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TRANSPORTATION MODELS
Introduction
All linear programming problems can be solved by simplex method, but certain special problems
lend themselves to easy solution by other methods. One such case is that of Transportation
Problems. Such problems involve a large number of variables and constraints and hence take a
long time to be solved by normal simplex method.
Transportation problems are encountered in physical distribution of goods.
The matrix of a typical transportation problem is as follows:
Cost of transporting one unit of the product from Supply Points to
Demand Points
Availability in
units at Supply
Points
D1 D2 D3 D4
S1 C11 C12 C13 C14 A1
S2 C21 C22 C23 C24 A2
S3 C31 C32 C33 C34 A3
Requirement in
units at demand
points
R1 R2 R3 R4
In the above matrix, S1, S2 and S3 are the supply points each supplying A1, A2 and A3 units and
D1, D2, D3 and D4 are the demand points each requiring R1, R2 and R3 units respectively.
C11 is the cost of transporting one unit of the product from S1 to D1;
C21 is the cost of transporting one unit of the product from S2 to D1 etc.
In general Cij is the cost of transporting one unit of the product from the supply point i to the
demand point j.
The supply point is also called as the origin, factory etc.
The demand point is also called as the destination, godown, warehouse etc.
Decision variables:
Calculating different Xij s, i.e., the quantities to be transported from various supply points to
various decision points.

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Objective
In a transportation models, the objective is to minimize the total transportation cost of
transporting the products from various supply points to the demand points.
Constraints
(a) Each of the supply points has an available quantity “Ai”
(b) Each of the demand points has a required quantity “Rj”
(c) All Ai and Rjs have non-negative values.
Formulation as an LPP Problem
The above example can be formulated as a linear programming problem.
Decision Variables: Find the values of X11, X12, X13, X14, X 21, X 22, X 23, X 24, X 31,
X32, X 33 and X34
Objective Function: MINIMIZE (C11 X11 + C12 X12 + C13 X13 + C14 X14 + C21 X21 +
C22 X22 + C23 X23 + C24 X24 + C31X31 + C32 X32 + C33 X33 + C34 X34).
Constraints:
(1) X11 + X12 + X13 + X14 = A1 (Availability at Supply point S1)
(4) X11 + X 21 + X31 + X41 = R1 (Requirement at demand point D1)
(7) X14+ X 24 + X34+ X44= R4 (Requirement at demand point D4).
Non-negativity constraints:
X11, X12, X13, X14, X 21, X 22, X 23, X 24, X 31, X32, X 33, X34 are all ≥ 0.
When formulated as an LP Problem, this 3 X 4 matrix problem (a small problem by realistic
situation) will have 12 decision variables, 7 artificial variables and 7 constraints besides the non-
negativity constraints. The solution would be a very complex problem if attempted as an LP
Problem.

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Thus, these distribution problems are generally solved by a special type of LP model known as a
Transportation Model.
Some useful information in Transportation Models
(1) In the transportation model, the two parameters i.e., supply and demand have the same
cumulative amount. Thus, it can be said that the material available for supply can be supplied
because the demand exists at the same level. This solution is a case of a balanced
transportation problem.
In real-life situations, the demand may exceed the supply or vice versa. Such a situation is
termed as an unbalanced transportation problem.
(2) Whenever there is a positive allocation to a particular cell, it is called an occupied cell. Other
cells, to which a positive allocation has not been made are treated as empty or unoccupied cells.
(3) The number of occupied cells should be equal to (m + n – 1) where m = number of rows and
n = number of columns.
If number of occupied cells < (m + n – 1) , the solution is said to be a degenerate solution.
STEPS IN TRANSPORTATION METHOD
Step 1: Formulate the problem and establish the transportation cost matrix, the cells indicating
the parameters’ values for various combinations of origins and destinations. The parameters may
be in terms of cost, time, distance etc.
Step 2: Check if the problem is balanced or not.
(a) If ∑Supply = ∑Demand, the problem is a balanced transportation problem.
(b) If ∑Supply ≠ ∑Demand, the problem is an unbalanced transportation problem.
(i) If ∑Supply > ∑ Demand, a dummy column is introduced ( with the difference as the
requirement). The cost for each cell in the column is assigned a zero value.
(ii) . If ∑Supply < ∑ Demand, a dummy row is introduced ( with the difference as the supply).
The cost for each cell in the row is assigned a zero value.
Step 3: Obtain an initial basic solution. This can be done in three different ways i.e., North-West
Corner Rule, Least Cost Method and Vogel’s Approximation Method.
The initial basic solution from any of the above named methods should satisfy the following
conditions:

