An optimization framework for the capacity allocation and admission control of MapReduce jobs in cloud systems

Abstract

Nowadays, we live in a Big Data world and many sectors of our economy are guided by data-driven decision processes. Big Data and Business Intelligence applications are facilitated by the MapReduce programming model, while, at infrastructural layer, cloud computing provides flexible and cost-effective solutions to provide on-demand large clusters. Capacity allocation in such systems, meant as the problem of providing computational power to support concurrent MapReduce applications in a cost-effective fashion, represents a challenge of paramount importance. In this paper we lay the foundation for a solution implementing admission control and capacity allocation for MapReduce jobs with a priori deadline guarantees. In particular, shared Hadoop 2.x clusters supporting batch and/or interactive jobs are targeted. We formulate a linear programming model able to minimize cloud resources costs and rejection penalties for the execution of jobs belonging to multiple classes with deadline guarantees. Scalability analyses demonstrated that the proposed method is able to determine the global optimal solution of the linear problem for systems including up to 10,000 classes in less than 1 s.

Appendices

Makespan bounds

In the following we report the results we presented in [32] providing an approximated formula for the estimation of MapReduce jobs. We consider a MapReduce system with up to $s_i^\text {M}$ and $s_i^\text {R}$ containers devoted for the Map and Reduce phase using the Capacity Scheduler.

Following the results in [14], the lower and upper bounds on the duration of the entire Map stage can be estimated as follows:

$$\begin{aligned} \begin{aligned} T_i^{\text {M}, \text {low}} =&\frac{N_i^\text {M} M_i^\text {avg}}{s_i^\text {M}} h_i \\ T_i^{\text {M}, \text {up}} =&\frac{N_i^\text {M} M_i^\text {avg} - 2M_i^\text {max}}{s_i^\text {M}} h_i+2M_i^\text {max} \end{aligned} \end{aligned}$$

where $M_i^\text {avg}$, $M_i^\text {max}$, $R_i^\text {avg}$, $R_i^\text {max}$, $S_i^{1, \text {avg}}$, $S_i^{1, \text {max}}$, $S_i^\text {avg}$, and $S_i^\text {max}$ denote the average and maximum duration of Map, Reduce, first Shuffle and typical Shuffle phases, respectively, while $N_i^\text {M}$ and $N_i^\text {R}$ are the number of Map and Reduce tasks (see Sect. 2). According to the results discussed in [14], we distinguish the non-overlapping portion of the first shuffle and the task durations in the typical shuffle. In the following bounds for the shuffle stage, this consideration affects the formula for $T_i^{\text {S}, \text {low}}$, where we subtract one wave:

$$\begin{aligned} \begin{aligned} T_i^{\text {S}, \text {low}} =&\left( \frac{N_i^\text {R}}{s_i^\text {R}} h_i - 1 \right) S_i^\text {avg} , \\ T_i^{\text {S}, \text {up}} =&\frac{N_i^\text {R} S_i^\text {avg} - 2 S_i^\text {max}}{s_i^\text {R}} h_i+2S_i^\text {max} . \end{aligned} \end{aligned}$$

$S_i^{1, \text {avg}}$ and $S_i^{1, \text {max}}$, the average and maximum execution time of the first shuffle phase, are estimated directly from the execution profile of class i.

Summing up all parts we get:

$$\begin{aligned} \begin{aligned} T_i^\text {low}&= T_i^{\text {M}, \text {low}} + S_i^{1, \text {avg}} + T_i^{\text {S}, \text {low}} + T_i^{\text {R}, \text {up}} , \\ T_i^\text {up}&= T_i^{\text {M}, \text {up}} + S_i^{1, \text {max}} + T_i^{\text {S}, \text {up}} + T_i^{\text {R}, \text {low}} . \end{aligned} \end{aligned}$$

$T_i^\text {low}$ and $T_i^\text {up}$ represent, respectively, an optimistic and a pessimistic prediction of class i job completion time. We also define $T_i^\text {avg} = \left( T_i^\text {up} + T_i^\text {low} \right) \!{/} 2$. Hence, the execution time of a class i job is at least:

$$\begin{aligned} T_i^\text {low} = \frac{\xi _i^{\text {M}, \text {low}} h_i}{s_i^\text {M}} + \frac{\xi _i^{\text {R}, \text {low}} h_i}{s_i^\text {R}} + \xi _i^{0, \text {low}} , \end{aligned}$$

