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Update: I will try to clarify the question: Let us say that the total harvest of the fish population at time t is $H_t$. Every harvest produce three types of fish: salmon ($f_1$), which is valuable and sold as a filee in salmon markets S; Pike ($f_2$), which is less valuable and sold to the fish soup markets P; and roach ($f_3$), which is solely used for fishsticks and sold at the markets R. The prices are endogenous and derived from the utility functions S,P,R, which are strictly concave and have a decreasing marginal utility.

If all the fish could be sold in any possible market, the constraints for the maximisation problem would be:

Total harvests at time t: $H_t=\sum_{i=1}^n v_{it}-\sum_{i=1}^n v_{i,t+1} $ , (1)

fish to markets at time t $H_t=\sum_{i=1}^n f_{it}$ , (2)

Where $v_{it} $ is the amount of fish $i$ in the lake at time $t$.

Now, if optimal, it should be possible to sell salmon to soup markets and salmon & pike to the fishstick markets: $S(f_1)$, $P(f_1, f_2)$ and $R(f_1,f_2,f_3)$. You cannot sell other fish but salmon in the filee markets, and roach cannot be sold at the soup markets.

Now the issue is: constraint (2) is pretty and you can easily see that the maximization problem is concave. However, it would allow any fish to enter any markets. If I use $S(f_1)$, $P(f_1, f_2)$ and $R(f_1,f_2,f_3)$ and (2), it would be optimal to sell all $f_1$ in $S$, all $f_1, f_2$ in $P$ and nothing in $R$, hence the double-counting. I struggling to come up with a constraint that would clearly satisfy the argument that fish that have been sold in some market cannot enter any other market. Thanks for your help!

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  • $\begingroup$ Thanks Giskard for your reply! I tried to clarify the question. $\endgroup$ – vonenzo Jun 11 at 8:42
  • $\begingroup$ Thank you for clarifying the question. Unfortunately I still do not understand it. Would the notation $f_{1,S},f_{1,P},f_{1,R}$ for salmon and $f_{2,P},f_{2,R}$ for pike sold at different markets not solve your double counting problem? If not, could you please explain what exactly you mean by double counting? $\endgroup$ – Giskard Jun 11 at 9:08
  • $\begingroup$ Perhaps it is more of a computation issue. As the shares of fish entering into markets are endogenously determined, I'm having trouble specifying the amounts of $f_1$ entering into each market. There is an exogenous share of each fish at every harvest, meaning that out of 10 fish harvested, lets say there are 30% salmon, 30% pike and 40% roach. Now, as the amounts entering to markets are endogenously determined, constraint (2) would lead to 3 fish entering in S markets, 6 fish entering in P markets and 1 fish entering in R markets. Thus, the salmon gets double-counted. $\endgroup$ – vonenzo Jun 11 at 9:46
  • $\begingroup$ Actually, if I add a constraint $H_{f_1t}=f_{1t,S},f_{1t,P},f_{1t,R}$ that might solve the issue. Thank you Giskard! $\endgroup$ – vonenzo Jun 11 at 9:49
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I think there is an approach problem in the constraint. (1) implies that:

\begin{align} v_{n,t} - v_{n,t+1}=H_t - (\sum_{i=1}^{n-1} v_{it}-\sum_{i=1}^{n-1} v_{i,t+1}) \end{align}

As the number of fish caught that are not the $ n $ species increases, the number of fish that belong to the $ n $ species decreases. This, i understand, may arise from a production decision: company 1 decides to produce more fish of species not $ n $ and reduces its production of type fish $ n $. But what exogenous force can create this phenomenon? it would make sense if they are species that compete for the same resources in a limited space, but this would affect future non-current populations, and it would be necessary to see if it is reasonable that the effect has a linear form, it is more likely that it affects according to a logistics function. The other option is for one species to be the predator of another, but in this case it appears that it is not.

I suggest either directly assuming that there is no inter-species resource struggle or choosing a different function that determines the resource struggle among fish. If you decide that there is no fight between species, you can separate the population dynamics of each species and you could choose a function like this:

\begin{align} P_{i,t} = P(v_{i,t} - v_{i,t+1},K) \\ \frac{\partial P_{t}}{\delta t}=\delta P_{ t}[1-\frac{P_{t,i}}{K}] - (v_{i,t} - v_{i,t+1}) \end{align}

The differential equation $ \delta P_ {t} [1- \frac {P_ {t}}{K}] $ is based on studies in biology on the dynamics of fish population, you can see a simple model in this link https://scholar.harvard.edu/stavins/publications/problem-commons-still-unsettled-after-100-years. $ P_{i, t} $ is the population of the species $ i $ in the period $ t $, and $ K $ is the "Carrying Capacity", this is the maximum level of fish population that can exist. You would only have to adapt that differential equation to the discrete case.

As for the second constraint, why make the quantity produced of one species of fish decrease with increasing another? This is something that arises endogenously as the producer's decision: it is optimal for company 1 to produce more of the type 1 fish and less of the type 2. This second restriction is not necessary, I think. You can use a production function like this:

\begin{align} f_{m}=f(v_{i}-v_{i,t+1},v_{-i}-v_{-i,t+1} ) \ \forall \ m \in M \end{align}

Each company endogenously chooses which fish to produce. It is clear that with this objective function, companies that cannot sell some type of fish will simply not sell that fish. In addition, you can analyze interesting phenomena such as the presence of economies of scope or simply choose a simpler function to find equilibrium in the long term. I think these observations would solve the problem you are having.

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  • $\begingroup$ Ha! i just realized that you solved the problem, but i hope that you may find my answer valuable anyway. $\endgroup$ – Samuel Cuevas Jun 28 at 7:39

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