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Is my modeling approach correct?

  1. Sep 30, 2013 #1
    Ecologists use the following model to represent the growth rate process of two competing species, x and y:



    x= number of blue whales (measured in 100,000's)
    y=number of fin whales (measured in 100,00's)
    t=time measured in centuries
    r=revenue obtained from harvesting (measured in 1,000,000 of dollars per year)

    A blue whale carcass is worth $12,000 and a fin whale carcass is worth $6,000. Assuming that controlled harvesting can be used to maintain x and y at any desired level, what population levels will produce the maximum revenue? (Once population reaches the desired levels, the population levels will be kept constant by harvesting at a rate equal to the growth rate).

    The equation without harvesting are

    I am a little unsure about my equation for revenue- more specifically, I don't know how to incorporate time.

    I have r= 12,000x + 6,000y, but that doesn't seem right because I neglected t. So I thought maybe this is more accurate:

    r=12,000 (x/100t) + 6,000 (y/100t) (since t is measured in centuries)

    Is this correct? Or am I completely missing the point?
  2. jcsd
  3. Sep 30, 2013 #2

    Simon Bridge

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    If the population is maintained - why would the harvest rate depend on time?
  4. Sep 30, 2013 #3


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    The variables 'x' and 'y' represent the total population of blue whales and fin whales, respectively.

    You cannot use 'x' and 'y' in your revenue equations without some modification, since you are not harvesting the entire whale population at any given time.

    What you are trying to do is manage the two whale populations so that, with harvesting, the number of whales in each population will not grow beyond certain figures, call them 'xp' and 'yp'.

    So, in a given year, if the population of the blue whales, x(t), exceeded the desired population, xp, then the excess would be harvested, yielding a revenue of [x(t) - xp]*12000/1000000 millions of dollars. The blue whale population would be reset to x(t+1/100) = xp for the following year. A similar analysis can be done for the fin whale population.

    Now, the trick is to select the values of xp and yp which will maximize the amount of revenue obtained from the harvests. According to the OP, the yearly rate of harvest should be equal to the growth rate of the population. For the blue whales, this harvest rate would be [5xp(1-xp/1.5)-xp*yp/10]/100 blue whales per year.
  5. Sep 30, 2013 #4


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    Wouldn't it be rather more simple and easier to get the big picture to split up the actual growth rate into the expression for the natural growth rate and the harvesting rate?
    That is, writing for dx/dt:
    Obviously, you can gain an equivalent formulation in your terms, but why do you want the harvesting rate to appear as a variable in the denominator in the expression in the parenthesis.

    I believe my approach will make the algebra simpler.
  6. Sep 30, 2013 #5


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    Agreeing with arildno. These are equations close to though not quite the same as classical Volterra-Lotka equation. They probably predict, like those, cycles around a fixed point. However the problem essentially is about the fixed point. Or rather the new fixed point when the animals are harvested.

    Firstly how do you find the fixed point? (There is a trivial one at 0,0 for the undisturbed system.)

    You don't need to find it for the undisturbed system, though interesting, only for the system where there is harvesting.

    So you have to add to your equations a term for the effect of harvesting.

    Arildno suggests one and it certainly simplifies the algebra.

    However you have to decide what is your model for the harvesting! Arildno's model is like when the animals are hard to catch and the yield is proportional to the effort and you catch all the animals you can so there is a term proportional to x. You'd have to make more complicated assumptions for 2 species. I'd think a better model is when you can just decide how many animals you catch. In that case the last term in each equation is a constant (two different).

    So you'd get the equilibrium populations of x and y as function of the two harvesting rates. I guess there is always only one relevant solution but that should emerge from the math. Then you'll have to determine what among the possible populations is the best $ yield.

    Interesting question, keep us posted when you have progressed.
  7. Sep 30, 2013 #6


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    Modelling harvesting as a constant makes the actual harvest insensitive to population fluctuations.
    Rather than the "hard to catch" idea, a term rx is the simplest way to model a case where you instantaneously adjust harvesting to balance the natural growth.
  8. Sep 30, 2013 #7
    Thank you. So would I need to define two new variables xp and yp?
  9. Sep 30, 2013 #8


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    I was going to argue about the considerations that introduces, but I have re-read the question and it says "Once population reaches the desired levels*, the population levels will be kept constant by harvesting at a rate equal to the growth rate". That is, the added term in each equation is a new constant (different for each) multiplied by dx/dt or dy/dt. The result is that the each RHS is the same for with as without harvesting except it is multipled by a constant when there is harvesting, so when you have solved for without harvesting you easily have the solutions for with.

    A slightly impractical model IMO but we are talking about principles. :biggrin:

    (*? However it does not do that by itself, so the question can only mean it is brought there by some unspecified means that is outside the question.)

    see ^^
    Last edited: Sep 30, 2013
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