OR

1. C h a p t e r Three
Transportation and Assignment Problem

2. 3.1.Transportation problem
 One important application of linear programming is the
area of physical distribution (transportation) of goods
from several supply centers (origins) to several demand
centers (destinations).
 Transportation problem involves a large number of
variables (transportation/shipping routes) and constraints,
it takes along time to solve it.
 Therefore, other methods (transportation algorithm)
havebeen developed for this purpose.

3. 



Objective:
amount of
the objective
commodities
is to determine the
which should be
transpor ted from several sources to different
destinations, atthe minimum transportation
cost and /or time.
Sources or origins are the places where goods
originate from (like plants,warehouses etc)
Destinations are placeswhere goods are to be shipped.
It can also be applied to the maximization of some total
value or utility,in such awaythatthe profit is maximized.

4. 3.1.1. GeneralTransportation
Problem Model
The transportation algorithm requires the
assumptions that:
All goods are homogeneous, so that anyoriginis
capableofsupplyingtoanydestination.
Transportation costs are alinear function of (or
directlyproportionalto) the quantity shipped over
any route.
Each sourcehas a fixedsupplyofunits,where this
entire supply must be distributed to the
destinations.
Similarly, each destination has a fixed demand for
units, where this entire demand must be received
from the sources.

5. A transportation problem model, which has ‘m’ sending
locations (origins) and ‘n’ receiving locations
(destinations), provides a framework for presenting all
relevant data. These are:
Quantity supply of each origin (SSi)
Quantity demand of each destination (DDi)
Unit transportation cost from each origin to each
destination (Cij)

6. Where:
SSi -istotalquantityofcommodityavailableatoriginI
(totalsupplyoforigini).
ddj-istotalquantityofcommoditiesneededat
destinations j(total demand of destination j).
Cij -measures thecostsoftransportingoneunitof
commodityfromsourceitodestinationj.
Xij -isthequantityofcommoditiestransportedfrom
ithorigintojthdestination.

7. The Linear ProgrammingRepresentation of a
Transportation Model

8. Minimize (total transportation cost)
Z=C11X11+C12X12+C13X13+….+CmnXmn
Subject to:
Capacity constraints (SS constraints)
X11+X12+----+X1n=SS1
X21+X22+----+X2n=SS2
. . . .
. . . .
Xm1+Xm2+.---+Xmn=SSm
Requirements constraints (DD constraint)
X11+X21+----+Xm1=SS1
X21+X22+----+Xm2=SS2
. . . .
. . .
. . .
.
.
Xn1+Xn2+.---+Xnm=SSn
Xij ≥ 0 for all i and j

9. Before applying the transportation techniques
(methods) to solve a specific problem, the
problem should satisfy the following
conditions.
Supplies(SS)and requirements(DD) must be
expressed in the same unit.
Thisconditionmeansthatshipmentsreceivedatany
destinationfromdifferentsources mustbe
indistinguishable.
In otherwords,allshipmentmustbemeasuredin
homogenous units.

10. T
otalsupplymustequaltototaldemand∑SS= ∑DD
Theproblemsatisfyingthisconditioniscalled
balanced transportation problem; otherwiseitis
knownas unbalanced transportation problem.
The condition ∑SS= ∑DD isthenecessary and
sufficient condition fortheexistenceoffeasible
solutiontothetransportation problem.

11. 5.1.2. Methods of Solving
Transportation Problem
Methods to get the initial solution: even if
there are different methods of such types, the
following three common methods can be used.
The North-West Corner Method
(NWC)
Minimum cost method (MCM), and
The Vogel’sApproximationMethod
(VAM).

12. 1.North West Corner method
1st. Balance the problem.That isseewhetherΣDD =
ΣSS.
 If not open a dummy column or dummy row
as the case maybe and balance the problem.
2nd. Allocate as much as possible to the selected
cell,and adjust the associated amounts of supply
and demand by subtracting the allocated amount.
Since :The method starts at the northwest-corner
cell (route) of the tableau (variable X11).

