2nd PUC Statistics Question Bank Chapter 8 Operations Research

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Karnataka 2nd PUC Statistics Question Bank Chapter 8 Operations Research

2nd PUC Statistics Linear Programming Exercise Problems

Question 1.
Define Linear Programming Problem.
Answer:
A general LPP deals with the optimization of a linear function of variables subject to a set of constraint.

Question 2.
Give the general form of LPP.
Answer:
Optimize the objective function
Z = c₁x₁ + c₂x₂ + ……… + c_nx_n
Subject to’m’ constraints

Question 3.
Write the general form of LPP in matrix notation?
Answer:
Optimize Z = CX
S.t. Ax(≤=≥)b and X ≥ 0
where c = (c₁c₂c₃ ………. c_n) is a row matrix of cost or profit coefficients

Question 4.
Mention the steps involved in formulation of an LPP.
Answer:

• Identify the decision variables of the problem.
• Identify the objective and express it as a linear function of the decision variables.
• Identify all the restrictions or constraints in the problems and express them as linear equations or inequalities of the unknown variables (decision variables)

Question 5.
Mention a method of finding a solution to an LPP with two variables.
Answer:

• Graphical method
• Simplex method

Question 6.
In an LPP define (i) Solution (ii) Feasible solution (iii) Optimal solution (iv)Objective function (v) Decision variables (vi) Unbounded solution (vii) Multiple solution.
Answer:
A set of real values x = (x₁x₂ ……… x_n) which satisfies the subject to constraints Ax(≤=≥)b is called a solution.

Feasible solution: A set of real values Y = (x₁x₂ ……… x_n ) which satisfies the

• Subject to constraints (Ax(≤=≥)b) and
• non negativity restriction (X ≥ 0)

Optimal solution: A set of real values X = (x₁x₂…x_n) which satisfies

• Subject to conditions (Ax(≤=≥)b)
• Non negativity restriction (X ≥ 0) and
• Which optimizes the objective function Z = CX is known as optimal solution to an LPP.

Objective function: The linear function Z = c₁x₁ + c₂x₂ + …………. + c_nx_n which is to be optimized is called the objective function.

Decision variables: The variables x₁x₂ ………. x_n whose values are to be determined are called as decision variables.

Unbounded solution: In case of maximization problem, when the LPP does not consist of a finite optimum i.e., value of Z, can be increased indefinitely then the LPP is said to have unbounded solution.

Multiple solutions: When there exists more than one feasible solution having the same optimal value for the objective function, then the LPP is said to have an alternative or multiple solution.

Question 7.
Write down the steps in the graphical method of solving an LPP.
Answer:
Step 1: Take each inequality in the constraints as equality. For each equation find two sets of points.

Step 2; Mark the obtained sets of points of each of the equations and join the points corresponding to each of the equations to obtain straight lines.

Step 3: Identify and shade the feasible region (Area which satisfies all the constraints (subject to and non negativity] by considering the inequalities.

Step 4: If the inequality is of ≤ then the FR will be towards the origin (shade below the line).

Step 5: If the inequality is of ≥ then the FR will be away from the origin (shade above the line).

Step 6: Mark the corner points of the feasible region and find values of the objective function at each of the corner points.

Step 7: The corner points which gives the optimal values of the objective function is the optimal solution to the given LPP.

Question 8.
The graphical solution to the LPP lies in the first quadrant give reason?
Answer:
Due to non negativity restrictions of the decision variables.

Question 9.
Which of the two feasible solutions (12,10) and (14, 4) of an L.P.P maximizes the objective function Z = 5x + 4y.
Answer:
Feasible solution (12,10), Z = 5(12) + 4(10) = 60 + 40 = 100 max
(14, 4) = 5(14) + 4(4) = 70 + 16 = 86
(12,10) maximizes the objective function Z = 5x + 4y

Question 10.
Consider the following LPP maximize Z = 5x + 3y x + 3y ≤ 10; x < 5 and x ≥ 0, y ≥ 0 Suppose x = -3 & y = 4 is it a feasible solution to given LPP? Give reason
Answer:
The value of x is not positive, so it is not a feasible solution.

Question 11.
A small manufacturer employs 5 skilled men and 10 semi skilled men for making a product in 2 qualities a deluxe material and ordinary material. The production of a deluxe model requires 2 hours by a skilled man and 3 hours by a semi skilled man. The ordinary model requires 1 hour by a skilled man and 2 hours by a semi skilled man. According to worker union rules, no man can work more than 8 hours per day. The profit of each deluxe model is Rs. 10 and that of each ordinary model is Rs. 8. Formulate a LPP such that the total profit is maximized.
Answer:
Let X be the no. of units of ordinary model to be manufactured.
Let Y be the no. of units of deluxe model to be manufactured.

The formulated LPP is
Max Z = 10x + 8y
st. 2x + y ≤ 40;
3x + 2y ≤ 80
and x, y, ≥ 0.

Question 12.
A company sells two different products A and B. The company makes a profit of 40 & ?30 per unit of products A and B respectively. The two products are produced in a common production process. The production process has a capacity of 30,000 man hours. It takes 3 hours to produce one unit of A and one hour to produce one unit of B. The company officials feel that the maximum number of units of A that can be sold is 8000 units and the maximum number of units of B that can be sold is 12000 units formulate the LPP.
Answer:
Let X denote the no. of units of product A to be produced.
Let Y denote the no. of units of product B to be produced.

The formulated LPP is
Maximize Z = 40x + 30y
st. 3x + y ≤ 30,000
x ≤ 8, 000
y ≤ 12, 000 and x, y ≥ 0

Question 13.
Graphically solve the following LPP
Max Z = 3x + 5y
S.t. x + 2y ≤ 200 ; y ≤ 60 and x, y ≥ 0
Answer:

Since the inequalities x + 2y ≤ 200 and y ≤ 60 are of ≤ type Shade the region towards the origin.
The common shaded region to all the constraints is ABCD which forms the feasible region.

Corner points z = 3x + 5y
A (0,0) Z = 3(0) + 5(0) = 0 + 0 ⇒ 0
B(0, 60) Z = 3(0) + 5(60) = 0 + 300 ⇒ 300
C (80, 60) Z = 3(80) + 5(60) = 240 + 300 ⇒ 540
D (200, 0) Z = 3(200) + 5(0) = 600 + 0 ⇒ 600 (max)
The max value of Z corresponds to the corner point D(200, 0)
The optimal solution is x = 200, y = 0 and Z = 600
The given LPP has unique optimal solution.

Question 14.
Solve graphically. Max Z = 20x + 80y
S.t. 2x + 6y ≤ 60, x + 4y ≤ 32
and x, y ≥ 0
Answer:
Considering the inequalities as equations
2x + 6y = 60 → (1)
x + 4y = 32 → (2)
x = 0, y = 0


The shaded region in the graph is the feasible region.
The corner points are A(0, 0), 3(0, 8), D(30, 0)
Solve eqns (1) and (2) to find C

Subs, y = 2 in (1)
2x + 6(2) = 60
2x = 60 – 12
2x = 48
x = \(\frac { 48 }{ 2 }\) = 24
= C(24, 2)
Corner points A, B, C and D.
Corner points Z = 20x+ 80y
A (0, 0) Z = 20(0) + 80(0) = 0 + 0 ⇒ 0
B (0, 8] Z = 20(0) + 80(8) = 0 + 640 ⇒ 640 Max
C (24, 2) Z = 20(24) + 80(2) = 480 + 160 ⇒ 640 Max
D (30, 0) Z = 20(30) + 80(0) = 600 + 0 ⇒ 600
The maximum value of Z corresponds to
x = 0, y = 8, x = 24, y = 2 and z = 640
The optimal solution is
X = 0, y = 8, x = 24, y = 2 and Z = 640
Thus LPP has multiple optimal solutions.

