59 population trait value will increase (decrease). When we achieve a 0 value of the partial derivative, then the second derivatives will be important as to whether there is a stable solution, or a branching into distinct values. 22.2. Using the standard matrix game formulae from Chapter 6 we have that (2, 3) and (1, 2, 3) are not ESS supports since 0 —d < 0 and there are no pure ESSs as no leading diagonal element is the biggest in its column. (1, 2) is an ESS support if e > c due to direct row domination, and (2, 3) is the support of an ESS if bc > be — dc ⇒ dc/be + c/e > 1. 22.3. As above, there is no internal ESS or ESS with support (2, 3) as — F D= —d < 0. There is a pure 1 if E < 0, C < 0, a pure 2 if A < 0, F < 0 (so there is no pure 2 unless θ < 0), a pure 3 if B < 0 < D, a pair (1, 2) if AC + EF < AE and a pair (1, 3) if BE — CD < BC. 22.4. In general varying θ does not affect the ESSs in a straightforward way and it depends upon the relative sizes of a, b, c, d and e. For instance if a > e sufficiently negative θ introduces a pure 2 ESS, whereas if a < e there is no pure 2 whatever the value of θ. We can see that A + E = a + e, B + C = b + c, F—D = d—so that the negative definiteness condition, which only depends upon these sums, is unaffected by the change in θ (and as noted in the above solution, there is never an internal ESS). 22.5. Here we need W (S) > W (RA) and W (S) > W (RB) for pA = pB = 0. For fixed dA, dB, X and Y this gives cA > dA + (1 — α)dB + X, cB > dB + (1 — β)dA + Y. The best treatment is the one that minimises W (S) subject to this, i.e. choose the maximum value of dA + dB for which the above hold. 22.6. For an equilibrium we need W (S) = W (RA) = W (RB) i.e. 1 — dA — dB = 1 — cA — dA — αdB + (1 — pA)X = 1 — cB — βdA + (1 — pB)Y. This rearranges to: pA = 1 — pB = 1 — cA — dA — (1 — α)dB , X 1 — cB — dB — (1 — β)dA Y . 22.7. The optimal strategy is the value of uc that maximises N ∗ from equation Solutions Manual for Game Theoretical Models in Biology, 2nd edition by Mark Broom.