13. 3rd. Cross out the row or column with zero supply
or demand to indicate that no further
assignments can be made in that row or column.
If both arow and acolumn net to zero
simultaneously, crossoutoneonly
,and leaveazero
supply (demand) in the uncrossed-out row
(column).
4th Ifexactlyonerow or column is leftuncrossed
out,stop.
Otherwise, move to the cell to the right ifacolumn
has just been crossed out or below ifarow has
been crossed out. Go to step 1.

14. 5th Make sure that allthe rim conditions are satisfied
and (m+n-1) cells are allocated.
6th Once allthe allocations are over
,i.e.,bothrim
requirement(columnandrowi.e.,availabilityand
requirement constraints) are satisfied, write
allocations and calculate the cost of
transportation.
Example 1.
Let us consider an example atthis juncture to
illustrate the application of NWC rule

15. To
From
D1 D2 D3 Total supply
S1 X11 5 X12 6 X13 7 70
S2 X21 4 X22 2 X32 5 30
S3 X31 1 X32 5 X33 4 50
Total Demand 65 42 43 150
150
ΣDD = ΣSS .The North West Cornercell X11 is chosen for
allocation..

16. The origin S1 has 70 items and the destination D1
requires only 65 items.
Hence it is enough to allot65 itemsfrom S1 toD1.
The origin S1 which is alivewith 5 more items can
supply to the destination to the right is alivewith
5 moreitemscan supplyto the destination to the
right of D1 namely D2 whose requirement is 42. So,
we supply 5 itemstoD2 therebytheoriginS1 is
exhausted.

17. D2 requires 37 items more. Now consider the
origin S2 that has 30 items to spare.We allot 30
itemstothecell(X22) so that the origin S2 is
exhausted.
Then, move to origin S3 and supply 7 more items
to the destination D2. Now the requirement of the
destination D2 is complete and S3 isleft with 43
items and the same can be allotted to the
destination D3.
Now the origin S3 is emptied and the requirement
atthe destination D3 is also complete.This
completes the initial solution to the problem

18. To
From
D1 D2 D3 Total supply
S1 65 5 5 6 7 70
S2 4 30 2 5 30
S3 1 7 5 43 4 50
Total Demand 65 42 43
150
150
The total cost of transportation by this method will be:
(65 × 5) + (5 × 6) + (30 × 2) + (7 × 5) + (43× 4) = 325 + 30 + 60 + 35
+ 172 = 622.

19. Example 2.
To
From
X Y Z Dummy Availability
A 7 4 3 3 2 0 10
B
5 8 6 1 0 8
C
6 1 4 4 3 0 5
D
3 5
1 4
5
0
6
Requirement 7 12 5 5 29
The total cost of transportation by this method will be:
(7x4)+(3x3)+(8x6)+(1x4)+(4x3)+(1x4)+(5x0) = 105

20. Tra
nsp
orta
tion
-20
Example 3
Destination
Supply
Source N S E W
A
16 13 22 17
200
100 100
B
14 13 19 15
350
40 300 10
C
9 20 23 10
150
150
Dummy
0 0 0 0
90
90
Demand 100 140 300 250
Z = 10770

21. 2.Least-Cost Method
Steps in Least-Cost Method:
Step 1: Determine the least cost among all the
rows of the transportation table.
Step 2: Identify the row and allocate the
maximum feasible quantity in the cell
corresponding to the least cost in the row.
Then eliminate that row (column) when an
allocation is made.

22. Step 3:Repeat steps 1 and 2 for the reduced
transportation table until allthe available quantities
are distributed to the required places.
If the minimum cost is not unique, the tie can be
broken arbitrarily.
Step 4: Make sure that allthe rim conditions are
satisfied and (m+n-1) cells are allocated.

