MAT9004 Suitable Optimization Method

Running head: MAT9004
MAT9004
Name of the Student
Name of the University
Author Note

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Q1:
a) Given, f ( x , y )= 10 x
x2 + 4 y2+ 9
f(x0,y0) = 0
 0 = 10x0/(x0^2 + 4*y0^2 + 9)
 x0 = 0
f(x1,y1) = 10x1/(x1^2 + 4y1^2 + 9)
 1 = 10x1/(x1^2 + 4y1^2 + 9)
 10x1 = x1^2 + 4y1^2 + 9
 x1^2 – 10x1 + 4y1^2 + 9 = 0
Minimize, (x1-x0)^2 + (y1-y0)^2
Or, Minimize x1^2 + (y1 – y0)^2
Subject to constraint,
 x1^2 – 10x1 + 4y1^2 + 9 = 0
Now, using suitable optimization method the value of objective function satisfying the
constraint can be found as 25.
Hence, the smallest possible distance = √25=5
Hence, option is D) 5.
b) By maximum rate of change theorem f(x,y) = x^2 + 2y^2 decreases most rapidly from
point (a,b) in the opposite direction of its gradient vector.
Here, (a,b) = (1,-1)

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Now, ∇ f ( x , y ) =
[ δf
δx
δf
δy ] = [ 2 x
4 y ]
Hence, ∇ f ( a , b ) = [ 2
−4 ]
Thus option (B) is correct.
c) Hessian matrix is given by,
H = [ fxx fxy
fyx fyy ] for two variables x and y.
Now, when f(x,y) = k then the Hessian is a null matrix.
Also, when the f(x,y) = 1st order function of x and y then also Hessian is a null matrix.
d) Let the binary string of length 6 is P.
Now, sample space = 2^6.
A = number of 1’s is even = zero 1’s + two 1’s + four 1’s + six 1’s
= 1 + 6P2/2! + 6P4/4! + 1 = 1 + 15 + 15 + 1 = 32.
Hence, Pr(A) = 32/2^6 = 0.5.
Now, B = string with no consecutive zeroes or ones.
Let, ai = bit strings with length i ending with 1.
bi = bit strings with length I ending with 0.
Thus at first for no consecutive zeroes for a string length of i+1, a zero or 1 can be appended
to the string ending with 1. However, only 1 can be appended to a string ending with 0. Thus
the recurrence relation is

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ai+1=ai+ bi
bi+1=ai
a1=1 ,b1=1 (string length of 1: 0/1)
a2=a1 +b1=1+1=2
b2=a1=1
a3=a2 +b2=3
b3=a2 =2
In similar manner,
a4=5 , b4=3
a5=8 , b5=5
a6=13 , b6=8
Total number of strings with consecutive zeroes = 13 + 8 = 21.
Similarly, total number of strings with consecutive ones = 21.
Thus the total favourable cases = 64 – 42 = 22.
Hence, the Pr(B) = 22/64 = 0.3438.
The probabilities of A and B are not independent as there can be even number of 1’s when
there is no consecutive 1’s in the string and hence same strings exist for both cases of A and
B.
Q2:
a) P = {(x,y) ∈ ℝ 2x^2 + y^2 <=12}

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Now, rightmost point of P can be found by,
f(x,0) = 2x^2 = 12
 x = sqrt(6) = 2.449.
Again, lowermost point is found by putting x = 0 in the constraint equation.
f(0,y) = y^2 = 12
 y = -3.464.
Hence, the rightmost point is (2.449,0) and lowermost point (0,-3.464).
Now, distance between (0,0) and (2.449,0) is 2.449.
Also, distance between (0,0) and (0,-3.464) is 3.464.
b)
T(x,y) = 2 x3 + x y2+5 x2 + y2.
∇ T ( x , y )=
[ δT
δx
δT
δy ]
δT
δx = 6 x2+ y2+10 x
δT
δy = 2xy + 2y
∇ T ( x , y )= [6 x2 + y2 +10 x
2 xy+ 2 y ]
c)

