Appendix
A.1. Optimal solution of the “Knapsack Continuos Problem” of Stage 3
$$ \matrix{{*{20}{c}} {{ \max }\,I{A_i} = \sum\limits_{{j = 1}}^m {{h_{{ij}}}{y_{{ij}}}}, i = 1,2} \hfill \\ {{\text{s}}{\text{.t}}{.}\,\sum\limits_{{j = 1}}^m {{y_{{ij}}}} \leqslant {Z_i}} \hfill \\ {{y_{{ij}}} \geqslant {r_i}{Z_i}\frac{{{n_{{ij}}}}}{{{N_i}}},\,\,\,\,\,\,j = 1,2, \ldots, m} \hfill \\ } $$
The result of Stage 3 is the function IA
i
(r
i
), the optimal solution of this “Knapsack Continuos Problem” is of the following form:
$$ \matrix{{*{20}{c}} {{y_{{i1}}} = {Z_i} - {r_i}{Z_i}\frac{{{n_{{i2}}}}}{{{N_i}}} - \ldots - {r_i}{Z_i}\frac{{{n_{{im}}}}}{{{N_i}}},i = 1,2} \hfill \\ {{y_{{ij}}} = {r_i}{Z_i}\frac{{{n_{{ij}}}}}{{{N_i}}},\,\,\,\,\,j = 2, \ldots, m} \hfill \\ } $$
The result of Stage 3 is the function IA
i
(r
i
),
$$ \matrix{{*{20}{c}} {I{A_i}\left( {{r_i}} \right) = {Z_i}\left( {{h_{{i1}}} - {r_i}\left( {{h_{{i1}}} - {h_{{i2}}}} \right)\frac{{{n_{{i2}}}}}{{{N_i}}} - \ldots - {r_i}\left( {{h_{{i1}}} - {h_{{im}}}} \right)\frac{{{n_{{im}}}}}{{{N_i}}}} \right)} \\ { = \left( {\frac{{\left( {1 - {r_i} + {r_{{-i }}}} \right)}}{2}\left( {1 - f} \right)B + \left( {\frac{{{N_i}}}{{{N_1} + {N_2}}}} \right)fB} \right)\left( {{h_{{i1}}} - {k_i}} \right),} \\ } $$
where \( {k_i} = \left( {{h_{{i1}}} - {h_{{i2}}}} \right)\frac{{{n_{{i2}}}}}{{{N_i}}} + \ldots + \left( {{h_{{i1}}} - {h_{{im}}}} \right)\frac{{{n_{{im}}}}}{{{N_i}}} < {h_{{i1}}},i = 1,2 \).
We define the following terms before presenting the proof:
$$ \matrix{{*{20}{c}} {{f_{{ti}}} = \frac{{\frac{{\left( {{h_{{i1}}} - {k_i}} \right)\left( {1 - 2{b_i}} \right) + {c_i}{k_i}}}{{{b_i}\left( {{h_{{i1}}} - {k_i}} \right) - {c_i}{k_i}}}}}{{\frac{{2{N_i}}}{{{N_1} + {N_2}}} + \frac{{\left( {{h_{{i1}}} - {k_i}} \right)\left( {1 - 2{b_i}} \right) + {c_i}{k_i}}}{{{b_i}\left( {{h_{{i1}}} - {k_i}} \right) - {c_i}{k_i}}}}},} \\ {{r_{{t - i}}} = \frac{{\left( {{c_i}{k_i} - {b_i}\left( {{h_{{i1}}} - {k_i}} \right)} \right)\frac{{fB{N_i}}}{{{N_1} + {N_2}}} + \left( {{h_{{i1}}} - {k_i}} \right)\left( {1 - f} \right)\frac{B}{2}\left( {1 - {b_i}} \right)}}{{\left( {1 - f} \right)\frac{B}{2}\left( {{b_i}\left( {{h_{{i1}}} - {k_i}} \right) - {c_i}{k_i}} \right)}}.} \\ } $$
A.2. Proposition 1: For problem L
i
with three the sub-populations:
-
1)
U
i
is a concave function of r
i
and therefore the optimal solution is either \( r_i^{*} = r_i^l \) or \( r_i^{*} = 1 \).
