Transportation problem

TRANSPORTATION PROBLEMS (TPs)

WHAT IS TRANSPORTATION PROBLEM?
A TRANSPORTATION PROBLEM (TP) CONSISTS OF
DETERMINING HOW TO ROUTE PRODUCTS IN A
SITUATION WHERE THERE ARE SEVERAL SUPPLY
LOCATIONS AND ALSO SEVERAL DESTINATIONS IN
ORDER THAT THE TOTAL COST OF TRANSPORTATION IS
MINIMISED

TRANSPORTATION PROBLEMS (TPs)

SUPPLY DEMAND

A D

B E

C F

MATHEMATICAL STATEMENT OF A
TRANSPORTAION PROBLEM
LET
ai = QUANTITY OF PRODUCT AVAILABLE AT SOURCE i,
bj = QUANTITY OF PRODUCT REQUIRED AT DESTINATION j,
cij = COST OF TRANSPORTATION OF ONE UNIT OF THE PRODUCT FROM
SOURCE i TO DESTINATION j,
xij = QUANTITY OF PRODUCT TRANSPORTED FROM SOURCE i TO
DESTINATION j.
ASSUMING THAT, TOTAL DEMAND = TOTAL SUPPLY, ie, Sai = Sbj THEN
THE PROBLEM CAN BE FORMULATED AS A LPP AS FOLLOWS.
m n
MINIMISE TOTAL COST Z = SS cij xij
i=1 j=1
n
SUBJECT TO Sxij = ai FOR i = 1,2,3…m
j=1
m
S xij = bj FOR j = 1,2,3…n
i=1
AND xij ≥ 0
FULL STATEMENT OF TP AS LPP

TRANSPORTATION MODEL –TABULAR FORM
.
SOURCES DESTINATIONS SUPPLY ai
1 2 3 … n

1 x11 x12 x13 x1n a1
c11 c12 c13 c1n
2 x21 x22 x23 x2n a2
c21 c22 c23 c2n
3 x31 x32 x33 x3n a3
c31 c32 c33 c3n
.
.
.
m xm1 xm2 xm3 xmn am
cm1 cm2 cm3 cmn
DEMAND bj b1 b2 b3 bn Sai= Sbj
MINIMISE TOTAL COST Z = SS cij xij FOR i=1 to m & j=1 to n

TRANSPORTAION PROBLEMS (TPs)
• TRANSPORTATION COST PER UNIT MATRIX
• TRANSPORTATION DECISION VARIABLE MATRIX
• SUPPLY COLUMN
• DEMAND ROW
• TOTAL TRANSPORTATION COST
• SOLUTION OF THE TRANSPORTATION PROBLEM
• OCCUPIED CELLS
• EMPTY CELLS
• CONSTRAINTS IN A TP
• VARIABLES IN A TP:

TRANSPORTATION PROBLEM–TABULAR FORM
.
SOURCES DESTINATIONS SUPPLY ai
1 2 3
1 x11 x12 x13 a1
c11 c12 c13
2 x21 x22 x23 a2
c21 c22 c23
3 x31 x32 x33 a3
c31 c32 c33
DEMAND bj b1 b2 b3 Sai= Sbj

MINIMISE TOTAL COST
Z = SS cij xij FOR i=1 to m & j=1 to n

EXAMPLE 1 OF TP
A STEEL COMPANY HAS THREE PLANTS P1,P2 AND P3 WITH ANNUAL
.
CAPACITIES OF 45,15 AND 40 THOUSAND TONNES OF CR COILS. THE
PRODUCT IS DISTRIBUTED FROM THREE WAREHOUSES W1,W2 AND
W3 WITH ANNUAL OFFTAKE OF 25,55 AND 20 THOUSAND TONNES OF
CR COILS. THE TRANSPORTATION COST (Rs LAKH PER THOUSAND
TONNES) IS AS PER FOLLOWING TABLE. FIND OPTIMUM
TRANSPORTATION SCHEDULE TO MINIMISE COST.
SOURCES DESTINATIONS SUPPLY
W1 W2 W3
P1
10 7 8 45
P2
15 12 9 15
P3
7 8 12 40
DEMAND 25 55 20 100

TP FOMULATED AS A LPP
Min Z =10x11+ 7x12+ 8x13+ 15x21+ 12x22+ 9x23+ 7x31+ 8x32+ 12x33
Subject to
x11+ x12+ x13= 45
x21+ x22+ x23= 15 SUPPLY CONSTRAINTS
x31+ x32+ x33= 40
x11+ x21+ x31= 25
x12+ x22+ x32= 55 DEMAND CONSTRAINTS
x13+ x23+ x33= 20
xij ≥ 0 FOR i = 1,2,3 AND j = 1,2,3

TP REFOMULATED AS A LPP
FOR SIMPLEX METHOD
Min Z =10x11+ 7x12+ 8x13+ 15x21+ 12x22+ 9x23+ 7x31+ 8x32+ 12x33
+ MA1 + MA2 + MA3 + MA4 + MA5 + MA6
Subject to
x11+ x12+ x13+A1= 45
x21+ x22+ x23+A2= 15 SUPPLY CONSTRAINTS
x31+ x32+ x33+A3= 40
x11+ x21+ x31+A4= 25
x12+ x22+ x32+A5= 55 DEMAND CONSTRAINTS
x13+ x23+ x33+A6= 20
xij ≥ 0 FOR i = 1,2,3 AND j = 1,2,3

DUAL OF A TP FORMULATED AS A LPP
Max G = u1+ u2+ u3+ v1+ v2+ v3+
Subject to
u 1+ v 1 +

METHODS OF SOLVING A TP
1. SIMPLEX METHOD
TP CAN BE STATED AS LPP AND THEN SOLVED BY THE
SIMPLEX METHOD

2. TRANSPORTATION METHOD
THIS INVOLVES THE FOLLOWING STEPS
i) OBTAIN THE INITIAL FEASIBLE SOLUTION USING
- NORTH WEST CORNER RULE
- VOGEL’S APPROXIMATION METHOD
ii) TEST FEASIBLE SOLUTION FOR OPTIMALITY USING
- STEPPING STONE METHOD
- MODIFIED DISTRIBUTION METHOD
iii) IMPROVE THE SOLUTION BY REPEATED ITERATION

NORTH WEST CORNER (NWC) RULE

1. START WITH THE NW CORNER OF TP TABLE
2. TAKE APPROPRIATE STEPS IF
a1 > b1
a1 < b1
a1 = b1
3. COMPLETE INITIAL FEASIBLE SOLUTION TABLE

