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)

A

B

C

D

E

F

SUPPLY DEMAND

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

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

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.

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

EXAMPLE 1 OF TP.

W1 W2 W3

P145

P215

P340

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

TP FOMULATED AS A LPPMin Z =10x11+ 7x12+ 8x13+ 15x21+ 12x22+ 9x23+ 7x31+ 8x32+ 12x33

Subject tox11+ x12+ x13= 45x21+ x22+ x23= 15 SUPPLY CONSTRAINTSx31+ x32+ x33= 40

x11+ x21+ x31= 25x12+ x22+ x32= 55 DEMAND CONSTRAINTSx13+ 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 tox11+ x12+ x13+A1= 45x21+ x22+ x23+A2= 15 SUPPLY CONSTRAINTSx31+ x32+ x33+A3= 40

x11+ x21+ x31+A4= 25x12+ x22+ x32+A5= 55 DEMAND CONSTRAINTSx13+ x23+ x33+A6= 20

xij ≥ 0 FOR i = 1,2,3 AND j = 1,2,3

DUAL OF A TP FORMULATED AS A LPPMax G = u1+ u2+ u3+ v1+ v2+ v3+

Subject to u1+ v1+

METHODS OF SOLVING A TP1. SIMPLEX METHOD

TP CAN BE STATED AS LPP AND THEN SOLVED BY THE SIMPLEX METHOD

2. TRANSPORTATION METHODTHIS INVOLVES THE FOLLOWING STEPSi) 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 TABLE2. 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


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 OPTIMALITYRULES 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 ONLY7. A LOOP MUST HAVE EVEN NUMBER OF CELLS 8. EACH UNOCCUPIED CELL CAN HAVE ONE AND ONLY ONE

LOOP9. ONLY OCCUPIED CELLS ARE TO BE ASSIGNED POSITIVE

OR NEGATIVE VALUES10. LOOP MAY NOT BE SQUARE OR RECTANGLE 11. ALL LOOPS MUST BE CONSISTENTLY CLOCKWISE OR

ANTICLOCKWISE

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.

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

EXAMPLE 1 OF TP.

W1 W2 W3

P145

P215

P340

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.


10 7

7

8

912

8 12

15

EXAMPLE 1 OF TP.

W1 W2 W3

P145

P215

P340

DEMAND 25 55 20 100

INITIAL BASIC FEASIBLE SOLUTION BY NORTH WEST CORNER METHOD


10 7

7

8

912

8 12

15

25

15

20

2020

EXAMPLE 1 OF TP.

1 2 3 41 1 3* X

3* X X X

1 1 1 1*

INTIAL BASIC FEASIBLE SOLUTION BY VOGEL APPROXIMATION METHOD

SOURCES DESTINATIONS SUPPLYW1 W2 W3

P110 7 8 45

P215 12 9 15

P37 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

EXAMPLE 1 OF TP.

W1 W2 W3

P145

P215

P340

DEMAND 25 55 20 100



10 7

7

8

912

8 12

15

25

15

20

2020

EXAMPLE 1 OF TP.

1 2 3 41 1 3* X

3* X X X

1 1 1 1*

INTIAL BASIC FEASIBLE SOLUTION BY VOGEL APPROXIMATION METHOD

SOURCES DESTINATIONS SUPPLYW1 W2 W3

P110 7 8 45

P215 12 9 15

P37 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


1. STEPPING STONE METHOD

2. MODIFIED DISTRIBUTION METHOD


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 OPTIMALITYRULES 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 ONLY7. A LOOP MUST HAVE EVEN NUMBER OF CELLS 8. EACH UNOCCUPIED CELL CAN HAVE ONE AND ONLY ONE

LOOP9. ONLY OCCUPIED CELLS ARE TO BE ASSIGNED POSITIVE

OR NEGATIVE VALUES10. LOOP MAY NOT BE SQUARE OR RECTANGLE 11. ALL LOOPS MUST BE CONSISTENTLY CLOCKWISE OR

ANTICLOCKWISE

TEST FOR OPTIMALITY & IMPROVEMENT OF SOLN

MODIFIED DISTRIBUTION METHOD

EXAMPLE 1 OF TP.

W1 W2 W3

P145

P215

P340

DEMAND 25 55 20 100



10 7

7

8

912

8 12

15

25

15

20

2020

UNBALANCED TRANSPORTATION PROBLEMS

• TOTAL SUPPLY EXCEEDS TOTAL DEMAND

• TOTAL DEMAND EXCEEDS TOTAL SUPPLY

EXAMPLE 2a OF TP.

