DATE
02/24/2023
MANAGEMENT SCIENCE
TRANSPORTATION, TRANSSHIPMENT
AND ASSIGNMENT MODEL
NAME OF GROUP PRESENTED BY
FSUU BLUE DRAGONS GROUP 1
INTRODUCTION:
In this chapter, we examine three special types of linear programming model formulations
transportation, transshipment, and assignment problems. They are part of a larger class of
linear programming problems known as network flow problems. We are considering these
problems in a separate chapter because they represent a popular group of linear
programming applications.
THE TRANSPORTATION MODEL
-is formulated for a class of problems with the following unique
characteristics: (1) A product is transported from a number of sources to a number of destinations
at the minimum possible cost; and (2) each source is able to supply a fixed number of
units of the product, and each destination has a fixed demand for the product. Although the
general transportation model can be applied to a wide variety of problems, it is
In a transportation
problem, items are
allocated from
sources to destinations at
a minimum cost.
TWO TYPES TRANSPORTATION
MODEL PROBLEM
A BALANCED UNBALANCED
TRANSPORTATION MODEL TRANSPORTATION MODEL,
IN WHICH SUPPLY SUPPLY IS GREATER THAN
EQUALS DEMAND, ALL DEMAND OR DEMAND IS
CONSTRAINTS ARE GREATER THAN SUPPLY
EQUALITIES.
SUPPLY = DEMAND SUPPLY ≠ DEMAND
TWO(2) PHASES OF SOLUTION
TRANSPORTATION PROBLEM:
1) PHASE 1: INITIAL BASIC FEASIBLE SOLUTION
3 METHODS :
A) NORTH- WEST CORNER RULE (NWCR)
B) LEAST COST METHOD (OR GREEDY METHOD)
C) VOGELS APPROXIMATION METHOD
2) PHASE 2: OPTIMAL BASIC SOLUTION
2 METHODS:
A) STEPPING STONE METHOD
B) MODIFIED DISTRIBUTION
METHOD (MODI)
1) PHASE 1: INITIAL BASIC FEASIBLE SOLUTION
NORTH- WEST CORNER RULE (NWCR)
-is a procedure for obtaining an initial feasible solution to a transportation
problem that starts with allocating units to the upper left-hand corner of any
transportation problem.
Steps:
1) Start in upper left-hand corner (source 1, destination A) and allocate as many units
as possible to satisfy demand at destination A.
2) Reduce the available supply at the current origin and unsatisfied demand at
the current destination by the amount of allocation.
3) Identify the first source with available supply. This is either the current source of the
one directly below it.
4)Identify the first destination with unsatisfied demand: This is either the current
destination or the one immediately to the right of it.
5) Allocate, as in Step 1, as many item as possible to the route associated with the
origin destination destination identified in Steps 3&4
6)Return to step 2
VOGELS APPROXIMATION METHOD (VAMS)
-is an algorithm that finds an initial feasible solution to a transportation problem by
considering the penalty cost of not using the cheapest available rate
Steps:
1) For each row and column in the transportation table identify the smallest and next to
smallest cost. The difference in the cost values is called. “PENALTIES”
2) Write those penalties in the right/bottom side of the transportation matrix.
3) Then identify the largest penalty among row/column penalties.
4) Select the largest penalty. Then select the lowest cost value correspond to that
row/column.
DO THE ALLOCATION
EXAMPLE:
PROBLEM:
Company A, B, and C have a total supply of cement amounting to 650 pcs. Each company
has a supply of 250, 250, and 150, respectively. These companies shipped and supplied
cement to the 4 factory in 4 different regions in the Philippines.
FACTORY 1
COMAPNY A
FACTORY 2
COMAPNY B
FACTORY 3
COMAPNY C
FACTORY 3
GIVEN TABLE:
GIVEN
TRANSPORTATION
COST FROM EACH
COMPANY TO EACH
FACTORY.
EACH FACTORY
DEMANDS THE
SUPPLY = DEMAND FOLLOWING
TOTAL OF 650 PCS OF CEMENT FOR SUPPLY NUMBER OF
AND DEMAND. CEMENT PER
MONTH:
TRANPORTATION PROBLEM MATRIX
DE
ST
IN
OR
IG
AT
IO
FACTORY 1 FACTORY 2 FACTORY 3 FACTORY 4 SUPPLY
IN N
COMPANY A 40 20 60 70 250
COMPANY B 20 30 15 80 250
COMPANY C 90 25 35 50 150
DEMAND 150 200 100 200 650 PCS
TRANPORTATION PROBLEM MATRIX
DE
ST
IN
OR
IG
AT
IO
FACTORY 1 FACTORY 2 FACTORY 3 FACTORY 4 SUPPLY
IN N
COMPANY A 40 20 60 70 250
COMPANY B 20 30 15 80 250
COMPANY C 90 25 35 50 150
DEMAND 150 200 100 200 650 PCS
WHERE MIN Z = 40XA1 + 20XA2 + 60XA3 + 70XA4 + 20XB1 + 30XB2 + 15XB3 + 80XB4 + 90XC1 +
25XC2 + 35XC3 + 50XC4
SUBJECT TO
XA1 + XA2 + XA3 + XA4 = 250 XA1 + XB1 + XC1 <= 150
XB1 + XB2 + XB3 + XB4 = 250 XA2 + XB2 + XC2 <= 200
XA3 + XB3 + XC3 <= 100
XIJ >= 0
XC1 + XC2 + XC3 + XC4 =150
XA4 + XB4 +XC4 <= 200
THE TRANSSHIPMENT MODEL
The transshipment model is an extension of the transportation model in which
intermediate transshipment points are added between the sources and
destinations. An example of a transshipment point is a distribution center or
warehouse located between plants and stores. In a transshipment
problem, items may be transported from sources through transshipment points on
to destinations, from one source to another, from one transshipment point to
another, from one destination to another, or directly from sources to destinations,
or some combination of these alternatives.
The transshipment
model includes
intermediate points
between sources and
destinations
TRANSSHIPMENT NETWORK
TRANSSHIPMENT
ALLOWS FOR THE
EFFICIENT TRANSFER OF
CARGO FROM LARGE
VESSELS TO SMALLER
ONES, ENSURING THAT
GOODS REACH THEIR
FINAL DESTINATION IN A
TIMELY AND COST-
EFFECTIVE MANNER.
NO DIRECT ROUTE:
DEPENDING ON THE
FINAL DESTINATION, A
SHIPMENT MAY BE
REQUIRED TO BE
SHIPPED BY TWO
MODES OF TRANSPORT.
TRANSSHIPMENT EXAMPLE
THE ASSIGNMENT MODEL
Is a special form of a linear programming model that is similar to the
transportation model. There are differences, however. In the
assignment model, the supply at each source and the demand at each
destination are each limited to one unit.
An assignment model
1) WHERE SUPPLY I = DEMAND J = 1
is for a special form
2) EACH DECISION VARIABLE IS A
of transportation
BINARY DECISION VARAIBLE (0,1)
problem in which all
3) ALL THE CONSTRAINTS ARE IN
supply and demand
FORM OF EQUATION (WITH “=”
values equal one.
SIGNS)
ASSIGNMENT
PROBLEM IS BALANCE
IS IT HAS AN EQUAL
ROW AND COLUMN JOB
1 2 3 4
PERSON
The assignment method is used A 10 12 19 11
to determine what resources are
assigned to which department, B 5 10 7 8
machine, or center of operation
C 12 14 13 11
in the production process. The
goal is to assign resources in
D 8 12 11 9
such a way to enhance
production efficiency, control
costs, and maximize profits.
