1. 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
3. 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.
MINIMISE TOTAL COST Z = SS cij xij
SUBJECT TO Sxij = ai FOR i = 1,2,3…m
S xij = bj FOR j = 1,2,3…n
AND xij ≥ 0
FULL STATEMENT OF TP AS LPP
i=1 j=1
m n
j=1
n
i=1
m
4. TRANSPORTATION MODEL –TABULAR FORM
.
1 2 3 … n
1 a1
2 a2
3 a3
...
m am
b1 b2 b3 bn Sai= Sbj
SOURCES DESTINATIONS SUPPLY ai
DEMAND bj
x11
x23x22x21
x13x12
x2n
x1n
x33x32x31 x3n
xmnxm3xm2xm1
c11 c12
cm1
c21
c31
c13 c1n
c23c22 c2n
c32 c33 c3n
cm2 cm3 cmn
MINIMISE TOTAL COST Z = SS cij xij FOR i=1 to m & j=1 to n
5. 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:
6. TRANSPORTATION PROBLEM–TABULAR FORM
.
1 2 3
1 a1
2 a2
3 a3
b1 b2 b3 Sai= Sbj
SOURCES DESTINATIONS SUPPLY ai
DEMAND bj
x11
x23x22x21
x13x12
x33x32x31
c11 c12
c21
c31
c13
c23c22
c32 c33
MINIMISE TOTAL COST
Z = SS cij xij FOR i=1 to m & j=1 to n
7. EXAMPLE 1 OF TP
.
W1 W2 W3
P1
45
P2
15
P3
40
DEMAND 25 55 20 100
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
10 7
7
8
912
8 12
15
8. 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
9. 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
10. DUAL OF A TP FORMULATED AS A LPP
Max G = u1+ u2+ u3+ v1+ v2+ v3+
Subject to
u1+ v1+
11. 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
12. 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
13. 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
15. 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
16. 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
17. TESTING FEASIBLE SOLUTION FOR OPTIMALITY
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.
18. TESTING FEASIBLE SOLUTION FOR OPTIMALITY
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
19. EXAMPLE 1 OF TP
.
W1 W2 W3
P1
45
P2
15
P3
40
DEMAND 25 55 20 100
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
10 7
7
8
912
8 12
15
20. EXAMPLE 1 OF TP
.
W1 W2 W3
P1
45
P2
15
P3
40
DEMAND 25 55 20 100
INITIAL BASIC FEASIBLE SOLUTION
BY NORTH WEST CORNER METHOD
SOURCES DESTINATIONS SUPPLY
10 7
7
8
912
8 12
15
25
15
20
2020
21. EXAMPLE 1 OF TP
.
1 2 3 4
1 1 3* X
3* X X X
1 1 1 1*
INTIAL BASIC FEASIBLE SOLUTION
BY VOGEL APPROXIMATION METHOD
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
40 5
1
3 1 1
3 1 4*
3 1 X
X X X
ITERATIONS
1525
15
1st ITERATION
2nd ITERATION
3rd ITERATION
4th ITERATION
22. EXAMPLE 1 OF TP
.
W1 W2 W3
P1
45
P2
15
P3
40
DEMAND 25 55 20 100
INITIAL BASIC FEASIBLE SOLUTION
BY NORTH WEST CORNER METHOD
SOURCES DESTINATIONS SUPPLY
10 7
7
8
912
8 12
15
25
15
20
2020
23. EXAMPLE 1 OF TP
.
1 2 3 4
1 1 3* X
3* X X X
1 1 1 1*
INTIAL BASIC FEASIBLE SOLUTION
BY VOGEL APPROXIMATION METHOD
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
40 5
1
3 1 1
3 1 4*
3 1 X
X X X
ITERATIONS
1525
15
1st ITERATION
2nd ITERATION
3rd ITERATION
4th ITERATION
25. 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
26. 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
30. EXAMPLE 2a OF TP
.
