WWW.PVPATILCLASS.IN 1

Unit -2

Transportation Problem

Introduction : Before 1941 when F.L.Hitchcook presented a study entitled “ The Distribution of a

Product from several sources to Numerous Localities “. This is first important contribution to the

solution of transportation problems. In 1947 T.C. Koopmans presented a study called “ Optimum

utilization of the transportation system “ . These contributions are mainly responsible for the

development of transportation model. It’s calculations are more simplified and less laborious than

that in simplex method it is a subclass of L.P.P and applied for machine assignment, plant location,

product mix problems, etc.

Definition: It is the special case of L.P.P in which all 𝑚 + 𝑛 − 1 constraints are equalities.

1] Transportation Problem as an L.P.P : Explain what do you mean by transportation problem give

it’s mathematical model and show that every T.P is a L.P.P.

Let there be m plants (source) manufacturing some product. These plants are situated at different

places. The artificial or goods produced are stored in n warehouses or go downs (destinations).

These warehouses are also at different places of transportation cost of one article depends the plant

and go down combination.

Let 𝑎1, 𝑎2, 𝑎3, . . . . 𝑎𝑛 be source and 𝑑1, 𝑑2, 𝑑3, . . . , 𝑑𝑛 be destinations. The capacity of the

plants (origin) are also differ and the requirement (demands) of warehouses are different.

Let 𝑎𝑖 = capacity of the ith plant

𝑏𝑗 = demand of the jth warehouse

𝑐𝑖𝑗 = cost of carrying one unit from ith plant to jth warehouse for 1 unit.

𝑥𝑖𝑗 = number of units carried from ith plant to jth warehouse.

In general matrix is

Now total cost of transportation = 𝑐11𝑥1 + 𝑐12𝑥2+ . . . . . +𝑐𝑚𝑛𝑥𝑚𝑛 = ∑ ∑ 𝑐𝑖𝑗𝑥𝑖𝑗𝑛𝑗=1

𝑚𝑖=1

This is to be minimised subject to ∑ 𝑥𝑖𝑗 = 𝑎𝑗𝑛𝑗=1 , 𝑖 = 1, 2, 3, . . . , 𝑚

∑ 𝑥𝑖𝑗 = 𝑏𝑖𝑚𝑖=1 , 𝑗 = 1, 2, 3, . . . , 𝑛 𝑎𝑛𝑑 𝑥𝑖𝑗 ≥ 0

WWW.PVPATILCLASS.IN 2

Thus mathematical formulation of T.P. is minimise 𝑧 = ∑ ∑ 𝑐𝑖𝑗𝑥𝑖𝑗𝑛𝑗=1

𝑚𝑖=1

subject to ∑ 𝑥𝑖𝑗 = 𝑎𝑗𝑛𝑗=1 , 𝑖 = 1, 2, 3, . . . , 𝑚

∑ 𝑥𝑖𝑗 = 𝑏𝑖𝑚𝑖=1 , 𝑗 = 1, 2, 3, . . . , 𝑛 𝑎𝑛𝑑 𝑥𝑖𝑗 ≥ 0

This is clearly as L.P.P. in which all m + n constants are in equalities.

∑ 𝑥𝑖𝑗 = 𝑎𝑗𝑛𝑗=1 , 𝑖 = 1, 2, 3, . . . , 𝑚 are known as supply condition.

∑ 𝑥𝑖𝑗 = 𝑏𝑖𝑚𝑖=1 , 𝑗 = 1, 2, 3, . . . , 𝑛 are known as demand condition.

Ex.1. A dairy farm has 3 plants located throughout a state. Daily milk production at each plant is as follows. Plant 1 → 6 million litres

Plant 2 → 1 million litres

Plant 3 → 10 million litres

Each day the firm must fulfil the needs of its 4 distribution centres. Minimum requirement to each centre as follows.

Distribution centre 1 → 7 million litres.

centre 2 → 5 million litres.

centre 3 → 3 million litres.

centre 4 → 2 million litres.

Cost of shipping 1 million litre of milk from each plant to each distribution centre is given in the following table.

Distribution 1 Distribution 2 Distribution 3 Distribution 4

Plant 1 2 3 11 7

Plant 2 1 0 6 1

Plant 3 5 8 15 9

Formulate as T.P.