23. To
From
D1 D2 D3 Total supply
S1 5 7 8 70
S2 4 4 6 30
S3 6 7 7 50
Total Demand 65 42 43 150
150

24. We examine the rows S1, S2 and S3, 4 istheleast
costelementinthecell(S2, D1)and
(S2, D2) and the tie can be brokenarbitrarily.Select
(S2,D1).
The originS2 can supply30 itemstoD1 andthus
originS2 is exhausted.
D1 stillrequires35 moreunits.Hence, shade the row
S2. Shading S2, we observe that 5 istheleastelement
inthecell(S1, D1) and examine the supply atS1 and
demand at D1.

25. The destination D1 requires 35 itemsand this
requirementissatisfiedfromS1 so that the column D1
is shaded next.
Next, we choose 7 as leastelementcorrespondingto
thecell(S1, D2).
We supply35 unitsfromS1 toD2. Now, only one row
is left behind.
Hence, weallow7 itemsfromS3 toD2 and43 itemsS3
to D3.

26. To
From
D1 D2 D3 Total supply
S1 35 5 35 7 8 70
S2 30 4 4 6 30
S3 6 7 7 43 7 50
Total Demand 65 42 43
150
150
The cost of the allocation by the least cost method is
(35 x 5) + (35 x 7) + (30 x 4) + (7 x 7) + (43 x 7) = 890

27. Example 2
T
o (A)
Albuquerque
(B)
Boston
(C)
Cleveland
(D) Des Moines
(E) Evansville
(F) Fort Lauder
Warehouse
requirement 300 200 200
Factory
capacity
300
300
100
700
$5
$5
$4
$4
$3
$3
$9
$8
$7
From
100
First, $3 is the lowest cost cell so ship 100 units fromDes
Moines to Cleveland and cross off the first row as Des
Moines is satisfied
Figure C.4
© 2011 Pearson Education

28. T
o (A)
Albuquerque
(B)
Boston
(C)
Cleveland
(D) Des Moines
(E) Evansville
(F) Fort Lauderdale
Warehouse
requirement 300 200 200
Factory
capacity
300
300
100
700
$5
$5
$4
$4
$3
$3
$9
$8
$7
From
100
100
Second, $3 is again the lowest cost cell so ship 100 units
from Evansville to Cleveland and cross off column C as
Cleveland is satisfied
Figure C.4
© 2011 Pearson Education

29. T
o (A)
Albuquerque
(B)
Boston
(C)
Cleveland
(D) Des Moines
(E) Evansville
(F) Fort Lauderdale
Warehouse
requirement 300 200 200
Factory
capacity
300
300
100
700
$5
$5
$4
$4
$3
$3
$9
$8
$7
From
100
100
200
Third, $4 is the lowest cost cell soship 200 units from
Evansville to Boston and cross off column B and row E as
Evansville and Boston are satisfied
Figure C.4
© 2011 Pearson Education

30. T
o (A)
Albuquerque
(B)
Boston
(C)
Cleveland
(D) Des Moines
(E) Evansville
(F) Fort Lauderdale
Warehouse
requirement 300 200 200
Factory
capacity
300
300
100
700
$5
$5
$4
$4
$3
$3
$9
$8
$7
From
100
100
200
300
Finally,ship 300 units fromAlbuquerque to Fort Lauder
as this is the only remaining cell to completethe
allocations
Figure C.4
© 2011 Pearson Education

31. T
o (A)
Albuquerque
(B)
Boston
(C)
Cleveland
(D) Des Moines
(E) Evansville
(F) Fort Lauderdale
Warehouse
requirement 300 200 200
Factory
capacity
300
300
100
700
$5
$5
$4
$4
$3
$3
$9
$8
$7
From
100
100
200
300
Total Cost
Figure C.4
© 2011 Pearson Education
= $3(100) + $3(100) + $4(200) + $9(300)
= $4,100

32. 3.Vogel Approximation Method
Steps inVAM:
Step 1: For each row (column), determine apenalty measure
by subtractingthesmallestunitcostelementintherow(column)
fromthenextsmallestunitcostelementinthesamerow
(column).
Step 2: Identify the row or column withthelargestpenalty.If
there isa tie (equal penalty) it can be broken by selectingthe
cellwheremaximumallocationcan bemade.
Allocate as much as possible to thevariablewiththeleastunit
costintheselectedrowor column.
Adjustthesupplyand demand,and crossoutthesatisfiedrowor
column.
If arow and acolumn are satisfied simultaneously, only one of
the two is crossed out, and the remaining row (column) is
assigned zero supply (demand).

