3. CONTENT
OBJECTIVES OFTRANSPORTATION PROBLEM
INTRODUCTION TO TRANSPORTATION PROBLEM
MATHEMATICALFORMULATION OFATRANSPORTATION PROBLEM
INITIALBASIC FEASIBLE SOLUTION
OPTIMALBASIC SOLUTION - MODI METHOD - MODIFIED
DISTRIBUTION METHOD
ASSIGNMENT QUESTIONS
MCQ QUESTIONS WITHANSWER
3
4. OBJECTIVES OF TRANSPORTATION PROBLEM
• Transportation problem works in a way of minimizing the cost function.
• The cost function is the amount of money spent to the logistics provider for
transporting the commodities from production or supplier place to the
demand place.
• It includes the distance between the two locations, the path followed, mode
of transport, the number of units that are transported, the speed of
transport, etc.
• To transport the commodities with minimum transportation cost without
any compromise in supply and demand.
4
5. INTRODUCTION TO TRANSPORTATION PROBLEM
• The transportation problem is a special type of linear programming problem, where the objective is to
minimize the cost of distributing a product from a number of sources to a number of destinations.
•Transportation deals with the transportation of a commodity (single product) from ‘m’ sources (origin or
supply or capacity centres) to ‘n’destinations (sinks or demand or requirement centres).
•It is assumed that, level of supply of each source and the amount of demand at each destination are
known . The unit transportation cost of commodity from each source to each destination are known.
•The objective is to determine the amount to be shifted from each source to each destination such that
the total transportation cost is minimum.
5
6. MATHEMATICAL FORMULATION OF A
TRANSPORTATION PROBLEM
m n
Let us assume that there are m sources and n destinations.
Let ai be the supply (capacity) at source i, bj be the
demand at destination j, cij be the unit transportation
cost from source i to destination j and xij be the number of
units shifted from source i to destination j.
Then the transportation problem can be expressed
mathematically as,
Minimize Z = ∑cij
∑xij subject to the constraints,
i =1 j =1
6
7. STANDARD TRANSPORTATION TABLE
Source
Transportation problem is explicitly represented by the following transportation table,
Destination
D1 D2 D3 … Dj … Dn Supply
S1
S2
S3
c11 c12 c13 … c1j … c1n a1
c21 c22 c23 … c2j … c2n a2
c31 c32 c33 … c3j … c3n a3
… …
ci1 ci2 ci3 … cij … cin ai
… …
cm1 cm2 cm3 … cmj … cmn am
b1 b2 b3 bj bn ai = bj
Si
Sm
Demand
The mn squares are called cells. The various a’s and b’s are called the constraints (rim
conditions)
7
8. BASIC DEFINITIONS
• Feasible Solution: A set of non-negative values xij, i = 1, 2 … m; j = 1, 2 …n that satisfies the
constraints (rim conditions) is called a feasible solution.
• Basic Feasible Solution: A feasible solution to a (m x n) transportation problem that contains no more
that m + n – 1 non-negative independent allocations is called a basic feasible solution(BFS) to the
transportation problem.
• Non-Degenerate Basic Feasible Solution: A basic feasible solution to a (m x n) transportation
problem is said to be a non-degenerate basic feasible solution if it contains exactly m + n – 1 non-
negative allocations in independent positions.
• Degenerate Basic Feasible Solution: A basic feasible solution that contains less than m + n – 1 non-
negative allocations is said to be a degenerate basic feasible solution.
• Optimal Solution: A feasible solution (not necessarily basic) is said to be an optimal solution if it
minimises the total transportation cost.
• The number of basic variables in an m x n balanced transportation problem is at the most m + n – 1.
8
9. TYPES OF TRANSPORTATIONPROBLEM
• Balanced Transportation Problem: where the total
supply equals to the total demand
• Unbalanced Transportation Problem: where the
total supply is not equal to the total demand
9
10. PHASES OF SOLUTION OF TRANSPORTATION PROBLEM
• Phase I- obtains the initial basic feasible solution
• Phase II-obtains the optimal basic solution
Optimality condition: m+n-1 no of allocation
10
Where m= number of rows and n=
number of columns
11. INITIAL BASIC FEASIBLE SOLUTION
The following methods are to find the initial basic feasible
solution for the transportation problems.
