## Info

a f 2(Ht(t + 1, 0)) = !1(1 — exp(—!2Ht(t + 1,0)))

where !1 and w2 are parameters. As total host population declines, the number that are inferior declines as well.

Next season's parasitoids emerge from hosts that have been attacked. If we assume that the parasitoid is solitary, then one offspring will emerge from a superior host, and one from an inferior host, but only with probability p. In that case, parasitoid numbers at the start of the next season are given by

P(t + 1)— H1 (t,0)-Hi(t,S) + /3[H2(t,0) -Htit,S)] (4 . 36)

in which the first two terms on the right hand side ofEq. (4.36) represent the number of superior hosts attacked and the second two terms represent the number of inferior hosts attacked.

We still need to find s*. If s* — 0, then we have the standard Nicholson-Bailey model and if s* — S, then we have a version of Taylor's stabilization of the Nicholson-Bailey dynamics because of a host refuge. To find s* in the more general situation, we will use an especially simple version of stochastic dynamic programming by assuming that the parasitoids never run out of eggs (a topic discussed in the next section). Given that there are a total of HT hosts for the parasitoid to attack, and the random search assumption of the Nicholson-Bailey model, we have

Prfencounter a host in a unit interval of time|HT hosts present}

so that if 1i(t) is the probability of encountering host type i within one unit of time in season t, we have m—H(fl)(1 - exp (— aHT(t, 0))) (4 . 38)

and we assume that hosts are sufficiently plentiful that the parasitoid never re-encounters a previously attacked host (see Exercise 4.8 below).

Now let F(s|t) denote the maximum expected accumulated reproduction between within-season time s and S, given host parameters of season t. We then have F(S|t) — 0 and if ms and mo are the probabilities of mortality during search and oviposition, the dynamic programming equation is

F(s|t) — (1 - l1(t)- l2(t))(1 - ms)F(s + 1|i)+ X1(t){1 + (1 - m0)F(s + 1|t)} + l2(t) max{P +(1 - ma)F(s + 1|t); (1 - ms)F(s + 1|t)}

which we solve backwards, as before.

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