VOGEL’S APPROXIMATION METHOD - STEPS

1. FIND DIFFERENCE IN TRANSPORTATION COSTS
BETWEEN TWO LEAST COST CELLS IN EACH ROW AND
COLUMN.
2. IDENTIFY THE ROW OR COLUMN THAT HAS THE
LARGEST DIFFERENCE.
3. DETERMINE THE CELL WITH THE MINIMUM
TRANSPORTATION COST IN THE ROW/COL
4. ASSIGN MAXIMUM POSSIBLE VALUE TO xij VARIABLE IN
THE CELL IDENTIFIED ABOVE
5. OMIT ROW IF SUPPLY EXHAUSTED AND OMIT COL IF
DEMAND MET
6. REPEAT STEPS 1 AND 5 ABOVE

TESTING FEASIBLE SOLUTION FOR
OPTIMALITY

1. STEPPING STONE METHOD

2. MODIFIED DISTRIBUTION METHOD

TESTING FEASIBLE SOLUTION FOR
OPTIMALITY

STEPPING STONE METHOD
1. IDENTIFY THE EMPTY CELLS
2. TRACE A CLOSED LOOP
3. DETERMINE NET COST CHANGE
4. DETERMINE THE NET OPPORTUNITY
5. IDENTIFY UNOCCUPIED CELL WITH THE
LARGEST POSITIVE NET OPPORTUNITY COST
6. REPEAT STEPS 1 TO 5 TO GET THE NEW
IMPROVED TABLES

TESTING FEASIBLE SOLUTION FOR OPTIMALITY
RULES FOR TRACING CLOSED LOOPS
1. ONLY HORIZONTAL OR VERTICAL MOVEMENT ALLOWED
2. MOVEMENT TO AN OCCUPIED CELL ONLY
3. STEPPING OVER ALLOWED
4. ASSIGN POSITIVE OR NEGATIVE SIGNS TO CELLS
5. LOOP MUST BE RIGHT ANGLED
6. A ROW OR COL MUST HAVE ONE CELL OF POSITIVE SIGN
AND ONE CELL OF NEGATIVE SIGN ONLY
7. A LOOP MUST HAVE EVEN NUMBER OF CELLS
8. EACH UNOCCUPIED CELL CAN HAVE ONE AND ONLY ONE
LOOP
9. ONLY OCCUPIED CELLS ARE TO BE ASSIGNED POSITIVE
OR NEGATIVE VALUES
10. LOOP MAY NOT BE SQUARE OR RECTANGLE
11. ALL LOOPS MUST BE CONSISTENTLY CLOCKWISE OR
ANTICLOCKWISE

MODIFIED DISTRIBUTION METHOD In case there are a large
number of rows and columns, then Modified distribution (MODI)
method would be more suitable than Stepping Stone method
Step 1
Add ui col and vj row: Add a column on the right hand side of the TP
table and title it ui. Also add a row at the bottom of the TP table and
title it vj.
Step 2 This step has four parts.
i) Assign value to ui=0 To any of the variable ui or vj, assign any
arbitrary value. Generally the variable in the first row i.e. u1 is
assigned the value equal to zero.
ii) Determine values of the vj in the first row using the value of u1 =
0 and the cij values of the occupied cells in the first row by applying
the formula ui + vj = cij
iii) Determine ui and vj values for other rows and columns with the
help of the formula ui + vj = cij using the ui and vj values already
obtained in steps a), b) above and cij values of each of the occupied
cells one by one.
iv) Check the solution for degeneracy. If the soln is degenerate [ie no.
of occupied cells is less than (m+n-1)], then this method will not be
applicable.

MODIFIED DISTRIBUTION METHOD
Step 3
Calculate the net opportunity cost for each of the unoccupied cells
using the formula δij = (ui + vj) - cij. If all unoccupied cells have
negative δij value, then, the solution is optimal.
Multiple optimality: If, however, one or more unoccupied cells have
δij value equal to zero, then the solution is optimal but not unique.
Non optimal solutionIf one or more unoccupied cells have positive δij
value, then the solution is not optimal.
Largest positive dj value: The unoccupied cell with the largest positive
δij value is identified.
Step 4
A closed loop is traced for the unoccupied cell with the largest δij
value. Appropriate quantity is shifted to the unoccupied cell and
also from and to the other cells in the loop so that the transportation
cost comes down.
Step 5
The resulting solution is once again tested for optimality.
If it is not optimal, then the steps from 1 to 4 are repeated, till an optimal
solution is obtained

EXAMPLE 1 OF TP
.

INITIAL BASIC FEASIBLE SOLUTION
BY NORTH WEST CORNER METHOD

W1 W2 W3
P1 25 20
10 7 8 45
P2 15
15 12 9 15
P3 20 20
7 8 12 40
DEMAND 25 55 20 100

EXAMPLE 1 OF TP
. INTIAL BASIC FEASIBLE SOLUTION
BY VOGEL APPROXIMATION METHOD
SOURCES DESTINATIONS SUPPLY ITERATIONS
W1 W2 W3 1 2 3 4
P1 40 5 1 1 3* X
10 7 8 45
P2 15 3* X X X
15 12 9 15
P3 25 15 1 1 1 1*
7 8 12 40
DEMAND 25 55 20 100
1st ITERATION
3 1 1
2nd ITERATION
3 1 4*
3rd ITERATION
3 1 X 1
4th ITERATION
X X X

EXAMPLE 1 OF TP
.
INITIAL BASIC FEASIBLE SOLUTION
BY NORTH WEST CORNER METHOD

W1 W2 W3
P1 25 20
10 7 8 45
P2 15
15 12 9 15
P3 20 20
7 8 12 40
DEMAND 25 55 20 100

TEST FOR OPTIMALITY & IMPROVEMENT OF SOLN


UNBALANCED
TRANSPORTATION PROBLEMS

• TOTAL SUPPLY EXCEEDS
TOTAL DEMAND

• TOTAL DEMAND EXCEEDS
TOTAL SUPPLY

EXAMPLE 2a OF TP
.
CAPACITIES OF 60, 20 AND 40 THOUSAND TONNES OF CR COILS. THE
W3 WITH ANNUAL OFFTAKE OF 25,55 AND 20 THOUSAND TONNES OF
W1 W2 W3
P1
10 7 8 60
P2
15 12 9 20
P3
7 8 12 40
120
DEMAND 25 55 20 100

SOLUTION OF EXAMPLE 2a OF TP
CREATE A DUMMY DESTINATION W4 WITH DEMAND = 20,000 TONNES

W1 W2 W3 W4

P1
10 7 8 0 60
P2
20
15 12 9 0
P3
40
7 8 12 0
120
DEMAND 25 55 20 20 120

EXAMPLE 2b OF TP
.
CAPACITIES OF 55, 15 AND 40 THOUSAND TONNES OF CR COILS. THE
W3 WITH ANNUAL OFFTAKE OF 30, 70 AND 20 THOUSAND TONNES OF
W1 W2 W3
P1
10 7 8 55
P2
15 12 9 15
P3
7 8 12 40
100
DEMAND 30 70 20 120

SOLUTION OF EXAMPLE 2b OF TP
CREATE A DUMMY SOURCE P4 WITH SUPPLY = 20,000 TONNES

W1 W2 W3

P1
10 7 8 55
P2
15
15 12 9
P3
40
7 8 12
P4
20
0 0 0
120
DEMAND 30 70 20 120

. EXAMPLE 3 OF TP (DEGENERACY)
AN ALUMINIUM MANUFACTURER HAS THREE PLANTS A, B, AND C WITH
ANNUAL CAPACITIES OF 60,100 AND 40 THOUSAND TONNES OF
ALUMINIUM INGOTS. THE PRODUCT IS DISTRIBUTED FROM FOUR
WAREHOUSES D, E, F, AND G WITH ANNUAL OFFTAKE OF 20, 50, 50,
AND 80 THOUSAND TONNES OF AL INGOTS. TRANSPORTATION COST
(Rs LAKH PER THOUSAND TONNES) IS AS PER FOLLOWING TABLE.
FIND OPTIMUM TRANSPORTATION SCHEDULE TO MINIMISE COST.
D E F G
A
7 3 8 6 60
B
4 2 5 10 100
C
2 6 5 1 40
DEMAND 20 50 50 80 200

TP TABLE 1 (NON OPTIMAL) dj IS NET OPPOR. AVAIL
D E F G ui (ui+vj)=cij EMPTY NET COST
A 20 40 u1 u1=0
ROW A CELL d =(u +v )-c dj
CHANGE
60 AD: v1= 7 ij i j ij
7 3 8 6 0 AE::v2= 3 AF 0+6-8=-2 -2
B 10 50 40 u2 ROW B AG 0+11-6=+5 +5
100 BE: u2= -1
BF: v3= 6 BD -1+7-4=+2 +2
4 2 5 10 -1 BG: v4= 11
CD -10+7-2=-5 -5
C 40 u3 ROW C
40 CG: u3= -10 CE -10+3-6=-13 -13
2 6 5 1 -10 CF -10+6-5=-9 -9
25 55 20 200 - SELECT THE CELL WITH THE LARGEST
POSITIVE dj VALUE (+5) ie CELL AG
v1 v2 v3 V4 - TRACE LOOP AG-BG-BE-AE
vj 7 3 6 11 - SHIFT 40 UNITS FROM HIGHER COST
CELL BG TO LOWER COST CELL AG
- SHIFT 40 UNITS FROM CELL AE TO BE
Z=7x20+3x40+2x10+5x50 SO THAT DEMAND SUPPLY
CONSTRAINTS ARE NOT AFFECTED
+10x40+1x40 = 970 -THIS GIVES US THE NEXT TABLE 2

TP TABLE 2 (NON OPTIMAL) dj IS NET OPPOR. AVAIL
A 20 e 40 u1 u1=0
CHANGE
60 AD: v1= 7 ij i j ij
7 3 8 6 0 AE::v2= 3 AF 0+6-8=-2 -2
B 50 50 u2 AG:v4= 6 BD -1+7-4=+2 +2
100 ROW B
BE: u2= -1 BG -1+6-10=-5 -5
4 2 5 10 -1 BF: v3= 6
CD -5+7-2=-0 0
C 40 u3 ROW C
40 CG: u3= -5 CE -5+3-6=-8 -8
2 6 5 1 -5 CF -5+6-5=-4 -4
25 55 20 200 -SOLN IS DEGENERATE SINCE NO. OF xij
VARIABLES (5) IS LESS THAN (m+n-1=6). TWO
v1 v2 v3 v3 RECENTLY VACATED CELLS ARE AE & BG
vj ASSIGN e VALUE TO AE SINCE IT HAS LOWER
7 3 6 6 cij VALUE. PROCEED LIKE EARLIER STEP 1
- CELL BD HAS LARGEST dj VALUE =+2
- TRACE LOOP BD-AD-AE-BE
Z=7x20+3xe+6x40+2x50 - SHIFT 20 UNITS FROM AD T0 BD
+5x50+1x40 = 770 (3xe=0) - SHIFT 20 UNITS FROM BE TO AE

TP TABLE 3 (OPTIMAL) dj IS NET OPPOR. AVAIL
A 20 40 u1 u1=0
CHANGE
60 AE::v2= 3 ij i j ij
7 3 8 6 0 AG: v4= 6 AD 0+5-7=-2 -2
B 20 30 50 u2 ROW B AF 0+6-8=-2 -2
100 BD: v1= 5
BE: u2= -1 BG -1+6-10=-5 -5
4 2 5 10 -1 BF: v3= 6
CD -5+5-2=-2 -5
C 40 u3 ROW C
40 - CG: u3= -5 CE -5+3-6=-8 -8
-5 CF -5+6-5=-4 -4
2 6 5 1
25 55 20 200 - SINCE ALL dj VALUE ARE NEGATIVE
v1 v2 v3 V4 THEREFORE THIS SOLUTION
vj 5 3 6 6 IS AN OPTIMAL SOLUTION

Z=3x20+6x40+4x20+3x30
+5x50+1x40 = 730

. EXAMPLE 4 OF TP (MAXIMISATION)
A FERTILIZER COMPANY HAS THREE FACTORIES A, B, AND C WITH
ANNUAL CAPACITIES OF 200, 500 AND 300 THOUSAND TONNES OF
UREA. THE PRODUCT IS DISTRIBUTED FROM FOUR WAREHOUSES
D, E, F, AND G WITH ANNUAL OFFTAKE OF 180, 320, 100,
AND 400 THOUSAND TONNES OF UREA. PROFIT
(Rs LAKH PER THOUSAND TONNES) IS AS PER FOMWING TABLE.
FIND OPTIMUM TRANSPORTATION SCHEDULE TO MAXIIMISE PROFIT.
D E F G
A
12 8 6 25 200
B
8 7 10 18 500
C
14 3 11 20 300
DEMAND 180 320 100 400 1000

