The document discusses linear programming models and the transportation problem. It defines linear programming as optimally allocating limited resources under certain assumptions. The transportation problem involves shipping goods from multiple origins to destinations in a way that minimizes costs. Basic concepts are defined, including feasible and basic feasible solutions. Methods for solving transportation problems are described, such as the North-West Corner Rule, Lowest Cost Entry Method, and Vogel's Approximation Method. Degeneracy in transportation problems and moving toward optimality are also covered.

1. QUANTITATIVE TECHNIQUES FOR
MANAGERS
UNIT-II RMB 207
DR. ZIAUL HASSAN BAKHSHI

2. Linear Programming Models
A model, which is used for optimum allocation of scarce or
limited resources to competing products or activities under
such assumptions as certainty, linearity, fixed technology, and
constant profit per unit, is linear programming.

3. PROPERTIES OF LINEAR PROGRAMMING MODEL
(a)The relationship between variables and
constraints must be linear.
(b) The model must have an objective function.
(c)The model must have structural constraints.
(d) The model must have non-negativity constraint.

4. Transportation Problem
The Transportation problem is to transport various amounts of
a single homogeneous commodity that are initially stored at various
origins, to different destinations in such a way that the total
transportation cost is a minimum.
It can also be defined as to ship goods from various origins to
various destinations in such a manner that the transportation cost is a
minimum
12/25/2016

5. Basic Definitions
• Feasible Solution
A set of non-negative individual allocations (xij ≥ 0) which
simultaneously removes deficiencies is called as feasible solution.
• Basic Feasible Solution
A feasible solution to ‘m’ origin, ‘n’ destination problem is said to be
basic if the number of positive allocations are m+n-1. If the number of
allocations is less than m+n-1 then it is called as Degenerate Basic
Feasible Solution. Otherwise it is called as Non-Degenerate Basic
Feasible Solution.
• Optimum Solution
A feasible solution is said to be optimal if it minimizes the total
transportation cost.
12/25/2016

6. Simple Network Representation
1
2
m
1
2
n
Sources Destinations
…
…
Supply s1
Supply s2
Supply sm
Demand d1
Demand d2
Demand dn
xij
Costs cij
12/25/2016

7. The Transportation Simplex Tableau
Destination
Supply ui
Source 1 2 … n
1
c11 c12
…
c1n
s1
2
c21 c22
…
c2n
s2
… … … … … …
m
cm1 cm2
…
cmn
sm
Demand d1 d2 … dn
Z =
vj
12/25/2016

8. Feasible Solutions
A transportation problem will have feasible
solutions if and only if
 

n
j
j
m
i
i ds
11

9. North-West Corner Rule
Step 1
The first assignment is made in the cell occupying the upper left-hand (north-west) corner of the
table.
The maximum possible amount is allocated here i.e. x11 = min (a1, b1). This value of x11 is then
entered in the cell (1,1) of the transportation table.
Step 2
If b1> a1, move vertically downwards to the second row and make the second allocation of
amount x21 = min (a2, b1- x11) in the cell (2, 1).
If b1< a1, move horizontally right side to the second column and make the second allocation
of amount x12 = min (a1 - x11, b2) in the cell (1, 2).
If b1 = a1, there is tie for the second allocation. One can make a second allocation of
magnitude x12 = min (a1 - a1, b2) in the cell (1, 2) or x21 = min (a2, b1 - b1) in the cell (2, 1)
Step 3
Start from the new north-west corner of the transportation table and repeat steps 1 and 2 until
all the requirements are satisfied.

10. Lowest Cost Entry Method (Matrix Minima Method)
Step 1
Determine the smallest cost in the cost matrix of the transportation table.
Allocate xij = min (ai, bj) in the cell (i, j)
Step 2
If xij = ai, cross out the ith row of the table and decrease bj by ai. Go to step
3.
If xij = bj, cross out the jth column of the table and decrease ai by bj. Go to
step 3.
If xij = ai= bj, cross out the ith row or jth column but not both.
Step 3
Repeat steps 1 and 2 for the resulting reduced transportation table until
all the requirements are satisfied. Whenever the minimum cost is not
unique, make an arbitrary choice among the minima.
12/25/2016

11. Vogel’s Approximation Method (Unit Cost Penalty Method)
Step1
For each row of the table, identify the smallest and the next to smallest
cost. Determine the difference between them for each row. These are
called penalties. Put them aside by enclosing them in the parenthesis
against the respective rows. Similarly compute penalties for each column.
Step 2
Identify the row or column with the largest penalty. If a tie occurs then use
an arbitrary choice. Let the largest penalty corresponding to the ith row
have the cost cij. Allocate the largest possible amount xij = min (ai, bj) in the
cell (i, j) and cross out either ith row or jth column in the usual manner.
Step 3
Again compute the row and column penalties for the reduced table and
then go to step 2. Repeat the procedure until all the requirements are
satisfied.
12/25/2016

12. MOVING TOWARDS OPTIMALITY
(1) DETERMINE THE NET-EVALUATIONS ( U-V METHOD)
Since the net evaluation is zero for all basic cells, it follows that
zij - cij = ui +vj - cij ,
for all basic cells (i, j). So we can make use of this relation to find the
values of ui and vj . Using the relation ui +vj = cij , for all i and j which (i,j)
is a basic cell, we can obtain the values of ui `s and vj `s. After getting the
values of ui `s and vj `s, we can compute the net-evaluation for each non-
basic cell and display them in parenthesis in the respective cells.
12/25/2016

13. MOVING TOWARDS OPTIMALITY
(2) SELECTION OF THE ELECTING VARIABLES
Choose the variable xrs to enter the basis for which the net evaluation
zrs - crs = max { zij - cij > 0} .
After identifying the entering variable xrs , form a loop which starts at the non-basic
cell (r,s) connecting only basic cells . Such a closed path exists and is unique for any non-
degenerate solution. Allocate a quantity  alternately to the cells of the loop starting + to the
entering cell. The value of  is the minimum value of allocations in the cells having -.
Now compute the net-evaluation for new transportation table and continue the above
process till all the net-evaluations are positive for non-basic cells.
12/25/2016

14. DEGENERACY IN TRANSPORTATION PROBLEM
Transportation with m-origins and n-destinations can
have m+n-1 positive basic variables, otherwise the basic
solution degenerates. So whenever the number of basic
cells is less than m + n-1, the transportation problem is
degenerate.
To resolve the degeneracy, the positive variables are
augmented by as many zero-valued variables as is
necessary to complete m +n –1 basic variables.
12/25/2016

15. W→
F
↓
W1 W2 W3 W4
Factory
Capacity
F1 19 30 50 10 7
F2 70 30 40 60 9
F3 40 8 70 20 18
Warehouse
Requirement
5 8 7 14 34
Find the initial basic feasible solution by using North-West
Corner Rule , Matrix Minima Method and Vogel’s
Approximation Method
12/25/2016