2. The Big-M Method is a technique, which is used in
removing artificial variables from the basis.
Assign coefficients to artificial variables, undesirable
from the objective function point of view.
If objective function Z is to be minimized, then a
very large positive price (called penalty) is assigned
to each artificial variable.
Similarly, if Z is to be maximized, then a very large
negative cost (also called penalty) is assigned to each
of these variables.
Following are the characteristics of Big-M Method
3. i. High penalty cost (or profit) is assumed as M
ii. M as a coefficient is assigned to artificial variable
A in the objective function Z(+M in
minimization and –M in maximization case).
iii. Coefficient of S (slack/surplus) takes zero values
in the objective function Z
iv. For minimization problem, the incoming variable
corresponds to the highest negative value of Cj-
Zj.
v. The solution is optimal when there is no negative
value bof Cj-Zj.
4. 1) Formulate the LP model, and express the in the standard form by introducing
surplus and artificial variables in the left hand side of the constraints.
Assign a 0 (zero) and +M as coefficient for surplus and
artificial variables respectively in the objective function.
M is considered a very large number so as to finally drive out the artificial
variables out of basic solution
2) Set up an initial solution.
Just to initiate the solution procedure, the initial basic
feasible solution is obtained by assigning zero value to decision variables.
3) Test for optimality of the solution.
If all the entries of Cj - Zj, row are positive, then the solution is optimum.
The solution can be further improved by removing one basic variable from
the basis and replacing it by some non-basic one.
Steps involved in using simplex method for
minimization problems
5. 4) Determine the variable to enter the basic solution
Identify the column with the largest negative value in the Cj - Zj
row of the table.
The entering variable is the one with the largest negative value
in the Cj-Zj row
The Leaving variable is the one with the smallest non negative
ratio.
5) Update the new solution
The optimal solution is obtained when the Cj-Zj row contains
entirely zeros and positive values.
6) Repeated Step (3—5) until an optimum solution is
obtained.
6. Min Z = 7X1+9X2
Subject to 3X1+6X2 >= 36
8X1+4X2 > = 64
X1, X2 > = 0
Solution
Step-1: write LP model in standard form
Min Z = 7X1+9X2+0S1+0S2+MA1+MA2
Subject to 3X1+6X2 –S1+A1 = 36
8X1+4X2-S2+A2 = 64
X1, X2 > = 0
Step-2: Construct tableaus for this problem are shown below.
Example
8. From above table largest negative value corresponding to X1 and 8
is pivot element. Do same thing similar to maximization case.
Third Simplex Tableau
The third tableau represents a final tableau since it is the optimal
solution with entirely zeros and non-negative values in the Cj-Zj
row.
Therefore, the optimal solution is: X1 = 20/3 and X2 = 8/3 and
value of objective function is 212/3.
11. 1) Maximize the revised objective function i.e multiply it by
negative sign if its positive and vice versa.
E.g Min 12.5X1 + 14.5X2 to -12.5X1 - 14.5X2 and then
maximize this objective function
2) Convert the inequalities into equality by introducing surplus
variables.
3) Introduce more additional variables a6 and a7 called as
artificial variable to facilitate the calculation of an initial
basic feasible solution.
In this method the calculation is carried out in two phases
hence tow phase method.
Step For Two phase method
12. Phase-I
In this phase we will consider the following linear programming
problem.
Max: A1-A2
Subject to: X1 + X2 –S1 +A1 = 2000
40X1 + 75X2 –S2 + A2 = 100000
75X1 + 100X2 +S3 = 200000
X1, X2, S1,S2,S3,A1,A2 ≥ 0
The initial basic feasible solution of the problem is A1 = 2000,
A2=100000 and S3 = 200000.
As the minimum value of the objective function of the Phase I
is zero at the end of the Phase I calculation both A1 and A2
become zero.
13. Cj 0 0 0 0 -1 -1
CB BS XB X1 X2 S1 S2 S3 A1 A2
-1 A1 2000 1 1 1 0 0 1 0
-1 A2 100000 40 75 0 -1 0 0 1
0 S3 200000 75 100 0 0 1 0 0
Zj-Cj -41 -71 1 1 0 0 0
Here X2 becomes a basic variable and A2 becomes non basic variable in the next
iteration. It is no longer considered for re-entry in the table.