W1 W2 W3

P160

P220

P340

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.


10 7

7

8

912

8 12

15

SOLUTION OF EXAMPLE 2a OF TP

CREATE A DUMMY DESTINATION W4 WITH DEMAND = 20,000 TONNES

W1 W2 W3 W4

P160

P220

P340

DEMAND 25 55 20 20 120

120


10 7

7

8

912

8 12

15 0

0

0

EXAMPLE 2b OF TP.

W1 W2 W3

P155

P215

P340

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.


10 7

7

8

912

8 12

15

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

W1 W2 W3

P155

P215

P340

P420

DEMAND 30 70 20 120

120


10 7

7

8

912

8 12

15

000

EXAMPLE 3 OF TP (DEGENERACY).

D E F G

A60

B100

C40

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.


7 3

2

8

52

6 5

4

6

10

1

TEST FOR OPTIMALITY & IMPROVEMENT OF SOLN MODIFIED DISTRIBUTION METHOD

D E F GA

60u1

0B

100u2

-1C

40u3

-10

25 55 20 200

v1

7v2

3v3

6V411

7 3

2

8

52

6 5

4

AF 0+6-8=-2 -2AG 0+11-6=+5 +5BD -1+7-4=+2 +2CD -10+7-2=-5 -5CE -10+3-6=-13 -13CF -10+6-5=-9 -9

djEMPTYCELL20

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=0ROW AAD: v1= 7AE::v2= 3ROW BBE: u2= -1BF: v3= 6BG: v4= 11ROW CCG: u3= -10

NET COST CHANGE

dij=(ui+vj)-cij

50 406

1

10


D E F GA

60u1

0B

100u2

-1C

40u3

-525 55 20 200

v1

7v2

3v3

6v3

6

7 3

2

8

52

6 5

4

AF 0+6-8=-2 -2BD -1+7-4=+2 +2BG -1+6-10=-5 -5CD -5+7-2=-0 0CE -5+3-6=-8 -8CF -5+6-5=-4 -4

djEMPTYCELL20

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=0ROW AAD: v1= 7AE::v2= 3AG:v4= 6ROW BBE: u2= -1BF: v3= 6ROW CCG: u3= -5

NET COST CHANGE

dij=(ui+vj)-cij

50

40

6

1

10


D E F GA

60u1

0B

100u2

-1C

40u3--5

25 55 20 200

v1

5v2

3v3

6V46

7 3

2

8

52

6 5

4

AD 0+5-7=-2 -2AF 0+6-8=-2 -2BG -1+6-10=-5 -5CD -5+5-2=-2 -5CE -5+3-6=-8 -8CF -5+6-5=-4 -4

djEMPTYCELL40

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=0ROW AAE::v2= 3AG: v4= 6ROW BBD: v1= 5BE: u2= -1BF: v3= 6ROW CCG: u3= -5

NET COST CHANGE

dij=(ui+vj)-cij

50306

1

10

EXAMPLE 4 OF TP (MAXIMISATION).

D E F G

A200

B500

C300

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.


12 8

14

6

107

3 11

8

25

18

20

TRANS SHIPMENT IN TPS I

PROHIBITED ROUTES INTPS I

EXAMPLE 5 OF TP.

U 1 2 3 4 50 1 1 1 X

1 3 3 3 X

1 1 4* X X

PROHIBITED ROUTES IN THE TPSOURCES DESTINATIONS SUPPLY

W1 W2 W3P1

M 407

58 45

P215 12

159 15

P3 257

158 12 40

DEMAND 25 55 20 100

25

V 6 7 81 M-7* 1 12 X 1 13 X 5* 14 X X 1*5 X

ITERATIONS

45

340

215

125

515

15

15

40 5

TP NUMERICALS S

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 FPRODN

A 16 19 22 14B 22 13 19 16C 14 28 8 12MONTHLY DEMAND 10 15 17DETERMINE THE OPTIMAL SHIPPING SCHEDULE SO THAT THE TRANSPORTATION COST IS MINIMIZED USING i) NWCR AND SSMii) VAM AND MODIFIED DISTRIBUTION METHOD

TP NUMERICALS

Q. NO. 2SOLVE 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. 3A 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.