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(i) The solution must be feasible, satisfying all supply and demand conditions.
(ii) The number of positive allocations must be equal to (m + n -1). Otherwise, the solution will
become degenerate. The degeneracy must then be resolved.
Step 4: Test the initial solution for optimality by Modified Distribution (MODI) method.
Step 5: Update the solution i.e., applying step 4 till the optimal feasible solution is obtained.
OBTAINING INITIAL BASIC SOLUTION
(1) North-West Corner Rule (NW Corner Rule)
Step 1: Identify the North-West Corner cell in the transport cost matrix.
Step 2: Allot the maximum possible units to that cell.
Step 3: Eliminate the row or column fully satisfied by this allocation. Identify the North-West
Corner cell in the new reduced cost matrix.
Step 4: Repeat Steps 1 to 3, until the South-East Corner is reached.
Note: The major advantage of the North–West Corner Rule method is that it is very simple
and easy to apply. Its major disadvantage, however, is that it is not sensitive to costs and
consequently yields poor initial solutions.
(2) Least Cost Method (LCM)
Step 1: Select the lowest cost cell in the cost matrix. (In case of a tie, select arbitrarily).
Step 2: Allot the maximum possible units (considering the supply as well as demand
considerations) to this selected lowest cost cell.
Step 3: Eliminate the row or column fully satisfied by this allocation.
Step 4: Repeat Steps 1 to 3, until all capacities and requirements are fully satisfied.
Note: Intuitively, this method promises the best solution as we select the least cost cell for
allocation and then go to the next least cost cell. However, in practice, the Vogel’s
approximation method usually yields a better solution.
(3) Vogel’s Approximation Method (VAM)
Step 1: For each row of the transportation table, identify the smallest and the next to-smallest
costs. Determine the difference between them for each row. Display them alongside the

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transportation table by enclosing them in parenthesis against the respective rows. Similarly
compute the differences for each column.
Step 2: Identify the row or column with the largest difference among all the rows and columns.
If a tie occurs, select any row or column arbitrarily. Now consider the cell with the minimum
cost in that row (or column as the case may be) and assign the maximum units possible,
considering the demand and supply positions corresponding to that cell. Assign only one cell at a
time.
Step 3: Delete the column or row that has been fully satisfied. Recompute the column and row
differences for the reduced transportation table and go to step 2. Repeat the procedure until all
units have been assigned..
Note: VAM is not quite as simple as the Northwest Corner Rule approach, but it facilitates
a very good initial solution—as a matter of fact, one that is often the optimal solution.
Vogel’s approximation method tackles the problem of finding a good initial solution by
taking into account the costs associatedwith each route alternative. This is something that
the Northwest Corner Rule did not do.
Degeneracy in transportation problems
Degeneracy exists in a transportation problem when the number of filled cells is less than the
number of rows plus the number of columns minus one (m + n - 1).
Degeneracy may be observed either during the initial allocation when the first entry in a row or
column satisfies both the row and column requirements or during the U-V method application,
when the added and subtracted values are equal.
Degeneracy requires some adjustment in the matrix to evaluate the solution achieved. The form
of this adjustment involves inserting some small value (denoted by €) in an empty cell so a
closed path can be developed to evaluate other empty cells. This value may be thought of as an
infinitely small amount, having no direct bearing on the cost of the solution. Procedurally, € is
used in exactly the same manner as a real number except that it may initially be placed in any
empty cell, even though row and column requirements have been met by real numbers.
While the choice of where to put an € is arbitrary, it saves time if it is placed where it may be
used to evaluate as many cells as possible without being shifted.

