table
contains the source of supply, while
each column is for the demand or
destination point.
2. The total available supply is written
on the right side of the table’s main
body.
3. The total requirements of each
destination are given in the lowest row
of the main body of the table.
4. Cells in the main body of the table
should correspond to the decision
variables.
5. At the corner of each table cell, the
“boxed-in” value contains the objective
function coefficient of the decision
variable.
STRUCTURE OF THE
TRANSPORTATION PROBLEM
Supply
Plant 1
Plant 2
Plant 3
Distributors
1 2 3 4
Availability
15 18 22 26
21 25 16 23
14 19 20 24
Requirements
X11 X12 X13 X14
X21 X22 X23 X24
X31 X32 X33 X34
Table 7.1
1,500 2,000 2,400 3,500
3,000
2,700
3,700
9,400

TRANSPORTATION
ALGORITHM
STEP 1:
Arrange the problem in
the transportation table.
STEP 2:
Obtain initial feasible
solution.
STEP 3:
Is the solution
optimal?
STEP 4:
Determine new
solution.
FINAL SOLUTION
Yes
No

NORTHWEST CORNER METHOD
IS THE MOST SIMPLE AND LOGICAL METHOD
OF FINDING THE INITIAL SOLUTION. THE
PROCEDURE BEGINS BY ALLOCATING UNITS
TO THE UPPER LEFT HAND CORNER AND
ENDS IN THE LOWER RIGHT CORNER OF THE
TRANSPORTATION PROBLEM
Lesson 7.4- OBTAINING INITIAL SOLUTION

MINIMUM COST METHOD
IS SYSTEMATIZED PROCEDURE THAT IS EASY TO
USE AND YIELDS AN INITIAL SOLUTION THAT IS
CLOSE TO THE OPTIMAL SOLUTION IN SMALL
PROBLEMS. IN THIS METHOD, THE ALLOCATION IS
MADE TO THE CELL WITH THE MOST POSSIBLE
LOWESR COST OR THE RIGHT HIGHEST PROFIT IN
A MAXIMIZATION CASE.

VOGEL'S APPROXIMATION METHOD
IS AN ALGORITHM THAT OBTAINS THE INITIAL
FEASIBLE SOLUTION BY DETERMINING "PENALTY
CAUSE" OF NOT USING THE LOWEST COST
ROUTE. THIS METHOD RESULTS IS AN OPTIMAL
OR NEAR OPTIMAL AS AN INITIAL SOLUTION

STEP
3
Column 1 has been removed
from further consideration.
Repeat Step 1

STEP
3
Column 3 has been removed.
Repeat Step 1

STEP
3
Row 2 has been removed.
Repeat Step 1

STEPPING STONE METHOD
IS GENERALLY THE EASIEST TO
VISUALIZE AND UNDERSTAND. ITS
PURPOSE IS TO DETERMINE THE EFFECR
ON THE TOTAL SHIPPING COST SHOULD
ONE UNIT OF GOODS ARE TO BE
DELIVERED THROUGH THE UNUSED
ROUTES.
LESSON 7.5 OPTIMAL FEASIBLE SOLUTION

STEP
2
There are two negative contribution
values (cell 2-1 and cell 2-4). This
means we will continue to Step 3

STEP
3
The new route will be at cell 2-4 as
it has the lowest net contribution
value.

MODIFIED DISTRIBUTION METHOD
IS A MORE EFFICIENT PROCEDURE IN
DETERMINING THE NET CONTRIBUTION OF
ROUTES THAT WERE NOT USED. IN THIS METHOD,
FOCUS IS ON THE EXTENSIVE USE OF THE COST
FACTORS ASSOCIATED WITH EACH CELL.

DISTRIBUTION
MODEL (PART 2)
Chapter 8

- The assignment model was developed and published in 1955 by
Harold Kuhn, an American Mathematician.
- This model is another type of transportation problem.
- The objective is to assign a number of origins to the equal
number of destinations at either minimum cost or maximum
profit.

Business situations that can utilized the
assignment model are the following:
(a). assigning machines to produce the purchase orders;
(b). assigning manpower to sales territories;
(c). asssigning contracts to bidders through a methodical bid selection process;
(d). asigning teaching loads to teachers;
(e). assigning account executives to different clientele; and so on.

