Full Text Transcript
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(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:2)(cid:9)(cid:2)(cid:4)(cid:10)
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LLLLLEEEEEAAAAARRRRRNNNNNIIIIINNNNNGGGGG OOOOOBBBBBJJJJJEEEEECCCCCTTTTTIIIIIVVVVVEEEEESSSSS
One of the widely used decision making problems, nowadays, is to decide on the optimal mix
of scarce resources in meeting the desired goal. In simplest form, it uses several linear inequations
in two variables derived from the description of the problem.
The objective in this section is to make a foundation of the working methodology for the above
by way of introduction of the idea of :
(cid:1) development of inequations from the descriptive problem;
(cid:1) graphing of linear inequations; and
(cid:1) determination of common region satisfying the inequations.
33333.....11111 IIIIINNNNNEEEEEQQQQQUUUUUAAAAALLLLLIIIIITTTTTIIIIIEEEEESSSSS
Inequalities are statements where two quantities are unequal but a relationship exists between
them. These type of inequalities occur in business whenever there is a limit on supply, demand,
sales etc. For example, if a producer requires a certain type of raw material for his factory and
there is an upper limit in the availability of that raw material, then any decision which he takes
about production should involve this constraint also. We will see in this chapter more about such
situations.
33333.....22222 LLLLLIIIIINNNNNEEEEEAAAAARRRRR IIIIINNNNNEEEEEQQQQQUUUUUAAAAALLLLLIIIIITTTTTIIIIIEEEEESSSSS IIIIINNNNN OOOOONNNNNEEEEE VVVVVAAAAARRRRRIIIIIAAAAABBBBBLLLLLEEEEE AAAAANNNNNDDDDD TTTTTHHHHHEEEEE
SSSSSOOOOOLLLLLUUUUUTTTTTIIIIIOOOOONNNNN SSSSSPPPPPAAAAACCCCCEEEEE
Any linear function that involves an inequality sign is a linear inequality. It may be of one
variable, or, of more than one variable. Simple example of linear inequalities are those of one
variable only; viz., x > 0, x < 0 etc.
x > 0 x ≤ 0
– 3 – 2 – 1 0 1 2 3 – 3 – 2 – 1 0 1 2 3
The values of the variables that satisfy an inequality are called the solution space, and is
abbreviated as S.S. The solution spaces for (i) x > 0, (ii) x ≤ 0 are shaded in the above diagrams,
by using deep lines.
Linear inequalities in two variables: Now we turn to linear inequalities in two variables x and
y and shade a few S.S.
y y y y
x>O x>O
x>O x>O
y>O y>O
x x x x
Let us now consider a linear inequality in two variables given by 3x + y < 6
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The inequality mentioned above is true for certain pairs A Y
of numbers (x, y) that satisfy 3x + y < 6. By trial, we may
arbitrarily find such a pair to be (1,1) because 3 × 1 +1=4, (0, 6)
and 4 < 6.
Linear inequalities in two variables may be solved easily
(x, y)
by extending our knowledge of straight lines.
(2, 0)
X
For this purpose, we replace the inequality by an equality O
B
and seek the pairs of number that satisfy 3x + y = 6. We
may write 3x + y = 6 as y = 6 – 3x, and draw the graph of
this linear function.
Let x = 0 so that y = 6. Let y = 0, so that x = 2.
Any pair of numbers (x, y) that satisfies the equation y = 6 – 3x falls on the line AB.
NNNNNooooottttteeeee::::: The pair of inequalities x ≥ 0, y ≥ 0 play an important role in linear programming problems.
Therefore, if y is to be less than 6 – 3x for the same value of x, it must assume a value that is less
than the ordinate of length 6 – 3x.
All such points (x, y) for which the ordinate is less than 6
– 3x lie below the line AB.
