1) Define the following : i) Objective function ii) Constraints
iii) Non – negative restrictions Ans. i) Objective function – If
, ,
1 2
C C Cn
are constants and, ,
x x 1 2 xn
are variable, then the linear function1 1 2 2
z c x c x c x n n
Which is to be maximized or minimized is called the objective function.
ii) Constraints – The inequalities or equation in the variables of a LPP which describes the conditions
under which the optimization (maximization or minimization) is to be accomplished are called constraints.
In the constraints given in the general form of LPP there may be one of the three sings
, ,
iii) Non – negative restrictions – These are the constraints which describes that the variable involved in a
LPP are non – negative.
2) A firm manufactures two types of products A and B and sells them at a profit of Rs. 2 on type A and Rs. 3 on B. Each product is processed on two machines G and H. Type A requires one minute of processing time on G and two minute on H. Type B requires one minute on G and one minute on H.
The machine G is available for not more than 6 hours 40 minutes, while machine H is available for 10 hours during any working day. Formulate the problem as a linear programming problem.
Ans. Let
x 1
be the number of products of type A andx 2
the number of products of type B. The given datamay be put in following tabular from.
Machine Time of product (Minutes) Available time Type A (
Similarly
3
x 2
will be the profit on sellingx 2
units of type B. Therefore total profit on sellingx 1
unit of Aand
x 2
units of B is given by2 3
1 2
Z x x
(objective function)Since machine G takes 1 minute time on type A and 1 minute time on type B, the total number by minutes
required on machine G is given by
1 2
Therefore
400
1 2
x x
Also the machine H is available for 10 hours only. There fore
2 600
1 2
x x
Since it is not possible to produce negative quantities so
1 0
x
and0
x 2
non – negative restrictions) Hence LPP of the given problem isMaximize
2 3
1 2
z x x
Subject to the constants
1 2 400
3) A resourceful home decorator manufactures two types of lamps say A and B.
Both lamps go through two technicians first a cutter, second a finisher. Lamp A requires 2 hours of the cutter’s time and 1 hour of finisher’s time. Lamb B requires 1 hour of cutter’s time and 2 hours of finisher’s time. The cutter has 104 hours and finisher has 76 hours of time available each month. Profit on lamp A is Rs. 6.00 and on one lamp B is Rs. 11.00. Assuming that he can sell all that he produces, how many of each type of lamps should he manufacture to obtain the best return.
Ans. The above information can be put in the following tabular form :-
Lamp Cutter’s time Finisher’s time Profit in Rs.
A 2 1 6
B 1 2 11
Maximum time available
104 76
Let the decorator manufacture x lamps of type A and y lamps of type B.
Total profit = Rs. (6x + 11y)
Total time taken lay the cutter in preparing x lamps of type A and y lamps of type B is (2x+y) hours. But
the cutter has 104 hours only for each month.
2 x y 104
Similarly, the total time taken by the finisher in preparing x lamps of type A and y lamps of type B is
(x+2y) hours. But the cutter has 76 hours only for each month.
2 76
x y
Since the number of lamps cannot be negative
x 0, y 0
So, the mathematical formulation of the given LPP is as follows : Maximize z = 6x + 11ySubject to
2 x y 104
x 2 y 76
4) Solve the following LPP graphically : maximize z= 5x+3y
Ans. Converting the given inequalities in to equations
3 x 5 y 15,5 x 2 y 10, x 0
And y=0The line 3x+5y=15 meets the coordinates axes at
5,0 A 1
and1 0,3
B
respectively. Join these points toget the line 3x+5y=15. Coordinate (0,0) satisfy the inequality
3 x 5 y 15
. So the region containingthe origin represents the solution set of the inequality
3 x 5 y 15
. The line 5x+2y =10 meets the coordinate axes at2,0
A 2
and2 0,5
B
respectively. Join these points toget the line 5x+2y =10. Coordinate (0,0) satisfy the inequality
5 x 2 y 10
. So the region containing theorigin represents the solution set of this inequality.
So the feasible region of given LPP is given by :-
The coordinates of vertices of the shaded feasible region are Y
20 45
0,0 , 2,0 , , 0,3
2 19 19 1
O A P B
Value of objective function at these points are
0,0 5 0 3 0 0
The optional value of
235
z 19
X’X
5) Define the followings :- O
i) Feasible solution Y’
ii) Infeasible solution
iii) Optimal feasible solution iv) Convex set
Ans. i) Feasible solution – A set of values of the variable
,
x x 1 2 xn
iscalled a feasible solution
of a LPP, if it satisfy the constraints and non – negativity restriction of the problem.
ii) Infeasible solution – A solution of LPP is an infeasible solution, if the system of constraints has no point
which satisfies all the constraints and non – negativity restrictions.
iii) Optional feasible solution – A feasible solution of a LPP is said to be an optional feasible solution, if it
also optimizes (Maximize or minimize) the objective function.
iv) Convex set – A set is a convex set, if every point on the line point on the line segment joining any two
points in it lies in it.
Ex.
Unit – VI
Probability
1) In how many ways can 3 girls and nine boys be seated in two vans, each having numbered seats, 3 in front row and 4 at the back ? How many seating arrangement are possible if 3 girls should sit together in a back row on adjacent seats ? Now if all the seating arrangements are equally likely, what is the probability of 3 girls sitting together in a back row on adjacent seats ? Ans. Each van has 7 seats, so there are 14 numbered seats in two vans. The total number
of ways in which 3
girls and 9 boys can sit on these seats is
14 12 C 12
So, total number of seating arrangements=
14 12
C 12
In a van 3 girls can choose adjacent seats in the back row in two ways (1,2,3 or 2, 3, 4)
So, the number of ways in which 3 girls can sit in the back row on adjacent seats in 2(3 ) ways. The numbe
Hence required probability
3! 9!
2) If the letters of the word ‘ATTRACTION’ are written down at random, find the probability that
I) All the T’s occur together
II) No two T’s occur together
Ans. The total number of arrangements of the cutter of the word ‘ATTRACTION’ is 10!
3! 2!
I) Consider all there T’s as one, then there are 8 letters consisting of two identical A’s. These 8 letters can be arranged is 8!
2!ways Hence required probability
8!
II) Other than 3 T’s there are 7 letters which can be arranged in 7!
2! ways. There are 8 places 6 between the 7 letters and one on extreme left and the other on extreme right. So 3 T’s can be arranged in 8 places
8
c 3
ways Number of ways in which no two T’s are together 7!2!
8 C 3
Hence required probability7! probability that either both are apples or both are good.
Ans. Out of 30 fruits two can be selected in
30
C 2
ways.Consider the following two events
A = Getting two apples, B= Getting two good fruits Required probability = P(AUB)
P A B P A P B P A B
There are 20 apples, out of which 2 can be selected in
20
C 2
ways.20
pieces, two can be selected in22
Since there are 15 pieces which are good apples, out of which two can be selected in
15
C 2
Ways. Therefore
15
4) A coin is tosses three times. Find P (E/F) in each of two following:
1) E= Head on the third toss, F= Heads in first two tosses