• North W
est Corner Rule (NWCR)
• Least Cost Method
• V
ogle’sApproximation Method(V
AM)
11
12. METHOD 1: NORTH-WEST CORNER RULE
Step 1: The first assignment is made in the cell occupying the upper left-hand (north-west) corner of
the transportation table. The maximum possible amount is allocated there. That is, x11 = min {a1, b1}.
i. If min {a1, b1} = a1, then put x11= a1, decrease b1 by a1 and move vertically to the
2nd row (to the cell (2, 1)) and cross out the first column.
ii. If min {a1, b1} = b1, then put x11= b1, decrease a1 by b1 and move horizontally
right (to the cell (1, 2)) and cross out the first column.
12
iii. If min {a1, b1} = a1 = b1, then put x11= a1 = b1 and move diagonally to the cell (2,
2) cross out the first row and the first column.
Step 2: Repeat the procedure until all the rim requirements are satisfied.
13. METHOD 2: LEAST COST METHOD (OR) MATRIX MINIMA METHOD
(OR) LOWEST COST ENTRY METHOD
13
.
Step 1: Balance the problem
Step 2: Select the lowest cost from the entire matrix
and allocate the minimum of supply or demand.
Step 3: Remove the row or column whose supply or
demand is fulfilled and prepare a new matrix
Step 4: Repeat the procedure until all the allocations
are over
Step 5: After all the allocations are over, write the
allocations and calculate the transportation cost
14. Example:1 A mobile phone
manufacturing company has three
branches located in three different
regions, say Jaipur, Udaipur and
Mumbai. The company has to transport
mobile phones to three destinations, say
Kanpur, Pune and Delhi. The availability
from Jaipur, Udaipur and Mumbai is 40,
60 and 70 units respectively. The
demand at Kanpur, Pune and Delhi are
70, 40 and 60 respectively. The
15.
16.
17.
18. METHOD 3:
VOGEL’S APPROXIMATION METHOD (OR)
UNIT COST PENALTY METHOD
Vogel’s Approximation Method (VAM) is one of the
methods used to calculate the initial basic feasible solution
to a transportation problem. However, VAM is an iterative
procedure such that in each step, we should find the
penalties for each available row and column by taking the
least cost and second least cost.
19. Procedure of Vogel’s Approximation Method
Step 1: Identify the two lowest costs in each row and column of the
given cost matrix and then write the absolute row and column difference.
These differences are called penalties.
Step 2: Identify the row or column with the maximum penalty and
assign the corresponding cell’s min(supply, demand). If two or more
columns or rows have the same maximum penalty, then we can choose
one among them as per our convenience.
Step 3: If the assignment in the previous satisfies the supply at the
origin, delete the corresponding row. If it satisfies the demand at that
destination, delete the corresponding column.
Step 4: Stop the procedure if supply at each origin is 0, i.e., every supply
is exhausted, and demand at each destination is 0, i.e., every demand is
satisfying. If not, repeat the above steps, i.e., from step 1.
21. Solution:
For the given cost matrix,
Total supply = 50 + 60 + 25 = 135
Total demand = 60 + 40 + 20 + 25 = 135
Thus, the given problem is balanced transportation problem.
Now, we can apply the Vogel’s approximation method to minimize the total cost of transportation.
Step 1: Identify the least and second least cost in each row and column and then write the
corresponding absolute differences of these values. For example, in the first row, 2 and 3 are the least
and second least values, their absolute difference is 1.
22. These row and column differences are called penalties.
Step 2: Now, identify the maximum penalty and choose the least value in
that corresponding row or column. Then, assign the min(supply, demand).
Here, the maximum penalty is 3 and the least value in the corresponding
column is 2. For this cell, min(supply, demand) = min(50, 40) = 40
Allocate 40 in that cell and strike the corresponding column since in this case
demand will be satisfied, i.e., 40 – 40 = 0.
23. Step 3: Now, find the absolute row and column differences for the remaining
rows and columns. Then repeat step 2.
Here, the maximum penalty is 3 and the least cost in that corresponding row is
3. Also, the min(supply, demand) = min(10, 60) = 10
Thus, allocate 10 for that cell and write down the new supply and demand for
the corresponding row and column.
Supply = 10 – 10 = 0
Demand = 60 – 10 = 50
As supply is 0, strike the corresponding row.
24. Step 4: Repeat the above step, i.e., step 3. This will give the below result.
In this step, the second column vanishes and the min(supply,
demand) = min(25, 50) = 25 is assigned for the cell with
value 2.
25. Step 5: Again repeat step 3, as we did for the
previous step.