33. Step 3:
(a)Ifexactlyoneroworcolumnwithzerosupply
ordemandremainsuncrossed out,stop.
(b)Ifonerow(column)withpositivesupply
(demand)remainsuncrossedout,determinethe
basicvariablesintherow(column)bytheleast-
costmethod. Stop.
(c)Ifalltheuncrossedoutrowsandcolumnshave
(remaining)zerosupplyanddemand,determine
thezerobasicvariablesbytheleast-costmethod.
Stop.
(d) Otherwise,gotostep 1.
Step 4: Make sure that alltherimconditionsare
satisfied and cellsareallocated.

34. To
From
D1 D2 D3 Total
supply
Row penalty
S1 65 5 5 7 8 70 2 2(ii) 1(iii) 0
S2 4 30 4 6 30 0 0 0 0
S3
6 7 7 43 7 50 1 1 0 0
Total
Demand
65 42 43 150
Column
penalty
1 3(i) 1
1 0 1
0 0 1
The cost of allocation (i.e., the associated objective value) by Vogel's
Approximation Method will be: (65× 5) + (5× 7) + (30 × 4) + (7 × 7) + (43× 7) =
325 + 35 +120 + 49 + 301 = 830.

35. The differencebetweenthesmallestand nexttothe
smallestelementineach rowand ineach columnis
calculated.
We choose the maximum from among the
differences.
The first individual allocation will be to the
smallest cost of a row or column with the
largest difference.
So we select the column D2 (penalty = 3) for the first
individual allocation, and allocate to (S2, D2) as much
as we can,since this cell has the least cost location.
Thus 30 units from S2 are allocated to D2.This exhausts
the supply from S2. However
, there is still a demand of
12 unitsfrom D2.

36. The allocations to other cells in that column are 0.The
next step is to cross out row S2 (as it is exhausted).
The next largest unit difference corresponds to the
row S1.This leads to an allocation in the corresponding
minimumcostlocationinrowS1, namelycell(S1, D1).
The maximum possible allocation is only 65 as required
byW1 fromS1 and allocation of 0 to others in the row
S1. Column D1 is thus crossed out.
Maximum difference is 1 in row S3 and in column D3.
Select arbitrarily S3 and allot the least cost cell (S1, D2)
5 units. Cross out row S1 for it is alreadyexhausted.
Now, we haveonly one row S3 and two columns D2
and D3 indicating that the entire availableamount from
S3 has to be moved to D2 and D3 as per their
requirements.

37. T
ra
Vogel’sMethod (1): calculate differences
Destination
nsp
orta
tion
-37
Supply diff
Source N S E W
A
16 13 22 17
200 3
B
14 13 19 15
350 1
C
9 20 23 10
150 1
Dummy
0 0 0 0
90 0
Demand 100 140 300 250
diff 9 13 19 10

38. T
ra
Vogel’
s Method (2):select xDummyE as basic
variable
Supply diff
Source N
Destination
S E W
Demand 100 140 300 250
diff 9 13 19 10
A
16 13 22 17
200 3
B
14 13 19 15
350 1
C
9 20 23 10
150 1
Dummy
0 0 0 0
90 0
90
nsp
orta
tion
-38

39. Vogel’sMethod (3): update supply,demand
and differences
Destination
Supply diff
Source N S E W
A
16 13 22 17
200 3
B
14 13 19 15
350 1
C
9 20 23 10
150 1
Dummy
0 0 0
90
0
--- ---
Demand 100 140 210 250
diff 5 0 3 5 Tra
nsp
orta
tion
-39

