Comments on: Tunnel-vision http://www.tangentspace.net/cz/archives/2007/01/tunnel-vision/ somewhere near the beginning. Mon, 01 Dec 2008 20:21:19 +0000 http://wordpress.org/?v=2.5 By: AnonEcon http://www.tangentspace.net/cz/archives/2007/01/tunnel-vision/#comment-88657 AnonEcon Thu, 01 Feb 2007 18:31:44 +0000 http://www.tangentspace.net/cz/archives/2007/01/tunnel-vision#comment-88657 For 0.25: I calculated the stationary probabilities for the markov chain when the value of flowers is [tex]x[/tex], then reasoned that since for most of the periods the transition probabilities would be arbitrary close to the stationary values and therefore any strategy that gave strictly superior per-period payoffs for the stationary probabilities could not do worse than any other strategy over an infinite horizon. For my variant: there are now two kinds of states. The state S_on when the singer sings and manager gets 0.75 and from which we transition with equal probability to S_on and S_off(0). And the states S_off(K) where the singer is not singing and has already received K worth of flowers. If managers chooses to send I worth of flowers then we transition with probability [tex]\sqrt{K+I}[/tex] to state S_on and with the remaining probability to S_off(K+I). If the manager's income/expenditure in period t is [tex]P_t[/tex] then his goal is to maximize the expected value of [tex]\sum_{t=0}^\infty {\beta^t P_t}[/tex] For 0.25: I calculated the stationary probabilities for the markov chain when the value of flowers is x, then reasoned that since for most of the periods the transition probabilities would be arbitrary close to the stationary values and therefore any strategy that gave strictly superior per-period payoffs for the stationary probabilities could not do worse than any other strategy over an infinite horizon.

For my variant: there are now two kinds of states. The state S_on when the singer sings and manager gets 0.75 and from which we transition with equal probability to S_on and S_off(0). And the states S_off(K) where the singer is not singing and has already received K worth of flowers. If managers chooses to send I worth of flowers then we transition with probability \sqrt{K+I} to state S_on and with the remaining probability to S_off(K+I). If the manager’s income/expenditure in period t is P_t then his goal is to maximize the expected value of
\sum_{t=0}^\infty {\beta^t P_t}

]]> By: Alex http://www.tangentspace.net/cz/archives/2007/01/tunnel-vision/#comment-88451 Alex Thu, 01 Feb 2007 07:49:16 +0000 http://www.tangentspace.net/cz/archives/2007/01/tunnel-vision#comment-88451 Yep, I got 0.25. What method did you use? Hmm, what do you mean by <blockquote> I assumed that it is the accumulated value of flowers that counts and that the manager discounts future incomes/expenditures at some rate [tex]\beta[/tex]. </blockquote> Yep, I got 0.25. What method did you use?

Hmm, what do you mean by

I assumed that it is the accumulated value of flowers that counts and that the manager discounts future incomes/expenditures at some rate \beta.

]]> By: AnonEcon http://www.tangentspace.net/cz/archives/2007/01/tunnel-vision/#comment-88342 AnonEcon Thu, 01 Feb 2007 05:36:25 +0000 http://www.tangentspace.net/cz/archives/2007/01/tunnel-vision#comment-88342 Thanks! This problem is very interesting too. ([tex]x=0.25[/tex]?) I was trying out a variant to teach myself how to solve dynamic programming problems numerically. I assumed that it is the accumulated value of flowers that counts and that the manager discounts future incomes/expenditures at some rate [tex]\beta[/tex]. I got a surprising result: the optimum policy for the manager is to just send one shipment of flowers (whose value depends on [tex]\beta[/tex]) when she stops singing and send nothing after that till she starts singing again. Do you have any idea why this might be happening? I would have expected the manager to send a smaller bouquet to begin with and then some more flowers if that didn't work and so on. Thanks! This problem is very interesting too. (x=0.25?)

I was trying out a variant to teach myself how to solve dynamic programming problems numerically. I assumed that it is the accumulated value of flowers that counts and that the manager discounts future incomes/expenditures at some rate \beta.

I got a surprising result: the optimum policy for the manager is to just send one shipment of flowers (whose value depends on \beta) when she stops singing and send nothing after that till she starts singing again.

Do you have any idea why this might be happening? I would have expected the manager to send a smaller bouquet to begin with and then some more flowers if that didn’t work and so on.

]]> By: Alex http://www.tangentspace.net/cz/archives/2007/01/tunnel-vision/#comment-88169 Alex Wed, 31 Jan 2007 18:36:26 +0000 http://www.tangentspace.net/cz/archives/2007/01/tunnel-vision#comment-88169 Each night that she refuses to sing, the promoter buys [tex] 10,000 x [/tex] worth of flowers. By the way, nice proof! Each night that she refuses to sing, the promoter buys 10,000 x worth of flowers.

By the way, nice proof!

]]> By: AnonEcon http://www.tangentspace.net/cz/archives/2007/01/tunnel-vision/#comment-87556 AnonEcon Tue, 30 Jan 2007 10:01:28 +0000 http://www.tangentspace.net/cz/archives/2007/01/tunnel-vision#comment-87556 <blockquote>...by sending her flowers every day...</blockquote> Is [tex]x[/tex] the accumulated value of the flowers sent to her since she last stopped singing, or the value of flowers sent to her tonight?

…by sending her flowers every day…

Is x the accumulated value of the flowers sent to her since she last stopped singing, or the value of flowers sent to her tonight?

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