.
EXAMPLE 5 OF TP
PROHIBITED ROUTES IN THE TP
SOURCES DESTINATIONS SUPPLY ITERATIONS
W1 W2 W3 U 1 2 3 4 5
P1 40
40 5
5 0 1 1 1 X
M 7 8 45
P2 15
15 1 3 3 3 X
15 12 9 15
P3 25
25 15
15 1 1 4* X X
7 8 12 40
DEMAND 25 55 20 100 1
25
V 6 7 8 2
15
1 M-7* 1 1
3
2 X 1 1 40
3 X 5* 1 4
4 X X 1* 5
5 X 5
15

TP NUMERICALS
Q. NO. 1. (NWC RULE,SSMI METHODS & VAM, MODI METHODS)
A PVC MANUFACTURING COMPANY HAS THREE FACTORIES
A, B, AND C AND THREE WAREHOUSES D, E, AND F. THE
MONTHLY DEMAND FROM THE WAREHOUSES AND THE
MONTHLY PRODUCTION OF THE FACTORIES, IN THOUSAND
OF TONNES OF PVC AND THE TRANSPORTATION COSTS PER
UNIT ARE GIVEN IN THE FOLLOWING TABLE.
WAREHOUSES MONTHLY
FACTORIES D E F PRODN
A 16 19 22 14
B 22 13 19 16
C 14 28 8 12
MONTHLY DEMAND 10 15 17
DETERMINE THE OPTIMAL SHIPPING SCHEDULE SO THAT
THE TRANSPORTATION COST IS MINIMIZED USING
i) NWCR AND SSM
ii) VAM AND MODIFIED DISTRIBUTION METHOD

TP NUMERICALS

Q. NO. 2
SOLVE Q. NO. 1, BY USING VAM AND
MODI DISTRIBUTION METHOD IF IT IS GIVEN
THAT, MONTHLY PRODUCTION OF FACTORIES A,
B AND C IS 16, 20 AND 12 THOUSAND TONNES
RESPECTIVELY MONTHLY DEMAND OF
WAREHOUSES D, E AND F IS 15, 15 AND 20
THOUSAND TONNES RESPECTIVELY.

TP NUMERICALS
Q. No. 3
A LIGHTING PRODUCTS COMPANY HAS FOUR FACTORIES F1, F2, F3,
AND F4, WHICH PRODUCE 125, 250, 175 AND 100 CASES OF 200-WATT
LAMPS EVERY MONTH.
THE COMPANY SUPPLIES THESE LAMPS TO FOUR WAREHOUSES W1,
W2, W3 AND W4 WHICH HAVE DEMAND OF 100, 400, 90 AND 60 CASES
PER MONTH RESPECTIVELY. THE PROFIT IN Rs PER CASE, AS CASES
ARE SUPPLIED FROM A PARTICULAR FACTORY TO A PARTICULAR
WAREHOUES, IS GIVEN IN THE FOLLOWING MATRIX.
WAREHOUSES
W1 W2 W3 W4
FACTORIES F1 90 100 120 110
F2 100 105 130 117
F3 111 109 110 120
F4 130 125 108 113
DETERMINE THE TRANSPORTATION SCHEDULE SO THAT PROFIT IS
MAXIMIZED GIVEN THE CONDITION THAT WARE HOUSE W1 MUST BE
SUPPLIED ITS FULL REQUIREMENT FROM FACTORY F1. USE VAM AND
MODIFIED DISTRIBUTION METHOD.
ALSO SOLVE THE TP WITHOUT THE CONDITION GIVEN ABOVE USING
NWCR AND STEPPING STONE METHOD.

TABLE 1 TP NUMERICALS ANS TO Q NO 3
W1 W2 W3 W4 Dj
F1W3 -5
F1 90 100 120 110 125
F1W4 -2
F2 100 105 130 117 250 F2W1 IGNORE
F3W1 IGNORE
F3 111 109 110 120 175
F4W1 IGNORE
F4 130 125 108 113 100 F3W3 -24
F3W4 -1
100 400 90 60
F4W3 -42
TABLE 2 OPTIMAL SOLN F4W4 -24

W1 W2 W3 W4 1 2 3 4 ui 1ST W1 GETS FULL
QTY FROM F1
F1 40 100 30 25 10 20 125 X 10 10 10 0
F2 30 25 100 0 90 13 60 250 X 13 13* 12* -5 2ND SUPPLY FROM
F4 EXHAUSTED
F3 19 21 175 20 10 175 X 10 10 11 -9
F4 0 5 100 22 17 100 X 12 X X -25 3RD DEMAND FROM
W3 MET
100 400 90 60
1 X X X X 4TH DEMAND FROM
W4 MET
2 X 16* 10 3 NOTE: In the first iteration for VOGEL
we put an X for all rows and columns
3 X 4 10 3 because the constraint is that warehouse 5th SUPPLY FROM
W1 is to be supplied entire quantity from F1 EXHAUSTED
4 X 4 X 3 factory F1
6th SUPPLY FROM
vj 40 30 5 18 F2 EXHAUSTED

7TH DEMAND FROM
W2 MET

TP NUMERICALS

Q. NO 4 ( DEGENERACY)
SOLVE THE FOLLOWING TRANSPORTATION PROBLEM.
D E F G SUPPLY
A 7 3 8 6 60
B 4 2 5 10 100
C 2 6 5 1 40
DEMAND 20 50 50 80

ANS FOR Q NO. 4
TABLE 1 TABLE 2
D E F G Sup Ui D E F G Sup Ui
A 20 40 -2 +5 60 0 A 20 e -2 40 60 0
B +2 10 50 40 100 -1 B +2 50 50 -5 100 -1
C -5 -13 -9 40 40 - C 0 -8 -4 40 40 -5
10 Dmd 20 50 50 80
Dmd 20 50 50 80
Vj 7 3 6 6
Vj 7 3 6 11
TABLE 3 OPTIMAL
D E F G Sup Ui
A -2 20 -2 40 60 0
B 20 30 50 -5 100 -1
C -2 -8 -4 40 40 -5
Dmd 20 50 50 80
Vj 5 3 6 6

TP NUMERICALS
Q. NO. 5
SOLVE THE FOLLOWING TRANSPORTATION PROBLEM
USING VOGEL’S APPROXIMATION METHOD. TEST THIS
SOLUTION FOR OPTIMALITY USING THE MODI METHOD.
DESTINATIONS SUPPLY
SOURCES D E F G
A 6 4 1 5 14
B 8 9 2 7 16
C 4 3 6 2 5
DEMAND 6 10 15 4

ANS FOR Q NO. 5 by VAM
TABLE 1 Optimal
D E F G Sup 1 2 3 Ui
A 4 10 -1 -1 14 3 1 2* 0
B 1 -3 15 -1 16 5* 1 1 2
C 1 -1 -8 4 5 1 1 1 -2
Dmd 6 10 15 4
1 2 1 1 3 1
15
2 2 1 X 3*
2
3 2 1 X X 4

Vj 6 4 0 4 3
10
4
1,4,1

TP NUMERICALS
Q. NO. 6
A COMPANY MANUFACTURING PUMPS FOR DESERT COOLERS SELLS
THEM TO ITS FIVE WHOLE-SELLERS A, B, C, D & E AT RS 250 EACH
AND THEIR DEMAND FOR THE NEXT MONTH IS 300,300, 1000, 500
AND 400 UNITS RESPECTIVELY. THE COMPANY MAKES THESE
PUMPS AT THREE FACTORIES F1, F2 & F3 WITH CAPACITIES OF
500, 1000 AND 1250 UNITS RESPECTIVELY. THE DIRECT COSTS OF
PRODUCTION OF A PUMP AT THE THREE FACTORIES F1, F2 & F3
ARE RS 100, 90 AND 80 RESPECTIVELY. THE COSTS OF
TRANSPORTATION FROM EACH FACTORY TO EACH WHOLE-
SELLER ARE AS GIVEN IN THE FOLLOWING TABLE.
WHOLESELLERS
FACTORIES A B C D E
F1 5 7 10 25 15
F2 8 6 9 12 14
F3 10 9 8 10 15
DETERMINE THE MAXIMUM PROFIT THAT THE COMPANY CAN MAKE
USING VOGEL APPROXIMATION METHOD AND MODI METHOD FOR
CHECKING OPTIMALITY.

ANS FOR Q NO. 6
PROFIT MATRIX
A B C D E

F1 250-100-5 250-100-7 250-100-10 250-100-25 250-100-15
145 143 140 125 135

F2 250-90-8 250-90-6 250-90-9 250-90-12 250-90-14
152 154 151 148 146

F3 250-80-10 250-80-9 250-80-8 250-80-10 250-80-15
160 161 162 160 155

A B C D E FDUMMY
F1 17 19 22 37 27 0 500
F2 10 8 11 14 16 0 1000
F3 2 1 0 2 7 0 1250
300 300 1000 500 400 250

TABLE 1 Optimal
1 A B C D E F Sup 1 2 3 4 5 6 Ui

2 F 250 -4 -4 -17 -4 250 500
-250
17* 2 2 2 5 X 0

3 1 17 19 22 37 27 0

4 F 50 300 250 400 -7 1000
-300
8 2 2 2 1 X -7
2 10 8 11 14 16 0 -250-400
F -3 -4 750 500 -2 -18 1250
-500
0 1 1 X X X -18
3 2 1 0 2 7 0 -750
D 300 300 1000 500 400 250
5
1 8 7 11 12 9 0 CIRCLED NUMERALS SHOW dj VALUES
2 8 7 11 12* 9 X
6 3 8 7 11* X 9 X
We choose B and not C or E because B has
4 7 11* 11 X 11 X lower cost cell (1) compared to C or E
5 7 X 11* X 11 X We choose C and not E because B has
lower cost cell (11) compared to E (16,27)
6 7 X X X 11* X
Vj 17 15 18 20 23 0

TP NUMERICALS
Q. No. 7
A COMPANY HAS FOUR FACTORIES F1, F2, F3, F4, MANUFACTURING
THE SAME PRODUCT. PRODUCTION COSTS AND RAW MATERIALS
COST DIFFER FORM FACTORY TO FACTORY AND ARE GIVEN IN
THE FOLLOWING TABLE (FIRST TWO ROWS).
THE TRANSPORTATION COSTS FROM THE FACTORIES TO SALES
DEPOTS S1, S2, S3 ARE ALSO GIVEN.
THE SALES PRICE PER UNIT AND REQUIREMENT AT EACH DEPOT ARE
GIVEN IN THE LAST TWO COLUMNS. THE LAST ROW IN THE TABLE
GIVES THE PRODUCTION CAPACITY AT EACH FACTORY.
DETERMINE THE MOST PROFITABLE PRODUCTION AND DISTRIBUTION
SCHEDULE AND THE CORRESPONDING PROFIT. THE SURPLUS
PRODUCTION SHOULD BE TAKEN TO YIELD ZERO PROFIT.
F1 F2 F3 F4 SALES REQUIRE
PRICE MENT
PRODN COST/UNIT 15 12 14 13 AT DIFF AT DIFF
RAW MATL COST 10 9 12 9 DEPOTS DEPOTS
TRANSPORT(TO S1) 3 9 5 4 34 80
-ATION (TO S2) 1 7 4 5 32 120
COSTS (TO S3) 5 8 3 6 31 150
PRODN. CAPACITY 100 150 50 100

ANS FOR Q NO. 7
PROFIT MATRIX
S1 S2 S3

F1 34-(15+10+3) 32-(15+10+1) 31-(15+10+5)
=6 =6 =1
F2 34-(12+9+9) 32-(12+9+7) 31-(12+9+8)
=4 =4 =2
F3 34-(14+12+5) 32-(14+12+4) 31-(14+12+3)
=3 =2 =2
F4 34-(13+9+4) 32-(13+9+5) 31-(13+9+6)
=8 =5 =3
NEGATIVE PROFIT MATRIX
S1 S2 S3 S4DUMMY) SUPPLY NOTE: SINCE
SURPLUS
F1 2 2 7 0 100 PRODUCTION
4 4 6 0 150 YIELDS ZERO
F2 PROFIT, THERE
F3 5 6 6 0 50 FORE, IN THE
PROFIT MATRIX
F4 0 3 5 0 100 S4 IS ASSIGNED
DEMAND 80 120 150 50 400 ZERO VALUE S
IN THE CELLS