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The stationary points can be found by solving the following equations.
6 x2+ y2+10 x = 0 (1)
2xy + 2y = 0 (2)
 2y(1+x) = 0
 1+ x = 0 => x = -1 and y = 0
Now, when x = -1
6 + y^2 -10 = 0 => y = +/- 2
When, y = 0
6x^2 + 10x = 0
=> 2x(3x + 5) = 0
=> x = 0 and x = -5/3
Hence, the stationary points are
(-1,-2), (-1,2), (0,0) and (-5/3,0)
d) T(x,y) = 2 x3 +x y2+5 x2 + y2.
H (x , y) = [fxx fxy
fyx fyy ]
fxx = ∂(6 x2+ y2+10 x )
∂ x = 12x + 10
fxy = ∂(2 xy+2 y )
∂ x = 2y
fyx = ∂(6 x2 + y2 +10 x)
∂ y = 2y

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fyy = ∂(2 xy+2 y )
∂ y = 2x
H ( x , y )= [12 x +10 2 y
2 y 2 x ]
e)
Stationary points: (-1,-2), (-1,2), (0,0) and (-5/3,0)
¿ H ( −1 ,−2 ) ∨¿ = (−12+10 ) (−2 )−4 (−2)2 = 4 – 16 = -12 <0 or (-1,-2) is saddle point.
T( −1 ,−2) = 2(−1)3 + (−1 )∗(−2)2+5 (−1)2 +(−2)2 = -2 -4 + 5 + 4 = 3
¿ H (−1,2 )∨¿ = (−12+10 ) (−2 )−4 (2)2 = -12 <0 or (-1,2) saddle point.
T( −1,2) = 2(−1)3 + (−1 )∗22 +5(−1)2 +22 = -2 – 4 + 5 + 4 = 3
¿ H ( 0,0 )∨¿ = ( 0+10 ) ( 2∗0 )−4 (0)2 = 0 test inconclusive.
It is found that T(0,0) = 0.
Hence, by comparison with other points it can be concluded that (0,0) is a local minimum.
¿ H ( −5
3 ,0 ) ∨¿ = ( 12∗( −5
3 )+10 ) ( 2∗( −5
3 ) )−4 (0)2 = -10 * (-10/3) = 100/3 >0
Now, fxx(-5/3,0) = 12∗(−5
3 )+10 = -10 < 0
Hence, the stationary point ( −5
3 ,0 )is a local maximum.
f) Now, all the stationary points are inside the ellipse and hence they do not fall on the
boundary of ellipse.
The boundary of P is defined by

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2x^2 + y^2 = 12
 y2=12 – 2 x2
Now, putting this boundary condition on T(x,y) gives
T(x,y) = 2x^3 + xy^2 + 3x^2 + (2x^2 + y^2) = 2x^3 + xy^2 + 3x^2 + 12
= 2x^3 + x(12-2x^2) + 3x^2 + 12 = 2x^3 +12x – 2x^3 + 3x^2 + 12 = 3x^2 + 12x + 12
This makes T as function of only one variable x.
Now, x cannot take any values as it is bounded by minimum and maximum x value in the
ellipse boundary. Hence, the two extreme points of x are -2.449 and 2.449.
Now, when x = -2.449 then y = 0 and when x = 2.449 then y = 0.
Thus the candidate points are (-2.449,0) and (2.449,0).
T(-2.449,0) = 3*(-2.449)^2 + 12*(-2.449) + 12 = 0.6048.
T(2.449,0) = 3*(2.449)^2 + 12*(2.449) + 12 = 59.38
g) Thus the global maxima of T(x,y) on P is at (2.449,0) which is 59.38 and global minima of
T(x,y) on P is at (-2.449,0) which is 0.6048.
Q3:
a) Given, there are 5 different flavours of ice-creams that are being served in two-scoop cones
and three-scoop sundaes. Repletion of flavours is allowed and the order in which they are
chosen does not matter.
Now, when served in two-scooped cones then, 2 items are chosen with repetition from 5
different items and irrespective of order.
Hence, possible cases are (5+2-1)C2 = 6C2 = 6!/(4!*2!) = 15.