Example:
The management of Metro
Utilities wants to assign three
service teams to Luzon,
Visayas, and Mindanao areas.
Each team has a certain
degree of familiarity with the
three geographical areas that
affect the team's level of
efficiency up as reflected in the
service cost. The management
of Metro Utilities wants to
come with an efficient
assignment schedule that will
minimize the total cost.
SERVICE
TEAMS
LUZON VISAYAS MINDANAO
Service
Team 1
20 15 30
Service
Team 2
16 13 33
Service
Team 3
18 19 27

Characteristics of the assignment
problem:
The problem of Metro Utilities is typical of a management situation with the following
characteristics:
1. The entities under consideration, such as service teams, jobs, employees,
account personnel, contracts, and projects, are finite in number.
2. The entities are assigned on a one-to-one basis to other objects.
3. The results of each assignment can be expressed as profits, payoffs, or costs.
4. The ultimate goal is to assign all entities in such a way that the total benefit is
maximized or minimized for cases involving total cost.

PRESENTATION OF THE ASSIGNMENT PROBLEM
Table 1
(METRO UTILITIES ASSIGNMENT PROBLEM)
SERVICE TEAMS LUZON
(z1)
VIASAYAS
(z2)
MINDANAO
(z3)
SUPPLY
Service Team 1
(s1)
20 15 30 1
Service Team 2
(s2)
16 13 33 1
Service Team 3
(s3)
18 19 27 1
DEMAND 1 1 1

Methods for Solving the Assignment Problem
• Complete Enumeration - The assignment problem is basically presented as a
balance problem with the number n of items equal to the number n of entities. Thus
there are n! (n factorial) different solutions to a given assignment problem. One way
of solving assignment problem is by comparing all possible solutions as presented in
Table 1. However, this method will not be practical when the number of possible
solutions is already unmanageable.
• Simplex Method - The simplex method can be used but not very efficient for solving
assignment problem.
• Transportation Method - Assignment problems can also be solved using the
transportation method. However, there are more efficient methods.
• Hungarian Method - The Hungarian method is the most efficient way of solving
largeor more complex assignment problems.

ASSIGNMENT ALTERNATIVE SOLUTION USING
COMPLETE ENUMERATION
Table 2
ALTERNATIVE
SOLUTIONS
COMBINATION TOTAL COST
1 S₁Z₁ S₂Z₂ S₃Z₃ 20 + 13 + 27 = 60
2 S₁Z₁ S₃Z₂ S₂Z₃ 20 + 19 + 33 = 72
3 S₂Z₁ S₁Z₂ S₃Z₃ 16 + 15 + 27 = 58 ← minimum
4 S₂Z₁ S₃Z₂ S₁Z₃ 16 + 19 + 30 = 65
5 S₃Z₁ S₂Z₂ S₁Z₃ 18 + 13 + 30 = 61
6 S₃Z₁ S₁Z₂ S₂Z₃ 18 + 15 + 33 = 66

The Hungarian Method
The Hungarian method or Flood's technique is an algorithm that provides an efficient
solution procedure for solving large, balanced assignment problems. It was
developed by Dénes König, a Hungarian mathematician and is based on the concept
of opportunity loss.
The procedure was based on the following theorem:
If one subtracts (or adds) a constant number from all entries in any row or column
of the assignment matrix, then the total cost of each of the n! possible
assignments is reduced (or increased) by the constant number subtracted (or
added).

