solution of problems on transportation

7/31/2019 Solution of Problems on Transportation

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1

QUANTITATIVEQUANTITATIVEMETHODSMETHODS

UNIVERSITY PROBLEMSUNIVERSITY PROBLEMS

ON TRANSPORTATIONON TRANSPORTATION



2

TRANSPORTATIONTRANSPORTATION

Q 1Q 1

The Punjab Transport Company has trucks availableat four different sites as stated below:

Site A B C D Trucks 5 10 7

3Customers W, X, and Y require trucks as shown

under:W: 5 Trucks, X: 8 Trucks, Y: 10 TrucksVariable costs of getting trucks to the customer are:From A to W - Rs.70, to X - Rs.30, to Y - Rs.60From B to W - Rs.40, to X - Rs.60, to Y - Rs.80

From C to W - Rs.50, to X - Rs.80, to Y - Rs.40



3


S 1S 1 Initial Solution Using VAM

W X Y DUMMY

A 70 30 (5) 60 0

B 40 (5) 60 (3) 80 (2) (-) 0 (+)

C 50 80 40 (5) (+) 0 (2) (-)

D 80 40 30 (3) 0

Ui + Vj to Test Initial Solution Using VAM

V1 = 10 V2 = 30 V3 = 50 V4 = 10

U1 = 0 10 30 50 10*

U2 = 30 40 60 80 40*

U3 = -10 0 20 40 0

U4 = -20 -10 10 30 -10



4


S 1S 1 1st Improvement of Initial Solution Using VAM [Degeneracy Exists]

W X Y DUMMY

A 70 30 (5) 60 0

B 40 (5) 60 (3) (-) 80 0 (2) (+)

C 50 80 40 (7) (+) 0 (€) (-)

D 80 40 (+) 30 (3) (-) 0

Ui + Vj to Test 1st Improvement of Initial Solution Using VAM

V1 = 10 V2 = 30 V3 = 10 V4 = -30

U1 = 0 10 30 10 -30

U2 = 30 40 60 40 0

U3 = 30 40 60 40 0

U4 = 20 30 50* 30 -10



5


S 1S 1 2nd improvement of Initial Solution Using VAM

W X Y DUMMY

A 70 30 (5) 60 0

B 40 (5) 60 (3) 80 0 (2)

C 50 80 40 (7) 0

D 80 40 (€) 30 (3) 0

Ui + Vj to Test 2nd improvement of Initial Solution Using VAM

V1 = 10 V2 = 30 V3 = 2 V4 = -30

U1 = 0 10 30 20 -30

U2 = 30 40 60 50 0

U3 = 20 30 50 40 -10

U4 = 10 20 40 30 -10



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S 1S 1

Optimal Cost = Rs.900/-



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TRANSPORTATIONTRANSPORTATIONQ 2Q 2

The following table shows all the necessaryinformation on the availability of supply toeach warehouse, the requirement of each

market and the unit transportation cost fromeach warehouse to each market.Warehouse Market Supply

P Q R S

A 6 3 5 4 22

B 5 9 2 7 15

C 5 7 8 6 8

Requirement 7 12 17 9



8


Q 2 Contd.Q 2 Contd.

The shipping clerk has worked out the followingschedule from experience.

12 units from A to Q, 1 unit from A to R,9 units from A to S, 15 units from B to R,7 units from C to P, 1 unit from C to R.a) Check and see if the clerk has the optimalschedule.b) Find the optimal schedule and minimumtransportation cost.c) If the clerk is approached by a carrier to routeC to Q, who offers to reduce rate in the hope of

getting some business, by how much the rate should



9


S 2S 2 Warehouse Market

P Q R S

A 6 3 (12) 5 (1) (+) 4 (9) (-)

B 5 9 2 (15) 7

C 5 (7) 7 8 (1) (-) 6 (+)

Ui + Vj

V1 = 2 V2 = 3 V3 = 5 V4 = 4

U1 =0 2 3 5 4

U2 = -3 -1 0 2 1

U3 = 3 5 6 8 7*



10



P Q R S

A 6 3 (12) 5 (2) 4 (8)

B 5 9 2 (15) 7

C 5 (7) 7 8 6 (1)

Ui + Vj

V1 = 3 V2 = 3 V3 = 5 V4 = 4

U1 =0 3 3 5 4

U2 = -3 0 0 2 1

U3 = 2 5 5 7 6



11

TRANSPORTATIONTRANSPORTATIONS 2S 2 (i) The clerk doesn’t have the optimal

schedule.(ii) Minimum transportation cost = Rs.149/-(iii) For having allocation from C to Q, the route

CQ must be economical, i.e., the opportunitycost for the route CQ should be negative orzero.