(7a)

where

$$\begin{aligned} \xi _i^{\text {M}, \text {low}}&= N_i^\text {M} M_i^\text {avg} , \end{aligned}$$

(7b)

$$\begin{aligned} \xi _i^{\text {R}, \text {low}}&= N_i^\text {R} \left( S_i^\text {avg} + R_i^\text {avg} \right) , \end{aligned}$$

(7c)

$$\begin{aligned} \xi _i^{0, \text {low}}&= S_i^{1, \text {avg}}- S_i^\text {avg} . \end{aligned}$$

(7d)

In the same way, the execution time of a job of class i is at most:

$$\begin{aligned} T_i^\text {up} = \frac{\xi _i^{\text {M}, \text {up}} h_i}{s_i^\text {M}} + \frac{\xi _i^{\text {R}, \text {up}} h_i}{s_i^\text {R}} + \xi _i^{0, \text {up}} , \end{aligned}$$

(8a)

where

$$\begin{aligned} \xi _i^{\text {M}, \text {up}}&= N_i^\text {M} M_i^\text {avg} - 2 M_i^\text {max} , \end{aligned}$$

(8b)

$$\begin{aligned} \xi _i^{\text {R}, \text {up}}&= N_i^\text {R} Sh^\mathrm{typ{i}}_\mathrm{avg} - 2 Sh^\mathrm{typ{i}}_\mathrm{max} + N_i^\text {R} R_i^\text {avg} - 2 R_i^\text {max} , \end{aligned}$$

(8c)

$$\begin{aligned} \xi _i^{0, \text {up}}&= 2S_i^\text {max} + S_i^{1, \text {max}} + 2 M_i^\text {max}+2R_i^\text {max}. \end{aligned}$$

(8d)

Depending on the guarantees required, it is possible to adopt either a conservative approach to meeting deadlines with $T_i^\text {up}$ or a less resource-demanding one with $T_i^\text {avg}$.

In both cases (upper bounds or approximation), the formulae above reduce to constraints (P1b) and (P2b), by defining $\zeta _i^{0, D} \triangleq \xi _i^0 - D_i < 0$:

$$\begin{aligned} T_i = \frac{\xi _i^\text {M} h_i}{s_i^\text {M}} + \frac{\xi _i^\text {R} h_i}{s_i^\text {R}} + \xi _i^0 \le D_i \Rightarrow T_i = \frac{\xi _i^\text {M} h_i}{s_i^\text {M}} + \frac{\xi _i^\text {R} h_i}{s_i^\text {R}} + \zeta _i^{0, D} \le 0. \end{aligned}$$

Proofs

Proof of Theorem 1

Since all the constraints of problem (P2) are convex and the Slater constraints qualification holds, the KKT conditions are necessary for optimality. The Lagrangian function of problem (P2) is given by:

$$\begin{aligned} L&= \delta d + \rho r - \sum _{i\in {\mathcal {U}}}\frac{p_i}{\varPsi _i} +\sum _{i\in {\mathcal {U}}} \phi _i \left( \frac{\xi _i^\text {M}}{s_i^\text {M} \varPsi _i } + \frac{\xi _i^\text {R}}{s_i^\text {R} \varPsi _i } + \zeta _i^{0, D} \right) \\&\quad + \mu _r (r - {\bar{r}}) +\nu \left[ \sum _{i\in {\mathcal {U}}} \left( \frac{s_i^\text {M}}{c_i^\text {M}} + \frac{s_i^\text {R}}{c_i^\text {R}}\right) -r-d\right] \\&\quad +\sum _{i\in {\mathcal {U}}} \left[ \mu _i(\varPsi _i - \varPsi ^\text {up}_i)+\lambda _i(-\varPsi _i + \varPsi ^\text {low}_i)\right] \\&\quad - \lambda _r r - \lambda _d d \nonumber - \sum _{i\in {\mathcal {U}}} \left( \omega _i\, s_i^\text {M} +\chi _i s_i^\text {R}\, \right) . \end{aligned}$$

Therefore, the KKT conditions are the following:

$$\begin{aligned} \delta - \nu - \lambda _d = 0,&\end{aligned}$$

(9)

$$\begin{aligned} \rho - \nu + \mu _r - \lambda _r = 0,&\end{aligned}$$

(10)

$$\begin{aligned} \frac{\nu }{c_i^\text {M}} - \phi _i \frac{\xi _i^\text {M}}{({s_i^\text {M}})^2 \varPsi _i } -\omega _i = 0,&\quad \forall i \in {\mathcal {U}}, \end{aligned}$$

