Exponential Decay
X Stock
O outflow
 TimeConstant
(sometimes called lifetime, decay time, or response time)
The above simple system flow diagram is the basic structure for exponential decay. It is
called exponential decay because if O is directly proportional to X (O=X/), then the
exact analytic solution of how the stock behaves over time is:
X  X 0 e t /
The time constant determines how fast the system approaches equilibrium which in the
above case is zero since there is no inflow. After a time =  the stock will drop by 63% ,
or to 37 % of its starting level. In the above system the time constant is 10 min. The time
to drop to half its starting value is 0.7 
t1/ 2  0.7
In this case since =10 min, t1/2=7 min. In the above graph notice that X drops from 100
to 50 after 7 min, to 25 after another 7 min, and to 12.5 after another 7 min. The this half
value time makes it easy to sketch the graph.
The above model structure of simple exponential decay is the most basic goal seeking
example, with a goal of X=0 in this case.
Goal seeking with inflow.
X Stock
F inflow
O outflow
 TimeConstant
(sometimes called lifetime, decay time, or
response time)
The above model is the same as before
except for a constant inflow F = 20
units/min. The stock X will stop changing
when F=O. That is when F=O the system
is in equilibrium. At equilibrium X=Xeq
and
F=O
F=Xeq/
Or
Xeq=F
Both  and t1/2 characterize a system’s response. Knowing the value of t1/2 makes is easy
to sketch a graph of the behavior of X. Knowing the value of  allows one to calculated
the outflow for any vaue of X (O=X/)and also calculated the equilibrium value of X,
Xeq Xeq=F. In this case since F=20 units/min and =10 min, Xeq=200 units.

The above graph shows this system’s approach to a new goal Xeq=200. It’s kind of
simple exponential decay inverted.
Notice that the gap between the value of X at
any time and the goal Xeq reduces by half
every 7 min (t1/2). That is at the beginning
the gap is 200 (X=0 and Xeq=200) after 7 min
X=100 and the gap close to 100, after
another 7 min X=150 and the gap closes to
50, and after another 7 min X=175 and the
gap closes to 25. This is goal seeking
behavior with the goal being Xeq.
The value of Xeq or the goal does not
depend on the initial value of X.
Stella Model (Cooling coffee cup)
The cooling rate depends on the difference
between the actual temperature and
ambient temperatures, and on the time
constant for the system.
The time constant, , characterizes the
cooling efficiency and would depend of the
size of the cup, insulation properties of the
cup and the wind speed.
CoffeeTemp(t) = CoffeeTemp(t - dt) + (CoolingRate) * dt
INIT CoffeeTemp = InitialTemp
OUTFLOWS:
CoolingRate = (CoffeeTempRoomTemp)/TimeConstant
InitialTemp = 40
RoomTemp = 20
TimeConstant = 20
After a long time the coffee will reach
ambient temperature. The time to cut the
gap between the coffee temperature and the
ambient temperature in half is called the
half value time T1/2 and is given by
T1/2= 0.7 * 

Note that the cooling flow icon was set to
biflow by clicking on the flow icon and
selecting the biflow option. A negative
cooling is a heating.
Set up the basic Stella model shown in the flow diagram below
Use the Stella model to explore the relation
Graph your results and also create a table.
T1/2= 0.7 * 
To Create a graph drag the graph pad
onto the workspace and then reposition to
where you want it. You’ll get a blank graph as shown below that can be enlarged by
using the resize workspace button
or dragging the graph’s bottom right hand corner.
Setup the interface page as shown on the next Page. Use the slider icon
to get slider
input for Initial Temp and AmbientTemp. To set the time duration of the run to 20
minutes use Run Specs under the Run menu.
Use the Stella model to explore the relation
Initial Temp (C)
80
60
40
30
0
T1/2= 0.7 * 
Time to halve gap, T1/2
Does the halving time depend on the initial temperature? _______________________
Explain what influences the halving time for exponential decay.
The instructions below and on the next page will help filling in this next
table easy.
Time constant, 
5
10
15
20
25
Time to halve T1/2
= 0.7 * 
% error
How much error is there between the halve time rule T1/2= 0.7 *  and your
experimentally determined values of T1/2? Calculate the % difference.
%error 
0.7 -T1/2
100
0.7
Use Sensi Specs from the Run menu to set up a sensitivity analysis.
Vary the time constant from 5 to 25 using 5 runs and define a graph and a table for this
sensitivity run. This graph (and table) will be page 2 of your graph (table) and page 1
will include a single graph (table column) representing your latest run. You can go
between pages by clicking in the lower left corner of you graph (table). Run your model
and print-out your graph showing all 5 runs on the one graph. You can use you table to
see what the halving time is for each run. To get a copy of your graph you can Save as
image under the file menu and then import this image into a Paint or a Word document.
Questions:
Q1: The temperature of a cooling coffe cup is shown in the figure below. What are:
The half-value time of this system? ______________________
The decay time constant of this system? ______________________
Q2: The cooling time constant for a coffee cup system is 10 minutes. The initial
temperature is 90 C and the ambient temperature is 10 C. Sketch the expected behavior
of the temperature over a 20 minute interval.
What is the half-value time for this system? ______________________
Questions:
Q1: The temperature of a cooling coffe cup is shown in the figure below. What are:
The half-value time of this system? ___3.5 minutes__(time to close the gap between
100 and 20 in half. i.e. time to drop to 60 C)_________________
The decay time constant of this system? ____about 5 minutes_____________
Q2: The cooling time constant for a coffee cup system is 10 minutes. The initial
temperature is 90 C and the ambient temperature is 10 C. Sketch the expected behavior
of the temperature over a 20 minute interval.
What is the half-value time for this system? __7 minutes____________________
time
0
7
14
21
T(C)
90
50
30
20