Question About Predator-Prey Relations
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Question About Predator-Prey Relations
During tonight's D&D session a rather interesting problem came up. For various reasons, my party is going to want to farm off a large population of goblins nearby, more or less for bounties and the like. It was asked just how many gobbos we could kill without hunting them to extinction. The Lotka-Volterra equations describe the cyclic nature of boom-bust in predators and prey, but this assumes the population of the predators is dependent on the population of the prey. How would these equations be altered to fit a predator population which is nearly constant and a population growth which is 0? Simply set dx/dt to be 0, and y to be constant, yielding 0 = ax - byx for the prey's population, where x is the prey population, a is the rate of growth in the population, and b is the rate of predation? This seems too obvious to be entirely correct.
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Ghetto Edit: In the equation, dx/dt is the growth of the prey population over time, a and b are the rates of population growth and predation, respectively, and x and y are the populations of prey and predator, respectively. I apologize if this caused confusion.
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- Alyrium Denryle
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You would not alter these equations, you would need to derive new ones. Or rather, use an exponential growth formula in which you periodically remove individuals from the population.
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- Kuroneko
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If the predator population is kept at a constant P, and if the prey resources are unlimited, then the prey population is subject to dp/dt = ap - bPp = p(a-bp), where a is the growth rate sans predation and b controls the probability of encounter with a predator. This is exactly the Lotka-Volterra condition with the predator equation removed, and the solutions have the form p(t) = p(0) exp[(a-bP)t], i.e., exponential growth with overall growth rate a-bP. If this parameter zero, there is no net growth. Yes, it really is that simple to keep the population constant, or it would be if this solution wasn't unstable.
On the other hand, if the prey is subject to a carrying capacity K, then we have a logistic form dp/dt = ap(1-p/K) - bPp, again with a the growth rate in the absense of predation. This experiences zero net growth if p = K[1-(b/a)P]. Note that at the level of predation b required to keep the population constant now depends on the current population, unlike the previous case.
On the other hand, if the prey is subject to a carrying capacity K, then we have a logistic form dp/dt = ap(1-p/K) - bPp, again with a the growth rate in the absense of predation. This experiences zero net growth if p = K[1-(b/a)P]. Note that at the level of predation b required to keep the population constant now depends on the current population, unlike the previous case.
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- Alyrium Denryle
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This will depend also on whether they are K or R selected. IE, whether the population reaches its carrying capacity and levels out, or whether or it oscillates around its carrying capacity.Kuroneko wrote:If the predator population is kept at a constant P, and if the prey resources are unlimited, then the prey population is subject to dp/dt = ap - bPp = p(a-bp), where a is the growth rate sans predation and b controls the probability of encounter with a predator. This is exactly the Lotka-Volterra condition with the predator equation removed, and the solutions have the form p(t) = p(0) exp[(a-bP)t], i.e., exponential growth with overall growth rate a-bP. If this parameter zero, there is no net growth. Yes, it really is that simple to keep the population constant, or it would be if this solution wasn't unstable.
On the other hand, if the prey is subject to a carrying capacity K, then we have a logistic form dp/dt = ap(1-p/K) - bPp, again with a the growth rate in the absense of predation. This experiences zero net growth if p = K[1-(b/a)P]. Note that at the level of predation b required to keep the population constant now depends on the current population, unlike the previous case.
How that works out mathematically I dont know, been a while since I have taken basic ecology (4 years...present work is in behavioral ecology and genetics where this stuff does not come into play)
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- Zixinus
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Infinite or till the DM gets bored of you lot farming goblins. This is fantasy after all. Also, goblins being somewhat intelligent, they may not be so simply subject to ecological rules as are animals.During tonight's D&D session a rather interesting problem came up. For various reasons, my party is going to want to farm off a large population of goblins nearby, more or less for bounties and the like. It was asked just how many gobbos we could kill without hunting them to extinction.
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Assuming we never reach the carrying capacity, and the growth stays exponential, all that must be done is make sure that the rate of death from predation is equivalent to the rate of growth, correct? a = bP will lead to zero growth?Kuroneko wrote:If the predator population is kept at a constant P, and if the prey resources are unlimited, then the prey population is subject to dp/dt = ap - bPp = p(a-bp), where a is the growth rate sans predation and b controls the probability of encounter with a predator. This is exactly the Lotka-Volterra condition with the predator equation removed, and the solutions have the form p(t) = p(0) exp[(a-bP)t], i.e., exponential growth with overall growth rate a-bP. If this parameter zero, there is no net growth. Yes, it really is that simple to keep the population constant, or it would be if this solution wasn't unstable.
On the other hand, if the prey is subject to a carrying capacity K, then we have a logistic form dp/dt = ap(1-p/K) - bPp, again with a the growth rate in the absense of predation. This experiences zero net growth if p = K[1-(b/a)P]. Note that at the level of predation b required to keep the population constant now depends on the current population, unlike the previous case.
For the second, we will obtain zero growth if the current population p = K*[1-(P * b/a)]? Or did I mix up the equation?
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- andrewgpaul
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Not just leads to zero growth; if a = bP, then the population of prey will remain constant forever. But the solution is unstable; any perturbation in a or b will cause exponential growth or decay.darthbob88 wrote:Assuming we never reach the carrying capacity, and the growth stays exponential, all that must be done is make sure that the rate of death from predation is equivalent to the rate of growth, correct? a = bP will lead to zero growth?
This is essentially what he said. You can arrive at this by setting the derivative dp/dt = 0 in the logistic equation given in Kuroneko's post and solving for p.For the second, we will obtain zero growth if the current population p = K*[1-(P * b/a)]? Or did I mix up the equation?
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Re: Question About Predator-Prey Relations
It assumes the population of the predators is dependent on the population of the prey because it assumes the predators will eat the prey, which makes sense because the population of anything will be limited by its food supply. You're not culling goblins for food, you're culling them for bounty. The equation is completely irrelevant to your scenario.darthbob88 wrote:During tonight's D&D session a rather interesting problem came up. For various reasons, my party is going to want to farm off a large population of goblins nearby, more or less for bounties and the like. It was asked just how many gobbos we could kill without hunting them to extinction. The Lotka-Volterra equations describe the cyclic nature of boom-bust in predators and prey, but this assumes the population of the predators is dependent on the population of the prey. How would these equations be altered to fit a predator population which is nearly constant and a population growth which is 0? Simply set dx/dt to be 0, and y to be constant, yielding 0 = ax - byx for the prey's population, where x is the prey population, a is the rate of growth in the population, and b is the rate of predation? This seems too obvious to be entirely correct.
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- Zixinus
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Actually, they are culling for EXP. Goblins usually drop shit.You're not culling goblins for food, you're culling them for bounty. The equation is completely irrelevant to your scenario.
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- Kuroneko
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You'd probably get oscillatory solutions in the full predator-prey equations modified to account for carrying capacity, but with the equation given above, there prey population cannot cross from below carrying capacity to above carrying capacity. Well, assuming the parameters are positive, anyway (b<0 means predators spawn more prey).Alyrium Denryle wrote:This will depend also on whether they are K or R selected. IE, whether the population reaches its carrying capacity and levels out, or whether or it oscillates around its carrying capacity.
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The XP would be shit too. The OP actually stated that they were farming them "for bounties and the like", which I took to mean that the local 'civilized' society is offering a bounty for each dead goblin.Zixinus wrote:Actually, they are culling for EXP. Goblins usually drop shit.You're not culling goblins for food, you're culling them for bounty. The equation is completely irrelevant to your scenario.
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