Question 15.
Solve graphically,
Max Z = x + y
S.t. x + y ≤ 1, 3x + y ≥ 3
and x, y ≥ 0
Answer:
Considering the inequalities as equations,
x + y = 1 → (1)
3x + y = 3 → (2)


There exists no feasible region.
∴The LPP has no solution

Question 16.
Solve graphically.
Max Z = 2x + y S.t. x – y ≤ 10, 2x – y ≤ 10
x, y ≥ 0
Answer:
Considering the inequalities as equations
x – y = 10 → (1)
2x – y = 10 → (2)

Question 17.
Solve graphicaIly.
Min Z = 5x + 4y S.t. 4x + y ≥ 40,
2x + 3y ≥ 60 and x, y ≥ 0
Answer:
Considering inequalities as equalities
4x + y = 40 → (1)
2x + 3y = 60 → (2)


The shaded region in the graph is the feasible region and is unbounded. Since the problem is to min Z, one of the existing corner points of the feasible region forms an optimal solution.

The corner points are A (0, 40), C(30, 0) and to find B
Solve eqns. (1) and (2)

Subs y = 16 in (1)
4x + 16 = 40
4x = 40 – 16
4x = 24
x = \(\frac { 24 }{ 4 }\) = 6
∴ B(6, 16)
Corner points
Z = 5x + 4y
A [0,40) Z= 5(0) + 4(0) =0 + 0 ⇒ 160
B(6,16) Z = 5(6) + 4(16) = 30 + 64 ⇒ 94
C (30, 0) Z = 5(30) + 4(0) = 150 + 0 ⇒ 150
LPP min Z value corresponds to x = 6, y = 16
∴ The optional solution is x = 6, y = 16 and Z = 94

Question 18.
Solve graphically.
Max Z = 3x + 5y
S.t. x + y ≤ 1500,
y ≤ 600
and x, y ≥ 0
Answer:
Converting inequalities to equalities
x + y = 1500 → (1)
y – 600 → (2)

The shaded region in the graph is the feasible region

The corner points are A(0,0)
B (0, 600)
C( )
D(1500, 0)
To find C solve eqns. (1) and (2)
x + y = 1500 → (1)
y = 600 → (2)
Subs. y = 600 in (1)
x + 600 = 1500
x= 1500 – 600
x = 900
C(900, 600)
corner points A, B, C and D

The max. value of Z correspond to C(900, 600)
∴ The optimal solution is x = 900, y = 600 and Z = 5,700
The given LPP has unique optimal solution.

2nd PUC Statistics Transportation Problems Exercise Problems

Question 1.
Describe the Transportation problem and give its mathematical formulation.
Answer:
T.P. is considered as a special type of LPP whose object is to be determine the quantity of units to be transported from various origins to different destinations at a minimum cost
Mathematical formulation

Question 2.
In a Transportation problem, Define the terms
(i) A feasible solution
(ii) Basic feasible solution
(iii) Optimal solution.
Answer:
Feasible solution: A set of non negative values x_ij where i = 1,2, …… m and j = 1,2, ………. n which satisfies the following
conditions:

and x_ij > 0 for all i & j is called a feasible solution to TP
Basic feasible solution: A feasible solution is said to be BFS if the no. of non zero allocations are m + n – 1.
Optimal solution: A feasible solution is said to be optimal if it minimises the total cost of transportation.

Question 3.
In a TP, when do you say that a solution is de generate?
Answer:
If the no. of positive allocations is any BFS is less than (m + n – 1), then the solution is degenerate.

Question 4.
What do you mean by a non degenerate solution in a TP?
Answer:
If the no. of positive allocations in any BFSis equal to (m + n – 1), then the solution is non-degenerate.

Question 5.
When is a TP balanced?
Answer:
If Σa_i = Σb_j then the TP is balanced or if total availability = total requirement.

Question 6.
What is an unbalanced TP?
Answer:
If Σa_i ≠ Σb_j then the TP is unbalanced

Question 7.
Mention any 2 methods of obtaining initial basic feasible solution for a transportation problem?
Answer:

• North west corner rule
• Matrix Minima method (Least cost)

Question 8.
Explain the (i) NWCR rule and (ii) MMM for obtaining an initial basic feasible solution to a TP.
Answer:
NWCR rule:
Step 1
Start with the cell at the upper left corner of the cost matrix of the transportation table allocate
x_ij = min (a_i, b_j)

Step 2

• If a_i > b_j move horizontally right and make the second allocation in the cell (1, 2)
• If a_i < b_j move vertically downward and make the next location point in the cell (2)
• If move diagonally to the cell (2, 2) and make the next location point here. Here x₂₂ = min(a_i, b_j)

Step 3:
The above procedure is repeated until all the availabilities in the various origins are allocated to cells as required in the different destinations.

Least Cost Method or Matrix Minima Method

Step 1
Find the least cost in the cost matrix. Let it be C_ij Allocate x_ij =min (a_i, b_j) in the cell (i, j)

Step 2

• If x_ij = a_i strike off the ith row of the transportation table and decrease b_j by a_i
• If x_ij = b_j strike off the jth column of the transportation table and decrease a_i by b_j
• If a_j = b_j = x_ij strike of both the i^th row and j^th column

Step 3
Steps 1 and 2 are repeated until all the availabilities in the origins and requirements in the destinations are satisfied.

Question 9.
Determine an initial basic feasible solution to the following TP by NWCR and MMM. Compute the transportation costs for the obtained solution by both the methods and write your conclusion.

Answer:
i. NWCR

Allocate x₁₁ = min(5, 7) = 5
Allocate x₂₁ = min(8, 2) = 2
Allocate x₂₂ = min(6, 9) = 6
Allocate x₃₂ = min(7, 3) = 3
Allocate x₃₃ = min(4, 18) = 4
Allocate x₄₃ = min(14, 14) = 14
x₁₁ = 5 x₂₁ = 2, x₂₂ = 6, x₃₂ = 3, x₃₃ = 4, x₄₃ = 14
Transportation cost = ΣΣc_ij . x_ij
= (2) (5) + 3(2) + 3(6) + 4(3) + 7(4) + 2(14)
=10 + 6 + 18 + 12 + 28 + 28 = 102

ii. MM

Here the least cost in the table is 1, it is in two cells (2, 3) and (4, 1), we make arbitrary choice and allocate in X₄₁ :
X₄₁ = Minimum (14,7), minimum is 7 and delete that column
For the next table in a_i 14 – 7 = 7

Here the least cost in the table is 1, it is in (2, 3) allocate X₂₃
X₂₃ = Minimum (8, 18), minimum is 8 and delete that row
For the next table in b_j 18 – 8 = 10

Here the least cost in the Lab1 is 2, it is in (4, 3) allocate X₄₃
X₄₃ = Minimum (7,10), minimum is 7 and delete that row
For the next table in b_j 10 – 7 = 3

Here the least cost in the table is 4, it is in two cells (3,2) and (1,3), we take through arbitrary choice and allocate X₁₃
X₁₃ = Minimum (5,3), minimum is 3 and delete that column
For the next table in a_i 5 – 3 = 2

Here the least cost in the table is 4, it is in (3,2) allocate X₃₂
X₃₂ = Minimum (7, 9), minimum is 7 and delete that row
For the next table in b_j 9 – 7 = 2

Here the final allocation is Minimum (1, 2), allocate
X₁₂ = (2, 2), minimum is 2
The obtained feasible solution is given below

Finally the calculated initial basic feasible solution by the least cost entry method or MMM is
X₁₂ = 2, X₁₃ = 3, X₃₂ = 7, X₄₁ = 7, X₂₃ = 8
The TP cost is
(2 × 7) + (3 × 4) + (8 × 1) + (7 × 4) + (7 × 1) + (7 × 2) = 83

Question 10.
Is the solution obtained by NWCR for the following TP degenerate?