-
2)
Table 1 specifies conditions for existence of Nash equilibria \( \left( {r_i^l,r_{{ - i}}^l} \right),\left( {r_i^l,1} \right),\left( {1,r_{{ - i}}^l} \right) \), and (1,1).
Proof of Proposition 1
-
(i)
The lower-level utility function is \( {\left( {{Z_i}} \right)^{{{a_i}}}}{\left( {{r_i}} \right)^{{{b_i}}}}{\left( {I{A_i}} \right)^{{{c_i}}}} \), where \( {Z_i} = \frac{{\left( {1 - {r_i} + {r_{{ - i}}}} \right)}}{2}\left( {1 - f} \right)B + \frac{{{N_i}}}{{{N_1} + {N_2}}}fB \) and \( I{A_i} = \left( {\frac{{\left( {1 - {r_i} + {r_{{ - i}}}} \right)}}{2}\left( {1 - f} \right)B + \left( {\frac{{{N_i}}}{{{N_1} + {N_2}}}} \right)fB} \right)\left( {{h_{{i1}}} - {r_i}\left( {{h_{{i1}}} - {h_{{i2}}}} \right)\frac{{{n_{{i2}}}}}{{{N_2}}} - {r_i}\left( {{h_{{i3}}} - {h_{{i3}}}} \right)\frac{{{n_{{i3}}}}}{{{N_3}}}} \right) \) are substituted to obtain:
$$ \matrix{{*{20}{c}} {{L_i}:\mathop{{\max }}\limits_{{{r_i}}} {U_i}\left( {{r_i}} \right) = {{\left( {{Z_i}} \right)}^{{{a_i}}}}{{\left( {{r_i}} \right)}^{{{b_i}}}}{{\left( {I{A_i}} \right)}^{{{c_i}}}},\,{\text{where}}\,{a_i} + {b_i} + {c_i} = 1} \hfill \\ {{\text{s}}{\text{.t}}{.}\,0 \leqslant {r_i} \leqslant 1} \hfill \\ } $$
The first derivative of ln U
i
(r
i
) with respect to r
i
is given by,
$$ \matrix{{*{20}{c}} {\frac{d}{{d{r_i}}}\left( {\ln {U_i}\left( {{r_i}} \right)} \right) = \frac{d}{{d{r_i}}}\left( {{a_i}\ln {Z_i} + {b_i}\ln {r_i} + {c_i}\ln I{A_i}} \right)} \hfill \\ { = - \frac{{{a_i}}}{{{Z_i}}}\left( {1 - f} \right)\frac{B}{2} + \frac{{{b_i}}}{{{r_i}}} - \frac{{{c_i}}}{{I{A_i}}}\left( {\left( {1 - f} \right)\frac{B}{2}\left( {{h_{{i1}}} - {r_i}{k_i}} \right) + {k_i}{Z_i}} \right)} \hfill \\ } $$
where, \( \frac{{d{Z_i}}}{{d{r_i}}} = - \left( {1 - f} \right)\frac{B}{2},\frac{{dI{A_i}}}{{d{r_i}}} = - \left( {1 - f} \right)\frac{B}{2}\left( {{h_{{i1}}} - {r_i}{k_i}} \right) - {k_i}{Z_i} \). The second derivative of ln U
i
(r
i
) with respect to r
i
is given by,
