We know that, for q < 1, the optimal level of monitoring is q = rL (12- k)rD. Substituting in
the value of rL above we get q = R (14ck)rD. We therefore have that
∂SW
∂k
rD RR +(1 - k)rD∖
4Ц-----2-----++ rD - rE = 0.
The solution we obtain is
re RrD + rD2 - 8c (rE - rD)
k g = min < -------D—2— -------,, 1 > .
rD2
From this expression, we obtain that kreg < 1 for c > 8(rRrD,c) ⇔ R < 8c (rEr-rD). Otherwise,
for R > 8c (rEr^-rD), we have that kreg = 1.
The previous solution assumed that q< 1. To get the bounds on when q =1, substitute
the solution for kreg, assuming kreg < 1, into
R — (1 — kreg)rD RrD — 4c (rE — rD)
4c 2crD
From here, we see that for c > ↑rR'2r ⇔ R < 2c2rEr-rD, q < 1. Otherwise, for R >
2c2rEr-rD, q = 1 and k should be set such that q(k) = 1 ⇔ kreg = 4c+rD R. Note, however,
that for R<4c this solution would imply that kreg > 1, which is not feasible. Therefore, for
R < 4c, we obtain that kreg = 1, which implies that q = R < 1.
One final point that needs to be verified is that, for R < 2c2rEr-rD, then q = R~(1-k)rD <
1, but that for R > 8c (rEr^-rD ), we have that kreg = 1, which would imply that q = R.
Note, however, that for both of these conditions to be true at the same time requires that
8c(rE-RD) < 2c2rEr-rD. This will be satisfied if and only if 4 (rE — rD) < 2rE — rD ⇔
rE < 3rD. We can now use this in the necessary condition for q < 1, which is R < 2c2rEr-rD.
Given the restriction on rE and rD, the right hand side must be less than 2c2(2 rrD) rD = 4c.
Therefore, the joint assumption that R < 2c2rEr-rD and R > 8c (rEr^-rD) implies that R < 4c,
and consequently that q = R < 1, as desired. □
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