(11)

$$\begin{aligned} \frac{\nu }{c_i^\text {R}} - \phi _i \frac{\xi _i^\text {R}}{({s_i^\text {R}})^2 \varPsi _i } -\chi _i = 0,&\quad \forall i \in {\mathcal {U}}, \end{aligned}$$

(12)

$$\begin{aligned} \frac{p_i}{\varPsi _i^2} - \frac{\phi _i}{\varPsi _i^2} (\frac{\xi _i^\text {M}}{s_i^\text {M} } + \frac{\xi _i^\text {R}}{s_i^\text {R}}) + \mu _i - \lambda _i = 0,&\quad \forall i \in {\mathcal {U}}, \end{aligned}$$

(13)

$$\begin{aligned} \phi _i\left( \frac{\xi _i^\text {M}}{s_i^\text {M} \varPsi _i} + \frac{\xi _i^\text {R}}{s_i^\text {R} \varPsi _i} + \zeta _i^{0, D}\right) =0,&\quad \phi _i \ge 0 , \quad \forall i \in {\mathcal {U}}, \end{aligned}$$

(14)

$$\begin{aligned} \mu _r(r-{\bar{r}}) = 0,&\quad \mu _r \ge 0, \end{aligned}$$

(15)

$$\begin{aligned} \nu \left[ \sum _{i\in {\mathcal {U}}} \left( \frac{s_i^\text {M}}{c_i^\text {M}} + \frac{s_i^\text {R}}{c_i^\text {R}} \right) - r - d \right] = 0,&\quad \nu \ge 0, \end{aligned}$$

(16)

$$\begin{aligned} \mu _i (\varPsi _i - \varPsi ^\text {up}_i) = 0,&\quad \mu _i \ge 0, \quad \forall i \in {\mathcal {U}}, \end{aligned}$$

(17)

$$\begin{aligned} \lambda _i ( -\varPsi _i + \varPsi ^\text {low}_i ) =0,&\quad \lambda _i \ge 0, \quad \forall i \in {\mathcal {U}}, \end{aligned}$$

(18)

$$\begin{aligned} \lambda _r\, r =0,&\quad \lambda _r \ge 0, \end{aligned}$$

(19)

$$\begin{aligned} \lambda _d\, d =0,&\quad \lambda _d \ge 0, \end{aligned}$$

(20)

$$\begin{aligned} \omega _i\, s_i^\text {M} =0,&\quad \omega _i \ge 0, \quad \forall i \in {\mathcal {U}}, \end{aligned}$$

(21)

$$\begin{aligned} \chi _i\, s_i^\text {R} =0,&\quad \chi _i \ge 0, \quad \forall i \in {\mathcal {U}}. \end{aligned}$$

(22)

Constraints (P2b) imply that $s_i^\text {M}$ and $s_i^\text {R}$ are positive; hence, multipliers $\omega _i$ and $\chi _i$ are equal to zero. As the adoption of reserved instances is favored, being cheaper than the on-demand ones, we obtain $r>0$ and $\lambda _r=0$. Furthermore, we have $\nu = \rho + \mu _r \ge \rho >0$. Therefore, Eq. (11) guarantees that $\phi _i > 0$ for all $i \in {\mathcal {U}}$; hence, constraints (P2b) hold as equalities.

Finally, we can use Eqs. (P2b), (11) and (12) to compute $s_i^\text {M}$ and $s_i^\text {R}$ as a function of $\varPsi _i$. First, we calculate the relation between $s_i^\text {M}$ and $s_i^\text {R}$ by using conditions (11) and (12) as follows:

$$\begin{aligned} \frac{\xi _i^\text {M}}{({s_i^\text {M}})^2 \varPsi _i } {c_i^\text {M}} =\frac{\xi _i^\text {R}}{({s_i^\text {R}})^2 \varPsi _i } {c_i^\text {R}} \ \ \Longleftrightarrow \ \ s_i^\text {M} = s_i^\text {R}\sqrt{\frac{\xi _i^\text {M}\, c_i^\text {M}}{\xi _i^\text {R}\, c_i^\text {R}}}. \end{aligned}$$