$$ \matrix{{*{20}{c}} {\frac{{{d^2}}}{{d{r_i}^2}}\left( {\ln {U_i}\left( {{r_i}} \right)} \right) = \frac{{{a_i}}}{{{Z_i}^2}}\left( {1 - f} \right)\frac{B}{2}\frac{d}{{d{r_i}}}\left( {{Z_i}} \right) - \frac{{{b_i}}}{{{r_i}^2}} - {c_i}\frac{d}{{d{r_i}}}\left( {\frac{{\left( {1 - f} \right)\frac{B}{2}\left( {{h_{{i1}}} - {r_i}{k_i}} \right) + {k_i}{Z_i}}}{{I{A_i}}}} \right)} \hfill \\ { = - \frac{{{a_i}}}{{{Z_i}^2}}{{\left( {1 - f} \right)}^2}\frac{{{B^2}}}{4} - \frac{{{b_i}}}{{{r_i}^2}} - {c_i}\frac{{ - \left( {1 - f} \right)\frac{B}{2}\left( {{h_{{i1}}} - {r_i}{k_i}} \right){k_i}{Z_i} + {{\left( {\left( {1 - f} \right)\frac{B}{2}\left( {{h_{{i1}}} - {r_i}{k_i}} \right) + {k_i}{Z_i}} \right)}^2}}}{{I{A_i}^2}}} \hfill \\ { = - \frac{{{a_i}}}{{{Z_i}^2}}{{\left( {1 - f} \right)}^2}\frac{{{B^2}}}{4} - \frac{{{b_i}}}{{{r_i}^2}} - {c_i}\frac{{{{\left( {1 - f} \right)}^2}\frac{{{B^2}}}{4}{{\left( {{h_{{i1}}} - {r_i}{k_i}} \right)}^2} + {k_i}^2{Z_i}^2}}{{I{A_i}^2}} \leqslant 0.} \hfill \\ } $$
Clearly, \( \ln \left( {{U_i}\left( {{r_i}} \right)} \right)\prime \prime \leqslant 0 \). Thus, U
i
(r
i
) is a concave function of r
i
and the optimal solution is either \( r_i^{*} = r_i^l \) or 1.
-
(ii)
To obtain specific conditions for existence of Nash equilibria, we solve for r
i
* by setting, \( \ln \left( {{U_i}\left( {{r_i}} \right)} \right)\prime = 0 \).
The first derivative of lnU
i
(r
i
) is
$$ \matrix{{*{20}{c}} { - \frac{{{a_i}}}{{{Z_i}}}\left( {1 - f} \right)\frac{B}{2} + \frac{{{b_i}}}{{{r_i}}} - \frac{{{c_i}}}{{I{A_i}}}\left( {\left( {1 - f} \right)\frac{B}{2}\left( {{h_{{i1}}} - {r_i}{k_i}} \right) + {k_i}{Z_i}} \right) = 0} \hfill \\ {\frac{{{b_i}}}{{{r_i}}} = \frac{{{a_i}}}{{{Z_i}}}\left( {1 - f} \right)\frac{B}{2} + \frac{{{c_i}}}{{{Z_i}\left( {{h_{{i1}}} - {r_i}{k_i}} \right)}}\left( {\left( {1 - f} \right)\frac{B}{2}\left( {{h_{{i1}}} - {r_i}{k_i}} \right) + {k_i}{Z_i}} \right)} \hfill \\ {{b_i}{Z_i} = \left( {{a_i} + {c_i}} \right){r_i}\left( {1 - f} \right)\frac{B}{2} + \frac{{{c_i}{r_i}}}{{\left( {{h_{{i1}}} - {r_i}{k_i}} \right)}}{k_i}{Z_i}} \hfill \\ } $$
On substituting value of Z
i
and rearranging terms in the above equation, we obtain
$$ r_i^2{a_q} + {r_i}{b_q} + {c_q} = 0 $$
(11)
where \( {a_q} = {k_i}\left( {1 + {c_i}} \right)\left( {1 - f} \right)\frac{B}{2} \), \( {b_q} = - \left( {{b_i} + {c_i}} \right){k_i}f\frac{{B{N_i}}}{{{N_1} + {N_2}}} - \left( {{h_{{i1}}} + \left( {{b_i} + {c_i}} \right)\left( {1 + {r_{{ - i}}}} \right){k_i}} \right)\left( {1 - f} \right)\frac{B}{2} \), \( {c_q} = {b_i}{h_{{i1}}}\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right) \).
Equation (11) has two positive real roots, since \( b_q^2 - 4{a_q}{c_q} > 0 \) as shown below.
$$ \matrix{{*{20}{c}} {b_q^2 - 4{a_q}{c_q} = {{\left( {\left( {{b_i} + {c_i}} \right){k_i}f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {{h_{{i1}}} + \left( {{b_i} + {c_i}} \right)\left( {1 + {r_{{ - i}}}} \right){k_i}} \right)\left( {1 - f} \right)\frac{B}{2}} \right)}^2} - } \hfill \\ {4\left( {{k_i}\left( {1 + {c_i}} \right)\left( {1 - f} \right)\frac{B}{2}} \right)\left( {{b_i}{h_{{i1}}}\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right)} \right)} \hfill \\ { = {{\left( {\left( {{b_i} + {c_i}} \right){k_i}\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right) + {h_{{i1}}}\left( {1 - f} \right)\frac{B}{2}} \right)}^2} - } \hfill \\ {4{k_i}\left( {1 - f} \right)\frac{B}{2}{b_i}{h_{{i1}}}\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right) - {c_i}4{k_i}\left( {1 - f} \right)\frac{B}{2}{b_i}{h_{{i1}}}\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right)} \hfill \\ { = {{\left( {{b_i}{k_i}\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right) + {c_i}{k_i}\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right) + {h_{{i1}}}\left( {1 - f} \right)\frac{B}{2}} \right)}^2} - } \hfill \\ {4{k_i}\left( {1 - f} \right)\frac{B}{2}{b_i}{h_{{i1}}}\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right) - {c_i}4{k_i}\left( {1 - f} \right)\frac{B}{2}{b_i}{h_{{i1}}}\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right)} \hfill \\ { = {{\left( {{b_i}{k_i}\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right) - {h_{{i1}}}\left( {1 - f} \right)\frac{B}{2}} \right)}^2} + {c_i}^2{k_i}^2{{\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right)}^2} + } \hfill \\ {2{c_i}{k_i}\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right){h_{{i1}}}\left( {1 - f} \right)\frac{B}{2} + 2{b_i}{c_i}{k_i}^2{{\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right)}^2} - } \hfill \\ {{c_i}4{k_i}\left( {1 - f} \right)\frac{B}{2}{b_i}{h_{{i1}}}\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right)} \hfill \\ } $$
On solving,
$$ \matrix{{*{20}{c}} { = {{\left( {{b_i}{k_i}\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right) - {h_{{i1}}}\left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right)}^2}} \hfill \\ { + {c_i}{k_i}\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right)\left( {\left( {{c_i}{k_i} + 2{b_i}{k_i}} \right)\left( {f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2}} \right) + 2{h_{{i1}}}\left( {1 - f} \right)\frac{B}{2}\left( {1 - 2{b_i}} \right)} \right) > 0} \hfill \\ } $$
Thus, Eq. (11) has two positive real roots, \( r_i^l \) and \( r_i^u \), defined by,
$$ r_i^j = \frac{{ - {b_q}\pm \sqrt {{{b_q}^2 - 4{a_q}{c_q}}} }}{{2{a_q}}},j = u,l,i = 1,2 $$
(12)
Since \( b_q^2 > b_q^2 - 4{a_q}{c_q} \), hence, we obtain, \( \left| {{b_q}} \right| > \sqrt {{b_q^2 - 4{a_q}{c_q}}} \) Since,
$$ \matrix{{*{20}{c}} {{b_q} < 0 \Rightarrow \left| {{b_q}} \right| = - {b_q}} \hfill \\ { - {b_q} > \sqrt {{b_q^2 - 4{a_q}{c_q}}} } \hfill \\ { - {b_q} - \sqrt {{b_q^2 - 4{a_q}{c_q}}} > 0} \hfill \\ {\frac{{ - {b_q} - \sqrt {{b_q^2 - 4{a_q}{c_q}}} }}{{2{a_q}}} > 0,\,{\text{because}},\,{a_q} > 0} \hfill \\ } $$
Thus, \( r_i^l > 0 \) and by definition \( 0 < r_i^l < r_i^u \). Further, \( r_i^u > 1 \) since \( - \frac{{{b_q}}}{{2{a_q}}} > 1 \) as shown below.Since h
i1 ≈ 0.001 and k
i
≈ .0001. Clearly, h
i1 > 2k
i
implies that \( - \frac{{{b_q}}}{{2{a_q}}} > 1 \).Therefore, clearly, \( r_i^u > 1 \).
