Samacheer Kalvi 12th Business Maths Guide Chapter 10 Operations Research Ex 10.1

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 10 Operations Research Ex 10.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Ex 10.1

Question 1.
What is transportation problem?
Solution:
The objective of transportation problem is to determine the amount to be transported from each origin to each destinations such that the total transportation cost is minimized.

Question 2.
Write mathematical form of transportation problem.
Solution:
Let there be m origins and n destinations. Let the amount of supply at th i th origin is a_i. Let the demand at j th destination is b_j.
The cost of transporting one unit of an item from origin i to destination j is C_ij and is known for all combination (i,j). Quantity transported from origin i to destination j be x_ij.

The objective is to determine the quantity x_ij to be transported over all routes (i,j) so as to minimize the total transportation cost. The supply limits at the origins and the demand requirements at the destinations must be satisfied.
The above transportation problem can be written in the following tabular form:

Now the linear programming model representing the transportation problem is given by
The objective function is Minimize Z = \(\sum_{\mathbf{i}=\mathbf{1}}^{\mathbf{m}}\), \(\sum_{\mathbf{j}=\mathbf{1}}^{\mathbf{n}}\) c_ij x_ij
Subject to the constraints
\(\sum_{\mathbf{j}=\mathbf{1}}^{\mathbf{n}}\) = x_ij = a_i, i = 1, 2 …….. m (supply constraints)
\(\sum_{\mathbf{i}=\mathbf{1}}^{\mathbf{m}}\) = x_ij = b_j, i = 1, 2 …….. n (demand constraints)
x_ij ≥ 0 for all i, j (non- negative restrictions)

Question 3.
What are feasible solution and non-degenerate solutions to the transportation problem?
Solution:
Feasible Solution: A feasible solution to a transportation problem is a set of non-negative values x_ij (i = 1, 2,.., m, j = 1, 2, …n) that satisfies the constraints.
Non-degenerate basic feasible Solution: If a basic feasible solution to a transportation problem contains exactly m + n – 1 allocation in independent positions, it is called a Non-degenerate basic feasible solution. Here m is the number of rows and n is the number of columns in a transportation problem.

Question 4.
What do you mean by a balanced transportation problem?
Solution:
In a transportation problem if the total supply equals the total demand (Σa_i = Σb_j) then it is said to be balanced transportation problem.

Question 5.
Find an intial basic feasible solution of the following problem using north west corner rule.

Solution:
Here total Supply 19 + 37 + 34 = 90
Total demand = 16 + 18 + 31 + 25 = 90
(i.e) Total supply = Total demand
∴ The given problem is balanced transportation problem
∴ we can final an initial basic feasible solution to due given problem.

From the above table we can choose the cell in the North west corner. Here the cell is (Q₁, D₁). Allocate as much as possible in this cell so that either the capacity of first row is exhausted or the destination requirement of the first column’s exhausted.
(i.e) x₁₁ = min (19, 16) = 16

Now the cell in the north west corner is (O₁, D₂) Allocate as much as possible in the first cell so that either the capacity of second row is exhausted or the destination requirement of the first column is exhausted.
(i.e) x₁₂ = min (3, 18) = 3

Reduced transportation table is

Now the cell in the north west corner is (O₂, D₂)
x₂₂ = min (37, 15) = 15

Now the cell in the north west corner is (O₂, D₃)
x₂₃ = min (22, 31) = 9

Now the cell in the north west corner is (O₃, D₃)
x₃₃ = min (34, 9) = 9

Thus we have the following allocations

Transportation schedule:
O₁ → D₁, O₁, → D₂, O₂ → D₂; O₂ → D₃ O₃ → D₃; O₃ → D₄
= (16 × 5) + (3 × 3) + (15 × 7) + (22 × 9) + (9 × 7) + (25 × 5)
= 80 + 9 + 105 + 198 + 63 + 125
= 580

Question 6.
Determine an intial basic feasible solution of the following transportation problem by north west corner method.

Solution:
Here total capacity (a_i) = 30 + 40 + 50 = 120
Total demand (b_j) = 35 + 28 + 32 + 25 = 120
(i.e) Total capacity = Total demand
∴ The given problem is balanced transportation.
∴ We can find an initial basic feasible solution to the given problem.

From the above table we can choose the cell in the North west corner. Here the cell is (chennai, Bangalore)
x₁₁ = min (30, 35) = 30
Reduced transportation table is

Now the cell in the North west corner is (Madurai, Bangalore)
x₂₁ = min(40, 5) = 5

From the above table we can choose the cell in the North west corner. Here the cell is (Nasik, Madurai)
x₂₂ = min (35, 28) = 28

From the above table we can choose the cell in the North west corner. Here the cell is (Bhopal, Madurai)
x₂₂ = min (7, 32) = 7

From the above table we can choose the cell in the North west corner. Here the cell is (Trichy, Bhopal)
x₃₃ = min (50, 25) = 25
Reduced transportation table is

x₃₄ = min (25, 25) = 25
Thus we have the following allocations

Transportation Schedule:
Chennai → Bangalore; Madurai → Bangalore;
Madurai → Naisk; Madurai → Bhopal
Trichy → Bhopal; Trichy → Delhi
The total transportation cost =
(30 × 6) + (5 × 5) + (28 × 11) + (7 × 9) + (25 × 7) + (25 × 13)
= 180 + 25 + 308 + 63 + 175 + 325
= 1076

Question 7.
Obtain an initial basic feasible solution to the following transportation problem by using least-cost method.

