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.