Ventana software support forum

Posted: **Fri Jul 12, 2002 7:37 am**

This is perhaps more semantic than literal, but I would tend to suggest
that considering a backlog a negative stock is more a convenience than
proper. A unit of backlog does not seem perfectly equivalent to a unit of
product. Suppose we define backlog is a positive stock of unfilled orders
(or a positive number of needed units to bring inventory to its desired
level). An inventory would be a positive stock of unsold product. I can
have both at the same time and receipt of a new product unit will raise
stock but doesnt reduce backlog until it is shipped.

My initial reaction after contemplating a variety of soft and hard models
is that negative stocks can only exist through definitions (usually to make
a model more convenient). Proper/formal formulation of a SD model should
technically use positive stocks. It seems that, as in the case of backlog,
little if any might be sacrificed by allowing stock to go negative,
but...forcing stocks to be positive may lead to deeper contemplation of the
assumptions underlying the model and lead to deeper insights.

Jay Forrest
From: Jay Forrest <jay@jayforrest.com>

Posted: **Fri Jul 12, 2002 6:40 pm**

Goeff,
Thanks for your suggestion.

I took yours and Bobs suggestion and it works nicely. The case I am
considering.is Similar to epidemic problem whre O1 is uncontrollable outflow
(e.g. recovery, death) and O2 is controllable (quarantine). So O1 has
higher precidence. The flow equations I ended up using are

O1 = MIN( S(n)/dt , f1(n))
O2 = MIN( Min( 0 , S(n)/dt - f2(n) ) , f2(n))
[ Hosts note the above equation for O2 is not actually quite right to
do what the author wants to do - the correct equation would be

O2(n) = MIN( MAX(0,S(n)/dt-f2(n)), f2(n))

I point this out because it is really easy to get mixed up on
MINs and MAXes in equations like this. The equation is the same as

O2(n) = MIN((S(n) - O1(n)*dt)/dt, ft(n))

The first term is the explicit Euler integration of S if only 1 flow
existed - a little easier to see the meaning, but it also means we
really are using O1 on the right.
]

I am sorry I am not familiar with DYNAMO language. Hope it is OK if I
continue using the difference eq. Besides, n helps keeping track of which
loop the values are taken from.

Also, I wouldnt put the rate O2 on the right hand side of O1 though, since
it is the value of O1(n+1) and not O(n) that is needed. So the infamous
rate to rate relation O1 ---> O2 is not really there.

chariya
From: "Chariya Peterson" <Chariya.Peterson@verizon.net>

Posted: **Sun Jul 14, 2002 9:54 am**

Niall,

It is fun to see others obsessed with systems!

Could you please clarify (by example?) your intent of nonlocality? Is
this something like one and the same system showing up in two different
places at the same time? As opposed to two analogous systems driven by
similar inputs.

I would also like to learn more about the idea of such a concept being
associated with physics only; how could something be associated one field
and not others?

Raymond T. Joseph, PE
rtjoseph@ev1.net
Aarden Control Engineering and Science

Posted: **Mon Jul 15, 2002 9:09 am**

geoff coyle wrote:

> Yes, I agree with this. A negative stock should not exist and, if necessary,
> a separate and explicit backlog should be modelled. Why? A backlog will
> probably trigger an entirely different set of policies and it is better to
> treat them explicitly.

I see that Jay Forrest made this comment, too. Id like to know
whether you see that as a firm rule or a useful heuristic. For
example,

I once had a bank account that had a line of credit attached. If I
wrote a check that made the balance go negative, it became a loan,
and I owed interest. (I forget whether they paid me interest for
positive balances.)

I think Id tend to model such an account as a stock that could assume
positive or negative values, but Im curious if youd split it into 2
stocks, both of which remained positive. I suspect we can think of
other such stocks that would seem to take on both polarities in the
real world.

IOW, is non-negativity of stocks a heuristic that ranks below ensuring
that stocks and flows represent real entities in the system that
create the problem to be studied?

That said, I agree that Id model a backlog separately as a positive
stock, if for no other reason than manufacturers tend to track backlog
as a separate entity.

> Incidentally, can anyone explain the origin of widgets?

I dont know if this is definitive, but its interesting: see
http://www.newscientist.com/lastword/an ... 66food.jsp

Widgets do sound worthy of more investigation and definitely on-topic
in a discussion on SD, given their apparent ties to the Beer Game.

(I suspect the OED would be a better reference, but I dont have a
copy.)

Thanks,

Bill
From: Bill Harris <bill_harris@facilitatedsystems.com>
--
Bill Harris 3217 102nd Place SE
Facilitated Systems Everett, WA 98208 USA
http://facilitatedsystems.com/

Posted: **Mon Jul 15, 2002 1:32 pm**

Geoff Coyle sparks interesting questions about widgets and dt.

First is the question of where the word "widget" comes from. According
to the American Heritage Dictionary, the word "widget" is "perhaps [an]
alteration of gadget". According to the same respected source the
origin of the word "gadget" is unknown. Just thought Id clear that up.