Cj 0 0 0 0 -1
CB BS XB X1 X2 S1 S2 S3 A1
-1 A1 2000/3 7/15 0 -1 1/75 0 1
0 X2 4000/3 8/15 1 0 -1/75 0 0
0 S3 200000/3 65/3 0 0 4/3 1 0
Zj-Cj -1/15 0 1 -1/75 0 0
14. Then X1 becomes a basic variable and A1 becomes a non
basic variable in the next iteration.
The calculation of Phase I end at this stage.
Note that, both the artificial variable have been removed
and also found a basic feasible solution of the problem.
The basic feasible solution is: X1 = 10000/7, X2 = 4000/2, S3= 250000/7.
Cj 0 0 0 0
CB BS XB X1 X2 S1 S2 S3
0 X1 10000/7 1 0 -15/7 1/35 0
0 X2 4000/7 0 1 8/7 -1/35 0
0 S3 250000/7 0 0 325/7 16/21 1
Zj-Cj 0 0 0 0 0
15. Phase II
The initial basic feasible solution obtained at the end of the Phase I
calculation is used as the initial basic feasible solution of the
problem.
In this Phase II calculation the original objective function is
introduced and the usual simplex procedure is applied to solve the
linear programming problem.
All zj-cj ≥ 0 the current solution maximizes the revised objective
function. Thus, the solution of the problem is: X1 = 10000/7 = 1428
and X2 = 4000/7 = 571.4 and Minimum Value of the objective
function is: 26135.3
Cj -12.5 -14.5 0 0 0
CB BS XB X1 X2 S1 S2 S3
-12.5 X1 10000/7 1 0 -15/7 1/35 0
-14.5 X2 4000/7 0 1 8/7 -1/35 0
0 S3 250000/7 0 0 325/7 5/7 1
Zj-Cj 0 0 143/14 2/35 0
16. 1. Non-feasible Solution/ Infeasibility
Optimality criteria is satisfied but still there exist an artificial
variable in the basis or solution mix, this is the indication of
infeasibility.
2. Unbounded Solution
The row with the smallest positive ratio is replaced. But if
the entire ratios turn out to be negative or undefined, it
indicates that the problem is unbounded.
No outgoing variable will exist.
3. Multiple Optimum Solutions
This situation can be recognized in a simplex method when
one of the non-basic variables in the Cj-Zj, row have a
value of zero.
SPECIAL ISSUES
18. Introduction: For every linear programming problem there is a
corresponding linear programming problem called the dual.
If the original problem is a maximization problem then the
dual problem is minimization problem and vice versa.
The solution of the dual problem is readily obtained from the
original problem solution if the simplex method is used.
The formulation of the dual problem also sometimes referred
as the concept of duality is helpful for the understanding of the
linear programming.
The variable of the dual problem is known as the dual
variables or shadow price of the various resources.
DUALITY
19. The shadow price is also defined as the rate of change in the
optimal objective function value with the respect to the unit
change in the availability of a resource.
Dual Problem Formulation
If the original problem is in the standard form then
the dual problem can be formulated using the
following rules:
The number of constraints in the original problem is equal to
the number of dual variables.
The original problem profit coefficients appear on the right
hand side of the dual problem constraints.
If the original problem is a maximization problem then the dual
problem is a minimization problem.
Similarly, if the original problem is a minimization problem
then the dual problem is a maximization problem.
20. The original problem has less than or equal to (≤) type of
constraints while the dual problem has greater than or equal
to (≥) type constraints.
The coefficients of the constraints of the original problem
which appear from left to right are placed from top to
bottom in the constraints of the dual problem and vice
versa.
Example: Three machine shops A, B, C produces three types
of products X, Y, Z respectively. Each product involves
operation of each of the machine shops. The time required for
each operation on various products is given as follows:
21. The available hours at the machine shops A, B, C are 100, 77,
and 80 only. The profit per unit of products X, Y, and Z is $12,
$3, and $1 respectively.
