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A SIMPLE INTRODUCTION TO DYNAMIC PROGRAMMING

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PROBLEM SET-UP Problem is arrayed as a set of decisions made over time. System has a discrete state Each decision results in some reward or cost, and results in the system being moved to another state. Usually has a finite number of transitions. Transitions can be probabilistic, as can the rewards. Solution is a decision strategy that maximizes summed reward (minimizes cost)

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Notation N = finite planning horizon S n (x) = cost of optimally operating from n to N given state x at time n. d n *(x) is the optimal policy at stage n given state x at time n. x(d n ) is the state resulting from deciding d at stage n. c(d n ) is the cost of taking decision d n

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EXAMPLE You have moved to Singapore, and you need to operate a car for 3 yrs. You plan to sell the car when you leave Your QOL is not affected by your wheels Cost/resale of cars and operating costs are below 0123 sale price1000800450150 op cost200400600

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MAPPING TO THE NOTATION State: Age of you car Stage: Years you have been in S-pore Policy: Car’s age you buy at the END of the year

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COST EXAMPLE you have a 2yr old car you operate for the year ($600) you sell your 3 yr old car (-$150) you buy a new (to you) 1 yr old used car ($800) TOTAL: $1250

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finish 0123 start0400200 1950750400 214501250900600

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car age"cost" end of yr 3 0-1000 1-800 2-450 3-150

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CONTINUED COST EXAMPLE It’s beginning yr 2, and you possess a 2 yr old car You can.... operate the car (600 + S 3 (3yr old car)) operate the car, sell it, buy new car (600 -150 + 1000 + S 3 (new)) operate the car, sell it, buy 1yr old car (600 -150 + 800 + S 3 (1 yr old car))...

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123 "cost" end of yr 3 01200-200-600-1000 11550350-50-800 21700850450-450 3 -150 1450 1250 900

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123 "cost" end of yr 3 01200-200-600-1000 11550350-50-800 21700850450-450 3 -150

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BELLMAN’S EQUATION Sometimes its easy to get your name on something!

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EXEMPLAR A specialized tool is available during the period 9am,..., 3pm Each hour, a bid for the asset is made according to the table below The asset is busy for 3 hr. if the bid is accepted 9101112123 100150160501754010

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0 000 100 150 160 9 1112 1 10 2 end 00 50175 10 40 0

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0 000 100 150 160 91112 1 10 2 end 00 50175 10 40 0 10

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0 000 100 150 160 91112 1 10 2 end 00 50175 10 40 0 10

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0 000 100 150 160 91112 1 10 2 end 00 50175 10 40 0 10 175

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0 000 100 150 160 91112 1 10 2 end 00 50175 10 40 0 10 175

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0 000 100 150 160 91112 1 10 2 end 00 50175 10 40 0 10 175

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0 000 100 150 160 91112 1 10 2 end 00 50175 10 40 0 10 175 325

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0 000 100 150 160 91112 1 10 2 end 00 50175 10 40 0 10 175 325 Note 1: Once the diagram is drawn, the problem can be solved by a shortest(longest) path algorithm Note 2: Dynamic Programming = Shortest Path

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PROBABILISTIC TRANSITIONS 1.c(d) is a random variable 2.x(d) is random 3.the “trial” takes place after the decision

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EXEMPLAR (Probabilistic) An “asset” is available during the period 8pm, 9pm,..., 3am Each hour, a bid for the asset is made according to the discrete probability density below The asset is busy for 3 hr. if the bid is accepted

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MANY APPROACHES TO FORMULATION N = 4am S n (x) = profit of optimally operating from n to N given state x at time n. d n *(x) is the optimal policy at stage n given state x at time n (ACCEPT, REJECT) c(d n ) is the profit of taking decision d n x(d n ) is the proposed bid (3,6,9) or the number of hours left in the remaining engagement (1hr, 2hr)

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RECURSION time hours before asset is available again See DP Example.xls

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UNLOCKING THE JARGON x(d) can be governed by a Markov Chain a different P i,j matrix for each decision d Result is a Markov Decision Process

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