W1 W2 W3
P1
60
P2
20
P3
40
DEMAND 25 55 20
120
100
A STEEL COMPANY HAS THREE PLANTS P1,P2 AND P3 WITH ANNUAL
CAPACITIES OF 60, 20 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
10 7
7
8
912
8 12
15
31. SOLUTION OF EXAMPLE 2a OF TP
CREATE A DUMMY DESTINATION W4 WITH DEMAND = 20,000 TONNES
W1 W2 W3 W4
P1
60
P2
20
P3
40
DEMAND 25 55 20 20
120
120
SOURCES DESTINATIONS SUPPLY
10 7
7
8
912
8 12
15 0
0
0
32. EXAMPLE 2b OF TP
.
W1 W2 W3
P1
55
P2
15
P3
40
DEMAND 30 70 20
100
120
A STEEL COMPANY HAS THREE PLANTS P1,P2 AND P3 WITH ANNUAL
CAPACITIES OF 55, 15 AND 40 THOUSAND TONNES OF CR COILS. THE
PRODUCT IS DISTRIBUTED FROM THREE WAREHOUSES W1,W2 AND
W3 WITH ANNUAL OFFTAKE OF 30, 70 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
10 7
7
8
912
8 12
15
33. SOLUTION OF EXAMPLE 2b OF TP
CREATE A DUMMY SOURCE P4 WITH SUPPLY = 20,000 TONNES
W1 W2 W3
P1
55
P2
15
P3
40
P4
20
DEMAND 30 70 20
120
120
SOURCES DESTINATIONS SUPPLY
10 7
7
8
912
8 12
15
000
34. EXAMPLE 3 OF TP (DEGENERACY).
D E F G
A
60
B
100
C
40
DEMAND 20 50 50 80 200
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.
SOURCES DESTINATIONS SUPPLY
7 3
2
8
52
6 5
4
6
10
1
35. TEST FOR OPTIMALITY & IMPROVEMENT OF SOLN
MODIFIED DISTRIBUTION METHOD
D E F G
A
60
u1
0
B
100
u2
-1
C
40
u3
-10
25 55 20 200
v1
7
v2
3
v3
6
V4
11
7 3
2
8
52
6 5
4
AF 0+6-8=-2 -2
AG 0+11-6=+5 +5
BD -1+7-4=+2 +2
CD -10+7-2=-5 -5
CE -10+3-6=-13 -13
CF -10+6-5=-9 -9
dj
EMPTY
CELL20
TP TABLE 1 (NON OPTIMAL) dj IS NET OPPOR. AVAIL
Z=7x20+3x40+2x10+5x50
+10x40+1x40 = 970
- SELECT THE CELL WITH THE LARGEST
POSITIVE dj VALUE (+5) ie CELL AG
- TRACE LOOP AG-BG-BE-AE
- SHIFT 40 UNITS FROM HIGHER COST
CELL BG TO LOWER COST CELL AG
- SHIFT 40 UNITS FROM CELL AE TO BE
SO THAT DEMAND SUPPLY
CONSTRAINTS ARE NOT AFFECTED
-THIS GIVES US THE NEXT TABLE 2
40
40
10
ui
vj
(ui+vj)=cij
u1=0
ROW A
AD: v1= 7
AE::v2= 3
ROW B
BE: u2= -1
BF: v3= 6
BG: v4= 11
ROW C
CG: u3= -10
NET COST
CHANGE
dij=(ui+vj)-cij
50 40
6
1
10
36. TEST FOR OPTIMALITY & IMPROVEMENT OF SOLN
MODIFIED DISTRIBUTION METHOD
D E F G
A
60
u1
0
B
100
u2
-1
C
40
u3
-5
25 55 20 200
v1
7
v2
3
v3
6
v3
6
7 3
2
8
52
6 5
4
AF 0+6-8=-2 -2
BD -1+7-4=+2 +2
BG -1+6-10=-5 -5
CD -5+7-2=-0 0
CE -5+3-6=-8 -8
CF -5+6-5=-4 -4
dj
EMPTY
CELL20
TP TABLE 2 (NON OPTIMAL) dj IS NET OPPOR. AVAIL
Z=7x20+3xe+6x40+2x50
+5x50+1x40 = 770 (3xe=0)
-SOLN IS DEGENERATE SINCE NO. OF xij
VARIABLES (5) IS LESS THAN (m+n-1=6). TWO
RECENTLY VACATED CELLS ARE AE & BG
ASSIGN e VALUE TO AE SINCE IT HAS LOWER
cij VALUE. PROCEED LIKE EARLIER STEP 1
- CELL BD HAS LARGEST dj VALUE =+2
- TRACE LOOP BD-AD-AE-BE
- SHIFT 20 UNITS FROM AD T0 BD
- SHIFT 20 UNITS FROM BE TO AE
e
40
50
ui
vj
(ui+vj)=cij
u1=0
ROW A
AD: v1= 7
AE::v2= 3
AG:v4= 6
ROW B
BE: u2= -1
BF: v3= 6
ROW C
CG: u3= -5
NET COST
CHANGE
dij=(ui+vj)-cij
50
40
6
1
10
37. TEST FOR OPTIMALITY & IMPROVEMENT OF SOLN
MODIFIED DISTRIBUTION METHOD
D E F G
A
60
u1
0
B
100
u2
-1
C
40
u3
-
-5
25 55 20 200
v1
5
v2
3
v3
6
V4
6
7 3
2
8
52
6 5
4
AD 0+5-7=-2 -2
AF 0+6-8=-2 -2
BG -1+6-10=-5 -5
CD -5+5-2=-2 -5
CE -5+3-6=-8 -8
CF -5+6-5=-4 -4
dj
EMPTY
CELL40
TP TABLE 3 (OPTIMAL) dj IS NET OPPOR. AVAIL
Z=3x20+6x40+4x20+3x30
+5x50+1x40 = 730
- SINCE ALL dj VALUE ARE NEGATIVE
THEREFORE THIS SOLUTION
IS AN OPTIMAL SOLUTION
20
40
20
ui
vj
(ui+vj)=cij
u1=0
ROW A
AE::v2= 3
AG: v4= 6
ROW B
BD: v1= 5
BE: u2= -1
BF: v3= 6
ROW C
CG: u3= -5
NET COST
CHANGE
dij=(ui+vj)-cij
5030
6
1
10
38. EXAMPLE 4 OF TP (MAXIMISATION).
D E F G
A
200
B
500
C
300
DEMAND 180 320 100 400 1000
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.
SOURCES DESTINATIONS SUPPLY
12 8
14
6
107
3 11
8
25
18
20
43. 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
44. 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.
45. 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.
46. TP NUMERICALS ANS TO Q NO 3TABLE 1
W1 W2 W3 W4
F1 40 30 10 20 125
F2 30 25 0 13 250
F3 19 21 20 10 175
F4 0 5 22 17 100
100 400 90 60
1 X X X X
2 X 16* 10 3
3 X 4 10 3
4 X 4 X 3
vj 40 30 5 18
Dj
F1W3 -5
F1W4 -2
F2W1 IGNORE
F3W1 IGNORE
F4W1 IGNORE
F3W3 -24
F3W4 -1
F4W3 -42
F4W4 -24
1 2 3 4 ui
X 10 10 10 0
X 13 13* 12* -5
X 10 10 11 -9
X 12 X X -25
100
W1 W2 W3 W4
F1 90 100 120 110 125
F2 100 105 130 117 250
F3 111 109 110 120 175
F4 130 125 108 113 100
100 400 90 60
TABLE 2 OPTIMAL SOLN
1ST W1 GETS FULL
QTY FROM F1
2ND SUPPLY FROM
F4 EXHAUSTED
3RD DEMAND FROM
W3 MET
100
6090
4TH DEMAND FROM
W4 MET
5th SUPPLY FROM
F1 EXHAUSTED
175
100
25
6th SUPPLY FROM
F2 EXHAUSTED
7TH DEMAND FROM
W2 MET
NOTE: In the first iteration for VOGEL
we put an X for all rows and columns
because the constraint is that warehouse
W1 is to be supplied entire quantity from
factory F1
47. 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
48. ANS FOR Q NO. 4
TABLE 1
D E F G Sup Ui
A 20 40 60 0
B 10 50 40 100 -1
C 40 40 -
10
Dmd 20 50 50 80
Vj 7 3 6 11
D E F G Sup Ui
A 20 e 40 60 0
B 50 50 100 -1
C 40 40 -5
Dmd 20 50 50 80
Vj 7 3 6 6
D E F G Sup Ui
A 20 40 60 0
B 20 30 50 100 -1
C 40 40 -5
Dmd 20 50 50 80
Vj 5 3 6 6
TABLE 3 OPTIMAL
TABLE 2
-9-13-5
+2
-2 +5
-4-80
-5+2
-2
-4-8-2
-5
-2-2
49. 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
50. ANS FOR Q NO. 5 by VAM
TABLE 1 Optimal
D E F G Sup 1 2 3 Ui
A 4 10 14 3 1 2* 0
B 1 15 16 5* 1 1 2
C 1 4 5 1 1 1 -2
Dmd 6 10 15 4
1 2 1 1 3
2 2 1 X 3*
3 2 1 X X
Vj 6 4 0 4
-8-1
-1-3
-1 -1
1
15
2
4
3
10
4
1,4,1
51. 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.
52. ANS FOR Q NO. 6
PROFIT MATRIX
A B C D E
F1 250-100-5
145
250-100-7
143
250-100-10
140
250-100-25
125
250-100-15
135
F2 250-90-8
152
250-90-6
154
250-90-9
151
250-90-12
148
250-90-14
146
F3 250-80-10
160
250-80-9
161
250-80-8
162
250-80-10
160
250-80-15
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
53. ANS FOR Q NO. 6 by VAM
TABLE 1 Optimal
A B C D E F Sup 1 2 3 4 5 6 Ui
F
1
250
17 19 22 37 27
250
0
500
-250
17* 2 2 2 5 X 0
F
2
50
10
300
8
250
11 14
400
16 0
1000
-300
-250-400
8 2 2 2 1 X -7
F
3 2 1
750
0
500
2 7 0
1250
-500
-750
0 1 1 X X X -18
D 300 300 1000 500 400 250
1 8 7 11 12 9 0
2 8 7 11 12* 9 X
3 8 7 11* X 9 X
4 7 11* 11 X 11 X
5 7 X 11* X 11 X
6 7 X X X 11* X
Vj 17 15 18 20 23 0
-7
-4-4 -4
-18-2-4-3
-17
1
2
3
4
5
We choose B and not C or E because B has
lower cost cell (1) compared to C or E
We choose C and not E because B has
lower cost cell (11) compared to E (16,27)
6
CIRCLED NUMERALS SHOW dj VALUES
54. 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
55. ANS FOR Q NO. 7
PROFIT MATRIX
S1 S2 S3
F1 34-(15+10+3)
=6
32-(15+10+1)
=6
31-(15+10+5)
=1
F2 34-(12+9+9)
=4
32-(12+9+7)
=4
31-(12+9+8)
=2
F3 34-(14+12+5)
=3
32-(14+12+4)
=2
31-(14+12+3)
=2
F4 34-(13+9+4)
=8
32-(13+9+5)
=5
31-(13+9+6)
=3
S1 S2 S3 S4DUMMY) SUPPLY
F1 2 2 7 0 100
F2 4 4 6 0 150
F3 5 6 6 0 50
F4 0 3 5 0 100
DEMAND 80 120 150 50 400
NEGATIVE PROFIT MATRIX
NOTE: SINCE
SURPLUS
PRODUCTION
YIELDS ZERO
PROFIT, THERE
FORE, IN THE
PROFIT MATRIX
S4 IS ASSIGNED
ZERO VALUE S
IN THE CELLS
56. TP NUMERICALS
.
D1 D2 D3 D4
R1 20
R2 25
R3 10
DEMAND 15 5 10 25 55
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.