We next show that if \( {b_i} > \frac{{{c_i}{k_i}}}{{{h_{{i1}}} - {k_i}}},r_i^l > {r_{{t - i}}} \), and f > f
ti
then \( r_i^l \geqslant 1 \), where \( 0 < {r_{{t - i}}} < 1 \).Let \( f > {f_{{ti}}} = \frac{{\frac{{\left( {{h_{{i1}}} - {k_i}} \right)\left( {1 - 2{b_i}} \right) + {c_i}{k_i}}}{{{b_i}\left( {{h_{{i1}}} - {k_i}} \right) - {c_i}{k_i}}}}}{{\frac{{2{N_i}}}{{{N_1} + {N_2}}} + \frac{{\left( {{h_{{i1}}} - {k_i}} \right)\left( {1 - 2{b_i}} \right) + {c_i}{k_i}}}{{{b_i}\left( {{h_{{i1}}} - {k_i}} \right) - {c_i}{k_i}}}}} \), where \( \left( {\frac{{2{N_i}}}{{{N_1} + {N_2}}} + \frac{{\left( {1 - 2{b_i}} \right)\left( {{h_{{i1}}} - {k_i}} \right) + {c_i}{k_i}}}{{{b_i}\left( {{h_{{i1}}} - {k_i}} \right) - {c_i}{k_i}}}} \right) > 0 \) since
$$ \matrix{{*{20}{c}} {{b_i} > \frac{{{c_i}{k_i}}}{{{h_{{i1}}} - {k_i}}}.} \hfill \\ { \Rightarrow f\left( {\frac{{2{N_i}}}{{{N_1} + {N_2}}} + \frac{{\left( {1 - 2{b_i}} \right)\left( {{h_{{i1}}} - {k_i}} \right) + {c_i}{k_i}}}{{{b_i}\left( {{h_{{i1}}} - {k_i}} \right) - {c_i}{k_i}}}} \right) > \frac{{\left( {1 - 2{b_i}} \right)\left( {{h_{{i1}}} - {k_i}} \right) + {c_i}{k_i}}}{{{b_i}\left( {{h_{{i1}}} - {k_i}} \right) - {c_i}{k_i}}}} \hfill \\ { \Rightarrow f\left( {\frac{{2{N_i}}}{{{N_1} + {N_2}}} + \frac{{\left( {{h_{{i1}}} - {k_i}} \right)\left( {1 - {b_i}} \right)}}{{{b_i}\left( {{h_{{i1}}} - {k_i}} \right) - {c_i}{k_i}}} - 1} \right) > \frac{{\left( {{h_{{i1}}} - {k_i}} \right)\left( {1 - {b_i}} \right)}}{{{b_i}\left( {{h_{{i1}}} - {k_i}} \right) - {c_i}{k_i}}} - 1} \hfill \\ { \Rightarrow \frac{{\left( {{h_{{i1}}} - {k_i}} \right)\left( {1 - {b_i}} \right)}}{{{b_i}\left( {{h_{{i1}}} - {k_i}} \right) - {c_i}{k_i}}} < 1 + \frac{{2f{N_i}}}{{\left( {{N_1} + {N_2}} \right)\left( {1 - f} \right)}}} \hfill \\ { \Rightarrow {r_{{t - i}}} < 1.} \hfill \\ } $$
(13)
Since, \( {b_i} > \frac{{{c_i}{k_i}}}{{{h_{{i1}}} - {k_i}}} \), clearly, \( {r_{{t - i}}} > 0 \).Let
$$ r_{{ - i}}^l \geqslant {r_{{t - i}}} = \frac{{\left( {{c_i}{k_i} - {b_i}\left( {{h_{{i1}}} - {k_i}} \right)} \right)\frac{{fB{N_i}}}{{{N_1} + {N_2}}} + \left( {{h_{{i1}}} - {k_i}} \right)\left( {1 - f} \right)\frac{B}{2}\left( {1 - {b_i}} \right)}}{{\left( {1 - f} \right)\frac{B}{2}\left( {{b_i}\left( {{h_{{i1}}} - {k_i}} \right) - {c_i}{k_i}} \right)}} $$
(14)
On rearranging the terms,