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Testing for Optimality:
MODIFIED DISTRIBUTION (MODI) Method
Step 1: Determine an initial basic feasible solution
Step 2: Determine the values of auxiliary variables, ui and vj, using ui + vj = cij for all allocated
cells.
Step 3: Compute the opportunity cost for all unallocated cells using cij – ( ui + vj ).
Step 4: Check the sign of each opportunity cost. If the opportunity costs of all the unoccupied
cells are either positive or zero, the given solution is the optimum solution. On the other hand, if
one or more unoccupied cell has negative opportunity cost, the given solution is not an optimum
solution and further savings in transportation cost are possible.
Step 5: Select the unoccupied cell with the smallest negative opportunity cost as the cell to be
included in the next solution.
Step 6: Draw a closed path or loop for the unoccupied cell selected in the previous step. Please
note that the right angle turn in this path is permitted only at occupied cells and at the original
unoccupied cell.
Step 7: Assign alternate plus and minus signs at the unoccupied cells on the corner points of the
closed path with a plus sign at the cell being evaluated.
Step 8: Determine the maximum number of units that should be shipped to this unoccupied cell.
The smallest value with a negative position on the closed path indicates the number of units that
can be shipped to the entering cell. Now, add this quantity to all the cells on the corner points of
the closed path marked with plus signs and subtract it from those cells marked with minus signs.
In this way an unoccupied cell becomes an occupied cell.
Step 9: Repeat the whole procedure until an optimum solution is obtained.
Unbalanced Transportation Problem
If the total supply is not equal to the total demand, then the problem is known as unbalanced
(a) If the total supply is more than the total demand, we introduce an additional column (called
dummy column), which will indicate the surplus supply with transportation cost zero.
(b) Similarly, if the total demand is more than the total supply, an additional row (called dummy
row) is introduced in the table, which represents unsatisfied demand with transportation cost
zero.

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ASSIGNMENT PROBLEMS
Introduction
 There are many situations where the assignment of people or machines etc. is called for.
Assignment of workers to machines, clerks to various counters, salesmen to different
sales areas, crews to different aircrafts are few such examples.
 The assignment becomes a problem because people possess varying abilities for
performing different jobs, and therefore, the cost of performing the jobs to different
people are different.
 The objective function would therefore be minimization of cost or time.
CHARACTERISTICS OF THE ASSIGNMENT PROBLEM
(1) The available resources (such as availability of workers, machines, project managers,
salesmen, jobs etc). are finite in number.
(2) These available resources can be assigned only on a one-to-one basis for example, a job can
be assigned to a particular employee and after this one-to-one assignment, neither the worker nor
the job so assigned is available for further consideration.
(3) The outcome or the results are expressed in terms of costs, time or profits.
(4) The assignment methods aim at either cost minimization or profit maximization, indicating
that the assignments are made with a specific purpose of either cost, time or distance reduction
(minimization) or profit or utility maximization.
(5) For one-to one assignment, the problem has to be of the balanced type, otherwise it has to be
converted into a balanced problem.
(6) The standard assignment problem is of the minimization type. A maximization assignment
problem has to be converted to the minimization type.
HUNGARIAN ASSIGNMENT METHOD
Step 1: Locate the smallest cost element in each row of the cost matrix. Then, subtract this
smallest element from each element in that row. As a result, there will be at least one zero in
each row of the revised matrix.
Step 2: Locate the smallest cost element in each column of the revised matrix. Then, subtract
this smallest element from each element in that column. As a result, there will be at least one
zero in each column of the second revised matrix.

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Step 3: (Making zero-assignment): Examine the rows successively until a row-wise exactly
single zero is found, mark this zero by to make zero assignment. Then mark a cross (X) over
all zeros if lying in the column of the marked zero, showing that they cannot be considered
for future assignments. Continue in this manner until all the rows have been examined. Repeat
the same procedure for columns also.
Step 4: Draw minimum number of horizontal and vertical lines to cover all zero elements. Let
the minimum number of lines be N. Now, there may be two possibilities:
(i) The number of lines drawn = the number of rows (or columns). Then an optimal assignment is
reached. So, make the zero assignments to get the required solution.
(ii) The number of lines drawn < the number of rows (or columns), then proceed to Step 5.
Step 5: Identify the smallest element uncovered by the horizontal and vertical lines. Subtract this
element from all uncovered elements and add it to the elements at the junction of two lines.
Step 6: Repeat Steps 4 and 5 till the optimal solution is obtained i.e., the minimum number of
lines drawn = the number of rows (or columns).
Step 7: Make feasible assignments on zero elements and work out the total cost/time.
A Rule for drawing minimum number of lines
Step 1: Tick (√) rows that do not have any marked () zero.
Step 2: Tick columns having marked ( ) zeros or zeros with (X) in ticked rows.
Step 3: Tick (√) rows having marked ( ) zeros in ticked columns.
Step 4: Repeat Steps 2 and 3 till the chain of ticking is complete.
Step 5: Draw lines through all unticked rows and ticked columns.
CONSTRAINED ASSIGNMENT PROBLEMS
(1) Unbalanced Assignment Problem
If the number of rows is not equal to the number of rows in an assignment problem, it is called
an unbalanced assignment problem.
An unbalanced assignment problem has to be converted into a balanced one as follows:
(i) If the number of rows > the number of columns, introduce dummy columns with cost
elements as zero.
(ii) If the number of rows < the number of columns, introduce dummy rows with cost
elements as zero.

