- fixed some typos in commentes

This commit is contained in:
Kaspar Fischer 2006-02-20 15:26:07 +00:00
parent 6fae29186c
commit 7a42e51754
1 changed files with 29 additions and 30 deletions

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@ -958,61 +958,60 @@ ratio_test_2( Tag_false)
}
}
// Idea here: At this point, the goal is to increase \mu_j until
// either we become optimal (\mu_j=0), or one of the variables in
// x^*_\hat{B} drops down to zero.
// Idea here: At this point, the goal is to increase \mu_j until either we
// become optimal (\mu_j=0), or one of the variables in x^*_\hat{B} drops
// down to zero.
//
// Technically, we do this as follows here. Eq. (2.11) in Sven's
// thesis holds, and by multlying it by $M_\hat{B}^{-1}$ we obtain
// an equation for \lambda and x^*_\hat{B}. The interesting
// equation (the one for x^*_\hat{B}) looks more or less as
// follows:
// Technically, we do this as follows here. Eq. (2.11) in Sven's thesis
// holds, and by multlying it by $M_\hat{B}^{-1}$ we obtain an equation
// for \lambda and x^*_\hat{B}. The interesting equation (the one for
// x^*_\hat{B}) looks more or less as follows:
//
// x(mu_j) = x(0) + mu_j q_it (1)
//
// where q_it is the vector from (2.12). In paritcular, for
// mu_j=mu_j(t_1) (i.e., if we plug the value of mu_j at the
// beginning of this ratio step 2 into (1)) we have
// mu_j=mu_j(t_1) (i.e., if we plug the value of mu_j at the beginning of
// this ratio step 2 into (1)) we have
//
// x(mu_j(t_1)) = x(0) + mu_j(t_1) q_it (2)
//
// where x(mu_j(t_1)) is the current solution of the solver at
// this point (i.e., at the beginning of ratio step 2).
// where x(mu_j(t_1)) is the current solution of the solver at this point
// (i.e., at the beginning of ratio step 2).
//
// By subtracting (2) from (1) we can thus eliminate the "unkown"
// x(0) (which is cheaper than computing it!):
// By subtracting (2) from (1) we can thus eliminate the "unkown" x(0)
// (which is cheaper than computing it):
//
// x(mu_j) = x(mu_j(t_1)) + (mu_j-mu_j(t_1)) q_it
// ----------------
// := delta
//
// In order to compute for each variable x_k in \hat{B} the value
// of mu_j for which x_k(mu_j) = 0, we thus evaluate
// In order to compute for each variable x_k in \hat{B} the value of mu_j
// for which x_k(mu_j) = 0, we thus evaluate
//
// x(mu_j(t_1))
// delta_k:= - ------------
// q_it
// x(mu_j(t_1))_k
// delta_k:= - --------------
// q_it_k
//
// The first variable in \hat{B} that hits zero "in the future" is
// then the one whose delta_k equals
// The first variable in \hat{B} that hits zero "in the future" is then
// the one whose delta_k equals
//
// delta_min:= min {delta_k | k in \hat{B} and (q_it)_k < 0 }
//
// Below we are thus going to compute this minimum. Once we have
// delta_min, we need to check whether we get optimal BEFORE a
// variable drwops to zero. As delta = mu_j - mu_j(t_1), the
// latter is precisely the case if delta_min >= -mu_j(t_1).
// delta_min, we need to check whether we get optimal BEFORE a variable
// drops to zero. As delta = mu_j - mu_j(t_1), the latter is precisely
// the case if delta_min >= -mu_j(t_1).
//
// (Note: please forget the crap identitiy between (2.11) and
// (2.12); the notation is misleading.)
// (Note: please forget the crap identitiy between (2.11) and (2.12); the
// notation is misleading.)
// By definition delta_min >= 0, such that initializing
// fw: By definition delta_min >= 0, such that initializing
// delta_min with -mu_j(t_1) has the desired effect that a basic variable
// is leaving only if 0 <= delta_min < -mu_j(t_1).
//
// The only initialization of delta_min as fraction x_i/q_i that works is
// x_i=mu_j(t_1); q_i=-1; (see below).
//
// Since mu_j(t_1) has been computed in ratio test step 1 we can
// reuse it.