40. Vogel’
s Method (4):select xCN as basic variable
Destination
Supply diff
Source N S E W
A
16 13 22 17
200 3
B
14 13 19 15
350 1
C
9
100
20 23 10
150 1
Dummy
0 0 0
90
0
--- ---
Demand 100 140 210 250
diff 5 0 3 5 Tra
nsp
orta
tion
-40

41. Vogel’sMethod (5): update supply,demand
and differences
Destination
Supply diff
Source N S E W
A
16 13 22 17
200 4
B
14 13 19 15
350 2
C
9
100
20 23 10
50 10
Dummy
0 0 0
90
0
--- ---
Demand --- 140 210 250
diff --- 0 3 5 Tra
nsp
orta
tion
-41

42. Vogel’
s Method (6):select xCW as basic
variable
Destination
Supply diff
Source N S E W
A
16 13 22 17
200 4
B
14 13 19 15
350 2
C
9
100
20 23 10
50
50 10
Dummy
0 0 0
90
0
--- ---
Demand --- 140 210 250
diff --- 0 3 5 Tra
nsp
orta
tion
-42

43. Vogel’sMethod (7): update supply,demand
and differences
Destination
Supply diff
Source N S E W
A
16 13 22 17
200 4
B
14 13 19 15
350 2
C
9
100
20 23 10
50
--- ---
Dummy
0 0 0
90
0
--- ---
Demand --- 140 210 200
diff --- 0 3 2 Tra
nsp
orta
tion
-43

44. Vogel’
s Method (8):select xAS as basic
variable
Destination
Supply diff
Source N S E W
A
16 13
140
22 17
200 4
B
14 13 19 15
350 2
C
9
100
20 23 10
50
--- ---
Dummy
0 0 0
90
0
--- ---
Demand --- 140 210 200
diff --- 0 3 2 Tra
nsp
orta
tion
-44

45. Vogel’sMethod (9): update supply,demand
and differences
Destination
Supply diff
Source N S E W
A
16 13
140
22 17
60 5
B
14 13 19 15
350 4
C
9
100
20 23 10
50
--- ---
Dummy
0 0 0
90
0
--- ---
Demand --- --- 210 200
diff --- --- 3 2 Tra
nsp
orta
tion
-45

46. Vogel’
s Method (10):select xAW as basic
variable
Destination
Supply diff
Source N S E W
A
16 13
140
22 17
60 60 5
B
14 13 19 15
350 4
C
9
100
20 23 10
50
--- ---
Dummy
0 0 0
90
0
--- ---
Demand --- --- 210 200
diff --- --- 3 2 Tra
nsp
orta
tion
-46

47. Vogel’sMethod (11): update supply,demand
and differences
Destination
Supply diff
Source N S E W
A
16 13
140
22 17
60 --- ---
B
14 13 19 15
350 4
C
9
100
20 23 10
50
--- ---
Dummy
0 0 0
90
0
--- ---
Demand --- --- 210 140
diff --- --- Tra
nsp
orta
tion
-47

48. Vogel’
s Method (12):select xBW and xBE as
basic variables
Destination
Supply diff
Source N S E W
A
16 13
140
22 17
60 --- ---
B
14 13 19
210
15
140 ---
C
9
100
20 23 10
50
--- ---
Dummy
0 0 0
90
0
--- ---
Demand --- --- --- ---
diff --- ---
Z = 10330
Tra
nsp
orta
tion
-48

49. 3.2.Assignment problem
In many business situations, management
needs to assign - personnel to jobs, - jobs
to machines, - machines to job locations,
or - salespersons to territories.
Consider the situation of assigning n jobston
machines.
When ajob i (=1,2,....,n)is assigned to
machine j(=1,2, .....n)that incurs acost Cij.
The objective is to assign the jobs to
machines at the least possible total cost.