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(2) Assignment Problem with Prohibitive Assignment
Sometimes, there might be a situation where a particular assignment is not desirable and cannot
be made.
In such a situation, the cost of assignment in that cell can be made prohibitive, writing it as ‘M’
(an excessively high value). The problem can then be solved in the usual manner.
MAXIMIZATION TYPE ASSIGNMENT PROBLEMS
The standard Hungarian Assignment method is for minimization problem. In certain situations,
maximization assignment problems, such as allotment of teachers to subjects or salesmen to
territories, are encountered.
Method 1: The maximization problem is converted into the minimization problem by deducting
all the elements from the highest element of the pay-off matrix. The resultant matrix is then used
for Hungarian Assignment method. In the final stage, after making all the assignments, the
values from the original matrix are used to find the optimal value of profit/rating etc.
Method 2: The maximization problem is converted into the minimization problem by deducting
the minimum value of the matrix from all the elements. The resultant matrix is then used for
Hungarian Assignment method. In the final stage, after making all the assignments, the values
from the original matrix are used to find the optimal value of profit/rating etc.
Method 3: All the elements of the matrix are multiplied by (-1). The resultant matrix is then used
for Hungarian Assignment method. In the final stage, after making all the assignments, the
values from the original matrix are used to find the optimal value of profit/rating etc.
THE TRAVELLING SALESMAN (ROUTING) PROBLEM
 The travelling salesman problem is one of the problems considered as puzzles by
mathematicians.
 Suppose a salesman wants to visit a certain number of cities allotted to him. He knows
the distance (or cost or time) of journey between every pair of cities, usually denoted by
Cij, i.e., from city i to city j. His problem is to select a route that starts from his home city,
passes through each city once and only once, and returns to his home city in the shortest
possible distance (or at the least cost or in least time).
 The problem may be classified in two forms:
(i) Symmetrical: The problem is said to be symmetrical if the distance (or cost or
time) between every pair of cities is independent of the direction of his journey.

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(ii) Asymmetrical: The problem is said to be asymmetrical, if for one or more
pair of cities, the distance (or cost or time) changes with the direction of his
journey.
 If the number of cities is only two, obviously there is no choice. If the number of cities
becomes three, say A, B and C and one of them (say A) is the home city, then there are
two possible routes:
(1) A- B –C –A and (2) A-C-B-A
 If the number of cities is four, say A, B, C and D and one of them (say A) is the home
city, then there are 3!= 6 possible routes:
(1) A- B –C –D-A (2) A-B-D-C-A (3) A-C-B-D-A
(4) A-C-D-B-A (5) A-D-B-C-A (6) A-D-C-B-A.
 If the number of cities is increased to 21, there are 20! = 2,402,902,008,176,640,000
different routes. Even a fast computer testing one route per micro-second and working 8
hours a dy 365 days a year, would take almost 3 lakh years. Thus, it is practically
impossible to find the best route by trying each one. In general, if there are n cities, there
are (n-1)! Possible routes.
 At present, the best procedure to solve a travelling salesman problem is to solve the
problem as if it were an assignment problem. It becomes necessary to formulate this type
of sequencing problem in the form of an assignment problem with the additional
restriction on his choice of route.
 The travelling salesman problem is formulated and solved as an assignment problem.
(The assignment from city i to city i, can be avoided by adopting the convention Cij = ∞
during the formulation of the problem).
 If the zero-value assignment yields a tour such that it starts from the home city, visits
each city only once and returns to the home city, the optimal solution is achieved.
However, if the zero-value assignment does not yield the complete tour but yields a number of
sub-tours, then the ‘next-best solution may have to be adopted for a few pair of cities to enable
the completion of the tour.

