Download presentation
Presentation is loading. Please wait.
Published byAngelina OβConnorβ Modified over 6 years ago
1
Quadratic Forms and Objective functions with two or more variables
2
Two Choice Variables Quadratic Forms Definiteness
Let π’=ππ₯, π£=ππ¦, π= π π₯π₯ , π= π π¦π¦ , β= π π₯π¦ = π π¦π₯ Then π 2 π= π π₯π₯ π π₯ 2 +2 π π₯π¦ ππ¦ππ₯+ π π¦π¦ π π¦ (πππ 1) can be written as π π π=π=π π π +ππππ+π π π (πππ 2)Β Definiteness Positive definite if q is invariably positive (q > 0) Positive semidefinite if q is invariably nonnegative (πβ₯0 ) Negative definite if q is invariably negative (q < 0) Negative semidefinite if q is invariably non positive (πβ€0 )
3
Rewriting the quadratic form using completing square
π π+ π π π π +ππβ π π π π π π ππ πππ ππ‘ππ£π πππππππ‘π πππ π>0 πππ ππβ β 2 >0 π ππ πππππ‘ππ£π πππππππ‘π ππ π<0 πππ ππβ β 2 >0 Writing in Matrix Form We have π’ π£ π β β π π’ π£ The determinant of the coefficient matrix π β β π is important in determining the sign of π π ππ ππππππππ π
πππππππ πππ β£πβ£>0 πππ π β β π >0 π ππ ππππππππ π
πππππππ πππ β£πβ£<0 πππ π β β π >0
4
Writing equation (2) in terms of equation (1) gives further insights
Since π= π π₯π₯ πππ ππβ β 2 ππ π π₯π₯ π π¦π¦ β π π₯π¦ . π π¦π₯ π π₯π₯ π π¦π¦ β π π₯π¦ . π π¦π₯ ππ ( π π₯π₯ π π¦π¦ β π π₯π¦ 2 ) Conditions for π 2 π πππ πππ ππ‘ππ£π πππππππ‘π ππ π π₯π₯ > 0 πππ π π₯π₯ π π₯π¦ π π₯π¦ π π¦π¦ >0 π 2 π πππ πππππ‘ππ£π πππππππ‘π ππ π π₯π₯ <0 πππ π π₯π₯ π π₯π¦ π π₯π¦ π π¦π¦ >0
5
Discriminant of a Quadratic form
In general the discriminant of a quadratic form π=π π’ 2 +2βπ’π£+π π£ 2 Is the symmetric determinant π β β π In the particular case of quadratic form π 2 π= π π₯π₯ π π₯ 2 +2 π π₯π¦ ππ¦ππ₯+ π π¦π¦ π π¦ 2 The discriminant is the determinant with second order partial derivatives as its elements.
6
Hessian Determinant β£π»β£= π ππ π ππ π ππ π ππ Determinant with all the second order partial derivatives is called Hessian Matrix.
7
Extremum Conditions for two choice variable
ππ π 2 π ππ πππ ππ‘ππ£π πππππππ‘π ππ‘ ππ ππππππ’π ππ π 2 π ππ πππππ‘ππ£π πππππππ‘π π‘βππ πππ₯πππ’π So, π 2 π πππ πππ ππ‘ππ£π πππππππ‘π(ππππππ’π) ππ π π₯π₯ > 0 πππ π π₯π₯ π π₯π¦ π π₯π¦ π π¦π¦ >0 π 2 π πππ πππππ‘ππ£π πππππππ‘π πππ₯πππ’π ππ π π₯π₯ < 0 πππ π π₯π₯ π π₯π¦ π π₯π¦ π π¦π¦ >0
8
Objective Functions with 3 choice Variables
π§=π π₯ 1 , π₯ 2 , π₯ 3 ππ§= π 1 π π₯ 1 + π 2 π π₯ 2 + π 3 π π₯ 3 First order condition: ππ§=0, π.π. π 1 = π 2 = π 3 =0 Second order condition : π 2 π π 2 π= π π π₯ 1 π 1 π π₯ 1 + π 2 π π₯ 2 + π 3 π π₯ 3 π π₯ 1 + π π π₯ 2 π 1 π π₯ 1 + π 2 π π₯ 2 + π 3 π π₯ π 1 π π₯ 1 + π 2 π π₯ 2 + π 3 π π₯ 3 π π₯ 2 + π π π₯ 3 π 1 π π₯ 1 + π 2 π π₯ 2 + π 3 π π₯ 3 π π₯ 3 = π 11 π π₯ π 12 π π₯ 1 π π₯ 2 + π 13 π π₯ 1 π π₯ 3 + π 21 π π₯ 2 π π₯ 1 + π 22 π π₯ π 23 π π₯ 2 π₯ 3 + π 31 π π₯ 3 π π₯ 1 + π 32 π π₯ 3 π π₯ 2 + π 33 π π₯ 3 2
9
Hessian Matrix for 3 choice Variables
π»= π 11 π 12 π 13 π 21 π 22 π 23 π 31 π 32 π 33 π€βπππ β£ π» 1 β£= π 11 , β£ π» 2 β£= π 11 π 12 π 21 π πππ β£ π» 3 β£= π 11 π 12 π 13 π 21 π 22 π 23 π 31 π 32 π 33 Optimization Conditions First Order Necessary condition: π π = π π = π π =π Second order Conditions πππ π β π‘π ππ πππ₯πππ’π β£ π» 1 β£<0, β£ π» 2 β£>0, β£ π» 3 β£<0 ; π 2 π(ππ·) πππ π β π‘π ππ ππππππ’π β£ π» 1 β£>0, β£ π» 2 β£>0, β£ π» 3 β£>0 ; π 2 π(ππ·)
10
π΅ Variable Case First order Necessary conditions
Optimization Conditions First order Necessary conditions π 1 = π 2 = π 3 = π 4 =β¦β¦β¦β¦β¦ π π =0 Second order Conditions πππ π β π‘π ππ πππ₯πππ’π β£ π» 1 β£<0, β£ π» 2 β£>0, β£ π» 3 β£< 0,β£ π» 4 β£>0β¦ β1 π β£ π» π β£> ; π 2 π(ππ·) πππ π β π‘π ππ ππππππ’π β£ π» 1 β£>0, β£ π» 2 β£>0, β£ π» 3 β£> 0 β¦.β£ π» π β£>0; π 2 π(ππ·)
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.