That is,CCQ – (V2 + U3) ≤ 0

i.e., CCQ ≤ 5.Hence, the rate should be reduced to Rs.5.



12

TRANSPORTATIONTRANSPORTATIONQ 3Q 3In the following transportation problem, thepenalty costs per unit of unsatisfied demandare 15, 13, and 12 for destinations 1, 2, and

3; then find the optimal solution. Further if,the demand at destination 3 must besatisfied exactly, obtain the revised optimalsolution.

Source Destination Supply

1 2 3

A 15 11 16 15

B 16 14 16 60

C 13 15 18 30

Demand 75 20 50



13


S 3S 3 Source Destination

1 2 3

A 15 11 16

B 16 14 16

C 13 15 18

DUMMY CD1

CD2

CD3

Difference between two lowest unit costs of the cells in Col.1, Col

and Col.3 must be 15, 13, and 12.

13 – CD1 = 15 CD1 = -211 – CD2 = 13 CD2 = -2

16 – CD3 = 12 CD3 = +4



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S 3S 3 Source Destination1 2 3

A 15 11 (15) 16

B 16 (5) 14 (5) 16 (50)

C 13 (30) 15 18DUMMY -2 (40) -2 4

Source Destination

V1 = 13 V2 = 11 V3 = 13

U1 = 0 13 11 13

U2 = 3 16 14 16

U3 = 0 13 11 13

U4 = -15 -2 -4 -2



15


S 3S 3




16

TRANSPORTATIONTRANSPORTATIONQ 4Q 4Solve the following transportation problem.


D1 D2 D3 D4

S1 10 20 5 7 10

S2 13 9 12 8 20

S3 4 15 7 9 30

S4 14 7 8 0 40

S5 3 12 5 19 50

Demand 60 60 20 10



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TRANSPORTATIONTRANSPORTATIONS 4S 4

Source Destination

D1 D2 D3 D4

S1 10 (10) 20 5 7

S2 13 9 (20) 12 8

S3 4 (30) 15 7 9

S4 14 7 (10) 1 (20) 0 (10)

S5 3 (20) 12 (30) 5 19



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S 4S 4

Source Destination

D1 D2 D3 D4

S1 10 (10) (-) 20 5 (+) 7

S2 13 9 (20) 12 8

S3 4 (30) 15 7 9

S4 14 7 (10) (+) 1 (20) (-) 0 (10)S5 3 (20) (+) 12 (30) (-) 5 19

Source Destination

V1 = 10 V2 = 19 V3 = 13 V4 = 12

U1 = 0 10 19 13* 12

U2 = -10 0 9 3 2

U3 = -6 4 13 7 6

U4 = -12 -2 7 1 0

U5 = -7 3 12 6 5



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S 4S 4

Source Destination

D1 D2 D3 D4

S1 10 20 5 (10) 7

S2 13 9 (20) 12 8

S3 4 (30) 15 7 9

S4 14 7 (20) (+) 1 (10) (-) 0 (10)

S5 3 (20) 12 (20) (-) 5 (+) 19

Source Destination

V1 = 2 V2 = 11 V3 = 5 V4 = 4

U1 = 0 2 11 5 4

U2 = -2 0 9 3 2

U3 = 2 4 13 7 6

U4 = -4 -2 7 1 0

U5 = 1 3 12 6* 5



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S 4S 4

Source Destination

D1 D2 D3 D4

S1 10 20 5 (10) 7

S2 13 9 (20) 12 8

S3 4 (30) 15 7 9

S4 14 7 (30) 1 0 (10)

S5 3 (20) 12 (10) 5 (10) 19

Source Destination

V1 = 3 V2 = 12 V3 = 5 V4 = 5

U1 = 0 3 12 5 5

U2 = -3 0 9 2 2

U3 = 1 4 13 6 6

U4 = -5 -2 7 0 0

U5 = 0 3 12 5 5



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S 4S 4




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Find the optimal solution for the followingcost matrix.