TP NUMERICALS
Q. NO. 8
.
AN OIL COMPANY HAS THREE REFINERIES R1, R2, R3 AND
FOUR REGIONAL OIL DEPOTS D1, D2, D3 D4. THE ANNUAL
SUPPLY AND DEMAND IN MILLION LITRES IS GIVEN BELOW
ALONG WITH THE TRANSPORTATION COSTS IN TERMS OF
RS THOUSANDS PER TANKER OF 10 KILOLITRES.
D1 D2 D3 D4
R1 5 10 5 20
5 7 2 4
R2 5 20 25
7 2 8 6
R3 10 10
4 5 10 5
DEMAND 15 5 10 25 55

TP NUMERICALS
Q. NO. 8 contd
ANSWER THE FOLLOWING QUESTIONS.
i. IS THE SOLUTION FEASIBLE?
ii. IS THE SOLUTION DEGENERATE?
iii. IS THE SOLUTION OPTIMAL?
iv. DOES THIS PROBLEM HAVE MULTIPLE OPTIMAL
SOLUTIONS? IF SO DETERMINE THEM.
v. IF THE TRANSPORTATION COST OF ROUTE R2 D1 IS
REDUCED FROM RS 7 TO RS 6, WILL THERE BE ANY
CHANGE IN THE SOLUTION?

ANS FOR Q NO. 8
ANSWER THE FOLLOWING QUESTIONS.
i) IS THE SOLUTION FEASIBLE?
Yes because it satisfies all supply and demand constraints.
x11+x13+x14 = 20; x11+x31=15 and so on.
ii) IS THE SOLUTION DEGENERATE?
No because No. of occupied cells = (m+n-1)
iii) IS THE SOLUTION OPTIMAL?
Yes soln is optimal since one dij value is zero and other all dij values are
negative. Z= 235 (SEE NEXT SLIDE)
iv) DOES THIS PROBLEM HAVE MULTIPLE OPTIMAL SOLUTIONS? IF
SO DETERMINE THEM.
Yes it has multipal optimal soludtions since one dij value is zero. Trace
the loop: R2D1-R1D1-R1D4-R2D4. Shift 5 units from R1D1to
R1D4. Shift 5 units from R2D4 to R2D1. The new solution has the
same Z value ie 235. (SEE SLIDE AFTER THE NEXT)
v) IF THE TRANSPORTATION COST OF ROUTE R2 D1 IS REDUCED FROM
RS 7 TO RS 6, WILL THERE BE ANY CHANGE IN THE SOLUTION?
Yes. The cost will come down by Rs 5 to Rs 230. (SEE THIRD SLIDE
FROM THIS)

ANS FOR Q NO. 8 OPTIMALITY CHECK BY MODI

Optimal Table
D1 D2 D3 D4 Supply Ui CIRCLED NUMERALS
SHOW dj VALUES
R 5 -7 10 5 0
20
1 5 7 2 4
R 0
5
-4
20 2
25
2 7 2 8 6 FOR FINDING THE SECOND
OPTIMAL SOLN, TRACE
R 10 -6 -9 -2 -1
LOOP FROM R2D1 AS
10
3 4 5 10 5 SHOWN AND SHIFT CELLS
AS SHOWN IN THE NEXT
Demand 15 5 10 25 55 SLIDE
Vj 5 0 2 4

Z = 235

ANS FOR Q NO. 8 MULTIPLE OPTIMALITY CHECK BY MODI

1st Optimal Table 2nd
D1 D2 D3
Optimal Soln D4 Supply Ui CIRCLED NUMERALS
SHOW dj VALUES
R 0 -7 10 10 0
20
1 5 7 2 4
R 5 5
-4
15 2
25
2 7 2 8 6
R 10 -6 -9 -2 -1
10
3 4 5 10 5
Demand 15 5 10 25 55
Vj 5 0 2 4

Z = 235

ANS FOR Q NO. 8 - TPT COST OF R2D1 CHANGED FROM 7 YO 6

Optimal Table
D1 D2 D3 D4 Supply Ui CIRCLED NUMERALS
SHOW dj VALUES
R 0 -7 10 10 0
20
1 5 7 2 4
R 5 5
-4
15 2
25
2 6 2 8 6
R 10 -6 -9 -2 -1
10
3 4 5 10 5
Demand 15 5 10 25 55
Vj 5 0 2 4

Z = 230

TP NUMERICALS
Q. NO.9
A LARGE BREAD-MANUFACTURING UNIT CAN PRODUCE SPECIAL
BREAD IN ITS TWO PLANTS P AND Q WITH MANUFACTURING CAPACITY
OF 5000 AND 4200 LOAVES OF BREAD PER DAY RESPECTIVELY AND
COST OF PRODUCTION OF Rs10 AND Rs 12 PER LOAF OF BREAD
RESPECTIVELY.
FOUR RETALING CHAINS A,B,C,AND D PURCHASE BREAD FROM THIS
COMPANY. THEIR DEMAND PER DAY IS RESPECTIVELY
3600,4600,1100,AND 3500 LOAVES OF BREAD AND THE PRICES THAT
THEY PAY PER LOAF OF BREAD ARE RESPECTIVELY Rs 19,17,20 AND
18.
THE COST OF TRANSPORTATION AND HANDLING IN Rs PER LOAF FOR
DELIVERY TO VARIOUS STORES OF THE RETAILING CHAINS IS AS
FOLLOWS.
PLANT RETAILING CHAINS
A B C D
P 1 2 3 2
Q 4 1 2 1
DETERMINE THE DELIVERY SCHEDULE FOR THE BREAD
MANUFACTURING COMPANY THAT WILL MAXIMIZE ITS PROFITS. WRITE
A DUAL OF THE TP

TP Q. NO. 9 – FOR INFO SUMMARY
Q. NO.9
A LARGE BREAD-MANUFACTURING UNIT CAN PRODUCE SPECIAL BREAD IN ITS TWO
PLANTS AS PER DETAILS GIVE BELOW.
PLANT Mfg CAP COST OF PRODN.
LOAVES/DAY Rs PER LOAF OF BREAD
P 5000 10
Q 4200 12
FOUR LARGE RETALING CHAINS PURCHASE BREAD FROM THIS COMPANY. THEIR
DEMAND AND THE PRICES THAT THEY PAY ARE GIVEN BELOW.
RETAILING MAX DEMAND PRICE
LOAVES/DAY RS PER LOAF
A 3600 19
B 4600 17
C 1100 20
D 3500 18
THE COST OF TRANSPORTATION AND HANDLING IN Rs PER LOAF FOR DELIVERY TO
VARIOUS STORES OF THE RETAILING CHAINS IS AS FOLLOWS.
PLANT RETAILING CHAINS
A B C D
P 1 2 3 2
Q 4 1 2 1
DETERMINE THE DELIVERY SCHEDULE FOR THE BREAD MANUFACTURING COMPANY
THAT WILL MAXIMIZE ITS PROFITS. WRITE A DUAL OF THE TP