WAREHOUSESW1 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.

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 -241 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 FROMF4 EXHAUSTED

3RD DEMAND FROM W3 MET

100

6090

4TH DEMAND FROM W4 MET

5th SUPPLY FROM F1 EXHAUSTED

17510025

6th SUPPLY FROM F2 EXHAUSTED

7TH DEMAND FROM W2 MET

NOTE: In the first iteration for VOGELwe put an X for all rows and columns because the constraint is that warehouse W1 is to be supplied entire quantity from factory F1

TP NUMERICALS

Q. NO 4 ( DEGENERACY)SOLVE THE FOLLOWING TRANSPORTATION PROBLEM.

D E F G SUPPLYA 7 3 8 6 60B 4 2 5 10 100C 2 6 5 1 40

DEMAND 20 50 50 80

ANS FOR Q NO. 4TABLE 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

TP NUMERICALS Q. NO. 5SOLVE THE FOLLOWING TRANSPORTATION PROBLEM USING VOGEL’S APPROXIMATION METHOD. TEST THIS SOLUTION FOR OPTIMALITY USING THE MODI METHOD.

DESTINATIONS SUPPLYSOURCES D E F GA 6 4 1 5 14B 8 9 2 7 16C 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 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

115

24

310

41,4,1

TP NUMERICALS Q. NO. 6A 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.

WHOLESELLERSFACTORIES A B C D EF1 5 7 10 25 15F2 8 6 9 12 14F3 10 9 8 10 15DETERMINE THE MAXIMUM PROFIT THAT THE COMPANY CAN MAKE

USING VOGEL APPROXIMATION METHOD AND MODI METHOD FOR CHECKING OPTIMALITY.

ANS FOR Q NO. 6PROFIT MATRIX

A B C D E

F1 250-100-5145

250-100-7143

250-100-10140

250-100-25125

250-100-15135

F2 250-90-8152

250-90-6154

250-90-9151

250-90-12148

250-90-14146

F3 250-80-10160

250-80-9161

250-80-8162

250-80-10160

250-80-15155

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

ANS FOR Q NO. 6 by VAMTABLE 1 Optimal

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

F1

250

17 19 22 37 27

250

0

500-250

17* 2 2 2 5 X 0

F2

50

10

300

8

250

11 14

400

16 0

1000-300-250-400

8 2 2 2 1 X -7

F3 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 X3 8 7 11* X 9 X

4 7 11* 11 X 11 X5 7 X 11* X 11 X

6 7 X X X 11* XVj 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

TP NUMERICALS Q. No. 7A 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 DIFFRAW MATL COST 10 9 12 9 DEPOTS DEPOTSTRANSPORT(TO S1)3 9 5 4 34 80-ATION (TO S2)1 7 4 5 32 120COSTS (TO S3) 5 8 3 6 31

150 PRODN. CAPACITY 100 150 50 100


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 MATRIXNOTE: SINCE SURPLUSPRODUCTION YIELDS ZERO PROFIT, THEREFORE, IN THE PROFIT MATRIX S4 IS ASSIGNEDZERO VALUE SIN THE CELLS

TP NUMERICALS.

D1 D2 D3 D4

R1 20

R2 25

R3 10

DEMAND 15 5 10 25 55

Q. NO. 8AN 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.


5

5

10

20

5

10

5 7

4

2 4

82 6

5 10 5

7

TP NUMERICALS Q. NO. 8 contdANSWER 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. 8ANSWER 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 TableD1 D2 D3 D4 Supply Ui