Solution:
The given transportation table is

Total supply = 25 + 35 + 40 = 100
Total demand = 30 + 25 + 45 = 100
(i.e) Total supply = Total demand
∴ The given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
The least cos is 4 corresponds to the cell (O₂, D₃)
Allocate min (35, 45) = 35 units to this cell

The least cost corresponds to the cell (O₁, D₃)
Allocate min (25, 10) = 10 units to this cell

The least cost is 6 corresponds to the cell (O₃, D₂)
Allocate min (40, 25) = 25 units to this cell

The least cost is 7 corresponds to the cell (O₃, D₁)
Allocate min (15, 30) = 15

Thus we have the following allocations

Transportation Schedule:
O₁ → D₁; O₁, → D₃; O₂ → D₃; O₃ → DI; O₃ → D₂
Total Transportation cost
= (15 × 9) + (10 × 5) + (35 × 4) + (15 × 7) + (25 × 6)
= 135 + 50 + 140 + 105 + 150
= 580

Question 8.
Explain vogel’s approximation method by obtaining initial feasible solution of the following transportation problem.

Solution:
Here Σ ai = 6 + 1 + 10 = 17
Σ bj = 7 + 5 + 3 + 2 = 17
Σ ai = Σ bj
(i.e) Total supply = Total Demand
∴ The given problem is balanced transportation problem
Hence there exists a feasible solution to the given problem.
First let us find the difference (penalty) between the first two smallest costs in each row and column and write them in brackets against the respective rows and columns.

Choose the largest difference. Here the largest difference is 6 which corresponds to column D₄
In this column choose the least cost. Here the least cost corresponds to (O₂, D₄)
Allocate min (1, 2) = 1 unit to this cell the reduced transportation table is

choose the largest difference 5 which corresponds to column D₂. Here the least cost corresponds to (O₁, D₂).
Allocate min (6, 5) = 5 units in this cell

Choose the largest difference 5 which corresponds to row O₁. Here the least cost corresponds to (O₁, D₁)
Allocate min (1, 7) = 1 unit in this cell

Choose the largest difference 4 which corresponds to row O₃. Here least cost corresponds to (10, 6) = 6 units in this cell.

Choose the largest difference 6 which corresponds to row O₃. Here the least cost corresponds to (O₃, D₄).
Allocate min (4, 1) = 1

Thus we have the following allocations

Transportation schedule
O₁ → D₁; O₁ → D₂; O₂ → D₄;
O₃ → D₁; O₃ → D₃; O₃ → D₄
Total transportation cost:
= (1 × 2) + (5 × 3) + (1 × 1) + (6 × 5) + (3 × 15) + (1 × 9)
= 2 + 15 + 1 + 30 + 45 + 9
= 102

Question 9.
Consider the following transportation problem.

Determine initital basic feasible solution by VAM
Solution:
Given Transportation problem is

Here Σ ai = Σ bj = 100
∴ The given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.
First let us find the difference (penalty) between the first two smallest costs in each row and column and write them in brackets against the respective rows and columns

Choose the largest difference. Here the difference is 3 which corresponds to D₂
In this column choose the least cost. Here the least cos corresponds to (O₃, D₂)
Allocate min (20, 40) = 20 units to this cell

Choose the largest difference is 4 which corresponds to column D₃. In this column choose the least cost. Here the least cost corresponds to (O₁, D₃).
Allocate min (30, 20) = 20 units to this cell

Choose the largest different is 3 which corresponds to column D₂. In this column choose the least cost. Here the least cost corresponds to (O₂, D₂)
Allocate min (50, 20) = 20 units to this cell

Choose the largest difference is 2 which corresponds to column D₄. In this column choose the least cost. Here the least cost corresponds to (O₂, D₄).
Allocate min (30, 10) = 10 units to this cell

Allocate min (20, 30) = 20 units to this cell

Here
x₁₁ = 10
x₁₃ = 20
x₂₁ = 20
x₂₂ = 20
x₂₄ = 10
x₃₂ = 20
Transportation schedule
O₁ → D₁; O₁ → D₃; O₂ → D₁;
O₂ → D₂; O₂ → D₄; O₃ → D₂
The transportation cost
= (10 × 5) + (20 × 3) + (20 × 4) + (20 × 5) + (10 × 4) + (20 × 2)
= 50 + 60 + 80 + 100 + 40 + 40
= 370

Question 10.
Determine basic feasible solution to the following transportation problem using North west Corner rule.