SOURCES DESTINATIONS SUPPLY
5
5
10
20
5
10
5 7
4
2 4
82 6
5 10 5
7
57. 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?
58. 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)
59. ANS FOR Q NO. 8 OPTIMALITY CHECK BY MODI
Optimal Table
D1 D2 D3 D4 Supply Ui
R
1
5
5 7
10
2
5
4
20
0
R
2 7
5
2 8
20
6
25
2
R
3
10
4 5 10 5
10
-1
Demand 15 5 10 25 55
Vj 5 0 2 4
-7
-2
-40
-9-6
CIRCLED NUMERALS
SHOW dj VALUES
FOR FINDING THE SECOND
OPTIMAL SOLN, TRACE
LOOP FROM R2D1 AS
SHOWN AND SHIFT CELLS
AS SHOWN IN THE NEXT
SLIDE
Z = 235
60. ANS FOR Q NO. 8 MULTIPLE OPTIMALITY CHECK BY MODI
1st Optimal Table 2nd
Optimal SolnD1 D2 D3 D4 Supply Ui
R
1 5 7
10
2
10
4
20
0
R
2
5
7
5
2 8
15
6
25
2
R
3
10
4 5 10 5
10
-1
Demand 15 5 10 25 55
Vj 5 0 2 4
-2
-4
-9-6
CIRCLED NUMERALS
SHOW dj VALUES
-70
Z = 235
61. ANS FOR Q NO. 8 - TPT COST OF R2D1 CHANGED FROM 7 YO 6
Optimal Table
D1 D2 D3 D4 Supply Ui
R
1 5 7
10
2
10
4
20
0
R
2
5
6
5
2 8
15
6
25
2
R
3
10
4 5 10 5
10
-1
Demand 15 5 10 25 55
Vj 5 0 2 4
-2
-4
-9-6
CIRCLED NUMERALS
SHOW dj VALUES
-70
Z = 230
62. 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
63. 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
64. ANS FOR Q NO. 9
PROFIT MATRIX
A B C D SUPPLY
P 19-10-1=
8
17-10-2=
5
20-10-3=
7
18-10-2=
6
5000
Q 19-12-4=
3
17-12-4=
4
20-12-2=
6
18-12-1=
5
4200
R DUMMY
SOURCE
0 0 0 0
3600
DEMAND 3600 4600 1100 3500 12800
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
NEGATIVE PROFIT MATRIX
65. ANS FOR Q NO. 9 by VAM
TABLE 1 Optimal
A B C D Sup 1 2 3 4 5 6 Ui
P
.
3600
0 3
1100
1
300
2
5000-
3600
1100-300
1 1* 1* X X X 0
Q
. 5
1000
4 2
3200
3
4200-
3200-
1000
1 1 1 1 X X 1
R
. 8
3600
8 8 8
3600 0 0 0 0 X X 5
D 3600 4600 1100 3500
1 5* 1 1 1
2 X 1 1 1
3 x 1 x 1
4 x 4 x 5*
5 X X X X
6 X X X X
Vj 0 3 1 2
-2
-4
0
-3 -1
0
1
3600
2
1100
3
300
4
3200
5
1000
In 2nd iteration, we choose P row and
not other row or cols because P has
lowest cost cell (1) compared to all others
6
3600
CIRCLED NUMERALS SHOW dj VALUES
In 3rd iteration, we choose P row and
not other row or cols because P has
lowest cost cell (2) compared to all others
3600
1000 3200
30011003600
66. 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.