Answer:
By following the steps of question no 11: Initial basic solution is

Allocate x₁₁ = min (7, 7) = 7
Allocate x₂₂ = min (9, 8) = 8
Allocate x₂₃ = min (1, 5) = 1
Allocate x₃₃ = min (18, 4) = 4
Allocate x₃₄ = min (14, 14) = 14
IBFS = x₁₁ = 7, x₂₂ = 8, x₂₃ = 1, x₃₃ = 4, x₃₄ = 14
Here no. of origins m = 3
no. of destinations n = 4
m + n – 1 = 3 + 4 – 1 = 6
No. of positive allocations = 5 < m + n – 1 = 6
∴ The solution is degenerate.

Question 11.
Obtain an initial basic feasible solution to following TP by MMM. Also obtain th transportation cost.

Answer:
The lowest cost corresponds to the cell (2, 2)

Allocate x₂₂ = min (3, 1) = 1
Delete 2^nd column
The reduced matrix is

Here the lowest cost corresponds to the cell (2, 1)
Allocate x₂₁ min. (2, 4) = 2
Delete the 2^nd row
The reduced cost matrix is

The least cost corresponds to the cell (3, 1)
Allocate = x₃₁ = min(5, 2) = 2
Delete 1st column
The reduced cost matrix is

The least cost corresponds to the cell (1, 3)
Allocate x₁₃ = min(2, 5) = 2
Delete 1st row
The reduced cost matrix is

Allocate x₃₃ = min(3, 3] = 3
IBFS is x₁₃ = 2, x₂₁ = 2, x₂₂ = 1, x₃₁ = 2 and x₃₃ = 3
Transportation cost Z = ΣΣc_ij . x_ij
= 1 × 1 + 2 × 2 + 3 × 2 + 4 × 2 + 6 × 3
= ₹37

Question 12.
Find an allocation of available sources by MMM and compute the transportation cost. Is the solution degenerate?

Answer:
The least cost corresponds to the cell (1, 3), (2, 1), (3, 2) choosing an arbitrary cell, allocate x₁₃ = min (60, 80) = 60

Delete 1^st row
The least cost corresponds to cell (2,1)
Allocate x₂₁ = min (70, 50) = 50
Delete 1st row.
The least cost corresponds to cell (3, 2)
Allocate x₃₂ = min (80, 80) = 80
Delete 3rd row and 2nd col.
The least cost corresponds to cell (2, 3)
Allocate x₂₃ = min(20, 20) = 20
IBFS is x₁₃ = 60, x₂₁ = 50, x₃₂ = 80, x₂₃ = 20
Transportation cost is Z = ΣΣc_ij . x_ij
= 3 × 60 + 3 × 50 + 3 × 80 + 9 × 20
= ₹750.

2nd PUC Statistics Game Theory Exercise Problems

Question 1.
What is n-person game?
Answer:
A game in which sum of the pay offs of the participants is zero, is called zero-sum game.

Question 2.
What is meant by two-person zero-sum game?
Answer:
1. There are finite no. of competitors called players or participants.

2. Each player has a finite number of courses of action. This list need not to be the same for each player.

3. A game is said to be played when each player selects one of his courses of action from the list of courses of action available to him. These choices are assumed to be made simultaneously so that no player knows the choice of the other until he has decided his own.

4. Every combination of courses of action is associated with an outcome/payment which may be positive, negative or zero to each player.

Question 3.
What do you mean by zero-sum game?
Answer:
A game in which sum of the pay offs of the participants is zero, is called zero-sum game.

Question 4.
What is meant by two-person zero-sum game?
Answer:
A game is a said be a two person zero sum game if
(a) 2 players participate.
(b) The game of one player is equal to the loss of the other player.

Question 5.
What are the properties of a rectangular game?
Answer:

• No. of players is 2
• Gain of one player will be the loss of other player
• Game plan is finite

Question 6.
Define (i) Strategy, (ii) Pure Strategy (iii) Mixed Strategy.
Answer:
Strategy: The strategy of a player is the pre-determined rule by which a player determines his course of action

Pure strategy: If a player decides to use only one particular course of action irrespective of the course of action chosen by his opponent during every play, this strategy is named as pure strategy

Mixed strategy: If a player decides to use all or some of his available courses of action in some fixed proportion, he is called to use mixed strategy.

Question 7.
Explain maximin and minimax of a game?
Answer:
Maximin: The maximum of the row minimums is known as maximin.
Minimax: The minimum of the column maximums is known as minimax.

Question 8.
What do you mean by saddle-point? When do you say that a game has saddle-point?
Answer:
Saddle point is the position in the pay-off matrix where maximin and minimax coincide.

Question 9.
What do you mean by value of game? For what value of the game is the game fair?
Answer:
The Pay off at the position of the saddle-point is the value of the game. If value of a game is zero its a fair game.

Question 10.
Explain the maximin-minimax principle of solving a rectangular game.
Answer:
The objective of finding the optimum Strategies for both the players which maximizes the gain of one player and minimizes the loss of the other plays under all situations will be obtained by applying maximin- minimax principle.

Step 1
Identify the minimum gain corresponding to each strategy of A. Write the row minimum on the right of each row. Find out the maximum of these, which called as maximin.

Step 2
Identify the maximum loss corresponding to each strategy .of B. Write the column maximum below each column. Find out the minimum among these, which is called as minimax.

Question 11.
Explain the principle of dominance of solving a rectangular game.
Answer:
If the strategy of a player dominates over another strategy then the latter strategy can be eliminated as the choice of such a strategy will not benefit the player in anyway.

1. If all the elements of a row in the pay off matrix say k^th row are greater than (or equal to) the corresponding elements of another row in the pay off matrix, say r^th row then k^th row dominates r^th row. Thus’the r^th row of pay off matrix can be deleted.

2. If all the elements of a column in the pay off matrix say p^th column are lesser than the corresponding elements of another column in the pay off matrix said to be q^th column, the p^th column dominates the q^th column. Thus the q^th column of the pay off matrix can be eliminated.

3. The process is repeated till optimal strategies are obtained.

Question 12.
In a two person zero sum game, if gain of one player is ‘3’ then what is the loss of the other player.
Answer:
The loss of the other player is (-3).

Question 13.
In a rectangular game, if saddle point exists and maximin is -4, what is the value of minimax?
Answer:
Value of minimax is -4

Question 14.
In a game the pay-off at saddle point is 4, what is the value of minimax?
Answer:
Value of minimax is 4

Question 15.
In a game of matching coins with two players, suppose A wins ₹ 10 when the coins show two heads coins, he wins ₹ 5 when coins show two tails and loses ₹ 4 when the coins show 1 head and 1 tail, write the pay off matrix of A. Does the game have saddle point, of so, write down the solution?
Answer:
The pay off matrix of A.

Here maximin ≠ minimax.
∴ Saddle point does not exist to the game and hence the suggested solution for both the players is mixed strategy.

Question 16.
If the value of the game is -6, is the game fair?
Answer:
If the value of the game is 0 then it is a fair game, but here the value is -6, hence The game is not fair

Question 17.
Solve the following game using maximin-minimax principle. Is the game fair?
Answer:

Maximin: From each column find the maximum value, and from that values choose the minimum value
Minimax: From each row find the minimum value, and from that values choose maximum value

From rows minimum values are -1, -1, -1, 0 out of these values maximum value is 0
From columns maximum values are 2, 0, 3 out of these values minimum value is 0
The saddle point is at (4, 2).
1. Best strategies for company X is S
2. Best strategies for company Y is B
3. V = 0
4. The game is fair (since V = 0)

Question 18.
The following is the pay-off matrix of player ‘A’, Write the pay-off matrix of player ‘B’.
Answer:

Question 19.
Solve the following using the principle of dominance.