In this case, we got 7 as the maximum penalty and 7 as
the least cost of the corresponding column.
26. Step 6: Now, again repeat step 3 by calculating the absolute
differences for the remaining rows and columns.
27. Step 7: In the previous step, except for one cell, every row and
column vanishes. Now, allocate the remaining supply or demand
value for that corresponding cell.
29. Practice problem : 1
Consider the transportation problem given below.
Solve this problem by Vogel’s approximation
method.
Origin Destination Supply
D1 D2 D3 D4
O1 3 1 7 4 300
O2 2 6 5 9 400
O3 8 3 3 2 500
Demand 250 350 400 200
30. Practice problem : 2
Find the solution for the following transportation
problem using VAM
From To Supply
D1 D2 D3
A1 6 8 10 150
A2 7 11 11 175
A3 4 5 12 275
Demand 200 100 350
.
31. Practice problem : 3
Use Vogel’s Approximation Method to find a basic
feasible solution for the following.
Sources Destinations Supply
D1 D2 D3
S1 4 5 1 40
S2 3 4 3 60
S3 6 2 8 70
Demand 70 40 60
32. PROBLEM 1: Obtain initial feasible solution for the following Transportation
table using (i) North West Corner rule. (ii) Least Cost Method. (iii) VAM
Method.
32
34. To find the IBFS using (i) North West Corner rule:
The number of allocations = 6 = 4+3-1 = m+n-1, is a non-degenerate solution.
The IBFS is = (2*5)+(3*2)+(3*6)+(4*3)+(7*4)+(2*14) = 153.
34
35. To find the IBFS using (ii) Least Cost Method
The number of allocations = 6 = 4+3-1 = m+n-1, is a non-degenerate solution.
The IBFS is = (7*2)+(4*3)+(1*8)+(4*7)+(1*7)+(2*7) = 83.
35
36. To find the IBFS using (iii) VAM
The number of allocations = 6 = 4+3-1 = m+n-1, is a non-degenerate solution.
The IBFS is = (2*5)+(3*2)+(1*6)+(4*7)+(1*2)+(2*12) = 76.
36
37. OPTIMAL BASIC SOLUTION - MODI METHOD - MODIFIED
DISTRIBUTION METHOD
Step 1: Find the initial basic feasible solution of the given problem by
Northwest Corner Rule (or) Least Cost Method (or) VAM.
Step 2: Check the number of occupied cells. If there are less than m + n – 1,
there exists degeneracy and we introduce a very small positive assignment of
in suitable independent positions, so that the number of occupied cells is
exactly equal to m + n – 1.
Step 3: Find the set values ui, vj (i = 1, 2, …m ; j = 1, 2, …n) from the relation
cij = ui + vj for each occupied cell (i, j), by starting initially with ui = 0 or vj =
0 preferably for which the corresponding row or column has maximum number
of individual allocations.
37
38. Step 4: Find ui + vj for each unoccupied cell (i, j) and enter at the upper right
corner of the
corresponding cell (i, j).
Step 5: Find the cell evaluations Pij = ui+ vj – Cij for each unoccupied cell (i,
j) and enter at the upper left corner of the corresponding cell (i, j).
Step 6: Examine the cell evaluations dij for all unoccupied cells (i, j) and
conclude that,
•If all dij<0, then the solution under the test is optimal and unique.
•If all dij> 0, with at least one dij = 0, then the solution under the test is
optimal and an alternative optimal solution exists.
•If at least one dij>0, then the solution is not optimal. Go to the next step. 38
39. Step 7: Form a new BFS by giving maximum allocation to the cell for which
dij is most
negative by making an occupied cell empty. For that draw a closed path
consisting of horizontal and vertical lines beginning and ending at the cell for
which dij is most negative and having its other corners at some allocated
cells. Along this closed loop indicate +θ and –θ alternatively at the corners.
Choose minimum of the allocations from the cells having –θ. Add this
minimum allocation to the cells with +θ and subtract this minimum allocation
from the allocation to the cells with –θ.
Step 8: Repeat steps (2) to (6) to test the optimality of this new basic feasible
solution.
Step 9: Continue the above procedure till an optimum solution is attained.
39
45. only for unallocated cells. We have two
unallocated cells in the first row, two in the
second row and two in the third row. Lets
compute this one by one.