50. This situation is aspecial case of the
transportation model and it is known as the
assignment problem.
Here, jobs represent “sources”and
machines represent“destinations.”
The supplyavailableateach sourceis1 unitand
demand at each destination is 1 unit.
Formulation/construction of the
model

51. The assignment model can be expressed mathematically as
follows:
Xij= 0, ifthejobjisnotassignedtomachine i
1,ifthejobjisassignedtomachine I
Now the problem is which work isto be assigned to whom
so that the cost of completion of work will be minimum.

52. To minimize z (cost)
where xij =
1; if ith person is assigned jthwork
0; if ith person is not assigned the jthwork
i.e., ith person will doonly one work.
i.e., jth work will be done only by oneperson.

53. Example
A farm produce different agricultural products and that
Products are manufactured on five different assembly lines
(1,2,3,4,5).
When manufacturing is finished, products are transported from
the assembly lines to one of the five different potential customers
(A,B,C,D,E).
Transporting products from five assembly lines to five inspection
areas requires different times (in minutes)
potential customers

54. Methods of solvingAP
(The Hungarian Method)
In order to find the proper assignment itis
essential for us to know the Hungarian method.
Step I
(A) Row reduction:
Select the smallest value in each row.
Subtract this value from each value in that row in
the cost matrix.
(B) Column reduction:
After completion of row reduction, subtract the
minimum entry of each column from all the
entries of the respective column.

55. Step II
Zero assignment:
(A) Starting with first row of the matrix received in
first step,examine the rows one byone until a
row containing exactly one zero is found.
 Then an experimental assignment indicated
by‘ ’ismarked to that zero.
 Now cross all the zeros in the column in which
the assignment is made.
 Cross out zero, if there are other zero in ether
column or row.
 This procedure should be adopted for each
row assignment.

56. (B) When the rows has been
completely an identical
procedure is
set of
examined,
applied successively to
columns.
Starting with column 1, examine all
columns until a column containing
exactly one zero is found.
Then makean experimentalassignmentinthat
position and
Cross otherzeros intherowinwhichthe
assignmentwas made.

57. Continue these successive operations on rows and
columns until all zero’shave either been
assigned or corssed-out.
Now there are two possibilities:
(a)Either all the zeros are assigned or crossed
out, i.e.,we get the maximal assignment.
or
(b)At least two zeros are remained by
assignment or bycrossing out in each row or
column.
In this situation we try to exclude some of the
zeros bytrial and error method.
This completes the second step.

58. After this step we can get two situations.
(i) Totalassigned zero’s= n
The assignment is optimal.
(ii)Totalassigned zero’s< n
Use step III and onwards.
Since n= # of assignments
Step III
Draw of minimum lines to cover zero’s :
In order to cover all the zero’sat least
once you may adopt the following
procedure.

59. (i)Marks (√) to allrows in which the assignment
has not been done.
(ii) See thepositionofzeroinmarked(√) rowand then
mark (√) to the corresponding column.
(iii)See the marked (√) column and find the
position of assigned zero’sand then mark (√) to
thecorrespondingrowswhicharenotmarked tillnow
.
(iv)Repeattheprocedure(ii)and (iii)tillthecompletion
of marking.
(v)Draw the lines through unmarkedrowsand
marked columns.
Note: If the above method does not work then
make an arbitraryassignmentandthenfollowstepIV
.