Then, we can replace $s_i^\text {M}$ by $\displaystyle s_i^\text {R}\sqrt{\frac{\xi _i^\text {M}\, c_i^\text {M}}{\xi _i^\text {R}\, c_i^\text {R}}}$ into (P2b) to derive an explicit formulation for $s_i^\text {R}$, as follows:

$$\begin{aligned} s_i^\text {R} = - \frac{1}{\zeta _i^{0, D} \varPsi _i} \left( {\sqrt{\frac{\xi _i^\text {M}\, \xi _i^\text {R} c_i^\text {R}}{c_i^\text {M}}} } + \xi _i^\text {R} \right) , \end{aligned}$$

and along the same lines we can express $s_i^\text {M}$ in closed form as follows:

$$\begin{aligned} s_i^\text {M} = - \frac{1}{\zeta _i^{0, D} \varPsi _i} \left( \sqrt{\frac{\xi _i^\text {M}\, \xi _i^\text {R}\, c_i^\text {M}}{c_i^\text {R}}} + \xi _i^\text {M} \right) . \end{aligned}$$

$\square$

Proof of Theorem 2

Theorem 1 implies that the variables $s_i^\text {M}$ and $s_i^\text {R}$ can be written as in Eqs. (3) and (4). Hence, constraint (P3b) is obtained by replacing $s_i^\text {M}$ and $s_i^\text {R}$ in constraint (P2d). Moreover, constraints (P2b) can be dropped since they have been used to derive the value of $s_i^\text {M}$ and $s_i^\text {R}$. Hence, problem (P3) is equivalent to problem (P2). $\square$

Proof of Theorem 3

First, we notice that the KKT conditions of problem (P3) are necessary and sufficient for optimality since the problem is linear. The KKT conditions can be written as follows:

$$\begin{aligned} \rho - \nu + \mu _r - \lambda _r = 0,&\end{aligned}$$

(23)

$$\begin{aligned} \delta - \nu - \lambda _d = 0,&\end{aligned}$$

(24)

$$\begin{aligned} -p_i + \gamma _i\,\nu + \mu _i - \lambda _i = 0,&\quad \forall \ i\ \in {\mathcal {U}}, \end{aligned}$$

(25)

$$\begin{aligned} \nu \, \left( \sum _{i\in {\mathcal {U}}} \gamma _i\,h^*_i - r^* - d^* \right) = 0,&\end{aligned}$$

(26)

$$\begin{aligned} \lambda _r \, r^* = 0,&\end{aligned}$$

(27)

$$\begin{aligned} \mu _r \, (r^* - {\bar{r}}) = 0,&\end{aligned}$$

(28)

$$\begin{aligned} \lambda _d \, d^* = 0,&\end{aligned}$$

(29)

$$\begin{aligned} \lambda _i \, (h^*_i-H_i^\text {low}) = 0,&\quad \forall \ i\ \in {\mathcal {U}}, \end{aligned}$$

(30)

$$\begin{aligned} \mu _i \, (h^*_i-H_i^\text {up}) = 0,&\quad \forall \ i\ \in {\mathcal {U}}, \end{aligned}$$

(31)

$$\begin{aligned} \nu , \lambda _r, \mu _r, \lambda _d \ge 0,&\end{aligned}$$

(32)

$$\begin{aligned} \lambda _i, \mu _i \ge 0,&\quad \forall \ i\ \in {\mathcal {U}}. \end{aligned}$$

(33)

• 1. Let us assume, by contradiction, that $r^* = 0$. Hence $d^* \ge \sum _{i\in {\mathcal {U}}} \gamma _i \, h^*_i \ge \sum _{i\in {\mathcal {U}}} \gamma _i \, H_i^\text {low} > 0,$ thus $\lambda _d=0$ and $\nu =\delta$. On the other hand, (28) implies that $\mu _r=0$ and $\lambda _r = \rho -\nu =\rho -\delta <0$, which is impossible.

• 2. Since $r^*>0$, we have $\lambda _r=0$, hence (23) implies $\nu = \rho + \mu _r \ge \rho >0$, thus constraint (P3b) is active at $(r^*,d^*,h^*)$.

• 3. It follows from (24) that $\nu =\delta - \lambda _d \le \delta$; hence, we have $\mu _i = \lambda _i + p_i - \gamma _i\,\nu \ge p_i - \gamma _i\,\nu \ge p_i - \gamma _i\,\delta > 0.$ Therefore $h^*_i=H_i^\text {up}$.