In other words, additions or subtractions to rows or columns are permissibles without changing the
ultimate optimal assignment. The total costs are changed but the relative ones remain the same. The
procedure involves four major step as shown in table 1.
Step 1. Construct “total
opportunity cost” matrix.
Step 2.
Evaluate if
optimal
assignment can
be achieved.
Step 3. Improve
the matrix.
Step 4. Establish the
final (optimal)
assignment.
↓
↓ No
←
Yes
←

● Step 1. Construct the "total opportunity cost" matrix. The first step concerns the
transformation of the cost matrix to a "total opportunity cost" matrix. Two operations
have to be done first:
First, the entry with the smallest value in each row is subtracted from all the other
elements in the same row. This step will eliminate the negative numbers. For example,
in the first row, the lowest element is 15. The first operation is illustrated below.
20 - 15 = 5 15 - 15 = 0 30 - 15 = 15
16 - 13 = 3 13 - 13 = 0 33 - 13 = 20
18 - 18 = 0 19 - 18 = 1 27 - 18 = 9
5 0 15
3 0 20
0 1 9
↑
New Matrix

● Second, the lowest entry in each column of the new matrix is subtracted from all elements in
the same column. The result is the "total opportunity cost" matrix as shown in Table 3.
5 - 0 = 5 0 - 0 = 0 15 - 9 = 6
3 - 0 = 3 0 - 0 = 0 20 - 9 = 11
0 - 0 = 0 1 - 0 = 1 9 - 9 = 0
Table 3
(TOTAL OPPORTUNITY COST)
5 0 6
3 0 11
0 1 0

Step 2. Evaluate optimality. All values in the "total opportunity cost" matrix
is non- negative, and the minimum possible cost will be at zero. Therefore,
a feasible assignment with a total opportunity cost value of zero is already
considered optimal. In testing for optimality, the matrix table must contain
enough zeros for each row and column. A simple procedure to test this is
by drawing straight lines, horizontally or vertically, to cover all zeros in the
matrix. Draw lines that will cover two or more zeros in a single line. To
illustrate this procedure, see the table below.
Z₁ Z₂ Z₃
S₁ 5 0 6
S₂ 3 0 11
S₃ 0 1 0

Step 3. Improve the "total opportunity cost" matrix. Step 3 requires three operations:
First, find the entry with the smallest value in the uncovered cells or cells without
a line. Then, subtract this entry from all entries in the uncovered cells. The smallest
entry in the "total opportunity cost" is 3.
Second, add the lowest entry to cells in which lines intersect. In our case, there
is only one, and that cell contains "1."
Third, cells that are with a single line through them, such as S₃ and Z₂, cells, to
be transferred with no change to the improved matrix table.
Z₁ Z₂ Z₃
S₁ 5 - 1 = 4 0 6 - 1 = 5
S₂ 3 - 1 = 2 0 11 - 1 = 10
S₃ 0 1 + 1 = 2 0
First improved total
opportunity cost
←

• Repeat step 2 to test optimality
Z₁ Z₂ Z₃
S₁ 4 0 5
S₂ 2 0 10
S₃ 0 2 0
• Optimality is not yet reached since there are only two lines. Thus, we repeat
step 3.
Z₁ Z₂ Z₃
S₁ 4 - 2 = 2 0 5 - 2 = 3
S₂ 2 - 2 = 0 0 10 - 2 = 8
S₃ 0 2 + 2 = 4 0
Second
improved total
opportuity cost
←

Table 4
OPTIMALITY TEST
Z₁ Z₂ Z₃
S₁ 2 0 3
S₂ 0 0 8
S₃ 0 2 0
Step 4. Establish an optimal assignment. The optimal assignment is
made using the values to the cell with a zero entry. The one-to-one
entry requirement must be maintained. If one solution is found, drop"
the row or column from the matrix and repeat the procedure. If there
will be two or more zeros in a row or column, then choose arbitrarily.

• The optimal assignment for Metro Utilities based on Table 4 are the cells with squares: S₂Z₁, S₁Z₂
and S₃Z₃ The values, therefore, are shown in the table below:
Table 5
(OPTIMAL ASSIGNMENT FOR METRO UTILITIES)
SERVICE TEAMS LUZON
(Z₁)
VIASAYAS
(Z₂)
MINDANAO
(Z₃)
SUPPLY
Service Team 1
(S₁)
20 15 30 1
Service Team 2
(S₂)
16 13 33 1
Service Team 3
(S₃)
18 19 27 1
DEMAND 1 1 1
Table 5 shows that the total cost is computed as 16 + 15 + 27 = 58 (thousand pesos) should there be more
than one optimal solution, a trial-and-error approach can be used to find the optimal solution.

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