# How to calculate the maximum zero fuel weight

## How do you calculate the maximum ellipsoid in a given polyhedron?

I am faced with the problem of finding the ellipsoid (is a symmetric positive definitive matrix) with maximum volume within a convex set which is considered to be a set of linear inequalities. I understood how it is formalized as a convex optimization problem as stated in "Convex Optimization, Stephen Boyd and Lieven Vandenberghe, Cambridge University Press, 2004" [pdf version]. My approach would be to use interior point methods and introduce an accuracy parameter t> 0BBCC = {x | aTix≤bi, i = 1,…, m}

minB, d [logdetB − 1] s.t.:||Bai||2+aTid≤bi,i=1,…,m

t> 0and integrate the constraints into the target using a logarithmic barrier function as explained in Chapter 11 of the book above, and try to solve the resulting unconstrained problem \ min_ {B, d} \ quad \ underbrace {\ left [\ log \ det B ^ to minimize {-1} - \ frac {1} {t} \ sum_ {i = 1} ^ m \ log (b_i- || Ba_i || _2-a_i ^ Td) \ right]} _ {= f (B, d)}. minB, d [logdetB − 1−1t∑i = 1mlog (bi− || Bai || 2 − aTid)] = f (B, d).

So I would use partial derivatives of f: ∂f∂B = B − 1 + 1t∑i = 1m⎛⎝⎜BaiaTi || Bai || bi− || Bai || 2 − aTid⎞⎠⎟

is a matrix and ∂f∂d = 1t∑i = 1m (aibi− || Bai || 2 − aTid)

is a vector. And then starting from an initial (feasible) point (B0, d0) I would iteratively update the actual solution (B ^ k, d ^ k) (Bk, dk) according to the negative partial derivatives: B ^ {k + 1} = B ^ k - s_B \ frac {\ partial f (B ^ k, d ^ k)} {\ partial B} \ d ^ {k + 1} = d ^ k - s_d \ frac {\ partial f (B ^ k, d ^ k)} {\ partial d} - Whoqol-bref questionnaire in Hindi pdf
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