Solution: The formulation of Linear Programming (original
problem) is as follows:
Max: 12x1 + 3x2 + x3
Subject to: 10x1 + 2x2 + x3 ≤ 100
7x1 + 3x2 + 2x3 ≤ 77
2x1 + 4x2 + x3 ≤ 80
x1, x2, x3 ≥ 0
Introduce slack variable:
Max: 12x1 + 3x2 + x3
Subject to: 10x1 + 2x2 + x3 +S1=100
7x1 + 3x2 + 2x3 +S2=77
2x1 + 4x2 + x3 +S3≤=80
x1, x2, x3 ≥ 0
22. Form the above equations, the first simplex table is obtained is
as follows:
Note that the basic variables are s1, s2 and s3. Therefore CB1
= 0, CB2 = 0, CB3 = 0.
The smallest negative element in the above table of z1 – c1 is
-12. Hence, x1 becomes a basic variable in the next iteration.
Determine the minimum ratios (100/10, 77/7, 80/2) =10, S1,
which is made as a non-basic variable.
Cj 12 3 1 0 0 0
CB BS XB X1 X2 X3 S1 S2 S3
0 S1 100 10 2 1 1 0 0
0 S2 77 7 3 2 0 1 0
0 S3 80 2 4 1 0 0 1
Zj-Cj -12 -3 -1 0 0 0
23. The next Table 2 is calculated using the following rules:
i. The revised basic variables are x1, s5, s6. Accordingly we
make CB1=22, CB2=0 and CB3=0.
ii. Since x1 is the incoming variable we make x1 coefficient
one by dividing each element of row 1 by 10. Thus the
numerical value of the element corresponding to x2 is
2/10, corresponding to x3 is 1/10, corresponding to S1 is
1/10, corresponding to S2 is 0/10 and corresponding to S3
is 0/10 in Table 2.
iii. The incoming basic variable should only appear in the
first row. So we multiply first row of Table 2 by 7 and
subtract if from the second row of Table 1 element by
element.
24. Cj 12 3 1 0 0 0
CB BS XB X1 X2 X3 S1 S2 S3
12 X1 10 1 2/10 1/10 1/10 0 0
0 S2 7 0 16/10 13/10 -7/10 1 0
0 S3 60 0 18/5 4/5 -7/5 0 1
Zj-Cj 0 -3/5 1/5 6/5 0 0
3. z2-c2 = -3/5. So X2 becomes a basic variable in the next iteration
4. Determine the minimum of the ratios, 70/16= third row, S2 will be a non
basic variable in the next iteration.
5. From Table 2, the Table 3 is calculated using the rules (i), (ii) and (iii)
mentioned above.
Cj 12 3 1 0 0 0
CB BS XB X1 X2 X3 S1 S2 S3
12 X1 73/8 1 0 -1/16 3/16 -1/8 0
3 X2 35/8 0 1 13/16 -7/16 5/8 0
0 S3 177/4 0 0 -17/8 11/8 -9/4 1
Zj-Cj 0 0 11/16 15/16 3/8 0
25. Since all the zi – cj ≥ 0, the optimum solution is as:
◦ x1 = 73/8 and x2 = 35/8 and
◦ The Maximum Profit is: $981/8 = $122.625
Suppose an investor is deciding to purchase the resources A, B, C.
What offers is he going to produce?
Let, assume that W1, W2 and W3 are the offers made per hour of machine
time A, B and C respectively.
Then these prices W1, W2 and W3 must satisfy the conditions given below:
1) W1, W2, W3 ≥ 0
2) Assume that the investor is behaving in a rational manner; he would try to
bargain as much as possible so that the total annual payable to the
produces would be as little as possible. This leads to the following
condition:
Minimize 100W1 + 77W2 + 80W3
3) The total amount offer by the investor to the three resources viz. A, B and
C required to produce one unit of each product must be at least as high as
the profit gained by the producer per unit.
26. Since, these resources enable the producer to earn the specified profit
corresponding to the product he would not like to sell it for anything
less assuming he is behaving rationally.
This leads to the following conditions:
10w1 + 7w2 + 2w3 ≥ 12
2w1 + 3w2 + 4w3 ≥ 3
w1 + 2w2 + w3 ≥ 1
Thus, in this case we have a linear problem to ascertain the values of the
variable w1, w2, w3. The variables w1, w2 and w3 are called as dual
variables.
Note:
The original (primal) problem illustrated in this example
a. considers the objective function maximization
b. contains ≤ type constraints
c. has non-negative constraints
This original problem is called as primal problem in the standard form.
27. Dual Problem Properties
a) If the original problem is in the standard form, then the
dual problem solution is obtained from the zj – cj values
of slack variables.