• 4. Since $\nu \ge \rho$, we get $\lambda _i = \mu _i + \gamma _i\,\nu - p_i \ge \gamma _i\,\nu - p_i \ge \gamma _i\,\rho - p_i >0,$ hence $h^*_i=H_i^\text {low}$.

• 5. We have $r^* = \sum _{i\in {\mathcal {U}}} \gamma _i\,h^*_i - d^* \le \sum _{i\in {\mathcal {U}}} \gamma _i\,H_i^\text {up} < {\bar{r}},$ thus $\mu _r=0$ and $\nu =\rho$. Therefore, $\lambda _d = \delta - \rho >0$ implies $d^*=0$.

• 6. We have $d^* = \sum _{i\in {\mathcal {U}}} \gamma _i\,h^*_i - r^* \ge \sum _{i\in {\mathcal {U}}} \gamma _i\,H_i^\text {low} - {\bar{r}} > 0,$ hence $\lambda _d=0$ and $\nu =\delta$. Therefore, $\mu _r = \delta - \rho >0$ implies $r^*={\bar{r}}$. $\square$

Proof of the closed-form optimal solution for the two-class case (Table 5)

First, we note that $\nu \in [\rho , \delta ]$, according to the properties (2) and (3) of Theorem 3. The following implications follow directly from the system (23)–(33):

$$\begin{aligned} \nu < \delta \quad \Longrightarrow \quad d^*=0, \\ \nu> \rho \quad \Longrightarrow \quad r^* = {\bar{r}}, \\ d^* > 0 \quad \Longrightarrow \quad r^* = {\bar{r}}. \end{aligned}$$

The proof is divided into six cases:

• 1. Let $p_1/\gamma _1< p_2/\gamma _2 <\rho$. Theorem 3 implies $h^*_1=H_1^\text {low}$ and $h^*_2=H_2^\text {low}$. If ${\bar{r}}<\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {low}$, then Theorem 3 guarantees that $r^*={\bar{r}}$ and $d^* = \gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {low} - {\bar{r}}$. If ${\bar{r}}>\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {low}$, then

  $$\begin{aligned} r^* = \gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {low} - d^* \le \gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {low} < {\bar{r}}, \end{aligned}$$

  thus $d^*=0$ and $r^*=\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {low}$.

• 2. Let $p_1/\gamma _1< \rho< p_2/\gamma _2 <\delta$. Theorem 3 implies $h^*_1=H_1^\text {low}$. We distinguish three cases.

  • (a) If $\nu \in [\rho , p_2/\gamma _2)$, then $\mu _2 \ge p_2-\gamma _2\,\nu >0$ hence $d^*=0$, $h^*_2=H_2^\text {up}$ and $r^*=\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {up} \le {\bar{r}}$.

  • (b) If $\nu =p_2/\gamma _2$, then $d^*=0$, $r^*={\bar{r}}$ and $h^*_2=\left( {\bar{r}}-\gamma _1\,H_1^\text {low}\right) /\gamma _2$. In particular, we have $\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {low} \le {\bar{r}} \le \gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {up}.$

  • (c) If $\nu \in (p_2/\gamma _2 , \delta ]$, then $\lambda _2 \ge \gamma _2\,\nu - p_2 >0$ thus $r^*={\bar{r}}$, $h^*_2=H_2^\text {low}$ and $d^*=\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {low} - {\bar{r}}$. In particular, we have ${\bar{r}} \le \gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {low}$.

• 3. Let $p_1/\gamma _1< \rho< \delta < p_2/\gamma _2$. It is similar to case 1. We have $h^*_1=H_1^\text {low}$ and $h^*_2=H_2^\text {up}$. If ${\bar{r}}<\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {up}$, then $d^* = \gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {up} - r^* >0$, thus $r^*={\bar{r}}$ and $d^*=\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {up} - {\bar{r}}$. If ${\bar{r}}>\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {up}$, then $r^* = \gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {up} - d^* < {\bar{r}}$, thus $d^*=0$ and $r^*=\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {up}$.

• 4. Let $\rho< p_1/\gamma _1< p_2/\gamma _2 < \delta$. We distinguish five cases.