D1 D2 D3 D4O1 20 22 17 4 120

O2 24 37 9 7 70

O3 32 37 20 15 50Demand 60 40 30 110



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S 5S 5 Source Destination

D1 D2 D3 D4

O1 20 (+) 22 (40) 17 4 (80) (-)

O2 24 (10) (-) 37 9 (30) 7 (30) (+)

O3 32 (50) 37 20 15

Source Destination

V1 = 21 V2 = 22 V3 = 6 V4 = 4

U1 = 0 21* 22 6 4

U2 = 3 24 25 9 7

U3 = 11 32 33 17 15



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S 5S 5Source Destination

D1 D2 D3 D4

O1 20 (10) (+) 22 (40) 17 4 (70) (-)

O2 24 37 9 (30) 7 (40)

O3 32 (50) (-) 37 20 15 (+)

Source Destination

V1 = 20 V2 = 22 V3 = 6 V4 = 4

U1 = 0 20 22 6 4

U2 = 3 23 25 9 7

U3 = 12 32 34 18 16*



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S 5S 5Source Destination

D1 D2 D3 D4

O1 20 (60) 22 (40) 17 4 (20)

O2 24 37 9 (30) 7 (40)

O3 32 37 20 15 (50)

Source Destination

V1 = 20 V2 = 22 V3 = 6 V4 = 4

U1 = 0 20 22 6 4

U2 = 3 23 25 9 7

U3 = 11 31 33 17 15



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S 5S 5




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Factory Warehouse Capacity

D1 D2 D3 D4F1 19 30 50 16 7

F2 70 30 40 60 9

F3 40 8 70 20 18Requirement 5 8 7 14



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S 6S 6 Factory Warehouse

D1 D2 D3 D4

F1 19 (5) 30 50 16 (2)

F2 70 30 (+) 40 (7) 60 (2) (-)

F3 40 8 (8) (-) 70 20 (10) (+)

Factory Warehouse

V1 = 19 V2 = 4 V3 = -4 V4 = 16

U1 = 0 19 4 -4 16

U2 = 44 63 48* 40 60

U3 = 4 23 8 0 20



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S 6S 6 Factory Warehouse

D1 D2 D3 D4

F1 19 (5) 30 50 16 (2)

F2 70 30 (2) 40 (7) 60

F3 40 8 (6) 70 20 (12)

Factory Warehouse

V1 = 19 V2 = 4 V3 = 14 V4 = 16

U1 = 0 19 4 14 16

U2 = 26 45 30 40 42

U3 = 4 23 8 18 20



30


S 6S 6




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Factory Destination Capacity

I II III IVA 20 5 25 15 50

B 17 13 16 17 50

C 5 21 19 23 100Requirement 30 40 60 70



32


S 7S 7 Factory Destination

I II III IV

A 20 5 (40) 25 15 (10)

B 17 13 16 17 (50)

C 5 (30) 21 19 (60) 23 (10)

Factory Destination

V1 = -3 V2 = 5 V3 = 11 V4 = 15

U1 = 0 -3 5 11 15

U2 = 2 -1 7 13 17

U3 = 8 5 13 19 23



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S 7S 7




34


Given the following transportation problem:

Warehouse Market Supply

A B CW1 10 12 7 180

W2 14 11 6 100

W3 9 5 13 160

W4 11 7 9 120

Demand 240 200 220



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TRANSPORTATIONTRANSPORTATIONQ 8 Contd.Q 8 Contd.It is known that currently no units can besent from warehouse W1 to market A andfrom warehouse W3 to market C. determine

the least cost transportation schedule. Is theoptimal solution obtained by you unique? If not, what is/are the other optimal solution/s?



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S 8S 8

Warehouse Market

A B C

W1 10 (60) 12 7 (120)

W2 14 11 6 (100)

W3 9 (+) 5 (160) (-) 13

W4 11 (80) (-) 7 (40) (+) 9DUMMY 0 (100) 0 0

Warehouse Market

V1 = 10 V2 = 6 V3 = 7

U1 = 0 10 6 7U2 = -1 9 5 6

U3 = -1 9* 5 6

U4 = 1 11 7 8

U5 = -10 0 -4 -3



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S 8S 8

Warehouse Market

A B C

W1 10 (60) 12 7 (120)

W2 14 11 6 (100)

W3 9 (80) 5 (80) 13W4 11 7 (120) 9

DUMMY 0 (100) 0 0

Alternate Solution will be

Optimal Solution: Rs.4000/-



38


Solve the following transportation problemfor maximum profit:

Warehouse Market Capacity

A B C D

X 12 18 6 25 200

Y 8 7 10 18 500

Z 14 3 11 20 300

Demand 180 320 100 400 1000



39

TRANSPORTATIONTRANSPORTATIONS 9S 9

Warehouse Market Capacity

A B C D

X 13 7 19 0 200

Y 17 18 15 7 500

Z 11 22 14 5 300

Demand 180 320 100 400 1000

n converting the given maximization problem to the minimization:



40



A B C D

X 13 7 (200) 19 0

Y 17 18 (100) (+) 15 7 (400) (-)

Z 11 (180) 22 (20) (-) 14 (100) 5 (+)

Warehouse Market

V1 = -4 V2 = 7 V3 = -1 V4 = -4

U1 = 0 -4 7 -1 -4

U2 = 11 7 18 10 7

U3 = 15 11 22 14 11*



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A B C D

X 13 7 (200) 19 0

Y 17 18 (120) 15 (+) 7 (380) (-)

Z 11 (180) 22 14 (100) (-) 5 (20) (+)

Warehouse Market

V1 = 2 V2 = 7 V3 = 5 V4 = -4

U1 = 0 2 7 5 -4

U2 = 11 13 18 16* 7

U3 = 9 11 16 14 5



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A B C D

X 13 7 (200) 19 0

Y 17 18 (120) 15 (100) 7 (280)

Z 11 (180) 22 14 5 (120)

Warehouse Market

V1 = 2 V2 = 7 V3 = 4 V4 = -4

U1 = 0 2 7 4 -4

U2 = 11 13 18 15 7

U3 = 9 11 16 13 5



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S 9S 9Hence, optimal allocation for maximizationWarehouse Market

A B C D

X 12 18 (200) 6 25

Y 8 7 (120) 10 (100) 18 (280)

Z 14 (180) 3 11 20 (120)

Maximum profit: Rs.15400/-



44

TRANSPORTATIONTRANSPORTATIONQ 10Q 10A manufacturer of jeans is interested indeveloping an advertising campaign that willreach 4 different age groups. Advertising

campaign can be conducted through T.V., radio,and magazines. The following table givesestimated cost per exposure for each group inappropriate units of money, according to

medium employed. The maximum exposurelevels possible in each of the media T.V., radio,and magazine are 40, 30, and 20 millionrespectively. Also desired exposures in each

age group 12 - 18, 19 - 25, 25 - 35, and 36 and



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Q 10 Contd.Q 10 Contd.

Formulate above problem as Transportation problemand find the optimal solution.

Media Age Group

12 – 18 19 – 25 26 – 35 36 and above

T.V. 12 7 10 10

Radio 10 9 12 10

Magazine 14 12 9 12



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S 10S 10 12 – 18 19 – 25 26 – 35 36 + DUMMY

TV 12 7 (25) 10 (5) (-) 10 (10) 0 (+)

RADIO 10 (30) 9 12 10 0 (€)

MAGAZINE 14 12 9 (10) (+) 12 0 (10) (-)

V1 = 11 V2 = 7 V3 = 10 V4 = 10 V5 =1

U1 = 0 11 7 10 10 1*

U2 = -1 10 6 9 9 0

U3 = -1 10 6 9 9 0



47


S 10S 10 12 – 18 19 – 25 26 – 35 36 and above DUMMY

TV 12 7 (25) 10 10 (10) (-) 0 (5) (+)

RADIO 10 (30) 9 12 10 (+) 0 (€) (-)

MAGAZINE 14 12 9 (15) 12 0 (5)

V1 = 10 V2 = 7 V3 = 9 V4 = 10 V5 =0

U1 = 0 10 7 9 10 0

U2 = 0 10 7 9 10* 0

U3 = 0 10 7 9 10 0



48


S 10S 10

Minimum Cost = Rs.710/-



49


A company has factories at A, B, and C which supplywarehouses at D, E, F, and G. monthly warehouserequirements are 80, 90, 110 and 160 unitsrespectively. Monthly capacities are 160, 150, and

190 units respectively. Unit shipping costs (Rs.) aregiven below:

D E F G

A 12 18 8 7

B 10 19 22 21

C 3 8 10 13

What is optimal shipping cost schedule?



50


S 11S 11 D E F G DUMMY

A 12 18 8 7 (160) 0 (€)

B 10 (80) 19 (+) 22 (10) (-) 21 0 (60)

C 3 8 (90) (-) 10 (100) (+) 13 0

V1 = 10 V2 = 20 V3 = 22 V4 = 7 V5 = 0

U1 = 0 10 20* 22* 7 0

U2 = 0 10 20* 22 7 0

U3 = -12 -2 8 10 -5 -12



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A 12 18 (+) 8 7 (160) 0 (€) (-)

B 10 (80) 19 (10) (-) 22 21 0 (60) (+)

C 3 8 (80) 10 (110) 13 0

V1 = 10 V2 = 19 V3 = 21 V4 = 7 V5 = 0

U1 = 0 10 19* 21* 7 0

U2 = 0 10 19 21 7 0

U3 = -11 -1 8 10 -4 -11



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A 12 18 (€) (-) 8 (+) 7 (160) 0