S1 S2 D1 D2
S1 0 18 5 10
S2 18 0 8 16
D1 5 8 0 3
D2 10 16 3 0
SOURCES DESTINATIONS
67. 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
S1 500
0 18
300
5 10
300+500 5 5 5* 5* X X 0
S2
18
500
0
200
8 16
200+500 8* 8* X X X X 2
D1
5 8
100
0
400
3
500 3 3 3 3 3* X -6
D2
10 16 3
500
0
500 3 3 3 3 3 X -9
D 500 500 100+500 400+500 2500
1 5 8 3 3
2 5 X 3 3
3 5 X 3 3
4 X X 3 3
5 X X 3 3
6 X X X 3*
Vj 0 -2 6 9
-5
-9-20
-16
-6
-11
1
500
2
200
3
500
4
300
5
100
6
500
500
100
500
200500
300
400
The interpretation of this is that S1 will
transport300 units to D1 and S2 will
transport 200 units to D1. D1 will trans
thip 400 units to D2. This means that
D2 will not get its packages from S1 or
S2 but will get 400 units trans shipped
trom D1.
The total number of units trans
ported from all the sources to
all the destinations is 500. This
qty is added to each supply and
each demand and TP is solved
-27-19
-16
CIRCLED NUMERALS SHOW dij VALUES
68. 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.
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
FACTORIES WAREHOUSES
70. 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.
71. ANS TO Q. NO. 12 PROHIBITED ROUTE TP.
WK1 WK2 WK3 WK4 DUMMY
DEMAND
SUPPLY
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
PRODUCTION WEEK COST OF PRODUCTION PER UNIT
72. ANS FOR Q NO. 10 TRANS SHIPMENT by VAM &MODI
A W1 W2 W3 W4 DUM
W
1
500
50
500
53 56 59 0
W
2 M
500
50
500
53 56 0
W
2OT M 55
100
58
200
61 0
W
3 M M
800
60 63
200
0
W
3OT M M 65 68
300
0
W
4 M M M
1000
60 0
-2
-4
1
500
2
300
3
200
4
800
5
1000
6
500
500
Sup Ui 1 2 3 4 5 6 7 8 9
1000
.
0 50 50 50 3 3
*
X X X
1000
.
-3 50 50 50 3 3 3 3 X
300
.
2 55 55 55 3 3 3 3 3
1000
.
4 60 60 60
*
M-
60
*
X X X X X
300
.
4 65 65
*
X X X X X X X
1000 1 60 60 60 M-
60
M-
60
*
X X X X
4600
TABLE 1 OPTIMAL SOLN.
D 500 1000 1400 1200 500
Vj 50 53 56 59 -4
1 M – 50* 3 3 3 0
2 X 3 3 3 0
3 X 3 3 3 0
4 X 3 3 3 X
5 X 3 3 3 X
6 X 3 3 3 X
7 X 5* 5 5 X
8 X X 5* 5 X
7
500
8
500
9
100
10
200
-M
+47
-M
+57
-M
+51
-M
+54
-M
+52
-M
+54
-M
+54
-M
+57
-M
+57 -3
-5 -5
-7
0
0
00
500
200
200100
500500
300
1000
800
Interpretation:
-Co. will make 1000 units in 1st week though dmd
is only 500 units. It will sell 500 of these in the
1st week and 500 in the 2nd week.
-I will produce 300 by running OT in the 2nd week
It will sell 100 of these in the 3rd week and 200 of
these in the 4th week.
-It will not run OT in the 3rd week.
CIRCLED NUMERALS SHOW dij VALUES
73. 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
74. TP NUMERICALS.
D1 D2 D3
W1 5 1 7 100
W2 6 4 6 800
W3 3 2 5 150
DEMAND
750 200 500
1050
1450
PENALTY 5 3 2
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.
SOURCES DESTINATIONS SUPPLY
75. ANS TO Q. NO. 14.
D1 D2 D3
W1
5
100
1 7
100
W2 600
6
100
4
100
6
800
W3 150
3 2 5
150
W4
DUMMY 5 3
400
2
400
DEMAND 750 200 500 1450
.
SOURCES DESTINATIONS SUP
There is a shortfall of 400 units. We
create a dummy warehouse (source)
with a supply capability of 400 units.
The penalty cost per unit payable to
the distributors is put in the cells in
row.
The transportation is Rs 5150.
The penalty cost is Rs 800
Sup Ui 1 2 3 4 5 6 7 8 9
V
1
2
3
4