Sol :

Types of T.P. :

1) Balanced T.P. : A T.P. is said to be balance if there is total supply = total demand in it. i.e. if ∑ 𝑎𝑖

𝑚𝑖=1 = ∑ 𝑏𝑗

𝑛𝑖=1 then it is balanced T.P.

WWW.PVPATILCLASS.IN 3

2) Unbalanced T.P. : A T.P. is said to be unbalance if there is total supply ≠ total demand in it.

i.e. if ∑ 𝑎𝑖𝑚𝑖=1 ≠ ∑ 𝑏𝑗

𝑛𝑖=1 then it is unbalanced T.P.

Before starting to find any solution to unbalanced problem we make it balanced by adding dummy warehouse (destination) or dummy plant (origin or source) with the required demand or capacity. For dummy destination or source all cost entries will be taken as 0.

Some useful definitions: 1] Solution of T.P. : Any set of 𝑥𝑖𝑗 satisfying the demand supply equation,

∑ 𝑥𝑖𝑗𝑛 𝑗=1

= 𝑎𝑖 … … … … … … … … 𝑖 = 1,2,3, … . , 𝑚

∑ 𝑥𝑖𝑗𝑛𝑗=1 = 𝑏𝑖 … … … … … … … … 𝑗 = 1,2,3, … . , 𝑛 Is known as solution of T.P.

2) Optimum Solution : A solution which gives minimum cost of transportation is known as optimum solution of T.P.

min ∑ ∑ 𝑐𝑖𝑗𝑥𝑖𝑗

𝑛

𝑗=1

𝑚

𝑖=1

3) Feasible solution : A set of non-negative values xij, i-1,2,3,…..,m , j=1,2,3,….,n that satisfies the constraints is called a feasible solution to the transportation problem.

4) Basic feasible solution : A set of values of xij satisfying demand supply equation in which number of occupied cells <= m+n-1 is known as basic feasible solution. If m+n-1 > occupied cell then it is degenerated solution. If occupied cells = m+n-1 then , it is non-degenerated basic solution.

5) Initial basic feasible solution : To given T.P. we first find a starting solution by any one of known methods. If it is basic then it is called as i.b.f.s.

Methods of finding Ibfs :

a) North West Corner Rule : Following are the steps in N.W. (North West)rule. 1) Start with north-west corner i.e. cell (1, 1) and then put their maximum possible

quantity according to demand supply equation of cell (1, 1) Thus 𝑖) if 𝑎1 < 𝑏1 then we take 𝑥11 = 𝑎1 𝑖𝑖) if 𝑎1 > 𝑏1 then we take 𝑥11 = 𝑏1 𝑖𝑖𝑖) if 𝑎1 = 𝑏1 then we take 𝑥11 = 𝑎1 = 𝑏1

2) Find new demand supply equation number and move to next cell For case 1 we move to cell (2,1) For case 2 we move to cell (1,2) For case 3 we move to cell (2,2):

3) Put maximum possible quantity according to new demand supply equations in cell selected in step 2. For this we follow step 1

4) Repeat the same procedure till all allocation are done.

Ex.1. Find the starting solution of T.P. by N-W corner method.

Origin Destination 𝑊1 𝑊2 𝑊3

Supply

𝐹1 𝐹2 𝐹3 𝐹4

2 7 4 3 3 1 5 4 7 1 6 2

5 8 7 14

Demand

7 9 18

34 34

Sol : Given T. P. is balanced.

∴ For starting solution

WWW.PVPATILCLASS.IN 4

Transportation cost = 5 × 2 + 2 × 3 + 6 × 3 + 3 × 4 + 4 × 7 + 2 × 14 = 102.

Ex.2.Find starting solution of T.P. by N-W rule.

Origin Destination 𝑊1 𝑊2 𝑊3 𝑊4

Supply

𝐹1 𝐹2 𝐹3

2 3 11 7 1 0 6 1 5 8 15 9

6 1 10

Demand

7 5 3 2

17 17

Sol : In given T.P. Total demand = Total supply, so T.P. is balanced

Transportation cost = 6 × 2 + 1 × 1 + 5 × 8 + 15 × 3 + 9 × 2 = 116

b) Matrix Minima Method ( Least cost method ) : Following are the steps in this method. 1) Observe all cost entries in transportation matrix and choose that cell whose cost entry

is minimum. In the case of tie we can choose any cell from the tie. 2) Observe the demand supply equation of the cell choosen in step one and put

maximum possible value of 𝑥𝑖𝑗 in that cell.

WWW.PVPATILCLASS.IN 5

3) Find new demand supply equation and repeat the process till all allocations are done.

Ex.1. Find the solution of T.P. by least cost method

Sol : Given T. P. is balanced because Total supply = Total demand.

Transportation cost = 7 × 2 + 3 × 4 + 8 × 1 + 7 × 4 + 7 × 2 + 7 × 1 = 83

Ex.2. Find solution for the following T.P. by least cost method.

Origin Destination 1 2 3 4

Supply

1 2 3

2 3 11 7 1 0 6 1 5 8 15 9

6 1 10

Demand 7 5 3 2

Sol : Given T. P. is balanced because Total supply = Total demand.

Origins 𝑊1 𝑊2 𝑊3 Supply

𝐹1 𝐹2 𝐹2 𝐹4

2 7 4 3 3 1 5 4 7 1 6 2

5 8 7 14

Demand 7 9 18

WWW.PVPATILCLASS.IN 6

Transportation cost = 6 × 2 + 1 × 1 + 5 × 1 + 8 × 5 + 15 × 3 + 9 × 1

= 12 + 1 + 5 + 40 + 45 + 9 = 112

c) Voggel Approximation Method : Following are the steps in this method

i) For each row find the difference between least cost and next to least cost. In that row in case of similar costs difference is 0. This is penalty for this row of element of that row is not chosen is solution say row penalty.

ii) For every column find the difference between least cost and next to least cost in that column. This is penalty for that column if element of that column is not chosen in solution say column penalty.

iii) Choose that row or column whose penalty is highest. iv) In the row or column choose in steps find that cell whose cost is least. v) For cell chosen in step iv observe demand supply equation put maximum possible

quantity in that cell. vi) Find new demand supply equation and repeat the same procedure. vii) Allow every step till all location are done.

Ex.1. Find the starting solution for following T.P. by V.A.M.

Sol : Given T.P. is balanced because, Total Supply = Total Demand

Transportation Cost = 5 × 2 + 8 × 1 + 7 × 4 + 2 × 1 + 2 × 6 + 2 × 10

= 10 + 8 + 28 + 2 + 12 + 20 = 80.

WWW.PVPATILCLASS.IN 7

Ex.2. Solve the following T.P. by V.A.M

Origin Destination 1 2 3 4

Supply

1 2 3

2 3 11 7 1 0 6 1 5 8 15 9

6 1 10

Demand 7 5 3 2

Sol : Given T.P. is balanced because, Total Supply = Total Demand

Transportation Cost = 1 × 2 + 5 × 3 + 5 × 6 + 15 × 3 + 9 × 1 + 1 × 1

= 2 + 15 + 30 + 45 + 9 + 1

= 102.

Ex.3. Find the ibts of following T.P. by 1)N-W rule , 2)Matrix Minama method , 3)Voggel’s method

Origin Destination 𝑤1 𝑤2 𝑤3 𝑤4

Supply

𝐹1 𝐹2 𝐹3

15 51 42 33 30 42 26 81 90 40 66 80

23 44 33

Demand 23 31 16 30

Solution: Given T.P is balanced because, Total Supply = Total Demand

i) By N-W corner rule

WWW.PVPATILCLASS.IN 8

Transportation cost = 23 × 15 + 31 × 42 + 13 × 26 + 3 × 66 + 30 × 80 = 345 + 1302 + 338 + 198 + 2400 = 4583

ii) Matrix Minima method

Transportation cost = 23 × 15 + 16 × 26 + 81 × 28 + 31 × 40 + 80 × 2 = 345 + 416 + 3268 + 1240 + 160 = 4389

iii) Voggel’s method

WWW.PVPATILCLASS.IN 9

Transportation cost = 33 × 23 + 23 × 30 + 5 × 42 + 16 × 26 + 40 × 26 + 7 × 60 = 759 + 690 + 210 + 416 + 1040 + 420 = 3535.