$$ \matrix{{*{20}{c}} { \Rightarrow \left( {{h_{{i1}}} + \left( {{b_i} + {c_i}} \right)\left( {1 + {r_{{ - i}}}} \right){k_i}} \right)\left( {1 - f} \right)\frac{B}{2} + {k_i}\left( {1 + {c_i}} \right)\left( {1 - f} \right)\frac{B}{2} + \left( {{b_i} + {c_i}} \right){k_i}f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + {b_i}{h_{{i1}}}\left( {\left( {1 + {r_{{ - i}}}} \right)\left( {1 - f} \right)\frac{B}{2} + f\frac{{B{N_i}}}{{{N_1} + {N_2}}}} \right) \geqslant 0} \hfill \\ { \Rightarrow {a_q} - \left( { - {b_q}} \right) + {c_q} \geqslant 0} \hfill \\ { \Rightarrow - 4a_q^2 + 4{a_q}\left( { - {b_q}} \right) + {{\left( { - {b_q}} \right)}^2} - 4{a_q}{c_q} \geqslant {{\left( { - {b_q}} \right)}^2}} \hfill \\ { \Rightarrow {{\left( { - {b_q} - 2{a_q}} \right)}^2} \geqslant {b_q}^2 - 4{a_q}{c_q}} \hfill \\ } $$
We next show that \( - {b_q} - {2}{a_q} > 0 \).
Clearly, \( 2\left( {1 + {c_i}} \right) < \frac{{{h_{{i1}}}}}{{{k_i}}} \) since \( {a_i} + {b_i} + {c_i} = {1} \) and h
i1 ≈ 0.001and k
i
≈ .0001.
$$ \left( {2\left( {1 + {c_i}} \right) - \frac{{{h_{{i1}}}}}{{{k_i}}}} \right)\frac{1}{{\left( {{b_i} + {c_i}} \right)}} - 1 - \frac{{2f{N_i}}}{{{N_1} + {N_2}}} < 0 $$
Clearly, \( r_{{ - i}}^l > \left( {2\left( {1 + {c_i}} \right) - \frac{{{h_{{i1}}}}}{{{k_i}}}} \right)\frac{1}{{\left( {{b_i} + {c_i}} \right)}} - 1 - \frac{{2f{N_i}}}{{{N_1} + {N_2}}} \)
$$ \matrix{{*{20}{c}} { \Rightarrow \left( {1 + {r_{{ - i}}}} \right)\left( {{b_i} + {c_i}} \right){k_i}\left( {1 - f} \right)\frac{B}{2} + {h_{{i1}}}\left( {1 - f} \right)\frac{B}{2} > {k_i}\left( {1 + {c_i}} \right)\left( {1 - f} \right)B - \left( {{b_i} + {c_i}} \right){k_i}f\frac{{B{N_i}}}{{{N_1} + {N_2}}}} \hfill \\ { \Rightarrow \left( {{b_i} + {c_i}} \right){k_i}f\frac{{B{N_i}}}{{{N_1} + {N_2}}} + \left( {{h_{{i1}}} + \left( {{b_i} + {c_i}} \right)\left( {1 + {r_{{ - i}}}} \right){k_i}} \right)\left( {1 - f} \right)\frac{B}{2} - {k_i}\left( {1 + {c_i}} \right)\left( {1 - f} \right)B > 0} \hfill \\ { \Rightarrow - {b_q} - 2{a_q} > 0.} \hfill \\ } $$
Thus, \( {\left( { - {b_q} - 2{a_q}} \right)^2} \geqslant {b_q}^2 - 4{a_q}{c_q} \)
$$ \matrix{{*{20}{c}} { \Rightarrow - {b_q} - \sqrt {{{b_q}^2 - 4{a_q}{c_q}}} \geqslant 2{a_q}} \hfill \\ { \Rightarrow r_i^l \geqslant 1} \hfill \\ } $$
(15)
Thus, if \( {b_i} > \frac{{{c_i}{k_i}}}{{{h_{{i1}}} - {k_i}}},r_i^l > {r_{{t - i}}} \), and f > f
ti
then \( r_i^l > = 1 \). On reversing the inequality in (13), if \( {b_i} > \frac{{{c_i}{k_i}}}{{{h_{{i1}}} - {k_i}}} \), and f ≤ f
ti
then \( r_i^l < 1 \). If \( {b_i} \leqslant \frac{{{c_i}{k_i}}}{{{h_{{i1}}} - {k_i}}} \) then r
t−i
< 0, therefore, clearly from (14) and (15), \( r_i^l < 1 \). Hence, we obtain conditions specified in Table 1.
