MAT9004 Suitable Optimization Method

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

MAT90042
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}