60. Step IV
Select the smallest element from the uncovered
elements :
(i)Subtract this smallest element from all those
elements which are not covered.
(ii)Add this smallest element to all those
elements which are at the intersection of two
lines.
Step V
Thus we haveincreasedthenumberof zero’s.
Now, modifythematrixwiththehelpofstepII and
findtherequired assignment.
Finally calculate the total minimized cost by
summing up numbers from the original table

61. Example 1.
Four persons A,B,C and D are to be assigned four
jobs I, II, III and IV
. The cost matrix is given as
under
, find the proper assignment.
Man
Jobs
A B C D
I 8 10 17 9
II 3 8 5 6
III 10 12 11 9
IV 6 13 9 7
Solution :
In order to find the proper assignment we apply the
Hungarian algorithm as follows:

62. I (A) Row reduction
Man
A B C D
Jobs
I 0 2 9 1
II 0 5 2 3
III 1 3 2 0
IV 0 7 3 1
Man
Jobs
A B C D
I 0 0 7 1
II 0 3 0 3
III 1 1 0 0
IV 0 5 1 1
I (B) Column reduction

63. II Zero assignment:
Man
Jobs
A B C D
I 0 0 7 1
II 0 3 0 3
III 1 1 0 0
IV 0 5 1 1
In this way allthe zero’s are either crossed out or assigned. Also
total assigned zero’s = 4 (i.e.,number of rows or columns).
Thus, the assignment is optimal.
From the table we get:
I B;
II. C:
III. D and
IV A. T
otal minimized cost = 10+5+9+6= 30

64. Example 2:
There are five machines and five jobs are to be
assigned and the associated cost matrix is as
follows. Find the proper assignment.
machines
Jobs
I II III IV V
A 6 12 3 11 15
B 4 2 7 1 10
C 8 11 10 7 11
D 16 19 12 23 21
E 9 5 7 6 10

65. Solution:
In order to find the proper assignment, we apply the
Hungarian method as follows:
I A (Rowreduction)
machines
Jobs
I II III IV V
A 3 9 0 8 12
B 3 1 6 0 9
C 1 4 3 0 4
D 4 7 0 11 9
E 4 0 2 1 5

66. machines
J
obs
I II III IV V
A 2 9 0 8 8
B 2 1 6 0 5
C 0 4 3 0 0
D 3 7 0 11 5
E 3 0 2 1 1
II (Zero assignment)
IB (Column reduction) machines
I II III IV V
A 2 9 0 8 8
Jobs
B 2 1 6 0 5
C 0 4 3 0 0
D 3 7 0 11 5
E 3 0 2 1 1

67. From the last table we see that allthe zeros are
either assigned or crossed out, butthetotalnumber
ofassignment,i.e.,4<5 (numberofjobstobeassigned
to machines).
Therefore, we haveto follow step III and onwards
as follows:
machines
I II III √ 2nd IV V
A 2 9 0 8 8 √3rd
Jobs
B 2 1 6 0 5
C 0 4 3 0 0
D 3 7 0 11 5 √ 1st
E 3 0 2 1 1

68. Step IV:Here, the smallest element among the uncovered
elements is 2.
(i)Subtract 2 from all those elements which are not
covered.
(ii)Add 2 to those entries which are at the
junction(intersection ) of two lines.
Make as it is rows that pass single line .
machines
Jobs
I II III IV V
A 0 7 0 6 6
B 2 1 8 0 5
C 0 4 5 0 0
D 1 5 0 9 3
E 3 0 4 1 1

69. Step V.using step II again
machines
Jobs
I II III IV V
A 0 7 0 6 6
B 2 1 8 0 5
C 0 4 5 0 0
D 1 5 0 9 3
E 3 0 4 1 1
Thus, we have got five assignments as required bythe
problem.
The assignment is as follows:
A I, B  IV, C V,D  III and E  II.
Thus from
the cost matrix the minimum cost = 6+1+11+12+5=35.

70. START
Construct the effectiveness matrix if not already given
Row reduction
Column reduction
Is zero assignment
possible
(i) Draw minimum number of lines to cover
all the zeros
(ii)Choose the least uncovered element
(iii)Subtract this from the uncovered
elements and add it to the elements at
intersection of the lines.
ASSIGNMENT
Put square over the zero and
cross out all zeros (if any) of
the corresponding column.
SOLUTION
Add the elements of the given
matrix correspond to each
square from original table
STOP
Is Zero
assignment
Possible
Y
es
No
No
Y
es