3] Optimity test and optimisation of solution to T.P by u.v. method : Modified Distribution Methode) : Following are the steps in u.v. method

1) Test whether T.P. is balanced or unbalanced. If unbalanced make it balanced. 2) Find ibfs to the T.P. by any one of method. 3) Check whether the starting solution is basic or not that i.e. check that number of occupied

cell = 𝑚 + 𝑛 − 1. If not, this is the case of degeneracy. Resolve the degeneracy first. 4) For occupied cell 𝑚 + 𝑛 − 1 cells solve the equations 𝑢𝑖 + 𝑣𝑗 = 𝑐𝑖𝑗 by taking 𝑢1 = 0 (or

some 𝑢𝑖 = 0 or 𝑣𝑗 = 0)

5) Compute net evaluations 𝛿𝑖𝑗 = 𝑢𝑖 + 𝑣𝑗 − 𝑐𝑖𝑗 .

6) If all 𝛿𝑖𝑗 are non-positive then the solution is optimized (test of optimity) and read the optimum solution but if not, then proceed further.

7) Select that cell whose net evaluation is positive and highest this we have to enter into basic of solution.

8) From the cell selected in previous step try a close path starting from that cell and ending in the same cell this path consist horizontal and vertical lines only and no two lines can intersect. We can take a turn at occupied cell only this mean that we can go through occupied cell such path is one and only one.

9) In above path make positive mark on the cell selected and then alternately negative and positive mark where we have taken turn.

10) Find the least quantity of the 𝑥𝑖𝑗 in cells marked by negative sign and add or subtract this

quantity as per mark of that cell. We get new and improved solution. 11) Apply the test of optimity for it and if it is not optimal then carry out complete process again

and again till we get optimum solution.

Ex.1. Following is the solution to a T.P. by V.A.M. optimise it.

Sol : Given solution is

WWW.PVPATILCLASS.IN 10

𝑖) 𝑢1 + 𝑣4 = 33 , 𝑖𝑓 𝑢1 = 0 ∴ 𝑣4 = 33 𝑖𝑖) 𝑢3 + 𝑣4 = 60 , ∴ 𝑢3 + 33 = 60 , ∴ 𝑢3 = 27 𝑖𝑖𝑖) 𝑢3 + 𝑣2 = 40 , ∴ 27 + 𝑣2 = 40 , ∴ 𝑣2 = 13 𝑖𝑣) 𝑢2 + 𝑣2 = 42 , ∴ 𝑢2 + 13 = 42 , ∴ 𝑢2 = 29 𝑣) 𝑢2 + 𝑣1 = 30 , ∴ 29 + 𝑣1 = 30 , ∴ 𝑣1 = 1 𝑣𝑖) 𝑢2 + 𝑣3 = 26 , ∴ 29 + 𝑣3 = 26 , ∴ 𝑣3 = −3 ∴ 𝑢1 = 0 , 𝑢2 = 30 , 𝑢3 = 27 𝑎𝑛𝑑 𝑣1 = 1 , 𝑣2 = 13 , 𝑣3 = −3 , 𝑣4 = 33 Now 𝛿𝑖𝑗 for unoccupied cells

𝛿11 = 𝑢1 + 𝑣1 − 𝐶11 = 0 + 1 − 15 = −14 𝛿12 = 𝑢1 + 𝑣2 − 𝐶12 = 0 + 13 − 51 = −38 𝛿13 = 𝑢1 + 𝑣3 − 𝐶13 = 0 − 3 − 42 = −45 𝛿24 = 𝑢2 + 𝑣4 − 𝐶14 = 30 − 33 − 81 = −84 𝛿31 = 𝑢3 + 𝑣1 − 𝐶31 = 27 + 13 − 60 = −32 𝛿33 = 𝑢3 + 𝑣3 − 𝐶33 = 27 − 3 − 66 = −42 All 𝛿𝑖𝑗 are negative, hence optimum solution is reached.

Transportation cost = 23 × 33 + 23 × 30 + 5 × 42 + 10 × 26 + 26 × 40 + 60 × 7 = 759 + 690 + 210 + 260 + 1040 + 420 = 3535

Next part will be upload in 4 days