A.3. Proposition 2: For problem IA with the three sub-populations, Table 2 specifies the optimal f and IA value for the conditions specified in Table 1
Proof
The upper-level utility function is \( IA = \sum\limits_{{i = 1}}^2 {I{A_i}} \)
$$ = B\sum\limits_{\begin{subarray} i, - i = 1 \\ i \ne - i \end{subarray} }^2 {\left[ {\frac{{\left( {1 - r_i^l + r_{{ - i}}^l} \right)\left( {1 - f} \right)B}}{2} + \frac{{fB{N_i}}}{{{N_i} + {N_{{ - i}}}}}} \right]\left( {{h_i}_1 - r_i^l{k_i}} \right)} $$
We next solve the upper-level problem.
$$ IA:\mathop{{\max }}\limits_f \sum\limits_{{i = 1}}^2 {I{A_i}\left( {r_i^l*(f)} \right)}, i = 1,2, $$
(16)
$$ {\text{s}}{\text{.t}}{.}\,{0} \leqslant f \leqslant 1 $$
(17)
$$ r_i^l*(f) = \arg \,\max \left( {{L_i}\left( {r_i^l} \right)} \right) $$
(18)
For the Nash equilibrium, (1,1), clearly, IA is a decreasing function in f. Thus, from Table 1, if \( {b_i} > \frac{{{c_i}{k_i}}}{{{h_{{i1}}} - {k_i}}},i = 1,2 \) then \( f* = \max \left\{ {{f_{{t1}}},{f_{{t2}}}} \right\} \) is optimal, resulting in \( {r_i}* = 1 \) at the lower level and \( IA_M^{*}\left( {r_1^{*},r_1^{*}} \right) = \frac{B}{2}\left( {{h_{{11}}} - {k_1} + {h_{{21}}} - {k_2}} \right) \).
For the Nash equilibrium, \( \left( {r_1^l,r_2^l} \right) \), first derivative of IA with respect to f is given below:
$$ \frac{{dIA}}{{df}} = \frac{{\partial IA}}{{\partial f}} + \frac{{\partial IA}}{{\partial r_1^l}} \cdot \frac{{\partial r_1^l}}{{\partial f}} + \frac{{\partial IA}}{{\partial r_2^l}} \cdot \frac{{\partial r_2^l}}{{\partial f}} \kern3pt \cdot$$
Clearly, the total number of infections prevented decrease with the increase in the preferences \( \left( {r_1^l,r_2^l} \right) \) of the LDs to allocate proportionally plus an increase in the fixed budget (budget to be allocated in proportion to the number of HIV/AIDS infections). That is, IA is decreasing in f, \( r_1^l \), and \( r_2^l,\frac{{\partial IA}}{{\partial f}} < 0,\frac{{\partial IA}}{{\partial r_1^l}} < 0,\,{\text{and}}\,\frac{{\partial IA}}{{\partial r_2^l}} < 0 \). Based on the literature, the preferences \( \left( {r_1^l,r_2^l} \right) \) of the LDs to allocate proportionally increase with the increase in the fixed budget (increase in f). That is, \( r_1^l,r_2^l \) is increasing in f. Therefore, IA is decreasing in f. Thus, we obtain optimal f * and IA* specified in Table 2, based on conditions of Table 1.