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lecture12

# lecture12 - Yinyu Ye MS&E Stanford MS&E211 Lecture Note#12...

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Yinyu Ye, MS&E, Stanford MS&E211 Lecture Note #12 1 Online Auction and Incentive Compatibility Yinyu Ye Department of Management Science and Engineering Stanford University Stanford, CA 94305, U.S.A. http://www.stanford.edu/˜yyye

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Yinyu Ye, MS&E, Stanford MS&E211 Lecture Note #12 2 Issues Consider the combinatorial auction problem discussed in the course. How to be less conservative? In real applications, data/information is revealed sequentially , and one has to make decisions sequentially based on what is known – cannot wait for solving the offline problem. All of the participants/palyers fare best when they truthfully reveal private information asked for by the mechanism. This is called incentive compatible .
Yinyu Ye, MS&E, Stanford MS&E211 Lecture Note #12 3 Combinatorial Auction: offline linear program Recall the auction offline linear program: max π T x y s.t. A T x 1 · y 0 , x q , x 0 , where 1 is the vector of all ones. π T x : the revenue amount can be collected. y : the worst-case cost (amount need to pay to the winners if the worst-case state is realized) when optimized.

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Yinyu Ye, MS&E, Stanford MS&E211 Lecture Note #12 4 Combinatorial Auction: offline utility formulation To value the uncertain revenue s i between the worst-case cost and the actual cost when state i is realized: max π T x y + U ( s ) s.t. A T x 1 · y + s = 0 , x q , x 0 . where U ( · ) is a concave (risk aversion) and increasing value function for the possible slack revenues s = 1 · y A T x . For example, U ( s ) = i u ( s i ) .
Yinyu Ye, MS&E, Stanford MS&E211 Lecture Note #12 5 Possible Concave Value Functions Exponential u ( s i ) = b · (1 exp( s i /b )) , for some positive constant b. Logarithmic u ( s i ) = b · log( s i ) , for some positive constant b. Quadratic u ( s i ) = b · (1 (1 s i /b ) 2 ) 0 s i b b s i b for some positive constant b.

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Yinyu Ye, MS&E, Stanford MS&E211 Lecture Note #12 6 Combinatorial Auction: online utility formulation Given the previous ( t 1) order-fills ¯ x 1 , . . . , ¯ x t 1 on input { π j , a j , q j } t 1 j =1 until time t , the t th order-fill decision is to choose x t such that, max π t x t y + i u ( s i ) + t 1 j =1 π j ¯ x j s.t. a t x t 1 · y + s + t 1 j =1 a j ¯ x j = 0 , x t q t , x t 0 . ( π t , a t , q t ) : the newly arrived bidding data. y : the new worst-case cost. t 1 j =1 π j ¯ x j : collected revenue before the new arrival. t 1 j =1 a j ¯ x j : outstanding shares in each state before the new arrival.
Yinyu Ye, MS&E, Stanford MS&E211 Lecture Note #12 7 Combinatorial Auction: initial solution max y + i u ( s i ) s.t. 1 · y + s = 0 , or max y + i u ( y ) KKT condition: let y 0 be the optimizer: u ( y 0 ) = 1 m , so that y 0 can be viewed as the initial outstanding shares in each state! p 0 i := u ( y 0 ) = 1 m can be seen as the initial price for state i .

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Yinyu Ye, MS&E, Stanford MS&E211 Lecture Note #12 8 Combinatorial Auction: simplified online formulation max π t x t y + i u ( s i ) s.t.
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