Second has to do with "dt". As a field were prejudiced against it,
saying things like "dt is artificial", "never use dt in an equation",
"dt is only a technical parameter". But this prejudice may have arisen
-- like many prejudices -- as a byproduct of the way we were taught as
opposed to anything substantive. Im referring to the fact that most
of us were taught calculus using the idea of limits.

Unfortunately for the unassuming dt, 19th century mathematicians used
the idea of limits to make the calculus rigorous. Using limits, you
view a net rate -- dStock/dt -- as a ratio taken to the limit as the
denominator approaches zero. As something thats going to a limit (but
never actually gets there) "dt" isnt really anything. And,
consequently, the poor little dt got a bad reputation as being
un-rigorous. Only those intrepid engineers and physicists continued to
use it heuristically -- thereby succeeding in solving their practical
problems and making obvious to mathematicians the superiority of
mathematicians.

But in the late 60s and 70s two developments in mathematics emerged
that each in its own way rehabilitated the infinitesimal dt. One was
non-standard analysis and the other was infinitesimal analysis.
Non-standard analysis assumes that around every real number there is an
infinite set of infinitesimal, hyper-real numbers. Infinitesimal
analysis assumes there are real, non-zero quantities, each so small that
its square is equal to zero. Non-standard analysis and infinitesimal
analysis each allows the derivation of the calculus using the
infinitesimal dt, rather than limits.

My point is just that dt can now be regarded as being just as rigorous
as any other real (or hyper-real) number -- zero, one, pi, etc. This
means that the Dynamo equations (and Euler integration in concept) are
correct **mathematically**. Of course, digital computers dont allow
infinitesimals, so we approximate dt as a small finite quantity -- much
as we approximate other quantities or relationships in our models, like
"the" net birth rate or the effect of fatigue on productivity.

In brief, "dt" (i.e. time-step) in our models actually does represent
something -- it represents an "instant" of time. The time step is not
exactly equal to the real infinitesimal dt, but its often close enough.

What do you think? Is it time to get over our general prejudice against
dt?

Jim Hines
jhines@Mit.edu

Posted: **Mon Jul 15, 2002 11:15 pm**

Stocks cant go negative? Dimensions have to be consistent.

I have a cylindrical water tank that is 10 meters tall. It is set on ground
that is 100 meters below sea level. When the tank is full, the water
elevation is -90 meters.

I have another cylindrical water tank that is 10 meters tall. It is set on
a hill such that the base of the tank is 100 meters above sea level. When
the tank is full, the water elevation is 110 meters. When the tank is
empty, the elevation is 100 meters.

What is the stock? It could be considered the volume of the water in the
tank, or the mass of the water in the tank, or the distance of the water
level above the base of the tank. In none of these cases was the stock
allowed to be a value with a meaningful negative value. But the actual
system measurement (variable) could have many different zero references (of
course not all at the same time!).

In any of these cases, the actual value of the stock is only meaningful over
a finite range. For example, for the tank on the hill, once it is full,
addition of water does not change the stock, the tank overflows any addition
beyond its rated capacity. What happens if an attempt is make to withdraw
stock from the tank if it is empty? If the attempt is to open a valve at
the base of the tank and let stock flow out onto the ground, the flow will
stop - there is no force to push the stock out, none the less, no stock to
push out. If the attempt is to drain the tank with a valve located below
the base of the tank, the tank empties and the stock goes to zero. But the
flow doesnt stop! The pipe connecting the tank to the discharge valve
begins to empty. Now the flow is produced not by the tank stock but the
pipe stock. Of course the model will only show this if the pipe stock has
been included in the model.

Now lets fill the tank, slowly. We add stock to the tank from a pipe above
the top of the tank so it free falls into the tank. Initially, very little
of the added water raise stock in the tank, it raises the stock in the pipe
first. But we poured the water into the tank. Does the tank have a
negative stock? When we look at it in detail we see that it does not have a
negative stock; its stock is bypassed (at first) to fill up the drain pipe.
But from an outside observer, the tank appears to have a negative stock.
Thus, to a loosely coupled observer, there may appear to be a negative
stock, but to a closely coupled observer, there are two stocks being
addressed.

Of course there is always the counter example to assure us that
non-negativity of stocks is not a law: We can look at the voltage on a
capacitor connected to an inductor. With an initial charge on the
capacitor, it will drain through the inductor but when the capacitor voltage
reaches zero, the inductor will not let its current fall to zero instantly -
current will continue to flow and force the capacitor voltage to become
negative. Any linear second order system can be driven through this bipolar
stock valuation.

Raymond T. Joseph, PE
rtjoseph@ev1.net
Aarden Control Engineering and Science

Bringing Control Science
Down to Earth

Ventana software support forum

Representing constraints

Representing constraints

Representing constraints

Representing constraints

Representing constraints

Representing constraints

Representing constraints