B 10 (80) 19 (10) 22 21 0 (60)

C 3 8 (80) (+) 10 (110) (-) 13 0

V1 = 9 V2 = 18 V3 = 20 V4 = 7 V5 = -1

U1 = 0 9 18 20* 7 -1

U2 = 1 10 19 21 8 0

U3 = -10 -1 8 10 -3 -11



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A 12 18 8 (€) 7 (160) 0

B 10 (80) 19 (10) 22 21 0 (60)

C 3 8 (80) 10 (110) 13 0

V1 = -3 V2 = 6 V3 = 8 V4 = 7 V5 = -13

U1 = 0 -3 6 8 7 -13

U2 = 13 10 19 21 20 0

U3 = 2 -1 8 10 9 -11



54


S 11S 11




55


Solve the following transportation problem given theunit transportation costs, demand and supply asbelow:

Sources Warehouse Supply

A B C

1 5 1 7 10

2 6 4 6 80

3 3 2 5 15

Demand 75 20 50



56


S 12S 12 Sources Warehouse

A B C

1 5 1 (10) 7

2 6 (60) (-) 4 (10) 6 (10) (+)

3 3 (15) 2 5DUMMY 0 (+) 0 0 (40) (-)

Sources Warehouse

V1 = 3 V2 = 1 V3 = 3

U1 = 0 3 1 3U2 = 3 6 4 6

U3 = 0 3 1 3

U4 = -3 0* -2 0



57


S 12S 12

Sources Warehouse

A B C

1 5 1 (10) 7

2 6 (20) 4 (10) 6 (50)

3 3 (15) 2 5DUMMY 0 (40) 0 0

Alternate Solution

Minimum Cost: Rs.515/-



58


A steel company has three open hearth furnaces andfive rolling mills. Transportation cost (Rs. per ton) forshipping steel from furnaces to rolling mills are asshown in the following table:

What is optimal shipping cost schedule?

Furnaces M1 M2 M3 M4 M5 Capacity

F1 4 2 3 2 6 8

F2 5 4 5 2 1 12

F3 6 5 4 7 3 14

Demand 4 4 6 8 8



59


S 13S 13 Furnaces M1 M2 M3 M4 M5 DUMMY

F1 4 2 (4) 3 (+) 2 (4) 6 0 (€) (-)

F2 5 4 5 2 (4) 1 (8) 0

F3 6 (4) 5 4 (6) (-) 7 3 0 (4) (+)

Furnaces V1 = 6 V2 = 2 V3 = 4 V4 = 2 V5 = 1 V6 = 0

U1 = 0 6* 2 4* 2 1 0

U2 = 0 6* 2 4 2 1 0

U3 = 0 6 2 4 2 1 0



60



F1 4 (+) 2 (4) 3 (€) (-) 2 (4) 6 0

F2 5 4 5 2 (4) 1 (8) 0

F3 6 (4) (-) 5 4 (6) (+) 7 3 0 (4)

Furnaces V1 = 5 V2 = 2 V3 = 3 V4 = 2 V5 = 1 V6 = -1

U1 = 0 5* 2 3 2 1 -1

U2 = 0 5 2 3 2 1 -1

U3 = 1 6 3 4 3 2 0



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F1 4 (€) 2 (4) 3 2 (4) 6 0

F2 5 4 5 2 (4) 1 (8) 0

F3 6 (4) 5 4 (6) 7 3 0 (4)

Furnaces V1 = 5 V2 = 2 V3 = 3 V4 = 2 V5 = 1 V6 = -1

U1 = 0 4 2 2 2 1 -2

U2 = 0 4 2 2 2 1 -2

U3 = 1 6 4 4 4 3 0



62


S 13S 13

Minimum Cost: Rs.80/-



63


Find the optimal solution for the following costmatrix.

D1 D2 D3 D4 Availability

O1 26 22 17 4 120

O2 24 37 9 7 70

O3 32 37 20 15 50

Demand 60 40 30 110 240



64


S 14S 14 D1 D2 D3 D4

O1 26 22 (40) 17 4 (80)

O2 24 (10) (+) 37 9 (30) 7 (30) (-)

O3 32 (50) (-) 37 20 15 (+)

V1 = 21 V2 = 22 V3 = 6 V4 = 4

U1 = 0 21 22 6 4

U2 = 3 24 25 9 7

U3 = 11 32 33 17 15*




S 14S 14

Minimum Cost = Rs.3520/

solution of problems on transportation

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