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SEQUENCING
 Operations scheduling is most important in job-shop processing.
 This is because in job-shop processing,
(i) The variety of items is large;
(ii) The number of items to be produced in a batch is less; and
(iii) The machines and workers are versatile and hence can perform different types of
jobs.
 In continuous and semi-continuous production processes (such as assembly lines),
scheduling is automatically done at the time of designing the facility.
 Scheduling in a job-shop involves the following activities:
(1) Assigning the job orders to different machines (or work centres).
(2) Deciding the sequence of processing on different machines on the basis of some
priority rule (called sequencing or prioritization).
(3) Planning the route of movement of material from one department to another (called
routing).
(4) Issuing dispatch lists to the various work centres containing information about:
(i) The work centres at which a customer’s order should be processed;
(ii) The customers orders to be processed first; and
(iii) The amount of time the processing should take.
(5) Tracking the progress of various scheduled jobs and the implementation of schedules,
revising the schedules in case of delays and accelerating the completion of certain
jobs (called expediting).
Importance of Proper Scheduling
Absence of proper scheduling may lead to the following problems:
(1) The workers may take up any job order at random and start processing it. It is likely that
this order has a much later due date compared to a job whose due date is near, This will
result in delays in meeting the due dates of customer orders.

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(2) If jobs at different work centres are chosen at random, so many jobs have to wait for a
long time, resulting in high work- in- process inventory.
(3) In the absence of scheduling, the overall set-up time increases resulting in high average
completion time of each job order.
(4) The operations manager will have no information about the current status of a particular
job, as the workers themselves will decide about processing a job randomly.
(5) At some point of time, at some work centres, workers and machines will be overburdened
with work, while at other work centres, they may be idle waiting for jobs to come to their
work centres. Hence, there will be low utilization of workers and machines leading to a
high cost of operation.
SEQUENCING AND PRIORITIZATION
Number of Jobs Number of
Machines
Jobs to be
processed in the
same sequence or
in different
sequences
Method
n 2 Same sequence Johnson’s Method
n 3 Same sequence Modified Johnson’s method
n M Same sequence Modified Johnson’s method
2 M Different
sequence
Akers’ method
Assigning n jobs to m machines Assignment Method
SEQUENCING N-JOBS ON 2 MACHINES
General Assumptions
(1) The processing times on different machines are independent of the order of the jobs in which
they are to be processed.
(2) Only one job can be processed on a machine at a given time.
(3) The time taken by the jobs in moving from one machine to another is very negligible and is
taken as equal to zero.

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(4) Each job, once started on a machine, is to be performed up to completion on that machine.
(5) The machines, on which the jobs are to be processed, are of different types.
(6) All jobs are known and are ready for processing before the period under consideration begins.
(7) Processing times are given and do not change.
JOHNSON’s ALGORITHM FOR N-JOBS ON 2 MACHINES
------------------------Processing Times--------------------------------------------
Jobs J1 J2 J3 J4 --------------------------------------------------- JN
Machine
A
A1 A2 A3 A4 ----------------------------------------------------- AN
Machine
B
B1 B2 B3 B4 ------------------------------------------------------ BN
The method:
Step (1) Identify the minimum of the processing times on both machines.
Step (2) (a) If the minimum processing time is for the kth job on Machine A, the job is to be
placed at the first available place in the sequence.
(b) If the minimum processing time is for the rth job on Machine B, the job is to be placed at
the last available place in the sequence.
(c) If the minimum processing time is simultaneously for the kth job on Machine A and for the
rth job on Machine B, then kth job is to be placed at the first available place in the sequence and
the rth job is to be placed at the last available place in the sequence.
(d) If the minimum processing time is the same for two or more jobs on Machine A, the job that
has the least time on Machine B is to be placed at the first available place in the sequence
followed by other jobs.
(e) If the minimum processing time is the same for two or more jobs on Machine B, the job that
has the least time on Machine A is to be placed at the last available place in the sequence
followed by other jobs.
Step (3) Remove this job from consideration and repeat Steps 1 and 2 till all jobs are placed in a
sequence.

