Ventana software support forum

I'd like to initialize each stock in an Nth order delay, assuming a known TOTAL of all stocks in the chain, a known average life L, and assuming an inflow that has been growing at rate R for since the beginning of the chain's growth. It seems to me there ought to be an analytical solution to this problem, but I haven 't figured it out yet. I've at least determined (I believe) that the ratio of any stock to the previous stock will be constant under such assumptions, but I haven't been able to determine what that ratio should be, although it must simply be a function of the rate R, the number of cohorts N, and the average age of the entire chain. Even if I can determine the ratio of the outflow to the inflow for a 1-stock model under such a situation (e.g., exponentially growing inflow at rate R with average stock lifetime L), I ought to be able to extend that to an Nth order chain to solve for the ratio of each stock to the total stock in the chain.

Any help on this?

If not, does anyone have any thoughts on initializing, with data, an Nth order delay? For example, vehicle stocks are well modeled using a 4th order delay (I've gotten excellent fit with estimated survival probabilities from Greenspan 1996). Thus, I'd like to initialize 4 stocks given an estimated total vehicle stock. But, even if I have an estimated stock of vehicles at 1-year intervals, it's not clear to me how this would translate into initializing a 4th order delay with this data.

Any advice from folks who may have come across something similar would be greatly appreciated.

Best,
Cory Welch

HI

There has been a discussion on the SD list mailing list this year I think on the average age of material in stock and the average time spent in the stock that are equal in the case of a simple exponential delay and the difference with a pipeline last in last out queuing process.

You could make a small model, and stard with N equal to one and eventually post it, and once you have found an answer add 1 to N.
I cannot help you very much, because in my profession (services) there are no stocks.
Regards.
JJ

I found a string of postings on a related topic (initializaton of aging chains) but it did not seem to be resolved. I've since calculated what I believe to be a fairly generic formula for initialization of each stock in an Nth order Erlang delay, assuming historic exponential growth at a constant fractional rate.

Let S_tot = Total Initial Stock
Let N = order of the delay (therefore, there will be N Stocks in the aging chain)
Let i be the index of each stock in the chain, starting at i=1 for the stock receiving the inflow
Let Si = the initial value of the ith stock in the chain
Let r = the historic fractional growth rate (inflow is assumed to have been growing at this rate for a sufficient time to ensure the proportions of each stock relative to the total have had time to equilibrate)
Let L = the average lifetime of the product (thus, the outflow of each Stock Si = Si/(L/N)

In such a case, define F as the ratio of each stock to its immediate upstream stock (upon reaching "exponential equilibrium" -- for lack of a better term, as I'm not sure "dynamic equilibrium" applies here). Then,

F = 1/(1+r * L/N) - {note this factor is independent of the stock index, L/N is the average life for each stock in the chain of N stocks}

Then, the initial value for each Stock Si will be:

Si = (S_tot * F^(i-1))/(SUM_over_all_i [F^(i-1)])

Where "SUM_over_all_i" means (F^0 + F^1 + F^2 + F^3 + .... F^N)

This formula should apply to both historic exponential growth and historic constant inflow (meaning the historic fractional growth rate is equal to zero, in which case the above formula would divide the total initial stock into N equal parts).

If anyone sees anything wrong with the above equation, please let me know. I've tested it and it seems to work, barring any transcription errors on my part.

That being said, I'm still thinking about how to translate initial stock data from annual vintages into the initial stock values of an Erlang delay of order N, as noted in the previous post. Any thoughts are welcome.

Best,

Cory Welch

Hi

Why not post a model with n stocks that you have tested, eventually with a little n?
Regards.
JJ