The region where these points fall is indicated by an
arrow and is shaded too in the adjoining diagram. Now
we consider two inequalities 3x + y ≤ 6 and x – y ≤ – 2
being satisfied simultaneously by x and y. The pairs of
numbers (x, y) that satisfy both the inequalities may be
found by drawing the graphs of the two lines y = 6 – 3x
and y = 2 + x, and determining the region where both the
inequalities hold. It is convenient to express each equality
with y on the left-side and the remaining terms in the
right side. The first inequality 3x + y ≤ 6 is equivalent to
y ≤ 6 – 3x and it requires the value of y for each x to be
less than or equal to that of and on 6 – 3x. The inequality
is therefore satisfied by all points lying below the line y
= 6 – 3x. The region where these points fall has been
shaded in the adjoining diagram.
We consider the second inequality x – y ≤ –2, and note that this is equivalent to y ≥ 2 + x. It
requires the value of y for each x to be larger than or equal to that of 2 + x. The inequality is,
therefore, satisfied by all points lying on and above the line y = 2 + x.
The region of interest is indicated by an arrow on the line y = 2 + x in the diagram below.
For x = 0, y = 2 + 0 = 2;
For y = 0, 0 = 2 + x i.e, x = –2.
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x3
–
6
{
Y
A
y
=
6
–
3
x
X
O
B
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y
y = 2 + x
(0, 2)
(-2, 0)
x
0
By superimposing the above two graphs we determine the common region ACD in which the
pairs (x, y) satisfy both inequalities.
Y
A
y = 2 + x
C
X
D O
y = 6 – 3x
We now consider the problem of drawing graphs of the following inequalities
x ≥ 0, y ≥ 0, x ≤ 6, y ≤ 7, x + y ≤ 12
and shading the common region.
NNNNNooooottttteeeee::::: [1] The inequalities 3x + y ≤ 6 and x – y ≤ 2 differ from the preceding ones in that these
also include equality signs. It means that the points lying on the corresponding lines
are also included in the region.
[2] The procedure may be extended to any number of inequalities.
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We note that the given inequalities may be grouped as follows :
x ≥ 0 y ≥ 0
x ≤ 6 y ≤ 7 x + y ≤ 12
Y Y
y =7
X
X
O
x > 0, x < 6 y > 0, y < 7
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:21)
x =
9
O
Y
x
+
y
=
12
X
x + y < 12
By superimposing the above three graphs, we determine the common region in the xy plane
where all the five inequalities are simultaneously satisfied.
Y
(0, 7) (5,7)
(6, 6)
0,0
X
O (6,0)
EEEEExxxxxaaaaammmmmpppppllllleeeee::::: A company produces two products A and B, each of which requires processing in
two machines. The first machine can be used at most for 60 hours, the second machine can be
used at most for 40 hours. The product A requires 2 hours on machine one and one hour on
machine two. The product B requires one hour on machine one and two hours on machine
two. Express above situation using linear inequalities.
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SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let the company produce, x number of product A and y number of product B. As
each of product A requires 2 hours in machine one and one hour in machine two, x number of
product A requires 2x hours in machine one and x hours in machine two. Similarly, y number
of product B requires y hours in machine one and 2y hours in machine two. But machine one
can be used for 60 hours and machine two for 40 hours. Hence 2x + y cannot exceed 60 and
x +2y cannot exceed 40. In other words,
2x + y ≤ 60 and x + 2y ≤ 40.
Thus, the conditions can be expressed using linear inequalities.
EEEEExxxxxaaaaammmmmpppppllllleeeee::::: A fertilizer company produces two types of fertilizers called grade I and grade II.
Each of these types is processed through two critical chemical plant units. Plant A has maximum
of 120 hours available in a week and plant B has maximum of 180 hours available in a week.
Manufacturing one bag of grade I fertilizer requires 6 hours in plant A and 4 hours in plant B.
Manufacturing one bag of grade II fertilizer requires 3 hours in plant A and 10 hours in plant B.