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Example
(1) In a machine shop, 8 different jobs are being manufactured, each requiring time on 2
machines A and B as given below:
JOBS Time (in Min) on Machine A Time (in Min) on Machine B
1 3 2
2 5 8
3 10 13
4 20 18
5 16 22
6 14 12
7 6 17
8 11 19
Decide the optimum sequence of processing of different jobs in order to minimize the total
manufacturing time for all the jobs. Calculate the total processing time.
Solution
The minimum of all processing times is 2 minutes. It corresponds to Job1 on Machine B. So, it is
to be placed at the end of the sequence.
J 1
Job 1 is removed from further consideration. The next minimum of processing time is 5 minutes.
It corresponds to Job 2 on Machine A. It is to be placed at the beginning of the sequence.
J2 J 1
Job 2 is removed from further consideration. The next minimum of processing time is 6 minutes.
It corresponds to Job 7 on Machine A. It is to be placed at the first available space in the
sequence.
J2 J7 J 1
Job 7 is removed from further consideration. The next minimum of processing time is 10
minutes. It corresponds to Job 3 on Machine A. It is to be placed at the first available space in the
sequence.

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J2 J7 J3 J 1
sequence.
J2 J7 J3 J8 J 1
minutes. It corresponds to Job 6 on Machine B. It is to be placed at the last available space in the
sequence.
J2 J7 J3 J8 J6 J 1
sequence.
J2 J7 J3 J8 J5 J6 J 1
The remaining job, Job 4 is placed in the only available space. The sequence in which the jobs
are to be processed is:
J2 J7 J3 J8 J5 J4 J6 J 1
MAKE SPAN
JOBS
MACHINE A MACHINE B
IN OUT IN OUT
J2 0 5 5 13
J7 5 11 13 30
J3 11 21 30 43
J8 21 32 43 62
J5 32 48 62 84
J4 48 68 84 102
J6 68 82 102 114
J1 82 85 114 116
The total processing time (MAKE-SPAN) = 116 minutes.

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SEQUENCING N-JOBS ON 3 MACHINES
------------------------Processing Times--------------------------------------------
Jobs J1 J2 J3 J4 --------------------------------------------------- JN
Machine
A
A1 A2 A3 A4 ----------------------------------------------------- AN
Machine
B
B1 B2 B3 B4 ------------------------------------------------------ BN
Machine
C
C1 C2 C3 C4 ------------------------------------------------------ CN
 There is no general procedure for solving the N-jobs, 3 Machines problem.
 However, if either one or both of the following conditions are satisfied, the problem can
be converted into a 2 machine-n jobs problem and modified Johnson’s method can be
used.
(i) Minimum of Processing times on Machine A ≥ Maximum of Processing times on Machine B.
(ii) Minimum of Processing times on Machine C ≥ Maximum of Processing times on Machine
B.
Modified Johnson’s Method
(1) Convert the three machines, n-jobs problem into a 2 machine-n jobs problem by creating two
fictitious machines G and H such that Gi = Ai + Bi and Hi = Bi + Ci
(2) Then solve the problem as a 2-machine, n jobs problem.

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Example
A company has six jobs which go through three machines X, Y, Z in the order X-> Y-> Z. The
processing time in minutes for each job on each machine is as follows:
PROCESSING TIME (MINUTES)
JOBS Machine X Machine Y Machine Z
1 18 7 19
2 15 12 12
3 29 11 23
4 35 2 47
5 43 8 28
6 37 15 38
What should be the sequence of the jobs that minimizes the total processing time? What is the
total processing time (MAKE SPAN)
Solution
Test
(i) Minimum of Xi s = 15 minutes. Maximum of Yi s = 15 minutes.
Minimum of Xi s = Maximum of Yi s
(ii) Minimum of Zi s = 12 minutes. Maximum of Yi s = 15 minutes.
Minimum of Zi s is NOTgreater than or equal to Maximum of Yi s.
Condition (1) is satisfied.
Hence, modified Johnson’s Method can be used.
The 3-machine, n jobs problem is converted into a 2 machine, n jobs problem by creating two
fictitious machines, G and H.
Processing Time (Minutes)
Job G H
1 25 26
2 27 24
3 40 34
4 37 49
5 51 36
6 52 55