Answer:
Since each of the pay offs in A₂ is greater than each of the pay offs in A₁
A₂ dominates A₁
∴ A₁ is deleted
The reduced pay off matrix x is

Similarly A₂, dominates A₃
∴ A₃ is deleted
The reduced pay off matrix is

Since the payoff in B₃ is less than payoff in B₁
B₃ dominates B₁
∴ B₁ is deleted
The reduced payoff matrix is

Similarly B₃ dominate B₂ and B₄
∴ B₂ and B₄ are deleted
The reduced payoff matrix is B is B₃
Saddle point is al (2, 3)
1. Strategy for player A is A₂
2. Strategy for player B is B₃
3. V = 3

Question 20.
Solve the following game by dominance property.

Answer:
Since each of the payoffs in A,₁is greater than 01 equal to each of the payoffs is A₂
A₁ dominates A₂
∴ A₂ is deleted
The reduced payoff matrix is

Since each of the payoffs in B₃ is less than each of the payoffs in B₁
B₃ dominates B₁
∴ B₁ is deleted
The reduced payoff matrix is

Similarly B₃ dominates B₂ and B₄
∴ B₂ and B₄ are deleted
The reduced payoff matrix x is

The payoff is A₁ is greater than the payoff in A₃
∴ A₁ dominates A₃
A₃ is deleted
The reduced payoff matrix is

Saddle point is at position (1, 3)
1. Strategy for player A is A₁
2. Strategy for player B is B₃
3. V = 1

Question 21.
In a rectangular game, pay off matrix of player A is

(i) Solve the game.
(ii) Write down the pay off matrix of B and then solve the game.
Answer:
Pay off matrix of player A is

Here maximin = minimax = 5
Saddle point exists at the position (3, 2)
1. Strategy for player A is (c)
2. Strategy for player B is (c)
3. V = 5

(ii) Payoff matrix of player B is

Here maximin = minimax = -5
Saddle point exists at the portion (2, 3)
1. Strategy for player A is (C)
2. Strategy for player B is (C)
3. V = -5

2nd PUC Statistics Replacement Theory Exercise Problems

Question 1.
What do you mean by replacement problem?
Answer:
Replacement theory handles with the problem of making decision at which age an item which deteriorates with time has to be replaced by a new one.

Question 2.
Give one example for equipments which deteriorate with age?
Answer:
A machine

Question 3.
What is the principle of equipments which deteriorate with age?
Answer
To date average annual cost stops decreasing and starts increasing.

Question 4.
Mention the needs for replacement of equipments?
Answer:

• As the machine ages the cost of its maintenance increases and it requires more of care.
• If the item does not depreciate with passage of time but suddenly stops functioning.

Question 5.
The following are the maintenance and depreciation

When should the truck be replaced?
Answer:

Here 6^th year is minimum, the truck should be replaced at the end of 6^th year.

Question 6.
The cost of a machine is 6600 and its resale value is 600. The maintenance costs is different years are as follows.

Determine the age at which the machine should be replaced?
Answer:
P = 6600 S_n = 600

Here 6^th year is minimum. The machine should be replaced at the end of 6^th year.

Question 7.
A machine cost 35000 and the operating cost is estimated to be 1500 for the first year and increases by 3000 every year for next 5 years. Determine the optimum period for replacement of the machine, assuming that the machine has no resale value.
Answer:
P = 35,000, s_n = 0

Here 5^th year is minimum.
The machine should be replaced at the end of 5^th year.

Question 8.
If the depreciation cost and the cumulative maintenance cost for an equipment for the third year is 10,100 and 10,400 respectively find the annual average cost.
Answer:
From the given data
(P – S_n for third year = 10,100
ΣC₁ for third year = 10,400
n = 3, A(n) = ?

Question 9.
The cumulative maintenance cost of a machine during fourth year is 13,500 its purchase cost is 8,500. Find the annual average cost assuming that the machine has no resale value?
Answer:
From the given data
(P – S_n) = 8,500
ΣC₁ for 4^th year = 13,500
Annual average cost for 4th year n n

Question 10.
A person wants to buy a machine. He has to choose between machine A and B. The purchase prices of these machines are 8,000 and 6,000 respectively. Their resale values and maintenance costs are as below.

Which machine should be person buy?
What is the annual average cost?
Answer:

Average annual cost of Machine A = 3740, machine a has to be replaced end of 5^th year

Machine B has to be replaced end of 6 years. Average annual cost = 2816.67
Conclusion: Machine A should be replaced at the end of 5^th year, Machine B should be replaced at the end of 6^th year and the minimum average annual cost of machine B is less than A. So It is profitable to purchase machine B.

2nd PUC Statistics Inventory Theory Exercise Problems

Question 1.
What is meant by inventory?
Answer:
Inventory is a physical stock of goods kept for future use.

Question 2.
What are advantages and disadvantages of in ventory?
Answer:
Advantages:

• Bulk production decreases the cost of production, transportation etc.,
• It reduces the set up cost.

Disadvantages:

• Interest on invested capital
• Labour for maintenance of inventory and records.

Question 3.
List the types of variables associated with an inventory?
Answer:

• Controlled variables
• Uncontrolled variables

Question 4.
Give one example for each controlled variable and uncontrolled variable.
Answer:
Controlled variables: The quantity of goods acquired to the inventory, frequency of replenishment.
Uncontrolled variables: Inventory costs, demand.

Question 5.
Mention different types of costs associated with an inventory?
Answer:

• Holding cost (or) Maintenance cost or storage cost or carrying cost.
• Capital cost
• Ordering cost (or) Setup cost
• Shortage cost (or) stock out cost or penalty cost.

Question 6.
Define:
(i) Holding cost
(ii) Ordering cost and up cost
Answer:
(i) Holding Cost: The cost associated with maintaining the inventory until it is sold or used in a production is called holding cost. It is denoted by C₁

(ii) Ordering Cost (Set up cost): The cost associated with setting up of machinery before starting production is called set up cost and is denoted by C₃.

Question 7.
What is meant by lead time?
Answer:
The time gap between the placing of order and arrival of goods at the inventory is called as lead time.

Question 8.
Mention any one objective of an inventory problem?
Answer:
To bring equilibrium between cost of inventory and cost of production.

Question 9.
Define EOQ?
Answer:
The order quantity which minimizes the total costs of carrying inventory and the ordering cost is the EOQ.

Question 10.
Under what conditions in EOQ model with shortage applicable?
Answer:

• Uniform demand
• Lead time zero.

Question 11.
The annual demand for an item is 3000 units. Capital cost is 7/- per unit. Inventory carrying cost is 20% of capital costperannum of setup cost is 150/- determine
(i) EOQ (ii) Number of orders per year (iii) Optimal cost.
Answer:
Given R = annual demand = 3000 units,
C₃ = setup cost = Rs. 150,
P = capital cost = Rs. 7 per unit, I = 20% 0.2
C₁ = P1 = 7 × \(\frac{20}{100}\) = 1.4peryear.