Express this using linear inequalities.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let us denote by x, the number of bags of fertilizers of grade I and by x, the number of
1 2
bags of fertilizers of grade II produced in a week. We are given that grade I fertilizer requires 6
hours in plant A and grade II fertilizer requires 3 hours in plant A and plant A has maximum of
120 hours available in a week. Thus 6x + 3x ≤ 120.
1 2
Similarly grade I fertilizer requires 4 hours in plant B and grade II fertilizer requires 10 hours in
Plant B and Plant B has maximum of 180 hours available in a week. Hence, we get the inequality
4x + 10x ≤ 180.
1 2
EEEEExxxxxaaaaammmmmpppppllllleeeee::::: Graph the inequalities 5x + 4x ≥ 9, x + x ≥ 3, x ≥ 0 and x ≥ 0 and mark the
1 2 1 2 1 2
common region.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: We draw the straight lines 5x + 4x = 9 and x + x = 3.
1 2 1 2
Table for 5x + 4x = 9 Table for x + x = 3
1 2 1 2
x 0 9/5 x 0 3
1 1
x 9/4 0 x 3 0
2 2
Now, if we take the point (4, 4), we find
5x + 4x ≥ 9
1 2
i.e., 5.4 + 4.4 ≥ 9
or, 36 ≥ 9 (True)
x + x ≥ 3
1 2
i.e., 4 + 4 ≥ 3
8 ≥ 3 (True)
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Hence (4, 4) is in the region which satisfies the x
2
inequalities.
We mark the region being satisfied by the
4
inequalities and note that the cross-hatched region
x
is satisfied by all the inequalities.
3 11111+++++x
22222=====33333
EEEEExxxxxaaaaammmmmpppppllllleeeee::::: Draw the graph of the solution set of the
2 55555x
11111+++++44444x
following inequality and equality:
x
2
= 0 1 22222=====99999
x + 2y = 4.
x – y ≤ 3.
Mark the common region.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: We draw the graph of both x + 2y = 4 and x – y ≤ 3 in the same plane.
The solution set of system is that portion of the graph of x + 2y = 4 that lies within the half-plane
representing the inequality x – y ≤ 3.
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0
=
x 1
x
0 1
1 2 3 4
y
For x + 2y = 4,
x 4 0
y 0 2
For x – y = 3, x
x 3 0
x+2y=4
x–y=3
y 0 –3
EEEEExxxxxaaaaammmmmpppppllllleeeee::::: Draw the graphs of the following inequalities:
x + y ≤ 4,
x – y ≤ 4,
x ≥ –2.
and mark the common region.
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For x – y = 4, y
x 4 0 x – y = 4
y 0 –4
For x + y = 4,
x 0 4
x
y 4 0
The common region is the one represented by overlapping x + y = 4
x = –2
of the shadings.
EEEEExxxxxaaaaammmmmpppppllllleeeee::::: Draw the graphs of the following linear
inequalities:
5x + 4y ≤100, 5x + y ≥ 40,
3x + 5y ≤ 75, x ≥ 0, y ≥ 0.
and mark the common region.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn:::::
x y
5x + 4y = 100 or, + =1
20 25
x y
3x + 5y = 75 or, + =1
25 15
x y
5x + y = 40 or, + =1
8 40
Plotting the straight lines on the graph paper we have the above diagram:
The common region of the given inequalities is shown by the shaded portion ABCD.
EEEEExxxxxaaaaammmmmpppppllllleeeee::::: Draw the graphs of the following linear inequalities:
5x + 8y ≤ 2000, x ≤ 175, x ≥ 0.
7x + 4y ≤ 1400, y ≤ 225, y ≥ 0.
and mark the common region:
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let us plot the line AB (5x +8y = 2,000) by joining
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the points A(400, 0) and B(0, 250).
x 400 0
y 0 250
Similarly, we plot the line CD (7x + 4y = 1400) by x 200 0
joining the points C(200, 0) and D(0, 350). y 0 350
Also, we draw the lines EF(x = 175)
and GH (y = 225).