For example: In the above example the variables S1, S2 and S3 are
the slack variables. Hence the dual problem solution is w1 = z4 – c4
= 15/16, w2 = z5 – c5 = 3/8 and w3 = z6 – c6 = 0.
b) The original problem objective function maximum value is the
minimum value of the dual problem objective function.
For example: From the above Example we know that the original
problem maximum values is 981/8 = 122.625. So that the minimum
value of the dual problem objective function is 100*15/16 + 77*3/8 +
80*0 = 981/8
c) Shadow Price: A resource shadow price is its unit cost, which is
equal to the increase in profit to be realized by one additional
unit of the resource.
28. ◦ For example: Let the minimum objective function value is expressed
as: 100*15/16 + 77*3/8 + 80*0
◦ If the first resource is increased by one unit the maximum profit also
increases by 15/16, which is the first dual variable of the optimum
solution.
◦ Therefore, the dual variables are also referred as the resource shadow
price or imputed price.
d) In the originals problem, if the number of constraints and
variables is m and n then the constraint and variables in the
dual problem is n and m respectively.
e) Suppose, the original problem is not in a standard form, then
the dual problem structure is unchanged. However, if a
constraint is greater than or equal to type, the corresponding
dual variable is negative or zero. Similarly, if a constraint in
the original problem is equal to type, then the corresponding
dual variables is unrestricted in sign.
29. Consider the following linear programming problem
Maximize: 22x1 + 25x2 +19x3
Subject to: 18x1 + 26x2 + 22x3 ≤ 350
14x1 + 18x2 + 20x3 ≥180
17x1 + 19x2 + 18x3 = 205
x1, x2, x3 ≥ 0
Note that this is a primal or original problem.
The corresponding dual problem for this problem is as follows:
Minimize: 250w1 + 80w2 +105w3
Subject to: 18w1 + 14w2 + 17w3 ≥ 22
26w1 + 18w2 + 19w3 ≥ 25
22w1 + 20w2 + 18w3 ≥ 19
w1 ≥ 0, w2 ≤, and w3 is unrestricted in sign (+ or -).
Now, we can solve this using simplex method as usual.
Example
30. Step 1: Set up the P-matrix and its transpose
Step 2: Form the constraint and objective function for dual
Step 3: Construct the initial simplex tableau for the dual
Step 4: solve the problem by constructing simplex tableau
Example: Minimize P = x1 + 2x2
Subject to: x1 + x2 ≥ 8
2x1 + x2≥ 12
x1 ≥ 1
P= =
z = 8w1 +12w2 + w3
Subject to: w1+2w2+w31
2w1+w2
Simple Way of Solving Dual Problem
31. Sensitivity analysis is the study of sensitivity of the optimal
solution of an LP problem due to discrete variations (changes) in
its parameters.
The degree of sensitivity of the solution due to these changes can
range from no change to all to a substantial change in the
optimal solution of the given LP problem.
Sensitivity analysis carries the LP analysis beyond the
determination of the optimal solution and begins with the final
simplex tableau.
Its purpose is to explore how changes in any of the parameters of
a problem, such as the coefficients of the constraints, coefficients
of the objective function, or the right hand side values.
3.2 Sensitivity Analysis
32. The process of studying the sensitivity of the optimal solution
of an LP problem is also called post optimality analysis.
Because it is done after an optimal solution, assuming a given
set of parameters has been obtained for the model.
To demonstrate sensitivity analysis, the microcomputer
example will again be used.
Resolving a problem can be very time consuming and as it
will be demonstrated below, it is unnecessary.
33. 3.2.1 A Change in the RHS of a constraint
The first step in determining how a change in the RHS (Right
Hand Side) of a constraint would influence the optimal
solution is to examine the shadow prices in the final simplex
tableau.
The final tableau for the microcomputer is shown in the
following table since sensitivity analysis starts from final
tableau.
34. A shadow price is a marginal value; it indicates the impact that a
one-unit change in the amount of a constraint would have on the
values of the objective function.
What shadow prices do not tell us?
o The extents to which the level of a scarce resource can be changed
and still have the same impact per unit.
o The ability to use additional amounts of a resource will disappear at
some point because of the fixed amounts of the other constraints.
o Therefore, we need to determine the range of feasibility, or the right
hand side range.
o The key to computing the range of feasibility for the constraint lies
in each slack column of the final simplex tableau.
o To compute the range for each constraint, the entries in the
associated slack column must be divided into the values in the
Quantity column.