Solution:
Here total supply = 4 + 8 + 9 = 21
Total demand = 3 + 3 + 4 + 5+ 6 = 21
(i.e) Total supply = Total demand
∴ The given problem is balanced transportation problem.
∴ we can find an initial basic feasible solution to the given problem.
From the above table we can choose the cell in the North west corner. Here the cell is (P, A)
Allocate as much as possible in this cell so that either the capacity of first row is exhausted or the destination requirement of the first column is exhausted.

Form the above table we can choose the cell in North west corner. Here the cell is (P,B)
x = min (1, 3) = 1

From the above table we can choose the cell in north west corner. Here the cell is (Q, B)
x = min (2, 8) = 2

From the above table, we can choose the cell in North west corner. Here the cell is (Q, C)

From the above table, we can choose the cell in North west corner. Here the cell is (Q, D)
x = min (2, 5) = 2

From the above table, we can choose the cell in North west corner. Here the cell is (R, D)
x = min (9, 3) = 3

Thus we have the following table

Transportation Schedule:
P → A; P → B; Q → B; Q → C; Q → D R → D; R → E
Total transportation cost:
= (3 × 2) + (1 × 11) + (2 × 4) + (4 × 7) + (2 × 2) + (3 × 8) + (6 × 2)
= 6 + 11 + 8 + 28 + 4 + 24 + 72
= 153

Question 11.
Find the initial basic feasible solution of the following transportation problem:

Using
(i) North West corner tule
(ii) Least Cost method
(iii) Vogel’s approximation method
Solution:
(i) North west corner rule:
Here the total supply = 10 + 10 + 10 = 30
Total demand = 7 + 12 + 11 = 30
(i.e) Total supply = Total demand
The given problem is balanced transportation problem.
∴ we can find an initial basic feasible solution to the given problem.
From the above table we can choose the cell in the North west corner. Here the cell is (A, I)
x₁₁ = min (7, 10) = 7

From the above table we can choose the cell in the north west corner. Here the cell is (B, I)
x = min (3, 12) = 3

From the above table we can choose the cell in the North west corner. Here the cell is (B, II)
x = min (9, 10) = 9

Here the cell in the North west corner is (C, II)
x = min (11, 1) = 1

Thus we have the following allocations

Transportation schedule:
A → I; B → I; B → II; C → II; C → III
Total transportation cost:
= (7 × 1) + (3 × 0) + (9 × 4) + (1 × 1) + (10 × 5)
= 7 + 0 + 36 + 1 + 50
= 94

(ii) Least cost method:
The given transportation table is

Here Total supply = Total demand = 30
∴ The given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
The least cost is 0 corresponds to the cell (B, I)
Allocate min (12, 10) = 10 units to this cell

The least cost 1 corresponds to the cell (C, II)
Allocate min (11, 10) = 10 units to this cell

Here the least cost 2 corresponds to the cell (B, III)
Allocate min (2, 10) = 2 units to this cell.

Here the least cost is 5 corresponds to the cell (C, III)

Transportation schedule:
A → III; B → I; B → III; C → II; C → III
Total transportation cost:
= (7 × 6) + (10 × 10) + (2 × 2) + (10 × 1) + (1 × 5)
= 42 + 0 + 4 + 10 + 5
= 61

(iii) Vogel’s approximation method:
Here Σ ai = Σ bj = 30
(i.e) Total supply = Total demand
∴ This given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.


A → I; B → I; B → III; C → I; C → II
Total transportation cost:
= (7 × 1) + (2 × 0) + (10 × 2) + (1 × 3) + (10 × 1)
= 7 + 0 + 20 + 3 + 10
= 40

Question 12.
Obtain an initial basic feasible solution to the following transportation problem by north west corner method.

Solution:
Here the total available = 250 + 300 + 400 = 950
Total Required = 200 + 225 + 275 + 250 = 950
(i.e) Total Available = total required
∴ The given problem is balanced transportation problem.
we can find an initial basic feasible solution to the given problem.
From the above table, we can choose the cell in the North west corner. Here the cell is (A, D).
x₁₁ = min (250, 200) = 200

From the above table we can choose the cell in North west corner. Here the cell is (A, E)
x = min (50, 225) = 50

From the above table, the north west corner cell is (B, E)
x = min (300, 175) = 175

From the above table, the north west corner cell is (B, F)
x = min (125, 275) = 125

Here the north west corner cell is (C, F)
x = min (400, 150) = 150

Transportation schedule:
A → D; A → E; B → E; B → F; C → G
Total Transportation cost:
= (200 × 11) + (50 × 13) + (175 × 18) + (125 × 14) + (150 × 13) + (250 × 10)
= 2200 + 650 + 3150 + 1750 + 1950 + 2500
= 12,200