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The sequence is
J1 J4 J6 J5 J3 J2
MAKE SPAN
MACHINE X MACHINE Y MACHINE Z
JOB In Out In Out In Out
J1 0 18 18 25 25 44
J4 18 53 53 55 55 102
J6 53 90 90 105 105 143
J5 90 133 133 141 143 171
J3 133 162 162 173 173 196
J2 162 177 177 189 196 208
Total Processing time = 208 minutes
______________________________________________________________________________
UNIVERSITY QUESTIONS FROM UNIT 2
I. TRANSPORTATION PROBLEM
Q (1) Discuss briefly the various methods for obtaining an initial basic feasible solution to a
Q (2) A company has four factories from which it ships its products to four warehouses, which
are the distribution centres.
Transportation costs per unit between various combinations of factories and warehouses are as
follows:
Factory
Warehouse
Availability
(Units)
1 2 3 4
1 48 60 56 58 140
2 45 55 53 60 260
3 50 65 60 62 360
4 52 64 55 61 220
Requirement (Units) 200 320 250 210

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Find the transportation schedule which minimizes the distribution cost.
Q (3) Solve the following transportation problem for minimum transportation cost. Also find the
alternate optimal solution if any.
From / To A B C Supply
X 4 8 8 76
Y 16 24 16 82
Z 8 16 24 77
Demand 72 102 41
Q (4) (a) Explain degeneracy in transportation problem.
(b) Solve the following transportation problem
Sources
Destinations
AvailabilityD1 D2 D3 D4
S1 19 30 50 10 7
S2 70 30 40 60 9
S3 40 8 70 20 18
Requirement 5 8 7 14
Q (5) The Reliable Company has three factories that ship products to four warehouses. The
shipping costs, requirements, capacities, are shown in table. Find the initial solution using
Vogel’s Approximation Method. What is the total cost of the optimal solution? Use the U-V
method to test optimality.
Cost of Transporting per case to
WarehouseFactories W1 W2 W3 W4 Supply
(in thousands)
F1 40 60 100 20 210
F2 140 60 80 120 270
F3 80 20 120 160 540
Demand
(in thousands)
150 240 210 420
II. ASSIGNMENT PROBLEM & TRAVELLING SALESMAN PROBLEM
Q (6) Solve the following Assignment Problem such that the total time required to complete all
jobs is minimum.
Time required to complete the job in minutes
Worker J1 J2 J3 J4 J5
W1 10 5 20 15 10
W2 20 10 10 30 5
W3 50 60 40 30 20
W4 30 40 40 20 60
W5 30 30 50 60 20

OR (2014-15) Unit 2
21
Q (7) The following table denotes the salesmen and the sales they can do if assigned to different
territories. Assign the salesmen to different territories such that the total sales are maximized.
Sales achieved if assigned to a territory (Rs Lakhs / month)
Salesman T1 T2 T3 T4 T5
S1 50 60 30 40 40
S2 70 60 80 50 60
S3 30 80 70 40 30
S4 40 40 50 70 80
S5 40 60 50 30 80
Q (8) A salesman starting at city P has to visit cities Q, R and S before returning to P. The
distances between pairs of cities in Kilometers are given below:
P Q R S
P ∞ 150 250 200
Q 220 ∞ 450 550
R 400 300 ∞ 250
S 200 260 380 ∞
Determine the route which enables the salesman to visit all the cities, at minimum total distance
travelled.
III. SEQUENCING
Q (9) Discuss in detail the optimal solution for processing each of n-jobs through three machines.
Q (10) (a) What is the difference between assignment problem and travelling salesman problem?
(b) Find the sequence that minimizes the total time in hours required to complete the jobs on
three machines.
Job 1 2 3 4 5 6 7
Machine A 5 7 3 4 6 7 12
Machine B 2 6 7 5 9 5 8
Machine C 10 12 11 13 12 10 11
What is the minimum elapsed time?
Q (11) Find the sequence that minimizes the total elapsed time required to complete the
following jobs:
Machine
Processing Time (in minutes) for
Job 1 Job 2 Job 3 Job 4 Job 5
A 2 5 4 3 2
B 6 8 1 2 3

OR (2014-15) Unit 2
23
Q (12) (a) Explain various methods to find Basic Feasible Solution to Transportation problems.
(b) Give mathematical definition of assignment problem.
(c) We have five jobs each of which must pass through the machines A, B and C in the order
ABC. Determine the sequence that minimizes the total elapsed time.
Job No -> 1 2 3 4 5
Machine A 5 7 6 9 5
Machine B 2 1 4 5 3
Machine C 3 7 5 6 7

OR 14 15-unit_2

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