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Similarly, when they are served in three-scoop sundaes then, 3 items chosen from 5 different
items with repetition and irrespective of order.
Hence, possible cases are (5+3-1)C3 = 7C3 = 7!/(4!*3!) = 35
Hence, different servings of two scoop-cones or three scoop sundaes = 35 + 15 = 50.
b) When exactly two flavours are served with repetition of flavours and irrespective of order
then, in two-scoop cones two flavours are picked from 5 flavours and assigned. Thus number
of different ways discarding order = 5C2 + 5*2 = 10 + 10 = 20. In this case three scoops
sundaes cannot be served as they cannot be filled with exactly two flavours.
c) In this case if A order two flavours then B should not order any of those two flavours. Thus
the number of ways when ordered pairing is considered is (5C2*2!)*(5-2)C2*2! =
5C2*2*3C2*2 = 10*2*3*2 = 120 ways.
d) In this case if A orders two flavours in three scoop cones then B must order from other
three flavours. Thus all possible cases =2*(5C2*3P2)*(3C3*3!) = 2*10*3*1*6 = 360 ways.
Q4:
a) X = event that MailX blocks M
S = event than M is spam
Given, P(X) = 25/29, P(X’) = 4/29
Also, given P(X | S’) = probability that M is blocked given M is not spam = 0.4/100 = 0.004.
Hence, P(X’|S’) = 1 - P(X | S’) = 1-0.004 = 0.996
P(X’|S) = probability that M is not blocked given M is spam = 30/100 = 0.3
P(X|S) = 1 - P(X’|S) = 1-0.3 = 0.7

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Now, P(S’) = P(X|S’) + P(X’|S) = 0.304
b) P(Y) = 0.999
P(Y|S’) = 0.14
P(S’|Y) = P(Y|S’)*P(S’)/P(Y) = (0.14*0.304)/0.999 = 0.0426 > 0.015 (Proved)
c) Given, MailX blocks 25/29 = 0.862 of the emails and MailY blocks 0.999 of the mails.
Then P(X∩Y) = 0.862
Now, if X and Y are independent then P(X)*P(Y) = P(X∩Y).
Now, P(X)*P(Y) = 0.862*0.999 ≠ 0.862.
Hence, the two events X and Y are not independent.
d) As given above, the error of second kind P(X’|S) = 0.3 and P(Y’|S) = 0.14.
Now, when combined or blocks the emails that MailX and MailY both consider to be spam
then
P(X’∩Y’|S) = 0.14.
This is lower than error of second kind for just MailX which is 0.3.
e) Now, the error of first kind P(X|S) = 0.7 and P(Y|S) = 1-0.0426 = 0.9574.
Hence, P(X∩Y|S) = 0.7.
This is lower than the error of first for just MailY which is 0.9574 and thus filtering system is
beneficial.
f) When MailX and MailY agrees then MailZ blocks, otherwise MailZ block with probability
½.
Thus Probability that MailZ block when not Spam

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= (½)*((1 – probability that MailX not blocks when not spam) + (1 – Probability that MailY
not blocks when not spam))
 P(Z|S’) = (½)(1-P(X’|S’) + 1-P(Y’|S’))
 P(Z|S’) = (½)(P(X|S’) + P(Y|S’)
Similarly,
Probability that MailZ not blocks when Spam = (½)(Probability that MailX not blocks when
Spam + Probability that MailY not blocks when Spam)
 P(Z’|S) = (½)(P(X’|S) + P(Y’|S))
g) P(S’|Z) = P(Z|S’)*P(S’)/P(Z) = 0.01*0.5/0.6 < 0.01
P(S|Z’) = P(Z’|S)*P(S)/P(Z’) = (0.15*0.5)/0.4 < 0.2 (Proved).