R1

5

5 7

10

2

5

4 200

R2 7

5

2 8

20

6 252

R3

10

4 5 10 5 10-1

Demand 15 5 10 25 55

Vj 5 0 2 4

-7

-2

-40

-9-6


FOR FINDING THE SECOND OPTIMAL SOLN, TRACE LOOP FROM R2D1 AS SHOWN AND SHIFT CELLS AS SHOWN IN THE NEXT SLIDE

Z = 235

ANS FOR Q NO. 8 MULTIPLE OPTIMALITY CHECK BY MODI

1st Optimal Table 2nd Optimal Soln D1 D2 D3 D4 Supply Ui

R1

5 7

10

2

10

4 200

R2

5

7

5

2 8

15

6 252

R3

10

4 5 10 5 10-1

Demand 15 5 10 25 55

Vj 5 0 2 4

-2

-4

-9-6


-70

Z = 235

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

Optimal Table D1 D2 D3 D4 Supply Ui

R1

5 7

10

2

10

4 200

R2

5

6

5

2 8

15

6 252

R3

10

4 5 10 5 10-1

Demand 15 5 10 25 55

Vj 5 0 2 4

-2

-4

-9-6


-70

Z = 230

TP NUMERICALS Q. NO.9A 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 DP 1 2 3 2Q 4 1 2 1DETERMINE 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.9A 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 BREADP 5000 10Q 4200 12FOUR 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 19B 4600 17C 1100 20D 3500 18THE 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 DP 1 2 3 2Q 4 1 2 1DETERMINE THE DELIVERY SCHEDULE FOR THE BREAD MANUFACTURING COMPANY THAT WILL MAXIMIZE ITS PROFITS. WRITE A DUAL OF THE TP


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 DUMMYSOURCE 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

ANS FOR Q NO. 9 by VAMTABLE 1 Optimal

A B C D Sup 1 2 3 4 5 6 Ui

P.

3600

0 3

1100

1

300

2

5000-36001100-300

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 13 x 1 x 1

4 x 4 x 5*5 X X X X

6 X X X XVj 0 3 1 2

-2

-4

0

-3 -1

0

13600

21100

3300

43200

51000

In 2nd iteration, we choose P row and not other row or cols because P has lowest cost cell (1) compared to all others

63600


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

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 D2S1 0 18 5 10S2 18 0 8 16D1 5 8 0 3D2 10 16 3 0

SOURCES DESTINATIONS

ANS FOR Q NO. 10 TRANS SHIPMENT by VAM &MODITABLE 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 25001 5 8 3 3

2 5 X 3 33 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

1500

2200

3500

4300

5100

6500

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 transthip 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 transported 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

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 W3F1 0 8 7 8 9F2 6 0 5 4 3W1 7 2 0 5 1W2 1 5 1 0 4W3 8 9 7 8 0

FACTORIES WAREHOUSES

ANS TO Q. NO. 11TRANSSHIPMENT PROBLEM).

F1 F2 W1 W2 W3 SUPPLYF1 0 8 7 8 9 200+500 =700F2 6 0 5 4 3 300+500 =800W1 7 2 0 5 1 500W2 1 5 1 0 4 500W3 8 9 7 8 0 500DEMAND 500 500 100+

500150+500

250+500

2500

FACTORIES WAREHOUSES

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.

WK1 WK2 WK3 WK4 DUMMYDEMAND

SUPPLY

WEEK1 (NORMAL) 50 53 56 59 0 1000WEEK2 (NORMAL) M 50 53 56 0 1000WEEK2 (OVERTIME) M 55 58 61 0 300WEEK3 (NORMAL) M M 60 63 0 1000WEEK3 (OVERTIME) M M 65 68 0 300WEEK4 (NORMAL) M M M 60 0 1000DEMAND 500 1000 1400 1200 500 4600

PRODUCTION WEEK COST OF PRODUCTION PER UNIT

ANS FOR Q NO. 10 TRANS SHIPMENT by VAM &MODI

A W1 W2 W3 W4 DUM

W1

50050

50053 56 59

0

W2 M

50050

50053 56

0

W2OT M 55

10058

20061 0

W3 M M

80060 63

2000

W3OT M M 65 68

3000

W4 M M M

100060 0

-2

-4

1500

2300

3200

4800

51000

6500

500

Sup Ui 1 2 3 4 5 6 7 8 91000.

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 500Vj 50 53 56 59 -41 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

7500

8500

9100

10200

-M+47

-M+57

-M+51

-M+54

-M+52

-M+54

-M+54

-M+57

-M+57 -3

-5 -5

-7

0

0

00500

200

200100

500500

300

1000

800

Interpretation: - Co. will make 1000 units in 1st week though dmdis 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 ofthese in the 4th week. -It will not run OT in the 3rd week.

CIRCLED NUMERALS SHOW dij VALUES

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 .

D1 D2 D3W1 5 1 7 100W2 6 4 6 800W3 3 2 5 150DEMAND

750 200 500 10501450

PENALTY 5 3 2

Q. NO. 14A 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.


ANS TO Q. NO. 14.

D1 D2 D3

W15

1001 7

100

W2 6006

1004

1006

800

W3 1503 2 5

150

W4 DUMMY 5 3

4002

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

V1234

Education

Transportation problem