35. For example:
Here, the smallest positive ratio indicates how much the
constraint level can be decreased before it reaches the lower
limit of its range of feasibility.
The smallest negative ratio (i.e. the negative ratio closest to 0)
indicates how much the storage constraint can be increased
before it reaches its upper limit of feasibility.
The general rule applies when computing the upper and lower
limits on the range of feasibility for maximization problem is:
◦ Allowable decrease: The smallest positive ratio
◦ Allowable increase: Negative ratio closest to zero.
Note: The reverse rule will be true for minimization problems.
36. 3.3.2 A Change in an Objective Function Coefficient
It is useful to know how much the contribution of a
given decision variable to the objective function can
change without changing the optimal solution.
There are two cases:
Changes for a variable that is not currently in the solution mix
Changes for a variable that is currently in the solution mix.
If a variable is not currently in the solution in a maximization
problem, its objective function coefficient would have to
increase by an amount that exceeds the C-Z value.
Range of Insignificance: the range over which a non-basic
variable’s objective function coefficient can change without
causing that variable to enter the solution mix.
37. Maximize
Subject to
In the above tableau we see that is a non- basic variable.
To determine the range over which the objective function
coefficient of would change without changing the optimal
solution, recall how a variable in a max problem enters the
solution mix.
38. Range of Optimality: the range over which objective
function coefficient of a variable that is in solution can
change with out changing the optimal values of the
decision variables.
Note, however, that such a change would change the
optimal value of the objective function.
Divide the values in row C-Z by the corresponding row
values of the variable in question and follow the following
rule for both maximization and minimization problems.
Allowable increase: The smallest positive ratio of C-Z value
and the variable’s substitution rate.
Allowable decrease: the smallest negative ratio of C-Z value
and the variable’s leave rate.
39. Note: If there is no positive ratio, it means that there is no
upper limit on that variable’s objective function coefficient.
Example: Determine the range of optimality for the decision
variables in the microcomputer problem.
40. The smallest positive ration is +40.Therefore, the
coefficient of X1 can be increased by Birr 40 without
changing the optimal solution.
The upper end of its range of optimality is this amount
added to its current (original) value.
Upper end is (Birr60 +Birr40) = Birr100
The smallest negative ratio is -10; therefore, the x1
coefficient can be decreased by as much as Birr10 from
its current value, making the lower end of the range
equal to (Birr60 - Birr10 )= Birr 50.
41. The smallest positive ration is +10.
This tells us that the x2 coefficient in the objective function
could be increased by Birr10 to (Birr50 + Birr10) = Birr 60.
The smallest negative ratio is -20, which tells us that the x 2
coefficient could be decreased by Birr20 to (Birr50 - Birr20) =
Birr30.
42. Consider a change for c1. This will change the c1 value from
c1=60 to c1=60+ .
△
This new value is not only included on the top cj row, but also
in the left-hand cj column.
Since 60+ is in
△ lefthand column, it becomes a multiple of
the column values when the new Zj row values and the
subsequent Cj – Zj row values.
The solution becomes optimal as long as the Cj –Zj row
values remain negative.
If Cj-Zj becomes positive, the product mix will change, if it
becomes zero, there will be an alternative solution.
43. Now recall that c1=60+ ; therefore, = c1-60. Now
△ △
substituting c1-60 for in the above inequalities, we get:
△
Therefore, the range of values for C1 over which the solution
basis remain optimal (although) the value of the objective
function may change) is:
45. The objectives of Transportation Problem is to minimize the
cost of distributing a product from a number of sources (e.g.
factories) to a number of destinations (e.g. warehouses)
while satisfying both the supply limits and the demand
requirement.
The model assumes that the distributing cost on a given rout
is directly proportional to the number of units distributed on
that route.
Generally, the transportation model can be extended to areas
other than the direct transportation of a commodity,
including among others, inventory control, employment
scheduling, and personnel assignment.
4.1 Introduction
46. Example 4.1:
Suppose a manufacturing company owns three factories
(sources) and distribute his products to five different retail
agencies (destinations).