The required graph is shown alongside
in which the common region is shaded.
EEEEExxxxxaaaaammmmmpppppllllleeeee::::: Draw the graphs of the following linear inequalities:
x + y ≥ 1, 7x + 9y ≤ 63,
y ≤ 5, x ≤ 6, x ≥ 0, y ≥ 0.
and mark the common region.
x 1 0 x 9 0
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: x + y = 1 ; ; 7x + 9y = 63, .
y 0 1 y 0 7
We plot the line AB (x + y = 1), CD (y = 5), EF (x = 6),
DE (7x + 9y = 63).
Given inequalities are shown by arrows.
Common region ABCDEF is the shaded region.
EEEEExxxxxaaaaammmmmpppppllllleeeee::::: Two machines (I and II) produce two grades of plywood, grade A and grade B. In
one hour of operation machine I produces two units of grade A and one unit of grade B, while
machine II, in one hour of operation produces three units of grade A and four units of grade B.
The machines are required to meet a production schedule of at least fourteen units of grade A
and twelve units of grade B. Express this using linear inequalities and draw the graph.
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:25)
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SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let the number of hours required on machine I be x and that on machine II be y.
Since in one hour, machine I can produce 2 units of grade A and one unit of grade B, in x hours
it will produce 2x and x units of grade A and B respectively. Similarly, machine II, in one hour,
can produce 3 units of grade A and 4 units of grade B. Hence, in y hours, it will produce 3y and
4y units Grade A & B respectively.
The given data can be expressed in the form of linear inequalities as follows:
2x + 3y ≥ 14 (Requirement of grade A)
x + 4y ≥ 12 (Requirement of grade B)
Moreover x and y cannot be negative, thus x ≥ 0 and y ≥ 0
Let us now draw the graphs of above inequalities. Since both x and y are positive, it is enough
to draw the graph only on the positive side.
The inequalities are drawn in the following graph:
For 2x + 3y = 14,
x 7 0 5 4.66
y 0 4.66 4
3
x+4y≥12
2
For x + 4y = 12,
1
x 0 12 2x+3y≥14
y 3 0 2 4 6 7 8 10 12
In the above graph we find that the shaded portion is moving towards infinity on the positive
side. Thus the result of these inequalities is unbounded.
EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee::::: 33333 (((((AAAAA)))))
CCCCChhhhhoooooooooossssseeeee ttttthhhhheeeee cccccooooorrrrrrrrrreeeeecccccttttt aaaaannnnnssssswwwwweeeeerrrrr/////aaaaannnnnssssswwwwweeeeerrrrrsssss
1 (i) An employer recruits experienced (x) and fresh workmen (y) for his firm under the
condition that he cannot employ more than 9 people. x and y can be related by the
inequality
(a) x + y ≠ 9 (b) x + y ≤ 9 (c) x + y ≥ 9 (d) none of these
(ii) On the average experienced person does 5 units of work while a fresh one 3 units of
work daily but the employer has to maintain an output of at least 30 units of work per
day. This situation can be expressed as
(a) 5x + 3y ≤ 30 (b) 5x + 3y >30 (c) 5x + 3y ≥ 30 (d) none of these
(iii) The rules and regulations demand that the employer should employ not more than 5
experienced hands to 1 fresh one and this fact can be expressed as
(a) y ≥ x/5 (b) 5y ≤ x (c) 5 y ≥ x (d) none of these
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(iv) The union however forbids him to employ less than 2 experienced person to each
