• Outlined in the paper "Using Markov Chains to Model Human Migration in a Network Equilibrium Framework", by J. Pan and A. Nagurney

• For the individual, each region is assumed to have its own utility function, which is a function of its population, and a cost of moving, which is a function of the number of people moving in the current iteration.

• The two regions have an initial population vector \(p = [p_1, p_2]\)

• The flow matrix, which we solve for, will be \(F = \left[ \begin{array}{cc} f_{11} & f_{12} \\ f_{21} & f_{22} \end{array} \right]\)

• This leaves us with the functions \(u_1 (p)\), \(u_2 (p)\), \(c_{12} (f)\) and \(c_{21} (f)\).

• People will move from locations of lower utility to locations of higher utility.

• At each iteration, we can solve the following for f:

\[ u_i (p) + c_{ij} (f) = u_j (p) \]

flowcalcpan <- function(p){
  if(u2pan(p) > u1pan(p)){
    f1 <- (p[1] - p[2] + 6)/4
    f <- c(f1,0)
  }else if(u1pan(p) > u2pan(p)){
    f2 <- (p[2] - p[1] - 6)/4
    f <- c(0,f2)
  }else{f <- c(0,0)}
  
  return(f)
}

p0 <- c(4,2)

u0 <- c(u1pan(p0), u2pan(p0))

pdatapan <- data.frame(p1 = 4, p2 = 2, u1 = u0[1], u2 = u0[2], 
                       totu = p0 %*% u0 )

iter <- 1:5

for(i in iter){
  p <- c(tail(pdatapan$p1,1),tail(pdatapan$p2,1))
  f <- flowcalcpan(p)
  p[1] <- p[1]-f[1]+f[2]
  p[2] <- p[2]+f[1]-f[2]
  u <- c(u1pan(p),u2pan(p))
  totu <- p %*% u

  pdatapan <- rbind(pdatapan, c(p[1], p[2], u, totu))
}

• In the last slide, you can see total social utility in the last column
• This is equal to the individual utility at each location times the population of each location
• or, \(p \times u^T\), with p as the population vector and u as the utility vector, whose lengths are the number of locations.
• In this model, it is not guarenteed that people acting in their own self interest will lead to a socially optimal distribution.

• This is solveable by linear programming

• \(u_{TOT} = p \times u\)

• \(u_{TOT} = p_1 (8-p_1) + p_2 (14-p_2)\)

• Subject to: \(p_1 + p_2 = p_{TOT} = 6\)

• Social Utility is Maximized at \(p_1 = 1.5\), and \(p_2 = 4.5\)

• How can a society achieve maximum social utility?

• Lets model the decision that taxes the residents in the higher utility region and redistributes the gains to the lower utility region

• This is somewhat like the prisoner's dilemma.

• If 5t < 2, moving is a strictly dominant strategy for player 1, so they will definitely do that. Given this, player 2 gets the same payoff whether or not they tax, so we'll assume they have a 50% chance of taxing.

• If 5t > 2, we still end up in the same place! Consider the situation where player 1 moves and player 2 taxes. Player 1 would be better off if they decided to stay.

• However, if player 1 decided to stay, player 2 would vote to not tax. Given that, it would make sense for player 1 to move. This suggests there might be a mixed strategy equilibrium.

• Lets assume Player 1 will move with probability \(p_m\), and Player 2 will tax with probability \(p_t\), and look at the decision facing player 2:

• Tax: \(8p_m + (1-p_m)(10-t) = 10 - 2p_m - t + p_m t\)

• No Tax: \(8p_m + 10(1-p_m) = 10 - 2p_m\)

• The p_m that would induce player 2 to mix between these two strategies would occur when these two payoffs are equal. So:

• \(10 - 2p_m - t + p_m t = 10 - 2p_m\)

• \(p_m = 1\). In a mixed strategy equilibrium, player 1 will move no matter what the tax is.