Intro to Optimization | Mathematical (MATLAB) Solutions

Running head: INTRO TO OPTIMIZATION
INTRO TO OPTIMIZATION
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INTRO TO OPTIMIZATION1
Question 1:
Q = [10 0
0 4 ] b = [−3
−2 ]
Conjugate direction method finds basis where the components of x minimizes
f(x) = (½)x^T*Q*x + b^T*x
= (½)*[x1 x2]*[ 10 0
0 4 ]*[ x 1
x 2 ] + [−3 −2 ]*[ x 1
x 2 ]
= [x1/2 x2/2] [10 x 1
4 x 2 ] -3x1 -2x2
= 5 x 12 +2 x 22 – 3 x 1 – 2 x 2
Hence, 1? = 5, 2? = 2, 3? = 3 and 4? = 2
Question 2:
f(x) = f(x1) + f(x2)
f(x1) = 5 x 12−3 x 1
f(x2) = 2 x 22−2 x 2
Now, f(x1) is minimum when f’(x1) = 0
 10x1 – 3 = 0
 x1 = 3/10
f’’(x1) = 10, f’’(3/10) = 10 > 0 (Hence, there is minimum of f(x1) at x1 = 3/10.
Thus optimal solution to min(f(x1)) is at x1 = 3/10 with f(3/10) = 5*(3/10)^2 – 3*(3/10)
= -0.45.

INTRO TO OPTIMIZATION2
Question 3:
Similarly, min(f(x2)) is found by
f’(x2) = 0
 4x2 – 2 = 0
 x2 = 2/4 = ½.
Then the optimal solution of f(x2) = 2∗0.52−2∗0.5 = -0.5 at x2 = ½.
Question 4:
Hence, the optimal value f(x1,x2) = 5∗( 0.3 )2 +2∗0.52 – 3∗0.3 – 2∗0.5 = -0.95.
At x1= 0.3
Question 5:
Min f(x1,x2) = -0.95
At x2 = 0.5.
Question 6:
Given, Q = [10 1
1 4 ] and d1 = [ 1
0 ] and first element of d2 is 1.
Let, d2 = [ 1
d ]
Hence, d 1T∗Q∗d 2=0
 [1¿ 0]∗
[ 10 1
1 4 ]∗
[ 1
d ] = 0
 [1¿ 0] [ 10+ d
1+4 d ] = 0

INTRO TO OPTIMIZATION3
 10+d = 0
 d=−10
Hence, the second element of d2 = -10.
Question 7:
Now, x = α1*d1 + α2*d2
= α 1 [1
0 ] + α2[ 1
−10 ] = [α 1+α 2
−10 α 2 ]
Hence, f(α1,α2) = (½)*[ α 1+ α 2 −10 α 2 ]*[ 10 1
1 4 ]∗
[ α 1+ α 2
−10 α 2 ] + [ −3 −2 ] [ α 1+α 2
−10 α 2 ]
= (1/2) [ α 1+ α 2 −10 α 2 ] [ 10 α 1
α 1−39 α 2 ] + −3 α 1+17 α 2
= (½)(10 α 12 +10 α 1 α 2−10 α 1 α 2+390 α 22 ) −3 α 1+17 α 2
= 5 α 12 +195 α 22−3 α 1+ 17 α 2
Hence, 1? = 5, 2? = 195, 3? = -3 and 4? = 17
Question 8:
f(α1,α2) = f(α1) + f(α 2)
f(α1) = 5 α 12−3 α 1
f’(α1) = 10α 1 – 3
f’(α1) = 0 => α 1 = 3/10
Hence, min f(α1) = 5∗0.32−3∗0.3 = -0.45.
Question 9: