The real system in the first graph is the refilling of the water barrel. There was a water problem in our unit a few years back. There were specific times in a day that we that there were no water. So we could have water in those times, we would fill up barrels with water empty or not.
The real system of the second graph is Strip Club. In the strip club, one pays a girl per hour to table them. You can extend for another hour but it will still cost the same. If one has lots of money, they keep on extending until they reach they’re limit. The graph will show the accumulation of money throughout time which is linear since the money being added is constant.
The behavior of the graph is analogous to the boiling of water in a pot or a kettle. The initial temperature slowly increases, but as the surrounding of the water (pot/kettle) absorbs heat, it speeds up the heating process thus giving it exponential growth.
Refrigeration of water. The same goes with the refrigeration of water. The temperature drops exponentially until it reaches a temperature that would cool it. This is the inverse of the third behavior and has the same auxiliary and could be a glass or container that releases heat and hastens up the cooling process.
The last behavior could be compared to acclimatization of an object to its surroundings, for example a hot liquid is being cooled at room temperature. The temperature of water slowly cools down but won’t cool down below the room temperature because the room temperature acts as the limiting factor for the water to cool down below it.
III.
That the program STELLA is easier to use than DYNAMO. Equations in STELLA are much easier to input than in DYNAMO. The stock flow diagrams are being developed simultaneously with the equations unlike DYNAMO where you need to develop the stock flow first then the equations.
Each system can be shown in a system dynamic model. The system can be as simple as the boiling of water or as difficult as climate change.
I.
1.
Equation:
Level_1(t) = Level_1(t - dt) + (Rate_1 - Rate_2) * dt
INIT Level_1 = 10
INFLOWS:
Rate_1 = 9
OUTFLOWS:
Rate_2 = 0
2.
Equation:
Level_1(t) = Level_1(t - dt) + (Rate_1 - Rate_2) * dt
INIT Level_1 = 100
INFLOWS:
Rate_1 = 1
OUTFLOWS:
Rate_2 = 10
3.
Equation:
Level_1(t) = Level_1(t - dt) + (Rate_1 - Rate_2) * dt
INIT Level_1 = 10
INFLOWS:
Rate_1 = Level_1/Auxilliary1
OUTFLOWS:
Rate_2 = 0
Auxilliary1 = 4.225
4.
Equation:
Level_1(t) = Level_1(t - dt) + (Rate_1 - Rate_2) * dt
INIT Level_1 = 100
INFLOWS:
Rate_1 = 0
OUTFLOWS:
Rate_2 = Level_1/Auxilliary1
Auxilliary1 = 4.47
5.
Equation:
Level_1(t) = Level_1(t - dt) + (Inflow_1) * dt
INIT Level_1 = 10
INFLOWS:
Inflow_1 = Auxilliary_2*(Auxilliary_1-Level_1)
Auxilliary_1 = 100
Auxilliary_2 = 1
II.
III.