Process Optimization
Assignment 4 Daniel Viju.V CH12B080
The Quadratic equation is modelled as, fx= 12xTHx+cTx Where, c is a (2x1) Parametric Vector
H is (2x2) is a symmetric Matrix x is (2x1) is the position vector
As seen in the demonstration the stationary point is x*=00
Stationary point as per definition is Hx*+c=0. Therefore we can conclude that the function is of the form, fx= 12xTHx
Now we know that the contours are ellipses. Therefore H is of the form,
H=a00b
Where a≠b.
The Eigen Vectors as seen from the demonstration is 10 and 01. We can also infer that, a=λ1 b=λ2
The aspect ratio relation as defined for ellipses is, e=λ1λ2 The H matrix can be rewritten as,
H=e2λ200λ2
Case 1: (e=0.3)
I set f(x)=20 to determine the eigen values. One point on the contour is 14.90880. From this,
H=0.18002
Now we determine the Objective value at (7,2) and is set correspondingly on the demonstration, f72=8.41 The Iterations are,
72→ 6.2872-0.1754→2.98810.857→2.688-0.0753→1.2990.371→1.14983-0.0323

Now we determine the Objective value at (-7,2) and is set correspondingly on the demonstration, f-72=8.41 -72→ -6.2872-0.1754→-2.98810.857→-2.688-0.0753→-1.2990.371→-1.14983-0.0323

Define,
Fx=12x-x*THx-x*= 12xTHx=f(x)
For 72 Iterate No. | 0 | 1 | 2 | 3 | 4 | 5 | F(x(k)) | 8.41 | 3.588 | 1.538 | 0.655 | 0.289 | 0.120 | F(x(k+1))/F(x(k)) | 0.4266 | 0.4286 | 0.4258 | 0.441 | 0.415 | |

For -72 Iterate No. | 0 | 1 | 2 | 3 | 4 | 5 | F(x(k)) | 8.41 | 3.588 | 1.538 | 0.655 | 0.289 | 0.120 | F(x(k+1))/F(x(k)) | 0.4266 | 0.4286 | 0.4258 | 0.441 | 0.415 | |

The upper bound is given by,
B=λmax-λminλmax+λmin2=0.697

For 0≤k≤4, we observe that, F(xk+1)F(xk ≤B

Case 2: (e=0.9)
I set f(x)=20 to determine the eigen values. One point on the contour is 4.9690. From this,
H=0.162002
Now we determine the Objective value at (7,2) and is set correspondingly on the demonstration, f72=43.69 The Iterations are,
72→ 0.252-0.462→0.041420.01489→7 E-4-0.0017→1.3 E-44 E-5→3.3 E-67.8 E-6

Now we determine the Objective value at (-7,2) and is set correspondingly on the demonstration, f-72=43.69 The Iterations are,
-72→ -0.252-0.462→-0.041420.01489→-7 E-4-0.0017→-1.3 E-44 E-5→-3.3 E-6-7.8 E-6

For 72 Iterate No. | 0 | 1 | 2 | 3 | 4 | 5 | F(x(k)) | 43.69 | 0.264 | 1.611 E - 3 | 3.286 E - 6 | 1.53 E -8 | 6.966 E -11 | F(x(k+1))/F(x(k)) | 6 E -4 | 0.0061 | 2.039 E -3 | 4.656 E -3 | 0.0045 | |

For -72 Iterate No. | 0 | 1 | 2 | 3 | 4 | 5 | F(x(k)) | 43.69 | 0.264 | 1.611 E - 3 | 3.286 E - 6 | 1.53 E -8 | 6.966 E -11 | F(x(k+1))/F(x(k)) | 6 E -4 | 0.0061 | 2.039 E -3 | 4.656 E -3 | 0.0045 | |
The upper bound is given by,
B=λmax-λminλmax+λmin2=0.011

For 0≤k≤4, we observe that, F(xk+1)F(xk ≤B

And it also converges faster than the earlier case

Case 3: (e=1.1)
I set f(x)=20 to determine the eigen values. One point on the contour is 4.06550. From this,
H=2.42002
Now we determine the Objective value at (6,2) and is set correspondingly on the demonstration, ( The point (7,2) is out of bounds in the demonstration) f62=47.56 62→ -0.0740.327→0.01430.00477→-1.77 E-47.79 E-4→3.39 E-51.14 E-5→-4.32 E-71.86 E-6

Now we determine the Objective value at (-6,2) and is set correspondingly on the demonstration, ( The point (-7,2) is out of bounds in the demonstration) f-62=47.56 -62→ 0.0740.327→-0.01430.00477→1.77 E-47.79 E-4→-3.39 E-51.14 E-5→4.32 E-71.86 E-6

For 62 Iterate No. | 0 | 1 | 2 | 3 | 4 | 5 | F(x(k)) | 47.56 | 0.114 | 2.70 E - 4 | 6.45 E - 7 | 1.53 E -8 | 3.69 E -12 | F(x(k+1))/F(x(k)) | 2.39 E -3 | 2.38 E -3 | 2.39 E -3 | 2.36 E -3 | 2.42 E -3 | |

For -62 Iterate No. | 0 | 1 | 2 | 3 | 4 | 5 | F(x(k)) | 47.56 | 0.114 | 2.70 E - 4 | 6.45 E - 7 | 1.53 E -8 | 3.69 E -12 | F(x(k+1))/F(x(k)) | 2.39 E -3 | 2.38 E -3 | 2.39 E -3 | 2.36 E -3 | 2.42 E -3 | |

The upper bound is given by,
B=λmax-λminλmax+λmin2=9.03 x 10-3

For 0≤k≤4, we observe that, F(xk+1)F(xk ≤B

Hence the theoretical validation has been validated for all the three cases