The following table shows the capacities of the three factories,
the quantity of products required by the various retail agencies
and the cost of shipping one unit of the product from each of
the three factories to each of the five retail agencies.
47. Usually the above table is referred as Transportation Table, which
provides the basic information regarding transportation problem.
The quantities inside the table are known as transportation cost per
unit of product.
A transportation problem can be formulated as LPP using variables
with two subscripts.
Let x11=Amount to be transported from factory 1 to retail agency 1
x12= Amount to be transported from factory 1 to retail agency
2 and so on
Let the transportation cost per unit be represented by C11, C12,
…..C35 that is C11=1, C12=9, and so on.
Let the capacities of the three factories be represented by a1=50,
a2=100, a3=150.
Let the requirement of the retail agencies are b1=100, b2=60,
b3=50, b4=50, and b5=40.
48. Thus, the problem can be formulated as:
Thus, the problem has 8 constraints and 15 variables.
So, it is not possible to solve such problem using simplex method. This is the
reason for the need of special computational procedure to solve transportation
problem.
49. 4.2 Transportation Algorithm
The steps of the transportation algorithm are exact parallels of
the simplex algorithm, they are:
Step 1: Determine a starting basic feasible solution, using any
one of the following three methods
1. North West Corner Method
2. Least Cost Method
3. Vogel Approximation Method
Step 2: Determine the optimal solution using the following
method
1. MODI (Modified Distribution Method) or UV Method.
50. Any basic feasible solution with m sources (such as factories) and n
destination (such as retail agency) has at most m + n -1 non-zero
Xij.
The special structure of the transportation problem allows securing a
non artificial basic feasible solution using one of three methods.
The difference among these three methods is the quality of the
initial basic feasible solution they produce, in the sense that a better
that a better initial solution yields a smaller objective value
Generally the Vogel Approximation Method produces the best initial
basic feasible solution, and the North West Corner Method produces
the worst, but the North West Corner Method involves least
computations.
4.3 Basic Feasible Solution of a Transportation
Problem
51. 1. North West Corner Method
The method starts at the North West (upper left) corner cell of the
tableau (variable x11).
Step -1: Allocate as much as possible to the selected cell, and adjust
the associated amounts of capacity (supply) and requirement
(demand) by subtracting the allocated amount.
Step -2: Cross out the row (column) with zero supply or demand to
indicate that no further assignments can be made in that row
(column).
If both the row and column becomes zero simultaneously, cross out
one of them only, and leave a zero supply or demand in the
uncrossed out row (column).
Step -3: If exactly one row (column) is left uncrossed out, then stop.
Otherwise, move to the cell to the right if a column has just been
crossed or the one below if a row has been crossed out. Go to step -1.
52. Consider the problem discussed in Example 4.1 to illustrate
the North West Corner Method of determining basic feasible
solution.
The arrows show the order in which the allocated (bolded)
amounts are generated. The starting basic solution is given
as
Retail agency
Factories 1 2 3 4 5 Capacity
1 50 50
2 50 50 100
3 0 10 50 50 40 150
Require
ment
100 60 50 50 40 300
1 9 13 36 51
24 16 20
12 1
14 33 1 23 26
53. The corresponding transportation cost is
It is clear that as soon as a value of Xij is determined, a row
(column) is eliminated from further consideration.
2. Least Cost Method
o The least cost method is also known as matrix minimum
method in the sense we look for the row and the column
corresponding to which Cij is minimum.
o This method finds a better initial basic feasible solution by
concentrating on the cheapest routes.
o We start by allocating as much as possible to the cell with the
smallest unit cost.
o If there are two or more minimum costs then we should select
the row and the column corresponding to the lower numbered
row.
54. o If they appear in the same row we should select the lower
numbered column.
o We then cross out the satisfied row or column, and adjust
the amounts of capacity and requirement accordingly
o If both a row and a column is satisfied simultaneously,
only one is crossed out.
o Next, we look for the uncrossed-out cell with the smallest
unit cost and repeat the process until we are left at the end
with exactly one uncrossed-out row or column.
55. The least cost method of determining initial basic feasible
solution.
So that the basic feasible solution developed by the Least Cost
Method has transportation cost is
Note that the minimum transportation cost obtained by the least
cost method is much lower than the corresponding cost of the
solution developed by using the north-west corner method
Retail agency
Factories 1 2 3 4 5 Capacity
1 50 1 9 13 36 51 50
2 24 60 12 16 20 40 1 100
3 50
14
33 50 1 50
23
26 150
Require
ment
100 60 50 50 40 300
56. VAM is an improved version of the least cost method that
generally produces better solutions.
The steps involved in this method are:
Step 1: For each row (column) with strictly positive capacity
(requirement), determine a penalty by subtracting the smallest
unit cost element in the row (column) from the next smallest
unit cost element in the same row (column).
Step 2: Identify the row or column with the largest penalty
among all the rows and columns.
If the penalties corresponding to two or more rows or columns
are equal we select the topmost row and the
extreme left column
Vogel Approximation Method (VAM)
57. Now, compute the penalty for various rows and columns which
is shown in the following table:
Origin
Destination
1
Destination
2
Destination
3
Destination
4
Supply (ai)
1 20 22 17 4 120
2 24 37 9 7 70
3 32 37 20 15 50
Demand
(bj)
60 40 30 110 240
Origin
Destinati
on 1
Destinatio
n 2
Destinatio
n 3
Destinati
on 4
Supply (ai)
1 20 22 17 4 120 17-4=13
2 24 37 9 7 70 9-7=2
3 32 37 20 15 50 20-15=5
Demand
(bj)
Row
penality
60
24-20=4
40
37-22=15
30
17-9=8
110
7-4= 3
240 Column
penality
58. Step 3: We select Xij as a basic variable if Cij is the minimum
cost in the row or column with largest penalty.
We choose the numerical value of Xij as high as possible subject
to the row and the column constraints. Depending upon whether
ai or bj is the smaller of the two ith row or jth column is crossed
out.
Step 4: The Step 2 is now performed on the uncrossed-out rows
and columns until all the basic variables have been satisfied.
Consider the following transportation problem
59. Look for the highest penalty in the row or column, the highest
penalty occurs in the second column and the minimum unit
cost i.e. cij in this column is c12=22.
Hence assign 40 to this cell i.e. x12=40 and cross out the
second column (since second column was satisfied).
Origin
Destinati
on 1
Destinati
on 2
Destinati
on 3
Destinati
on 4
Supply
(ai)
Column
penality
1 20 22 40 17 4
120 80
13
2 24 37 9 7 70 2
3 32 37 20 15 50 5
Demand
(bj)
60 40
30 110 240
Row
penality
4 15 8 3
60. The next highest penalty in the uncrossed-out rows and columns is
13 which occur in the first row and the minimum unit cost in this
row is c14=4, hence x14=80 and cross out the first row.
The next highest penalty in the uncrossed-out rows and columns is
8 which occurs in the third column and the minimum cost in this
column is c23=9, hence x23=30 and cross out the third column
with adjusted capacity, requirement and penalty values.
Origin
Destin
ation 1
Destinati
on 2
Destin
ation
3
Destin
ation 4
Supply
(ai)
Column
penality
1 20 22 40 17 4 80 120 0 13
2 24 37 9 7 70 2
3 32 37 20 15 50 5
Demand (bj) 60 40 30 110 30 240
Row penality 4 15 8 3
62. The next highest penalty in the uncrossed-out rows and
columns is 17 which occurs in the third row and the smallest
cost in this row is c31=32, hence xi31=50 and cross out the third
row or first column.
The transportation cost corresponding to this choice of basic
variables is
Origin
Destinati
on 1
Destinati
on 2
Destin
ation 3
Destin
ation 4
Supply
(ai)
Column
penality
1 20 22 40 17 4 80
120 0
13
2 24 10 37 9 30 7 30 70 0 17
3 32 50 37 20 15 50 0 17
Demand
(bj)
60 40 30 110 240
Row
penality
8 15 8 8
63. Step 2: Modified Distribution Method
The Modified Distribution Method, also known as MODI method or u-
v method, which provides a minimum cost solution (optimal solution)
to the transportation problem.
The following are the steps involved in this method.
Step 1: Find out the basic feasible solution of the transportation problem
using any one of the three methods discussed in the previous section.
Step 2: Introduce dual variables corresponding to the row constraints and
the column constraints.
If there are m origins and n destinations then there will be m+n dual
variables.
The dual variables corresponding to the row constraints are represented
by , i=1,2,…..m where as the dual variables corresponding to the
column constraints are represented by j=1,2,…..n.
