Shuzhen
(Shuzhen Liu)
August 22, 2023, 7:25am
1
For N_1 , is it right in CVX?
Expression E
Expression N1
E = inv_pos(A) + inv_pos(B);
N1 = 0.5[ pow_pos(E,2) + 2A/(A0)^3 - 3/(A0)^2+ 2B/(B0)^3 - 3/(B0)^2 ] ;
Expression N_1 :
A_0,B_0 are the feasible point of A,B ,respectively.
1/(AB)<N_1 = 0.5[ (1/A+1/B)^2 + 2A/A_0^3-3/A_0^2 + 2B/B_0^3-3/B_0^2]
If A
and B
are both variables, that inequality is not convex, because
1/(A*B) - 0.5*(1/A+1/B)^2
is concave in A
and `B, and the other terms in the inequality are affine.
if the direction in the inequality were reversed, the constraint would be convex.
Shuzhen
(Shuzhen Liu)
August 22, 2023, 12:58pm
3
I think you misread my expression.
I just need to write the right of the inequality:
0.5[ (1/A+1/B)^2 + 2A/A_0^3-3/A_0^2+2B/B_0^3-3/B_0^2 ]
I interpreted that as an inequality, because that’s what you wrote. If you just wanted N1
, you shouldn’t have included the inequality.
Presuming A and B > 0,
(1/A+1/B)^2
can be entered as square_pos(inv_pos(A)+inv_pos(B))
or equivalently as pow_pos(inv_pos(A)+inv_pos(B),2)
Shuzhen
(Shuzhen Liu)
August 22, 2023, 1:12pm
5
Yes,you are right.
When the expression is like this : A+B-0.5*(1/A+1/B)^2 >T , it is convex .
A + B - 0.5( pow_pos( inv_pos(A) + inv_pos(B) , 2) ) > T;
And thank you for your reply very much.
Shuzhen
(Shuzhen Liu)
August 22, 2023, 1:15pm
6
Thank you very much for your reply again! I get it.