ANS FOR Q NO. 9
PROFIT MATRIX
A B C D SUPPLY
P 19-10-1= 17-10-2= 20-10-3= 18-10-2= 5000
8 5 7 6
Q 19-12-4= 17-12-4= 20-12-2= 18-12-1= 4200
3 4 6 5
R DUMMY 3600
SOURCE
0 0 0 0
DEMAND 3600 4600 1100 3500 12800

NEGATIVE PROFIT MATRIX
A B C D Supply
P 0 3 1 2 5000
Q 5 4 2 3 4200
R 8 8 8 8 3600
Demand 3600 4600 1100 3500 12800k

TABLE 1 Optimal
A B C D Sup 1 2 3 4 5 6 Ui 1
3600
P 3600
3600 0 1100
1100 300 5000-
300
3600
1 1* 1* X X X 0
. 0 3 1 2 1100-300 2
1100
Q -4
1000
1000 0 3200 4200-
3200
3200-
1 1 1 1 X X 1
. 5 4 2 3 1000
3
R -3 3600
3600 -2 -1 3600 0 0 0 0 X X 5 300
. 8 8 8 8
D 3600 4600 1100 3500 In 2nd iteration, we choose P row and 4
not other row or cols because P has 3200
1 5* 1 1 1 lowest cost cell (1) compared to all others
2 X 1 1 1 5
In 3rd iteration, we choose P row and 1000
3 x 1 x 1
not other row or cols because P has
4 x 4 x 5* lowest cost cell (2) compared to all others
6
5 X X X X 3600
6 X X X X
Vj 0 3 1 2 CIRCLED NUMERALS SHOW dj VALUES

TP NUMERICALS
Q. NO. 10 (TRANSSHIPMENT PROBLEM)
A TRANSPORTER HAS DETERMINED THE COST OF TRANSPORTATION
PER PACKAGE FOR A CUSTOMER’S PRODUCT IS AS PER TABLE GIVEN
BELOW. EVERY WEEK HE HAS TO PICK UP300 PACKAGES FROM
SOURCE S1 AND 200 PACKAGES FROM SOURCE S2 AND DELIVER 100
PACKAGES TO DESTINATION D1 AND 400 PACKAGES TO DESTINATION
D2. THE TRANSPORTER HAS THE OPTION OF EITHER SHIPPING
DIRECTLY FROM THE SOURCES TO THE DESTINATIOS OR TO
TRANSSHIP IF ECONOMICAL. DETERMINE THE OPTIMUM SHIPPING
SCHEDULE, WITH WOULD MINIMISE COST OF TRANSPORTATION.

SOURCES DESTINATIONS
S1 S2 D1 D2
S1 0 18 5 10
S2 18 0 8 16
D1 5 8 0 3
D2 10 16 3 0

ANS FOR Q NO. 10 TRANS SHIPMENT by VAM &MODI
TABLE 1 Optimal
S1 S2 D1 D2 Sup 1 2 3 4 5 6 Ui 1
500
S1 500
500 -20 300
300 -9 300+500 5 5 5* 5* X X 0
0 18 5 10 2
S2 500
500 200
200 -5 200+500 8* 8* X X X X 2 200
-16
18 0 8 16
D1 -16 100
100 400 500
400 3 3 3 3 3* X -6 3
-11 500
5 8 0 3
D2 500 500
500 3 3 3 3 3 X -9
-19 -27 -6
CIRCLED NUMERALS SHOW dij VALUES
4
10 16 3 0 The total number of units trans
300
D 500 500 100+500 400+500 2500 ported from all the sources to
1 5 8 3 3
all the destinations is 500. This 5
qty is added to each supply and 100
2 5 X 3 3 each demand and TP is solved
3 5 X 3 3 The interpretation of this is that S1 will
4 X X 3 3 transport300 units to D1 and S2 will 6
transport 200 units to D1. D1 will trans 500
5 X X 3 3 thip 400 units to D2. This means that
6 X X X 3* D2 will not get its packages from S1 or
S2 but will get 400 units trans shipped
Vj 0 -2 6 9 trom D1.

TP NUMERICALS
Q. NO. 11 (TRANSSHIPMENT PROBLEM)
A COMPANY HAS TWO FACTORIES F1 AND F2 HAVING PRODUCTION
CAPACITY OF 200 AND 300 UNITS RESPECTIVELY. IT HAS THREE
WAREHOUSES W1,W2 AND W3, HAVING DEMAND EQUAL TO 100, 150
AND 250 RESPECTIVELY. THE COMPANY HAS THE OPTION OF EITHER
SHIPPING DIRECTLY FROM THE FACTORIES TO THE WAREHOUSES OR
TO TRANSSHIP IF ECONOMICAL. DETERMINE THE OPTIMUM SHIPPING
SCHEDULE, WITH MINIMUM COST OF TRANSPORTATION.
FACTORIES WAREHOUSES
F1 F2 W1 W2 W3
F1 0 8 7 8 9
F2 6 0 5 4 3
W1 7 2 0 5 1
W2 1 5 1 0 4
W3 8 9 7 8 0

. ANS TO Q. NO. 11TRANSSHIPMENT PROBLEM)

FACTORIES WAREHOUSES
F1 F2 W1 W2 W3 SUPPLY
F1 0 8 7 8 9 200+500 =700
F2 6 0 5 4 3 300+500 =800
W1 7 2 0 5 1 500
W2 1 5 1 0 4 500
W3 8 9 7 8 0 500
DEMAND 500 500 100+ 150+ 250+ 2500
500 500 500

TP NUMERICALS
Q. NO. 12 (PROHIBITED ROUTES)
A TOY MANUFACTURER HAS DETERMINED THAT DEMAND FOR A
PARTICULAR DESIGN OF TOY CAR FROM VARIOUS DISRIBUTORS
IS 500, 1000, 1400 AND 1200 FOR THE 1ST , 2ND , 3RD , AND 4TH WEEK
OF THE NEXT MONTH WHICH MUST BE SATISFIED.
THE PRODUCTION COST PER UNIT IS RS 50 FOR THE FIRST TWO
WEEKS AND RS 60 PER UNIT FOR THE NEXT TWO WEEKS DUE TO
EXPECTED INCREASE IN COST OF PLASITIC. THE PLANT CAN
PRODUCE MAXIMUM OF 1000 UNITS PER WEEK. THE
MANUFACTURER CAN ASK EMPLOYEES TO WORK OVER TIME
DURING THE 2ND AND THE 3RD WEEK WHICH INCREASES THE
PRODUCTION BY ADDITIONAL 300 UNITS BUT ALSO IT INCREASES
THE COST BY RS 5 PER UNIT. EXCESS PRODUCTION CAN BE
STORED AT A COST OF RS 3 PER UNIT PER WEEK.
DETERMINE THE PRODUCTION SCEHEDULE SO THAT TOTAL COST IS
MINIMISED.

. ANS TO Q. NO. 12 PROHIBITED ROUTE TP
PRODUCTION WEEK COST OF PRODUCTION PER UNIT
WK1 WK2 WK3 WK4 DUMMY SUPPLY
DEMAND
WEEK1 (NORMAL) 50 53 56 59 0 1000
WEEK2 (NORMAL) M 50 53 56 0 1000
WEEK2 (OVERTIME) M 55 58 61 0 300
WEEK3 (NORMAL) M M 60 63 0 1000
WEEK3 (OVERTIME) M M 65 68 0 300
WEEK4 (NORMAL) M M M 60 0 1000
DEMAND 500 1000 1400 1200 500 4600

ANS FOR Q NO. 10 TRANS SHIPMENT by VAM &MODI 10
TABLE 1 OPTIMAL SOLN.
200
A W1 W2 W3 W4 DUM Sup Ui 1 2 3 4 5 6 7 8 9 1
W 500 500
500 500 1000 0 50 50 50 3 3 X X X 500
0 0 -4 . *
1 50 53 56 59 0
W -M 500 500
500 500 1000 -3 50 50 50 3 3 3 3 X 2
0 -7
2 M +47 50 53 56 0 . 300
W -M 100 200
100 200 300 2 55 55 55 3 3 3 3 3
2OT M +52 55 58 61 0
-2
. 3
200
W -M -M 800
800 200
200 1000 4 60 60 60 M-
60
X X X X X
3 M +54 M +57 60 63 0 0 . * * 4
W -M -M
-5 -5 300
300 300 4 65 65 X X X X X X X 800
3OT M +54 M +57 65 68 0 . *

W -M -M -M 1000
1000 1000 1 60 60 60 M- M- X X X X 5
-3 60 60
4 M +51 M +54 M +57 60 0 *
1000
4600 6
D 500 1000 1400 1200 500 CIRCLED NUMERALS SHOW dij VALUES 500
Vj 50 53 56 59 -4 7
1 M – 50* 3 3 3 0
Interpretation:
-Co. will make 1000 units in 1 st week though dmd
2 X 3 3 3 0 500
3 X 3 3 3 0
is only 500 units. It will sell 500 of these in the
4 X 3 3 3 X 1st week and 500 in the 2nd week. 8
-I will produce 300 by running OT in the 2 nd week
5 X 3 3 3 X 500
6 X 3 3 3 X It will sell 100 of these in the 3rd week and 200 of
7 X 5* 5 5 X these in the 4th week. 9
-It will not run OT in the 3 rd week.
8 X X 5* 5 X 100

TP NUMERICALS
Q. NO. 13 (PROHIBITED ROUTES)
A COMPANY IS PLANNING ITS NEXT FOUR WEEKS’
PRODUCTION. THE PER UNIT PRODUCTION COST IS RS
10 FOR THE FIRST TWO WEEKS AND RS 15 FOR THE
NEXT TWO WEEKS. DEMAND IS 300, 700, 900 AND 800 FOR
THE 1ST, 2ND, 3RD, AND 4TH WEEK, WHICH MUST BE
MET.
THE PLANT CAN PRODUCE MAXIMUM OF 700 UNITS PER
WEEK. THE COMPANY CAN ASK EMPLOYEES TO WORK
OVER TIME DURING THE 2ND AND THE 3RD WEEK WHICH
INCREASES THE PRODUCTION BY ADDITIONAL 200 UNITS
BUT ALSO IT INCREASES THE COST BY RS 5 PER UNIT.
EXCESS PRODUCTION CAN BE STORED AT A COST OF
RS 3 PER UNIT PER WEEK

.
TP NUMERICALS
Q. NO. 14
A FMCG COMPANY HAS THREE WARE HOUSES W1,W2 AND W3 AND SUPPLIES
PRODUCTS FROM THESE WAREHOUSES TO THREE DISTRIBUTORS D1,D2 AND D3.
FMCG COMPANY HAS DETERMINED THAT DURING THE NEXT MONTH, THERE WILL
BE A SHORT FALL IN SUPPLY AGAINST THE PROJECTED DEMAND. IT HAS AGREED
TO PAY A PENALTY PER UNIT AS PER THE TABLE GIVEN BELOW TO DISTRIBUTORS
FOR DEMAND THAT IS NOT MET. FIND THE DELIVERY SCHEDULE THAT THE
COMPANY SHOULD FOLLOW TO MINIMISE TRANSPORTATION COSTS AND PENALTY
COST AND DETERMINE VALUES OF BOTH COSTS.

D1 D2 D3
W1 5 1 7 100
W2 6 4 6 800
W3 3 2 5 150
DEMAND 1050
750 200 500 1450
PENALTY 5 3 2

. ANS TO Q. NO. 14

SOURCES DESTINATIONS SUP
D1 D2 D3 Sup Ui 1 2 3 4 5 6 7 8 9

W1 100 100
5 1 7
W2 600 100 100 800
6 4 6
W3 150 150
3 2 5
W4 400 400
DUMMY 5 3 2
DEMAND 750 200 500 1450
The transportation is Rs 5150.
The penalty cost is Rs 800
V
1 There is a shortfall of 400 units. We
2 create a dummy warehouse (source)
3
with a supply capability of 400 units.
The penalty cost per unit payable to
4
the distributors is put in the cells in
row.

Transportation problem

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