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16 Nov 2020: PB(x, y) = P(x, y for x, y B)) is irreducible, in thesense that for all x, y B, there exists n 0 such that P nB(x, y) > 0. ... PπB (τA > t) =k. i=1. aiγti,. where πB(x) = π(x)/π(B) for all x B.

16 Nov 2020: P(Y = j) c, for all j > 0 and P(Y = j) is decreasing in j,. ... Prove that for all ε < 1/2 we have. tmix(ε) D/2.

7 May 2020: 3) Choose the pivot row. Look within the pivot column j and findthe i B which minimises xi,0/Γi,j over all i B such thatΓi,j > 0. ... Remark 4. Suppose Γi,j 0 for all i B in step (3).

19 May 2020: of. transport to zero for all suppliers i. Mike Tehranchi IB Optimisation: Lecture 12. ... Theset B is the basis for the b.f.s x0. I We choose λi,0 and µi,0 such that λi,0 µj,0 = dij for all(i, j) B.

4 May 2020: P : minimise f (x) subject to g(x) b. where f and gi are convex for all i. ... Thisshows c>x c>x for all feasible x and b>λ b>λ for allfeasible λ, proving the optimality of x and λ.

24 Apr 2020: f (y) f (x) λ(x)>(y x). for all y X. Mike Tehranchi IB Optimisation: Lecture 1. ... Hence. f (y) f (x) λ(x)>(y x). for all x,y X , where λ(x) = Df (x).

30 Apr 2020: L(x,λ) L(x,λ). for all x X. In particular, we have. infxX, g(x)=b. ... 3. the functional constraint is g(x) b4. and gj is convex for all 1 j m.

6 May 2020: objective function c>x = c>B xB. I Repeat this procedure for all(nm. ... exists). We may be unlucky and have to search through all possible b.f.s.However, by choosing the sequence of b.f.s.

29 Apr 2020: L(x,λ) for all x X by assumption= f (x) for all feasible x. ... i=1. pi = 1, pi > 0 for all i. Mike Tehranchi IB Optimisation: Lecture 3.

19 Dec 2020: p.245, in display (3.253), all six occurrences of ‘f’ should replaced by ‘H’. ... p.621, in Theorem 8.2.5, it should read ‘for τ large enough depending only on an upper bound S > s,all 0 < s < S,B > 0 and every.’.