For C13, P13 = 0 + 0 – 7 = -7
(here C13 = 7, u1 = 0 and v3 = 0)
For C14, P14 = 0 + (-1) -4 = -5
For C21, P21 = 5 + 3 – 2 = 6
For C24, P24 = 5 + (-1) – 9 = -5
For C31, P31 = 3 + 3 – 8 = -2
For C32, P32 = 3 + 1 – 3 = 1
46. If we get all the penalties value as zero or
negative values that mean the optimality is
reached and this answer is the final answer.
But if we get any positive value means we need
to proceed with the sum in the next step. Now
find the maximum positive penalty.
Here the maximum value is 6 which
corresponds to
C21
47.
48. The rule for drawing closed-path or loop.
Starting from the new basic cell draw a closed-path in such
a way that the right angle turn is done only at the allocated
cell or at the new basic cell. See the below images:
49. Consider the cells with a negative sign. Compare the allocated value
(i.e. 200 and 250 in this case) and select the minimum (i.e. select 200
in this case). Now subtract 200 from the cells with a minus sign and
add 200 to the cells with a plus sign. And draw a new iteration. The
work of the loop is over and the new solution looks as shown below.
53. For C11, P11 = 0 + (-3) – 3 = -6
For C13, P13 = 0 + 0 – 7 = -7
For C14, P14 = 0 + (-1) – 4 = -5
For C24, P24 = 5 + (-1) – 9 = -5
For C31, P31 = 0 + (-3) – 8 = -11
For C32, P32 = 3 + 1 – 3 = 1
There is one positive value i.e. 1 for C32
54. Now draw a loop starting from the new basic cell(C32). Assign
alternate plus and minus sign with new basic cell assigned as a plus
55.
56. Select the minimum value from allocated values to the cell with a
minus sign. Subtract this value from the cell with a minus sign and
add to the cell with a plus sign. Now the solution looks as shown in
the image below:
58. Check if the total number of allocated cells is
equal to (m + n – 1). Find u and v values as
above.
Now again find the penalties for the unallocated cells as
59. For P11 = 0 + (-2) – 3 = -5
For P13 = 0 + 1 – 7 = -6
For P14= 0 + 0 – 4 = -4
For P22= 4 + 1 – 6 = -1
For P24= 4 + 0 – 9 = -5
For P31= 2 + (-2) – 8 = -8
All the penalty values are negative values. So
the optimality is reached. Now, find the total
cost i.e.
(250 * 1) + (200 * 2) + (150 * 5) + (50 * 3) +
(200 * 3) + (150 * 2) = 2450
60. Practice PROBLEM 1: Solve the
Transportation Problem and identify
OPTIMAL SOLUTION Using MODI Method.
Source
Destination
Supply
A B C D
1 19 30 50 10 7
2 70 30 40 60 9
3 40 8 70 20 18
Demand 5 8 7 14 34
60
61. Source Destination Supply
A B C
1 2 2 3 10
2 4 1 2 15
3 1 3 1 40
Demand 20 15 30 61
Practice PROBLEM 2: Solve the
Transportation problem and identify
OPTIMAL SOLUTION Using MODI Method.
62. 𝑖
𝑚
𝑖=1 𝑗
𝑛
𝑗 =1 ,
62
The given problem is balanced
ie.,Total Supply = Total Demand
63. To find the IBFS using VAM
The number of allocations = 5 = 3+3-1 = m+n-1, is a non-degenerate solution.
The IBFS is = (2*10)+(1*5)+(2*10)+(1*20)+(1*20) = 85.
63
64. Tofind the optimalsolutionusingMODIMethod:
• Find the set values ui, vj(i = 1, 2, …m; j = 1, 2, …n) from the relation cij= ui+ vjfor each occupied cell (i, j), by starting
initially with ui= 0 or vj= 0 preferably for which the corresponding row or column has maximum number of individual
allocations.
• Findui +vj for each unoccupied cell(i, j) and enter at the upperrightcorner of thecorrespondingcell (i, j).
• Find the cell evaluations dij = cij – (ui + vj) for each unoccupied cell (i, j) and enter at the upper left corner of the
corresponding cell (i, j).
64
65. Form a new BFS by giving maximum allocation to the cell for which dij is most negative by
making an occupied cell empty. For that draw a closed path consisting of horizontal and
vertical lines beginning and ending at the cell for which dij is most negative and having its
other corners at some allocated cells. Along this closed loop indicate +θ and –θ alternatively at
the corners. Choose minimum of the allocations from the cells having –θ. Add this minimum
allocation to the cells with +θ and subtract this minimum allocation from the allocation to the
cells with –θ.
65
66. Since all dij > 0 the above allocation is optimal unique solution.
The Optimal solution = (2*10)+(1*15)+(2*0)+(1*30)+(1*10) = 75.
66
67. A S S I G N M E N T Q U E S T I O N S
1 . O b t a i n i n i t i a l f e a s i b l e s o l u t i o n f o r t h e f o l l o w i n g T r a n s p o r t a t i o n t a b l e u s i n g ( i )
N o r t h W e s t C o r n e r r u l e . ( i i ) L e a s t C o s t M e t h o d . ( i i i ) V A M M e t h o d .
S o u r c e D e s t i n a t i o n S u p p l y
A B C
1 7 3 4 2
2 2 1 3 3
3 3 4 6 3
D e m a n d
4 1 3
2 . S o l v e t h e T r a n s p o r t a t i o n t a b l e .
30
S o u r c e
D e s t i n a t i o n
S u p p l y
A B C D
1 5 4 2 6 2 0
2 8 3 5 7 3 0
3 5 9 4 6 5 0
D e m a n d
1 0 4 0 2 0
3 0
68. Meaning of Assignment Problem:
An assignment problem is a particular case of
transportation problem where the objective is to
assign a number of resources to an equal number of
activities so as to minimise total cost or maximize
total profit of allocation.
The problem of assignment arises because available
resources such as men, machines etc. have varying
degrees of efficiency for performing different
activities, therefore, cost, profit or loss of performing
the different activities is different.
II ASSIGNMENT PROBLEMS
69. An assignment problem is a particular case of
transportation problem.
The objective is to assign a number of resources to
an equal number of activities .
So as to minimize total cost or maximize total
profit of allocation. The problem of assignment
arises because available resources such as men,
machines etc. have varying degrees of efficiency
for performing different activities, therefore, cost,
profit or loss of performing the different activities
is different
70. ASSIGMENT ALGORITHM (OR) HUNGARIAN
METHOD
Step:1
First check whether the number of rows is equal to number of
columns, if it is so, the assignment problem is said to be balanced.
Then proceed to step 1. If it is not balanced, then it should be
balanced before applying the algorithm.
Step 2:
Subtract the smallest cost element of each row from all the
elements in the row of the given cost matrix. See that each row
contains at least one zero.
Step 3:
Subtract the smallest cost element of each column from all the
elements in the column of the resulting cost matrix obtained by
step 1 and make sure each column contains at least one zero.
71. Step 4:
(Assigning the zeros)
(a)Examine the rows successively until a row with exactly one
unmarked zero is found. Make an assignment to this single
unmarked zero by encircling it. Cross all other zeros in the column of
this encircled zero, as these will not be considered for any future
assignment. Continue in this way until all the rows have been
examined.
(b)Examine the columns successively until a column with exactly one
unmarked zero is found. Make an assignment to this single
unmarked zero by encircling it and cross any other zero in its row.
Continue until all the columns have been examined.
Step 5:
(Apply Optimal Test) (a) If each row and each column contain exactly
one encircled zero, then the current assignment is optimal.
72. Step 6: Cover all the zeros by drawing a minimum
number of straight lines as follows:
(a) Mark the rows that do not have assignment.
(b)Mark the columns (not already marked) that have
zeros in marked rows.
(c) Mark the rows (not already marked) that have
assignments in marked columns.
(d) Repeat (b) and (c) until no more marking is
required.
(e) Draw lines through all unmarked rows and marked
columns. If the number of these lines is equal to the
order of the matrix then it is an optimum solution
otherwise not
73. Step 7:
Determine the smallest cost element not
covered by the straight lines. Subtract this
smallest cost element from all the uncovered
elements and add this to all those elements
which are lying in the intersection of these
straight lines and do not change the remaining
elements which lie on the straight lines.
Step 8: Repeat steps (1) to (6), until an optimum
assignment is obtained.
74.
75.
76.
77.
78.
79.
80.
81.
82. Since each row and each column contain
exactly one encircled zero, then the
current assignment is optimal.
Where the optimal assignment is as
1 to IV ,
2 to I ,
3 to II ,
4 to III
and 5 to V.
The optimal z = 11 + 43 + 28 + 27 + 25 =
134 hours.
83. M C Q Q U E S T I O N S W I T H A N S W E R
( 1 ) T o f i n d i ni t i al f e a s i b l e s o l u t i o n o f a t r a n s p o r t a t i o n p r o b l e m t h e m e t h o d w h i c h st art s
a l l o c a t i o n f r o m t h e l o w e s t c o s t i s c a l l e d m e t h o d .
