2. Simple Consumer Location Model (from R. Braid) © Allen C. Goodman, 2009

3. Consumer Equilibrium U = U (z, q) q = size of land plot for consumer z = consumption of other goods k = distance from urban center R(k) = rent per acre (or per sq. mile) 1 = price of composite good z t = commuting cost per mile to center of urban area

4. Assume all consumers work at center (k = 0) W = z + R(k)q + tk W = budget constraint of consumer. Consumer does this problem: Max U (z, q) s.t. W = z + R(k)q + tk Or: L = U (z, q) - (W - z - R(k)q - tk)  L/  z = U z – = 0.  L/  q = U q – R(k) = 0.

5. This gives MRSqz U q /U z = MRS = R(k). In spatial equilibrium:  L/  k = (-q R k –t) = 0.  R k = -t/q What does this mean? At k = 5? z q W = z + R(5)q + t*5 At k = 6? W = z + R(5)q + t*6 W = z + R(6)q + t*6

6. Cobb-Douglas Utility Function U (z, q) = Az 1-  q  Figure out what happens at a particular value of k. R(k) = MRS = U q /U z = A  z 1-  q  -1 / A(1-  ) z –  q  = [  /(1-  )] [z/q]. z = [(1-  )/  ] q R(k).

7. Continuing … z = [(1-  )/  ] q R(k). Substituting: W = z + R(k)q +tk W = [(1-  )/  ] q R(k) + R(k)q +tk W = (1/  ) q R(k) + tk.  q =  [(W – tk)/R(k)] q(k) =  [(w – tk)/R(k)]

8. We recall that in eq’m: q(k) =  [(w – tk)/R(k)]  R/  k = -t/q(k). Substituting for q(k), we get:  R/  k = -tR(k)/[  (W-tk)] or: dR/R = -(t/  ) dk/(W-tk)

9. Differential Equation! dR/R = -(t/  ) dk/(W-tk) From math, we remember that d ln R = dR/R. ln R = log C + (1/  ) ln (W –tk) = ln [C (W – tk) 1/  ]. R(k) = C (W – tk) 1/ .  What happens to rent as k  ? We can now address some other questions.

10. How big is the city?

11. Solving for R(k)

12. Solving for q(k)

13. Integrating the expression

16. What does this tell us? Why rent is high at small distances Why density is high at small distances Why the rich live further away.

17. Rich and poor R k = -t/q Look at poor, p and rich r. q r > q p, so: R k p is steeper than R k r Distance k Rent RkpRkp RkrRkr kpkp krkr