Homework 2 Solutions - University of California, Los Angeles LIFESCIENC 30B
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LIFESCIENC 30B
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Homework 2 Solutions
Exercise 4.2.FE 1 Some people have difficulty maintaining a stable weight. Instead, they
gain a lot of weight, go on a diet, lose the weight, but then gain it back. This pattern is
sometimes referred to as yo-yo dieting.
a) What kind of feedback loop is involved in this sit...
Exercise 4.2.FE 1 Some people have difficulty maintaining a stable weight. Instead, they
gain a lot of weight, go on a diet, lose the weight, but then gain it back. This pattern is
sometimes referred to as yo-yo dieting.
a) What kind of feedback loop is involved in this situation?
b) Use your understanding of feedback loops and oscillations to suggest what might help
such a person to stabilize their weight.
Part a:
Negative Feedback Loop (since a gain in weight will ultimately leads to the decrease in weight, and
in reverse)
Part b:
The oscillations are caused by:
1. High sensitivity (steep reaction by the person): too much weight → drastic diet → too low
weight → eat a lot → too much weight → repeat
2. Time delay by the body because it takes time for body weight to change
It is not as feasible to control the time delay since it is a physiological process. Thus, the person
can only stabilize their weight by not changing their diet too drastically. Just adjust a little. Let
the weight stabilize around a new value. If that is still not the desired weight, then make additional
small change to the diet.
Exercise 4.2.FE 2 While traveling, you find yourself in a hotel room in which using the
thermostat leads to large oscillations in the room’s temperature. The thermostat responds to
the room’s air temperature by turning on an air conditioner on the other side of the room if
the temperature near the thermostat gets too warm. Similarly, when the temperature near the
thermostat gets cold, the air conditioner switches off. What could the builder of the hotel have
done to prevent the oscillations you are experiencing?
Oscillations are caused by:
1. Steep reaction (high sensitivity)
2. Time delay
In this case, the reaction is steep (on/off) but it is not controllable.
Thus, we can only control the time delay.
To reduce drastic oscillations, we need to reduce time delay. In this case, it can be argued that the
time delay is at maximum because the distance between the air conditioner and the thermostat is
the greatest. Thus, the distance the air has to travel is also the greatest.
To reduce time delay, they can put the thermostat closer to the AC (or AC closer to the thermostat,
either way works).
1
, LS 30B Homework 2 Solutions
Exercise 4.2.FE 4 Meerkats are highly social small carnivores that live in southern Africa.
They rely on each other to raise their young. Use the following assumptions to model the
number of adult meerkats, M, in a population. You can invent parameters as necessary.
• The per capita rate at which meerkats give birth to babies who survive to adulthood
is a steep sigmoid function of the adult population, with higher reproductive success at
higher populations.
• Meerkats die of natural causes at a constant per capita rate d.
• Meerkats are preyed upon by eagles and jackals. These predators have many other prey,
so their population does not depend on the meerkat population.
• The rate at which jackals prey on meerkats is a nonsigmoid saturating function of the
meerkat population.
• The rate at which eagles prey on meerkats is a sigmoid function of the meerkat population.
The sigmoid is not very steep.
Bullet #1:
For this one, it is up to interpretation, so this answer is acceptable:
M0 Mn 0 Mn
=k· → M = k · ·M
M (a1 )n + M n (a1 )n + M n
However, one can argue that the number of adults now is dependent on the number of babies born
τ years back. And we know that the number of babies born is dependent on the number of adults
AT THAT TIME, not at this moment. As a result,
M0 [M (t − τ)]n 0 [M (t − τ)]n
=k· → M = k · · M (t − τ)
M (t − τ) (a1 )n + [M (t − τ)]n (a1 )n + [M (t − τ)]n
Bullet #2:
M0
= −d → M 0 = −d · M
M
Bullet #4:
c1 · M
M0 = − ·J
a2 + M
Bullet #5:
c2 · M s
M0 = − ·E (s is small)
(a3 )s + M s
2
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