User:PAR/Work3

Pricing Algorithm

Sales of the object is a Poisson process. The probability of a sale occurring in the time interval t to t+dt is

:

dP=\lambda(p)dt\,

where p is the price of the object and \lambda(p) is the average rate of sales at some fixed price p via the demand curve. A simple linear demand curve will be assumed:

:

\lambda(p)=2\lambda_0\left(1-\frac{p}{2p_0}\right)\,\,\,\,(0 \le p<2p_0)

:

\lambda(p)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(p \ge 2p_0)

where \lambda_0 is a constant equal to the rate of sales at optimum price p_0. The optimum price p_0 is the price at which the rate of income p\lambda is maximum. For prices above 2 p_0 the sales rate will be zero.

The pricing algorithm will be to have a linearly decreasing price, decreasing to zero at time \tau or until a sale is made, at which point the price jumps to 1+\alpha times the sale price, and again begins a linear decline. That is, if p_n=p_s(1+\alpha) where p_s is the sale price, then the price p as a function of time t after that sale is

:p(t)=p_n\left(1-\frac{t}{\tau}\right)\,\,\,\,\,\, for t\le\tau

:p(t)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, for t\ge\tau

\lambda(t) divides into two cases. When p_n is greater than 2p_0, then \lambda(t) remains zero until p_n=2p_0, at which point it begins to rise linearly. It does so until t=\tau, at which point it remains at 2\lambda_0

:\lambda(t)=0\, for 0\le t \le \tau(1-2p_0/p_n)

:\lambda(t)=2\lambda_0\left(1-\frac{p_n}{2p_0}\left(1-\frac{t}{\tau}\right)\right) for \tau(1-2p_0/p_n)\le t \le \tau

:\lambda(t)=2\lambda_0\, for t \ge \tau

When p_n is less than 2p_0, \lambda(t) rises linearly until t=\tau, at which point it remains at 2\lambda_0.

:\lambda(t)=2\lambda_0\left(1-\frac{p_n}{2p_0}\left(1-\frac{t}{\tau}\right)\right) for 0\le t \le \tau

:\lambda(t)=2\lambda_0\, for t \ge \tau

Approximate Equilibrium

Depending on the initial price, the price function will take a certain amount of time to equilibrate. (This does not mean it is constant, of course, only that its average behavior gives no clue as to the amount of time elapsed since time zero.)

An approximate equilibrium condition is that the average time between sales (T=1/\overline{\lambda}(t)) is such that the price after a sale decays to the price before the sale.

:

p_s=p_s(1+\alpha)(1-T/\tau)\,

:

\frac{1}{T}=?

These are two equations in two unknowns (p_s and t). Solving:

:

T=\tau\frac{\alpha}{1+\alpha}\,

:

\frac{p_s}{p_o}=?

Note the problems when 2\lambda_0T<1

Exact equilibrium

The probability that the price is p at time t+dt is the probability that the price was pe^{t/\tau} at time t and a sale was not made, plus the probability that the price was p/(1+\alpha) at time t and that a sale was made. Normalizing to unity p_0 and \lambda_0

:

P(p,t+dt)

=(1-\lambda(pe^{dt/\tau})dt)P(pe^{dt/\tau},t) + \lambda(p/(1+\alpha)) P(p/(1+\alpha),t)dt\,

:

=(1-\lambda(p)dt)P(p(1+dt/\tau),t) + \lambda(p/(1+\alpha)) P(p/(1+\alpha),t)dt\,

:

=(1-\lambda(p)dt)P(p+pdt/\tau,t) + \lambda(p/(1+\alpha)) P(p/(1+\alpha),t)dt\,

or

:

P(p,t)+\frac{\partial P}{\partial t}dt=\left(1-\lambda(p)dt\right)\left(P(p,t)+\frac{\partial P}{\partial p}\frac{p}{\tau}dt\right) + \lambda(p/(1+\alpha)) P(p/(1+\alpha),t)dt\,

or

:

\frac{\partial P}{\partial t}=\lambda(p)P(p,t)+\frac{p}{\tau}\frac{\partial P}{\partial p}+ \lambda\left(\frac{p}{1+\alpha}\right) P\left(\frac{p}{1+\alpha},t\right)\,

The sale price

The sale price is a random variable, but its not a Poisson process. The sale price probability is dependent on the previous sale price.

Given that the last sale price was p_i at time t_i the probability that the next sale will occur between time t and t+dt is

:

P(t)dt = e^{-\overline{\lambda}t}\lambda(t)dt

:

\lambda(t)=2\lambda_0\left(1-\frac{p_i(1+\alpha)e^{-t/\tau}}{2p_0}\right) \,\,\,\,\,\,\,\mathrm{for}\,\,p_i(1+\alpha)e^{-t/\tau}\le 2p_0\mathrm{\,\,zero\,\,otherwise}

\lambda(t) will be zero when

:

p_i(1+\alpha)e^{-t/\tau} \ge 2p_0

or, equivalently,

:

t \le \tau\ln\left(\frac{p_i(1+\alpha)}{2p_0}\right)

as long as that t>0. For p_i(1+\alpha)e^{-t/\tau}\le 2p_0 then, we have:

:

\overline{\lambda}=\frac{1}{t}\int_0^t \lambda(t')dt' =

2\lambda_0\left(1+\frac{\tau}{t}\,\frac{ p_i(1+\alpha)\left(e^{-t/\tau}-1\right)}{2p_0}\right)

and for p_i(1+\alpha)e^{-t/\tau} > 2p_0

:

\overline{\lambda}=\frac{1}{t}\int_{t_x}^t \lambda(t')dt' =

2\lambda_0\left(1-\frac{t_x}{t}+\frac{\tau}{t}\,\frac{ p_i(1+\alpha)\left(e^{-t/\tau}-e^{-t_x/\tau}\right)}{2p_0}\right)

and that sale price will be

:p_{i+1}(t)=p_i(1+\alpha)e^{-t/\tau}\,

The expected value of p_{i+1} is

:

\langle p_{i+1}\rangle = \int_0^\infty P(t)p_{i+1}(t)dt = \int_0^\infty p_i(1+\alpha)e^{-t/\tau}e^{-\overline{\lambda}t}\lambda(t)dt

Given a sale at [p,0], and given that there is a sale at time t, what is the probability distribution for that sale price? The sale is not necessarily the first sale after the original.

References

  • [http://www.springerlink.com/content/q870032g57u05715/]
  • [http://portal.acm.org/citation.cfm?id=1018409.1018759&coll=GUIDE&dl=GUIDE&type=series&idx=SERIES134&part=series&WantType=Proceedings&title=AGENTS&CFID=70591674&CFTOKEN=97332196]
  • [http://ecommerce.hostip.info/pages/865/Pricing-DYNAMIC-PRICING.html]
  • [http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?tp=&arnumber=1243848&isnumber=27871]
  • [http://www.vjolt.net/vol6/issue2/v6i2-a11-Weiss.html]
  • [http://www.managingchange.com/dynamic/overview.htm]
  • [http://nymag.com/nymetro/news/bizfinance/columns/bottomline/1778/]
  • [http://www.meiss.com/download/RM-Maglaras-Meissner.pdf]
  • [http://www.jstor.org/pss/2661464]
  • [http://cat.inist.fr/?aModele=afficheN&cpsidt=17941386 Dynamic pricing with real-time demand learning K.Y. Lin]