The values of the dual variables are calculated from the equation given
below
64. Step 3: Any basic feasible solution has m + n -1 xij > 0. Thus, there
will be m + n -1 equation to determine m + n dual variables. One of
the dual variables can be chosen arbitrarily.
It is also to be noted that as the primal constraints are equations,
the dual variables are unrestricted in sign.
Step 4: If , the dual variables calculated in Step 3 are compared
with the cij values of this allocation as .
If all ≥ 0, then by the theorem of complementary slackness it can
be shown that the corresponding solution of the transportation
problem is optimum.
If one or more < 0, we select the cell with the least value of and
allocate as much as possible subject to the row and column
constraints.
The allocations of the number of adjacent cell are adjusted so that a
basic variable becomes non-basic.
Step 5: A fresh set of dual variables are calculated and repeat the
entire procedure from Step 1 to Step 5.
65. Step 1: Finding the basic feasible solution using three of
them for e.g using least cost method.
Step 2: The dual variables and can be calculated from
the corresponding cij , that is
Based on Example 4.2
Factdest 1 2 3 4 5 Supply
1 1 9 13 36 51 50
2 24 12 16 20 1 100
3 14 33 1 23 26 150
Demand 100 70 50 40 40 300
66. Step 3: Choose one of the dual variables arbitrarily is zero that
is as it occurs most often in the above equations.
The values of the variables calculated are
Step 4: Now we calculate values for all the cells where xij=0
(.e. unallocated cell by the basic feasible solution)
That is:
67. Note that in the above calculation all the except for cell
(1, 2) where = 9+13-33 = -11
Thus in the next iteration x12 will be a basic variable
changing one of the present basic variables non-basic.
We also observe that for allocating one unit in cell (1,
2) we have to reduce one unit in cells (3, 2) and (1, 1)
and increase one unit in cell (3, 1).
The net transportation cost for each unit of such
reallocation is -33 -1 + 9 +14 = -11
The maximum that can be allocated to cell (1, 2) is 10
otherwise the allocation in the cell (3, 2) will be
negative.
Thus, the revised basic feasible solution is
69. A flow of customers from finite or infinite population
towards the service facility forms a queue (waiting
line) an account of lack of capability to serve them all
at a time.
In general, the queueing system consists of one or
more queues and one or more servers and operates
under a set of procedures.
Depending upon the server status, the incoming
customer either waits at the queue or gets the turn to be
served.
Introduction:
70. Queueing System
A queueing system can be completely described by
1. The input (arrival pattern)
2. The service mechanism (service pattern)
3. The queue discipline and
4. Customer’s behaviour
1. The input (arrival pattern)
The input described the way in which the customers arrive and
join the system.
Generally, customers arrive in a more or less random manner
which is not possible for prediction.
The arrival pattern can be described in terms of probabilities and
consequently the probability distribution for inter-arrival times
(the time between two successive arrivals) must be defined.
We deal with those Queueing system in which the customers
arrive in poisson process.
The mean arrival rate is denoted by λ.
71. 2. The Service Mechanism
This means the arrangement of service facility to serve
customers.
If there is infinite number of servers, then all the
customers are served instantaneously or arrival and there
will be no queue.
If the number of servers is finite then the customers are
served according to a specific order with service time a
constant or a random variable.
Distribution of service time follows ‘Exponential
distribution’ defined by
The mean Service rate is E(t) = 1/λ
72. 3. Queueing Discipline
It is a rule according to which the customers
are selected for service when a queue has been
formed.
The most common disciplines are
1. First come first served – (FCFS)
2. First in first out – (FIFO)
3. Last in first out – (LIFO)
4. Selection for service in random order
(SIRO)
73. 4. Customer’s behaviour
1. Generally, it is assumed that the customers arrive into the
system one by one. But in some cases, customers may arrive in
groups.
Such arrival is called Bulk arrival.
2. If there is more than one queue, the customers from one
queue may be tempted to join another queue because of
its smaller size.
This behaviour of customers is known as jockeying.
3. If the queue length appears very large to a customer, he/she
may not join the queue.
This property is known as Balking of customers.
4. Sometimes, a customer who is already in a queue will leave
the queue in anticipation of longer waiting line.
This kind of departare is known as reneging.