( a ) n o r t h w e s t c o r n e r
( b ) l e a s t c o s t
( c ) s o u t h e a s t c o r n e r
( d ) Vo g e l ’s a p p r o x i m a t i o n
( 2 ) I n a t r a n s p o r t a t i o n p r o b l e m , t h e m e t h o d o f p e n a l t i e s i s c a l l e d m e t h o d .
( a ) l e a s t c o s t
( b ) s o u t h e a s t c o r n e r
( c ) Vo g e l ’ s a p p r o x i m a t i o n
( d ) n o r t h w e s t c o r n e r
( 3 ) W h e n t h e t o t a l o f a l l o c a t i o n s o f a t r a n s p o r t a t i o n p r o b l e m m a t c h w i t h s u p p l y a n d d e m a n d
v a l u e s , t h e s o l u t i o n i s c a l l e d s o l u t i o n .
( a ) n o n - d e g e n e r a t e
( b ) d e g e n e r a t e
( c ) f e a s i b l e
( d ) i n f e a s i b l e
( 4 ) W h e n t h e a l l o c a t i o n s o f a t r a n s p o r t a t i o n p r o b l e m s a t i s f y t h e r i m c o n d i t i o n ( m + n – 1 ) t h e
s o l u t i o n i s c a l l e d s o l u t i o n .
( a ) d e g e n e r a t e
( b ) i n f e a s i b l e
( c ) u n b o u n d e d
( d ) n o n - d e g e n e r a t e
( 5 ) W h e n t h e r e i s a d e g e n e r a c y i n t h e t r a n s p o r t a t i o n p r o b l e m , w e a d d a n i m a g i n a r y a l l o c a t i o n
c a l l e d i n t h e s o l u t i o n .
( a ) d u m m y
( b ) p e n a l t y
( c ) e p s i l o n
( d ) r e g r e t
31
84. (6) If M + N – 1 = N u m b e r o f allocatio n s in tran s p o rta tio n , it m ean s . (W h ere ‘M ’ is
n u m b e r of r o w s a n d ‘ N ’ is n u m b e r o f c o l u m n s )
( a ) T h e r e is n o d e g e n e r a c y
( b ) P r o b l e m is u n b a l a n c e d
( c ) P r o b l e m is d egen er a t e
( d ) S ol ut i on is o p t i m a l
( 7 ) W h i c h of t he fo l l o wi n g consi ders di fference b e t w e e n t w o least co s t s for e a c h r o w a n d
c o l u m n w h i l e fi ndi ng initial basi c feasible sol ut i on in t ransport at i on?
( a ) N o r t h w e s t co rn er rule
( b ) Leas t cost m e t h o d
( c ) Vo g e l ’s a p p r o x i m a t i o n m e t h o d
( d ) R o w m i n i m a m e t h o d
( 8 ) T h e sol ut i on to a transportation p r o b l e m w i t h m - s o u r c e s a n d n-dest i nat i ons is feasible if t he
n u m b e r s of allocations are _ .
(a)
(b )
(c)
(d )
m + n
m n
m - n
m + n -1
( 9 ) W h e n t he total d e m a n d is equal to s u p p l y t h en t he transportation p r o b l e m is sai d to b e
( a ) b a l a n c e d
( b ) u n b a l a n c e d
( c ) m a x i m i z a t i o n
( d ) m i n i m i z a t i o n
( 1 0 ) W h e n the total d e m a n d is not equal to s u p p l y then it is said to b e .
( a ) b a l a n c e d
( b ) u n b a l a n c e d
( c ) m a x i m i z a t i o n
( d ) m i n i m i z a t i o n
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85. (11) The allocation cells in the transportation table will be called cell
(a) occupied
(b) unoccupied
(c) no
(d) finite
(12) In the transportation table, empty cells will be called .
(a) occupied
(b) unoccupied
(c) basic
(d) non-basic
(13) In a transportation table, an ordered set of or more cells is said to form a loop
(a) 2
(b) 3
(c) 4
(d) 5
(14) To resolve degeneracy at the initial solution, a very small quantity is allocated in
cell
(a) occupied
(b) basic
(c) non-basic
(d) unoccupied
(16) For finding an optimum solution in transportation problem method is used.
( a ) Modi
(b) Hungarian
(c) Graphical
(d) simplex
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