Question 12.
The demand for commodity is at a constant rate of 200 units per year. There is an inventory in which setup cost is 800/- per production run, holding cost is 10/- per unit per year and shortage cost is 12/- per unit per year. Find the EOQ and maximum shortage level.
Answer:
Annual demand = R = 200 units per year
Holding cost = C₁ = 10/- per year
Shortage cost = C₂ = 12/- per unit per year C₂
Setup cost = 800 per production run = C₃

Question 13.
A Stockist has to supply 400 units of a product every Monday to his customers. He gets the product at ?50 per unit from the manufacturer. The cost of ordering and transportation from the manufacturer is 75/- per order. The cost of carrying inventory is 75/- per order the cost of carrying inventory is 75% per year of the cost of the product. Find
(i) Economic lot size
(ii) Minimum average cost.
Answer:
R = 400/- week,
R = 400 × 52 ⇒ 20,800
C₃ = ₹ 75/order
P = 50, I = 7.5%
C₁ = PI = 50 × 7.5% ⇒ 375

Question 14.
There is a demand for 10,000 items per year. The replenishment cost is 200/- and the maintenance cost is 10/- per item per year. Replenishment is instantaneous and shortage of are not allowed. Find (i) Optimal lot size (ii) The optimum time between orders (iii) The optimum annual average cost, (iv) The optimum no. of orders.
Answer:
R = 10,000, C₃ = 200, C₁ = 10

Question 15.
The demand for an item is 700 units per year. The cost of placing an order is 7/- and holding cost is 10/- per year. The cost of shortage is 3/- per unit. Find (i) EOQ (ii) Time between two orders.
Answer:
R = 700, C₃ = 7, C₁ = 10, C₂ = 3

Question 16.
A manufacturer has to supply 25,000 units of his product every year to a customer. This space demand is uniform. The customer has no storage and so the manufacturer has to ship requirements daily. Inability to meet the the demand in time costs ₹2 per unit per day. Inventory carrying cost is ₹0.50 per unit per day. Set up cost is ₹500 per production run. Find the frequency of replenishment which would minimize cost. What is the maximum shortage level?
Answer:
R = 25,000/- Units
C₂ = ₹2/un it/day
C₁ = ₹0.50/unit/day
C₁ = ₹500/production run

Maximum inventory level
Q° – S°= 7905.454 – 6324.36 = 1581.094

2nd PUC Statistics Operations Research LPP Pratical Assignments

Question 1.
A Tailor gets a profit of ₹ 100 from a shirt and ₹ 170 from pant. In a week of 56 hours, he uses 36 hours for cutting and 20 hours for stiching. He requires 2 hours to cut a shirt and 3 hours to cut a pant for stiching he requires 1 hour to a shirt and 2 hours to a pant. Formulate LPP.
Answer:
The statement can be written in the following table as below:

The L.P.P. is: Max, z = 100x + 170y
Subject to constraints: 2x + 3y ≤ 36
x + 2y ≤ 20 and
Non-negativity: x ≥ 0, y ≥ 0.

Question 2.
A manufacturer produces 2 products A and B, which needs two machine P and Q. Product A requires 6 hours on machine P and 2 hours on machine Q. Product B requires 4 hours on machine P and 4 hours on machine Q. There are 100 hours of time available on machine P and 80 hours on machine Q. Profit earned by the manufacturer on selling one unit of A is Rs. 10 and on selling one unit of B is Rs. 15. Formulate LPP.
Answer:
The given statement can be written the following table:

The L.P.P is: Maximize Z = 10x + 15y
Subject to constrants : 6x + 4y ≤100
2x + 4y ≤ 80 and
Non-negativity: x ≥ 0, y ≥ 0.

Question 3.
A company owns two four mills A and B, which have different Production capacities of high, medium and low grade flour. This company has entered into a contract to supply flour to a firm every week with 12, 8 and 24 quintals of high (H), medium (M) and low (L) grade flour respectively. It costs the company Rs. 1000 and Rs. 800 per day to run Mill A and Mill B respectively. On a day mill A products 6, 2,4 quintals of H, M, L respectively. How many days per week should each mill be operated in order to meet the contract order most economically?
Answer:

The L.P.P is : minimize z = 1000 x + 800y
Subject to constraints: 6x + 2y ≥ 12
2x + 2y ≥ 8
4x + 12y ≥ 24 and
Non-negativity: x ≥ 0, y ≥ 0; Number of days per week can be determined using graphical method.

Question 4.
Graphically solve the following LPP
Max Z = 8x + 3y
s.t. 2x+ 5y ≥ 20
x + y ≤ 7 and x, y ≥ 0
Answer:
Converting the inequalities to equalities
2x + 5y = 20 …(1)
x + y = 7 …(2)
x = 0, y = 0


The corner points of the feasible region are A(0, 7) B(0, 4) and C.
To find C solve eqn.(1) and (2)

Subs y = 2 in Eqn. (1) (2)
x + 2 = 7
x = 7 – 2 = 5
∴ C (5, 2)

The maximum value of Z corresponds to corner point C(5, 2)
∴ x = 5, y = 2 and z = 46 forms an optimal solution.

Question 5.
Solve graphically
Minz = 45x + 38y
s.t 3x + 5y ≤ 30
2x – 4y ≥ 8 And x, y ≥ 0
Answer:
Converting inequalities to equalities
3x + 5y = 30 …(1)
2x – 4y = 8 …(2)
x = 0, y = 0

x = 0
y = 0

The corner points of the feasible region are A(.) B(4, 0) C (10, 0)
Solving Equation (1) and (2) to find A

Subs y = 1.6 in Equation (1)
3x + 5(1.6) = 30
3x + 8.18 = 30
3x = 30 – 8.18
x = \(\frac{21.82}{3}\) x = 7.2 ∴ A(7.2, 1.6)
Corner point Z = 45x + 38y
A(7.2, 1.6) z = 45(7.2) + 38(1.6)
z = 384.8
B(4, 0) z = 45(4) + 38(0)
z = 180
C(10, 0) z = 45(10) + 38(0)
z = 450
The minimum value of z corresponds to the corner point
B(4,0)
∴ x = 4y = 0 and z = 180
forms an optimal solution to the gain LPP.

Question 6.
Consider the LPP
Max Z = 3x + 5y
s.t x + 2y ≤ 9
x ≤ 3 and x,y ≥ 0
lf,x = – 2 and y = 4 is a solution to LPP. Is it a feasible solution? Give reason to your answer
Answer:
x = -2 y = 4 is not a feasible solution
(For a feasible solution x and y ≥ 0)
Here x = -2 is negative.

Question 7.
Max Z = x₁ + 5X₂
s.t x₁ + 5X₂ ≤ 16
5x₁ ≤ 30 And x₁, y ≥ 0
Answer:
Converting the inequalities to equalities
x₁ + 5X₂ = 16 …(1)
5x₁ = 30 …(2)
x₁ = 0 x₂ = 0

x₁ = 0 x₂ = 0

The corner points of the feasible region are A(0, 3.2) B(0, 0) C(6, 0) and D( )
Solve eqn. (1) and (2) to find D
x₁ + 5X₂ = 16 …(1)
5X₂ = 30 …(2)
From eqn. (2), 5x₁ = 30 x₁ = \(\frac { 30 }{ 5 }\) = 6
Subs x₁ = 6 in …(1)
6 + 5X₂ = 16
5X₂ = 16 – 6
5X₂ = 10
x₂ = \(\frac { 11 }{ 5 }\) = 2
∴ D(6, 2.2)

The maximum value of Z corresponds to the corner point A(0, 3.2) and D(6, 2)
∴ x = 0 y = 32
And x = 6, y = 2 and z = 16 forms the optimal solutions.
The given LPP has multiple optimal solutions

Question 8.
Solve Max Z = x + y
s.t x + y ≥ 1
3x + y ≤ 3
And x, y ≥ 0
Answer:
Converting the inequalities to equalities
x + y = 1 :..(l)
3x + y = 3 …(2)
x = 0, y = 0


The corner points of the feasible region are A(0, 3) B(0,1) and C(1, 0)

Corner point	Z = x + y
A(0, 3)	Z = 0 + 3= 3
B(0,1)	Z=0+1=1
C(1,0)	Z =1 + 0 = 1

The max value of Z correspond to the corner point A(0, 3)
∴ x = 0 y = 3 and z = 3 forms an optimal solutions.

