sol2_1 - ORIE 4350 (Shmoys 1:25): Problem Set #2 1. 1....

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ORIE 4350 (Shmoys 1:25): Problem Set #2 1. 1. (Binmore, Section 2.12, Number 19) Notice that ( s,t 0 ) is in the column corresponding to t 0 and the row corresponding to s ; ( s 0 ,t ) is in the column corresponding to t and the row corresponding to s 0 . By the definition of saddle point , [ 10pt ] ( s,t ) 1 ( s 0 ,t ) , ( s,t ) 1 ( s,t 0 ) ( s 0 ,t 0 ) 1 ( s,t 0 ) , ( s 0 ,t 0 ) 1 ( s 0 ,t ) which results in ( s,t ) 1 ( s 0 ,t ) 1 ( s 0 ,t 0 ) 1 ( s,t 0 ) 1 ( s,t ) Hence it has to be the case that [ 5pt ] ( s,t ) 1 ( s 0 ,t ) 1 ( s 0 ,t 0 ) 1 ( s,t 0 ) 1 ( s,t ) To show that( s,t 0 ) is also a saddle point, consider any outcome (ˆ s,t 0 ) in the column corresponding to t 0 and any outcome ( s, ˆ t ) in the row corresponding to s . By the fact that ( s,t ) and ( s 0 ,t 0 ) are saddle points, [ 5pt ] ( s,t 0 ) ( s 0 ,t 0 ) s,t 0 ) ( s,t 0 ) ( s,t ) ( s, ˆ t ) which says exactly that ( s,t 0 ) is a saddle point. [ 5pt ] The Same argument applies to ( s 0 ,t ). 2.
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This note was uploaded on 02/06/2011 for the course ORIE 4350 at Cornell.

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sol2_1 - ORIE 4350 (Shmoys 1:25): Problem Set #2 1. 1....

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