fresh person. This situation can be expressed as
(a) x ≤ y/2 (b) y ≤ x/2 (c) y ≥ x /2 (d) x ≥ 2y
(v) The graph to express the inequality x + y ≤ 9 is
(a) (b)
o o
(c) (d) none of these
o
(vi) The graph to express the inequality 5x + 3y ≥ 30 is
(a) (b)
o o
(c) (d) none of these
o
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:26)(cid:26)
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(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:1)(cid:8)(cid:1)(cid:3)(cid:9)
(vii) The graph to express the inequality y ≤ 1/2x is indicated by
(a) (b)
o
o
(c) (d)
o o
(viii)
L1 : 5x + 3y = 30L2 : x+y = 9 L3 : y = x/3 L4 : y = x/2
The common region (shaded part) shown in the diagram refers to
(a) 5x + 3y ≤ 30 (b) 5x + 3y ≥ 30 (c) 5x + 3y ≥ 30 (d) 5x + 3y > 30 (e)None of these
x + y ≤ 9 x + y ≤ 9 x + y ≥ 9 x + y < 9
y ≤ 1/5 x y ≥ x/3 y ≤ x/3 y ≥ 9
y ≤ x/2 y ≤ x/2 y ≥ x/2 y ≤ x/2
x ≥ 0, y≥ 0 x ≥ 0, y≥ 0 x ≥ 0, y≥ 0
(cid:10)(cid:11)(cid:26)(cid:12) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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2. A dietitian wishes to mix together two kinds of food so that the vitamin content of the
mixture is at least 9 units of vitamin A, 7 units of vitamin B, 10 units of vitamin C and 12
units of vitamin D. The vitamin content per Kg. of each food is shown below:
A B C D
Food I : 2 1 1 2
Food II: 1 1 2 3
Assuming x units of food I is to be mixed with y units of food II the situation can be expressed
as
(a) 2x + y ≤ 9 (b) 2x + y ≥ 30 (c) 2x + y ≥ 9 (d) 2x + y ≥ 9
x + y ≤ 7 x + y ≤ 7 x + y ≥ 7 x + y ≥ 7
x + 2y ≤ 10 x + 2y ≥ 10 x + y ≤ 10 x +2 y ≥ 10
2x +3 y ≤ 12 x + 3y ≥ 12 x + 3y ≥ 12 2x +3 y ≥ 12
x > 0, y > 0 x ≥ 0, y ≥ 0,
3. Graphs of the inequations are drawn below :
o
L1 : 2x +y = 9 L2 : x + y = 7 L3 : x+2y= 10 L4 : x + 3y = 12
The common region (shaded part) indicated on the diagram is expressed by the set of
inequalities
(a) 2x + y ≤ 9 (b) 2x + y ≥ 9 (c) 2x + y ≥ 9 (d) none of these
x + y ≥ 7 x + y ≤ 7 x + y ≥ 7
x + 2y ≥ 10 x +2 y ≥ 10 x +2y ≥ 10
x +3 y ≥ 12 x + 3y ≥ 12 x +3 y ≥ 12
x ≥ 0, y≥ 0
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:26)(cid:10)
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(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:1)(cid:8)(cid:1)(cid:3)(cid:9)
4. The common region satisfied by the inequalities L1: 3x + y ≥ 6, L2: x + y ≥ 4, L3: x +3y ≥ 6,
and L4: x + y ≤ 6 is indicated by
(a) (b)
o o
(c) (d) none of these
o
5. The region indicated by the shading in the graph is expressed by inequalities
o
(cid:10)(cid:11)(cid:26)(cid:20) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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(a) x + x ≤ 2 (b) x + x ≤ 2 (c) x + x ≥ 2 (d) x + x ≤ 2
1 2 1 2 1 2 1 2
2x + 2x ≥8 x x + x ≤ 4 2x + 2x ≥ 8 2x + 2x > 8
1 2 2 1 2 1 2 1 2
x ≥ 0 , x ≥ 0,
1 2
6. (i)The inequalities x ≥ 0, x ≥ 0, are represented by one of the graphs shown below:
1 2
(a) (b)
o o
(c) (d)
o
o
(ii) The region is expressed as
(a) x – x ≥ 1
1 2
(b) x + x ≤ 1
1 2
(c) x + x ≥ 1
1 2
(d) none of these
o
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:26)(cid:21)
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(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:1)(cid:8)(cid:1)(cid:3)(cid:9)
(iii) The inequality –x + 2x ≤ 0 is indicated on the graph as
1 2
(a) (b)
o
o
(c) (d) none of these
o
7.