Question 9.
Solve graphically
Max z = 20x + 10y
s.t x ≥ 2
y ≤ 5 And x, y ≥ 0
Answer:
Converting inequalities to equalities
x = 2 …(1)
y = 5 …(2)

x = 0, y = 0

Feasible region exists and is unbounded
The objective of the given LPP is to maximize Z.
Therefore the LPP is said to have unbounded solution.

Question 10.
Max Z = 20x + 10y
s.t x + y ≥ 50
20x + 40y ≤ 800 and x, y ≥ 0
Answer:
Converting the inequalities to equalities x + y = 50 …(1)
20x + 40y = 800 …(2)
x = 0, y = 0

x = 0, y = 0

There exists no feasible region and hence no solution to the given LPP

Question 11.
Min Z = 3x + 2
s.t x + 3y ≥ 6
2x + y ≥ 8 And x, y ≥ 0
Answer:
Converting inequalities to equalities
x + 3y = 6
2x + y = 8
x = 0 y = 0

x = 0 y = 0

The feasible region is unbounded since the objective is to minimize Z, one of the corner points form an optimal solution.
Corner points are A(0, 8) C(6, 0) and B( ).
To find B solve eqn. (1) and (2)

Subs y = 0.8 in eqn. …. (1)
x + 3(0.8) = 6
x = 6 – 2.4
x = 3.6
∴ B(3.6, 0.8)

Minimum value of Z corresponds to the corner points B(3.6, 0.8)
∴ x = 3, y = 0.8 and Z = 12.4 forms an optimal solution

Question 12.
Solve graphically
Max z = x + y
s.t 5x + 3y ≤ 15
3x + 5y ≤ 15 And x, y ≥ 0
(a) If objective function is Max Z = x + 2y what will be the optimal solution?
(b) Reverse the inequalities in the two constraints and solve graphically
Answer:
(a) Converting the inequalities to equalities 5x + 3y = 15 …(1)
3x + 5y = 15 …(2) and x = 0 y = 0

x = 0, y = 0

The corner points of the feasible region are
A(0, 3) B(0, 0) C(3, 0) and D( )
To find D solve eqn. (1) and (2)

Subs y =1.8 in equation (1)
5x + 3(1.8) = 15
5x + 5.4 = 15
5x =9.6
x = \(\frac { 9.6 }{ 5 }\) = 1.92 ∴ D(1.92, 1.8)

The max value of z corresponds to the corner point A(0,3)
∴ x = 0, y = 3 and Z = 3.72 forms an optimal solution

(a) If objective function is Max Z = x + 2y

The Max value of Z corresponds to the corner point A(0, 3)
∴ x = 0, y = 3 forms an optimal solution

(b) Reversing the inequalities
5x + 3y ≥ 15
3x + 5y ≥ 15
From the point (x,y) for the respective equations,

Feasible region exists and is unbounded
Since the objective is to max Z the given LPP is said to have unbounded solution.

2nd PUC Statistics Transportation Problems : I Pratical Assignments

Question 1.
Obtain an initial B.F.S for the following T.P by NWCR method. Also obtain the transportation cost.

Answer:


Here Σ^a₁ – supply = Σ^b₁ – Demand = 120 the T.P is balanced
By N.W.C.R first allocation is made at the cell (1,1) as
X₁₁ = min (a₁, b₁) = min (27,46) = 27.
‘A’ is satisfied, and replace b₁ = 46 by (46 – 27) = 19
Next allocate in the same column in the cell (2, 1) / (B, x) as:
X₂₁ = min (33 , 19) = 19
‘X’ is satisfied and replace 33 by (33 – 19) = 14
Next allocation is made in the same row in the cell (2 ,2) (B,y)as:
X₂₂ = min (14,44) = 14,
‘B’ is satisfied and replace 44 by (44 – 14) = 30
Next allocation is made in the same column in the cell (3 ,2) / (c ,y) as:
X₃₂ = min (38 ,30) = 30
Next allocate in the same row at (3,3) / (C , Z) as
X₃₃ = min (8.30) = 8,
‘C’ is satisfied and replace 30 by (30 – 8) = 22
Next allocation is made in same column at (4 ,3) / (D , Z) as
X₄₃ = min (22 , 22) = 22.
The Initial Basic Feasible solution (1 .B.F.S) and the Total transportation cost is

Question 2.
For the following T.P, obtain an initial B.F.S by NWCR and show that it is degenerate.

Answer:

T.P. is balanced
By N.W.C.R First allocations is made at the north – west corner cell (1,1) as:
X₁₁ = min (a_i, b_j) = min (35,25) = 25
A is satisfied and replace a₁ = 35 by (35 – 25) = 10
Next allocate in the same row at (1,2) / (II, B) as:
X₁₂ = min (10, 45) = 10,
I’ satisfied and replace 45 by [45 – (0)] = 35
Next allocate in the same column at (2 ,2) / (II ,B) as:
X₂₂ = min (50,35) = 35
‘B’ is satisfied and replace 50 by (50 – 35) = 15
Next allocate in the same row at (2,3)/ (II ,c) as:
X₂₃ = min = 15
Both II and ‘C’ are satisfied next allocate in the diagonal cell (3,4)/ (II, D) as:
X₃₄ = min (15,15) = 15
The I.B.F.S and the total transportation cost is:

Here Total allocations = 5 < (m + n -1) = (3 + 4 -1) = 6, The solution is degenerate

Question 3.
Verify whether the following solution is non – degenerate.

Answer:
No. of allocations = 6 = (m + n – 1) = 3 + 4 – 1 = 6, The solution is non – degenerate.

Question 4.
Obtain an initial basic feasible solution to following T.P by matrix minima method. Also obtain the transportation cost.

Answer:

T.P. is balanced
By M.M.M . the least cost is 1 at (2 2). Allocate as:
X₂₂ = min(a₂, b₂) = min(3, 1) = 1,
2^nd col, is satisfied, so delete and replace a₂ = 3 by (3 – 1) = 2, and prepare the reduced table.

The least cost in the reduced cost matrix is 2, allocate as:
X₂₁ min (2 ,4) = 2,
2^nd row is satisfied, so delete and replace b₁ = 4 by (4 – 2) = 2 prepare the reduced table.

The least cost in the reduced cost matrix is 3. Allocate as:
X₃₁ = min(5,2) = 2
First column is satisfied, so delete and replace a₃ = 5 by (5 – 2) = 3. Prepare the reduced table.

The least cost is 4 allocate as: X₁₃ = min (2 , 5) = 2.
First row is satisfied so delete and replace 5 by (5 – 2) = 3 and prepare the reduced table as:

Last allocation is made as:
X₃₃ = min (3 .3) = 3.
The I.B.F.S and the total transportation cost is:

Question 5.
Obtain an initial B.F.S by Matrix minima methods for T.P given below:

Answer:

T.P is balanced.
By M.M.M. the least cost is ‘10’ at (2 ,4) allocate as:
X₂₄ = min (a₂, b₄) = min (250,, 250) = 250.
Here O₂ and D₄ are satisfied so delete prepare the reduced cost matric.

The least cost in the reduced cost matrix is ’14’ at two cells, so cheese any one cell, allocate as:
X₃₃ = min (150 ,150) = 150
Both O₃ and D₃ are satisfied so delete, and prepare the reduced table.

Allocate at (1,2) as:
X₁₂ = min (300 ,200) = 200. .
D₂ is satisfied so delete and replace 300 by (300 – 200) = 100 Prepare the reduced Table:

Last allocations is made at (1,1) as:
X₁₁ = min (100,100) = 100
The I.B.F.S and the total cost is:

Question 6.
A company has 4 warehouses and six stores. warehouses altogether have a surplus of 22 units given commodity divided among them as follows:

The six stores altogether need 22 units of the commodity and the individual requirement are as follows,

Cost (Rs) of shipping one unit of a commodity from i^th warehou to j^th is as given below:

Find a Feasible solution using NWCR technique.
Answer:
The given availabilities in the warehouses and requirement of the stores with respective cost matrix can be written as below.