  • (a) If $\nu \in [\rho , p_1/\gamma _1)$, then $d^*=0$. Furthermore, $\mu _1 \ge p_1-\gamma _1\,\nu >0$ and $\mu _2 \ge p_2-\gamma _2\,\nu >0$, hence $h^*_1=H_1^\text {up}$ and $h^*_2=H_2^\text {up}$. Thus, $r^*=\gamma _1\,H_1^\text {up}+\gamma _2\,H_2^\text {up} \le {\bar{r}}$.

  • (b) If $\nu =p_1/\gamma _1$, then $d^*=0$ and $r^*={\bar{r}}$. Since $\mu _2 \ge p_2-\gamma _2\,\nu >0$, we get $h^*_2=H_2^\text {up}$ and $h^*_1 = [{\bar{r}}-\gamma _2\,H_2^\text {up}]/\gamma _1$. In particular, we have $\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {up} \le {\bar{r}} \le \gamma _1\,H_1^\text {up}+\gamma _2\,H_2^\text {up}$.

  • (c) If $\nu \in (p_1/\gamma _1 ,p_2/\gamma _2)$, then $d^*=0$ and $r^*={\bar{r}}$. Since $\lambda _1>0$ and $\mu _2>0$, we have $h^*_1=H_1^\text {low}$ and $h^*_2=H_2^\text {up}$. Thus, ${\bar{r}}=\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {up}$.

  • (d) If $\nu =p_2/\gamma _2$, then $d^*=0$ and $r^*={\bar{r}}$. Since $\lambda _1 > 0$, we get $h^*_1=H_1^\text {low}$ and $h^*_2 = \left( {\bar{r}}-\gamma _1\,H_1^\text {low}\right) /\gamma _2$. In particular, we have $\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {low} \le {\bar{r}} \le \gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {up}$.

  • (e) If $\nu \in (p_2/\gamma _2 , \delta ]$, then $r^*={\bar{r}}$. Furthermore, $\lambda _1 >0$ and $\lambda _2 >0$ thus $h^*_1=H_1^\text {low}$, $h^*_2=H_2^\text {low}$ and $d^*=\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {low} - {\bar{r}}$. In particular, we have ${\bar{r}} \le \gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {low}$.

• 5. Let $\rho< p_1/\gamma _1< \delta < p_2/\gamma _2$. It is similar to case B. We have $h^*_2=H_2^\text {up}$ and distinguish three cases:

  • (a) If $\nu \in [\rho , p_1/\gamma _1)$, then $d^*=0$, $h^*_1=H_1^\text {up}$ and $r^*=\gamma _1\,H_1^\text {up}+\gamma _2\,H_2^\text {up} \le {\bar{r}}$.

  • (b) If $\nu =p_1/\gamma _1$, then $d^*=0$, $r^*={\bar{r}}$ and $h^*_1=\left( {\bar{r}}-\gamma _2\,H_2^\text {up}\right) /\gamma _1$. In particular, we have $\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {up} \le {\bar{r}} \le \gamma _1\,H_1^\text {up}+\gamma _2\,H_2^\text {up}$.

  • (c) If $\nu \in (p_1/\gamma _1 , \delta ]$, then $r^*={\bar{r}}$, $h^*_1=H_1^\text {low}$ and $d^*=\gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {up} - {\bar{r}}$. In particular, we have ${\bar{r}} \le \gamma _1\,H_1^\text {low}+\gamma _2\,H_2^\text {up}$.

• 6. Let $\delta< p_1/\gamma _1 < p_2/\gamma _2$. It is similar to case 1. We have $h^*_1=H_1^\text {up}$ and $h^*_2=H_2^\text {up}$. If ${\bar{r}}<\gamma _1\,H_1^\text {up}+\gamma _2\,H_2^\text {up}$, then $d^* = \gamma _1\,H_1^\text {up}+\gamma _2\,H_2^\text {up} - r^* >0$, thus $r^*={\bar{r}}$ and $d^*=\gamma _1\,H_1^\text {up}+\gamma _2\,H_2^\text {up}-{\bar{r}}$. If ${\bar{r}}>\gamma _1\,H_1^\text {up}+\gamma _2\,H_2^\text {up}$, then $r^* = \gamma _1\,H_1^\text {up}+\gamma _2\,H_2^\text {up} - d^* < {\bar{r}}$, thus $d^*=0$ and $r^*=\gamma _1\,H_1^\text {up}+\gamma _2\,H_2^\text {up}$. $\square$