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The common region indicated on the graph is expressed by the set of five inequalities
(a) L1 ::::: x ≥ 0 (b) L1 ::::: x ≥ 0 (c) L1 ::::: x ≤ 0 (d)None of these
1 1 1
L2 ::::: x ≥ 0 L2 ::::: x ≥ 0 L2 ::::: x ≤ 0
2 2 2
L3 ::::: x + x ≤ 1 L3 ::::: x +x ≥ 1 L3 ::::: x + x ≥ 1
1 2 1 2 1 2
L4 ::::: x – x ≥ 1 L4 ::::: x –x ≥ 1 L4 ::::: x –x ≥ 1
1 2 1 2 1 2
L5 ::::: –––––x + 2x ≤ 0 L5 :::::––––– x +2x ≤ 0 L5 :::::––––– x +2x ≤ 0
1 2 1 2 1 2
8. A firm makes two types of products : Type A and Type B. The profit on product A is Rs. 20
each and that on product B is Rs. 30 each. Both types are processed on three machines M1,
M2 and M3. The time required in hours by each product and total time available in hours
per week on each machine are as follows:
Machine Product A Product B Available Time
M1 3 3 36
M2 5 2 50
M3 2 6 60
The constraints can be formulated taking x = number of units A and x = number of unit of
1 2
B as
(a) x + x ≤ 12 (b) 3x + 3x ≥ 36 (c) 3x + 3x ≤ 36 (d)none of these
1 2 1 2 1 2
5x + 2x ≤ 50 5x + 2x ≤ 50 5x + 2x ≤ 50
1 2 1 2 1 2
2x + 6x ≤ 60 2x + 6x ≥ 60 2x + 6x ≤ 60
1 2 1 2 1 2
x ≥ 0, x ≥ 0 x ≥ 0, x ≥ 0 x ≥ 0, x ≥ 0
1 2 1 2 1 2
9. The set of inequalities L1: x + x ≤ 12, L2: 5x + 2x ≤ 50, L3: x + 3x ≤ 30, x ≥ 0, and x
1 2 1 2 1 2 1 2
≥ 0 is represented by
(a) (b)
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(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:1)(cid:8)(cid:1)(cid:3)(cid:9)
(c) (d) none of these
10. The common region satisfying the set of inequalities x ≥ 0, y ≥ 0, L1: x+y ≤ 5, L2: x +2y ≤ 8
and L3: 4x +3y ≥ 12 is indicated by
(a) (b)
(c) (d) none of these
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77777..... b 88888..... c 99999..... b 1111100000..... a
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1. On solving the inequalities 2x+5y≤20, 3x+2y≤12, x≥0,y≥0, we get the following
situation
( ) ( )
(A) (0, 0), (0, 4), (4, 0) and 20 ,36 (B) (0, 0), (10, 0), (0, 6) and 20 ,36
11 11 11 11
(C) (0, 0), (0, 4), (4, 0) and (2, 3) (D) (0, 0), (10, 0), (0, 6) and (2, 3)
2. On solving the inequalities 6x+ y≥18, x+4y≥12, 2x+ y≥10, , we get the following situation
(A) (0, 18), (12, 0), , (4, 2) and (7, 6)
(B) (3, 0), (0, 3),, (4, 2) and (7, 6)
(C) (5, 0), (0, 10), , (4, 2) and (7, 6)
(D) (0, 18), (12, 0), (4, 2), (0, 0) and (7, 6)
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1) A
2) A
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:26)(cid:25)
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