By N.W.C.R first allocation is made in the North- west corner cell (1,1) as:
X₁₁ = min (a₁, b₁) = min (5,4) = 4
‘I’ is satisfied and replace 5 by (5,4)= 1, and Next allocation is made in the same row in the cell (1,2) as:
X₁₂ = min (1,4) = 1
‘A’ is satisfied and replace b2 = 4 by (4 – 1) = 3. Next allocate in the same row at (2,2) as:
X₂₂ = min (6,3) = 3
‘II’ is satisfied and replace a2 = 6 by (6 – 3)=3. Next allocate in the same row (2,3) as:
X₂₃ = min (3,6) = 3
‘B’ is satisfied and replace b3 = 6 by (6 – 3)=3. Next allocate in the same row (3,3) as:
X₃₃ = min (2,3) = 2
‘C’ is satisfied and replace b3 = 3 by (3 – 2)=l. Next allocate in the same row (4,3) as:
X₄₃ = min (9,1) = 1
III’ is satisfied and replace a4 = 9 by (9 – 1)=8. Next allocate in the same row (4,4) as:
X₄₄ = min (8,2) = 2
‘IV’ is satisfied and replace 8 by (8 – 2)=6. Next allocate in the same row (4,5) as:
X₄₅ = min (6,4) = 4
‘V’ is satisfied and replace 6 by (6 – 4)=2. Next allocate in the same row (4,6) as:
X₄₆ = min (2,2) = 2.
The Feasible solution and the total transportation cost is:

2nd PUC Statistics Game Theory Pratical Assignments

Question 1.
A and B roll a die each. If both the dice show odd numbers, then A pays Rs. 10 to B, if both the dice show even numbers, then A gets Rs. 10 from B. gets Rs. 5 from A. Write the pay off matrix of A. Does the game have saddle point?
Answer:
The pay off Matrix of player A is:

By M.M.P
(1) Row minimum is circled
(ii) Column maximum is boxed
(iii) The payoff – 5 at (2, 1) is circled and boxed, is the saddle point.

Question 2.
Player A has 3 marbles – a white, a black and a green marble. Player B has 2 marbles a white and a black marble. Each player selects a marble simultaneously. If they are of matching a colour A loses Rs. 10, otherwise he gains Rs. 5. Write B’s payoff matrix.
Answer:
The payoff Matrix of player B is:

Question 3.
Find the value of the game using maximin – minimax principle from the following data:

Answer:

By Maximin – Minimax principle
(i) Minimum payoff in each row is / are circled
(ii) Maximum payoff in each column is / are boxed.
(iii) The payoff 7′ at (2, 3) is circled as well as boxed, is the saddle point.
(iv) The suggested optimal solution for the players are : A₂,B₃.
(v) Value of the game : v = 7.

Question 4.
Solve the following game using dominance principle :

Answer:

By Dominance principle. All the payoff of A₃ are more than all the payoff of A₂
A₃ dominates A₂, so delete A₂.

All the payoff of B₃ are less than all the pay off of B₁, B₂ and
B₄. B₃ dominates B₁, B₂, B₄
So, delete B₁, B₂, B₄

The pay off of A₃ is more than A₁, A₃ dominates A₁, so delete A₁

(i) Saddle points occurs at (3, 3)
(ii) The suggested optimal strategies for the players are : A₃. B₃
(iii) The value of the game V = 4.

Question 5.
Given pay off matrix of A is below write the payoff matrix of B and then solve the game:

Answer:
The pay off matrix of player B is:
Player A

By M.M.P:

• Minimum payoff in each row is / are circle
• Maximum payoff in each column is / are boxed
• The payoff 7’ is circled as well as boxed at (2,2) is the saddle point.
• The suggested optimal strategies for the players are B₂, A₂.
• Value of the game v = -7.

Question 6.
Solve the following game by dominance principle. Is the game fair?

Answer:

By Dominance principle, All pay off of A₁ are more than all pay off of A₂ and A₃.
A₁ dominate A₂ and A₃. so delete A₂, A₁

The payoff of B₃ is less than the payoff’s of B₁, B₂ and B₄. B₃ dominates B₁, B₂, B₄. So delete B₁, B₂, B₄

saddle point occurs at (1, 3).
The suggested strategies for respective players are : A₁, B₃. The value of the game v = 0 yes the game is fair.

2nd PUC Statistics Replacement Theory Pratical Assignments

Question 1.
The cost of scooter is Rs. 36,000. Its maintenance cost and resale value at different age given below:

Determine the optimal age for replacement of the scooter:
Answer:
Given: p = 36,000, Let C₁, S_n,
be the Maintenance cost and resale value and n be the years

From the above table the Annual average maintenance cost A(n) is Minimum for the 5^th year. The optimum period for the replacement of the scooter is n = 5^th year. The optimum cost A(5) = Rs. 5,920.

Question 2.
The capital cost of a machine is Rs. 10,500. Its resale value is Rs. 500. The maintenance cost are as follows:

Find out when the machine should be replaced?
Answer:
Given: p = 10, 500, S_n = 500, ∴ (p – s_n) = 10,000

From the above table the A(n) is minimum for the year n = 5 is the optimum period for replacement of the machine : The optimum cost is A(S) = Rs. 3600.

Question 3.
The Purchase price of machine A is Rs. 10000. Its salvage rates and maintenance cost are as below:

What should be the optimum replacement period? What would be the annual cost?
Answer:
Given : p = 10,000; salvage rate – s_n

From the above table A(n) is minimum for the 4^th year is the optimum replacement period. The Annual average cost. A(4) = Rs. 2950.

Question 4.
The purchase price of machine B is Rs. 8000. Its salvage rates and maintenance cost are as below:

What would be the optimum replacement period?
What would be the annual average cost?
Answer:
Given : p = 8000, Salvage rate – S_n

From the above table A(n) is minimum for the 4^th years is the optimum period for replacement of the machine. The Annual Average cost: A(4) = Rs. 2375.

Question 5.
Example 3 and 4 describe about machines A and B. After looking at these examples, which machine does the consumer prefer to buy and why?
Answer:
Annual Average cost of machine A : A (n) = A(4) = Rs. 2950 and of machine B : A(4) = Rs. 2375. Here A(n) of machine B < A(n) of machine A.
∴ machine B is preferred.

Question 6.
If the depreciation costand the cumulative maintenance cost for an equipment for the third year Rs. 6000 and Rs 6200 respectively. What is the annual cost?
Answer:
Given : Depreciation cost: (p – S_n) = (p – S₃) = 6000
Cumulative maintenance cost: Σci = 6200 for n = 3
∴ Annual average cost A(n) = A(3)
\(=\frac{\left(p-s_{n}\right)+\Sigma c_{C}}{3}=\frac{6000+6200}{3}=R s .4066 .67\)

Question 7.
An equipment costs Rs.5000 the running cost is Rs. 500 for the first two years and increased by Rs. 2000 from third year onwards. The scrap cost of the machine at all times is Rs. 300. Find the optimal replacement age.
Answer:
P = 5000, Running cost: c₁; scrap cost – S_n

From the above table A(n) is minimum for the 3^rd year, n = 3 is the optimum period for the replacement of the equipment optimum cost A(3) = Rs. 2,733.33

2nd PUC Statistics Replacement Theory Pratical Assignments

Question 1.
Given:
Setup cost = Rs. 50 / cycle.
Storage cost = Rs. 3 / items / year
Demand = 5000 items / year
Calculate minimum average inventory cost.
Answer:
Given:
C₃ = 50/cycle.C₁ = 3/year, R = 5000/year.
C₂ – shortage not allowed use EOQ model I
Minimum average inventory cost: C(Q⁶)
= \(\sqrt{2 C_{3} R C_{1}}=\sqrt{2 \times 50 \times 5000 \times 3}\) = Rs. 1224.74

Question 2.
A company has to supply 12,000 units of a product / year to its customers. The shortage cost is infinite. The holding cost is Rs. 0.2/ unit / month and setup cost is Rs. 350. Determine (i) Number of orders per year. (ii) Minimum average cost / month.
Answer:
Given:
R = 12,000 / year, C₁ = 0.2 1 month.
∴ C₁ = 0.2 × 12 = 2.4,C₃ = 350, C₂ = 0.
Shortages not allowed use EOQ model I.
(i) Number of orders per year : n° = \(\frac{I}{t^{0}}=\frac{R}{Q^{0}}\)
Here
Q° = \(\sqrt{\frac{2 C_{3} R}{C_{1}}}=\sqrt{\frac{2 \times 350 \times 12,000}{2.4}}\)
= 1870.83 units / year
∴ n° = \(\frac{\mathrm{R}}{\mathrm{Q}} \quad \frac{12,000}{1870.83}\)
= 6.414 orders/ years.

(ii) Min. Avg cost / month
= C(q°) = \(\sqrt{2 c_{3} R c_{1}}=\sqrt{2 \times 350 \times 12,000 \times 2.4}\)
= 4489.98 = 4490 / year
∴ \(\frac{4490}{12}\) = Rs. 74.17 /month 12

Question 3.
There is demand for 8000 ¡tenis / year. The replenishment cost is loo and the maintaince cost Rs. 2 / item / year. Replenishment is instantaneous and shortages are not allowed. Find (I) EOQ (ii) to (iii) no (iv) C(Q°).
Answer:
Given: R =8000/year, C₃ = 100,C₁ = 2/year.
Shortage – C₂ not allowed use EOQ model I.
(i) EOQ:
Q° = \(\sqrt{\frac{2 C_{3} R}{C_{1}}}=\sqrt{\frac{2 \times 100 \times 8,000}{2}}\)
= 894.43 items/ year

Question 4.
Maruthi udyog company purchase 10,000 rear mirror for its cars annually. Each mirror costs Rs. 18. The ordering cost per order is Rs. 12 and the annual inventory carrying cost is 12% of capital cost. Compute EOQ and the min. average inventory cost.
Answer:
Given: R = 10,000/year. P = 18, C₃ 12,1 = 12% /year
∴ C₁ = P₁ = 18 × 12% = 2.16/year.
Shortage not allowed use EOQ model I:
EOQ: Q° = \(\sqrt{\frac{2 C_{3} R}{C_{1}}}=\sqrt{\frac{2 \times 12 \times 10,000}{2.16}}\)
= 3 33.33 mirrors / year.
Min. Avg. In cost:
C(Q°) = \(\sqrt{2 \mathrm{C}_{3} \mathrm{RC}_{1}}=\sqrt{2 \times 12 \times 10,000 \times 2.16}\)
= Rs. 720 / year.

Question 5.
The annual demand for an item is 3000 units, capital cost is Rs. 7/ unit. Inventory carrying cost per year is 20% of the capital cost. If setup cost per year is 20% of the capital cost. IF setup cost is Rs. 150. Determine (I) EOQ (ii) no. of orders per year (iii) Optimal cost per year.
Answer:
Given: R = 3000/year,p = 7,I = 20% year
∴ C₁ =P₁ = 7 × 20% = 1.4/year C₃ = 150
Shortage not allowed tise EOQ mode I
(i) EOQ:
Q° = \(\sqrt{\frac{2 C_{3} R}{C_{1}}}=\sqrt{\frac{2 \times 150 \times 3,000}{1.4}}\)
= 801.78 iternš per year.

(ii) No. of orders:
n° = \(\frac{1}{t^{0}}=\frac{R}{Q^{0}}=\frac{3,000}{801.78}\)
= 3.74 orders / year

(iii) Optimal cost per year: C(Q°)
= \(\sqrt{2 c_{3} R c_{1}}=\sqrt{2 \times 150 \times 3,000 \times 1.4}\)
= Rs. 1122.49 /year

Question 6.
The demand for motor cycle tyres is 500 / year. The cost of placing an order is Rs. 250. Holding cost is Rs. 25 p.a. The penalty cost for not supplying on deman is Rs. 10 per month. Find the optimal lot size and maximum shortage level.
Answer:
Given: R = 500/year, C₃ = 250,C₁ = 25/year
C₂ =10/month.
∴ C₂ = 10 × 12=120/year.
C₂ = shortages allowed use EOQ model Il.
Optimal lot size:
Q° = \(\sqrt{\frac{2 C_{2} R}{C_{1}}} \times \sqrt{\frac{C_{1}+C_{2}}{C_{2}}}=\sqrt{\frac{2 \times 250 \times 500}{25}} \times \sqrt{\frac{25+120}{120}}\)
= 100 × 1.0992 = 109.92 tyres/years.
Maximum shortage level = (Q° – s°)
Here S° = \(\frac{C_{2} Q_{0}}{C_{1}+C_{2}}=\frac{120 \times 109.92}{25+120}=90.968\) = 90.968
∴ (Q°- s°) = 109.92 – 90.968 = 19.952 tyre/year.

Question 7.
The following data gives various costs and other factors for the production of inventory systems of gears:
Demand / year = 5,000 gears, Set Lip cost = Rs. 360
Holding cost / year = Rs. 40, Shortage cost / unit / year = Rs. loo
Find (ï) Optimum lot size, (ii) Number of orders / year and (iii) Re – order time.
Answer:
Given: R = 5000 / year, C₃ = 360, C₁ = 40 / year
C₂ = 1oo / year. Use EOQ model Il

i. Optimum lot size:
Q° = \(\sqrt{\frac{2 C_{3} R}{C_{1}}} \times \sqrt{\frac{C_{1}+C_{2}}{C_{2}}}=\sqrt{\frac{2 \times 360 \times 5,000}{40}} \times \sqrt{\frac{40+100}{100}}\)
= 300 × 1.18232 = 35.496

ii. Frequency of ordering:
n° = n° = \(\frac{1}{t^{\circ}}=\frac{R}{Q^{\circ}}=\frac{5,000}{354.96}\) = 14.086 orders / year.

iii. Re-order time:
t° = \(\frac{Q^{0}}{R}=\frac{354.96}{5,000}\) = 0.07099 years.

Question 8.
The demand for an item is at a rate of 5 units/month. There is a inventory in which the setup cost is Rs. 1800 / Production run, holding cost Is Rs. 10 unit / year and shortage cost is Rs. 12 / unit / year. Suggest a suitable inventory policy.
Answer:
Given: R =5/month.\R = 5 × 12 = 60/year
C₃ = 1800,C₁ = 10 year, C₂ = 12/year.
Use EOQ model II.
(i) Q° = \(\sqrt{\frac{2 C_{3} R}{C_{1}}} \times \sqrt{\frac{C_{1} \times C_{2}}{C_{2}}}\)
= \(\sqrt{\frac{2 \times 1800 \times 60}{10}} \times \sqrt{\frac{10+12}{12}}\)
= 198.99 items / year.

(ii) Recorder time = t° = \(\frac{198.99}{60}\)
= 3.316 years.

(iii) Frequency or ordering:
n° = \(\frac{1}{t^{0}}=\frac{1}{3.316}\)
= 0.3 orders

(iv) Annual Avg. mv.
Cost = C (Q°, S°)
= \(\sqrt{2 C_{9} R C_{1}} \times \sqrt{\frac{C_{3}}{C_{1}+C_{2}}=\sqrt{2 \times 1800 \times 60 \times 10} \times \sqrt{\frac{12}{10+12}}}\)
= Rs. 1085.44/year.