Worst-case analysis of non-cooperative load balancing

Abstract

We investigate the impact of heterogeneity in the amount of incoming traffic routed by dispatchers in a non-cooperative load balancing game. For a fixed amount of total incoming traffic, we show that for a broad class of cost functions the worst-case social cost occurs when each dispatcher routes the same amount of traffic, that is, the game is symmetric. Using this result, we give lower bounds on the Price of Anarchy for (i) cost functions that are polynomial on server loads; and (ii) cost functions representing the mean delay of the shortest remaining processing time service discipline.

Appendices

Proof of results in Sect. 3.1

1.1 Proof of Proposition 1

We first prove a series of technical lemmata before proving Proposition 1.

Lemma 7

${\mathcal {S}}_i\cap {\mathcal {S}}_k\ne \emptyset$.

Proof

Assume the contrary, i.e., if $m\in {\mathcal {S}}_i$ then $m\notin {\mathcal {S}}_k$, and if $n\in {\mathcal {S}}_k$ then $n\notin {\mathcal {S}}_i$. For one such pair $m$ and $n$, from (4), we can conclude that $\mu _i > \psi ^\prime _m(0,y_m) \ge \mu _k$ and $\mu _k > \psi ^\prime _n(0,y_n) \ge \mu _i$, which is a contradiction. $\square$

Since ${\mathcal {S}}_i\cap {\mathcal {S}}_k\ne \emptyset$, from (2), we have

$$\begin{aligned} \mu _i - \mu _k = \psi ^\prime _j(x_{i,j},y_j) - \psi ^\prime _j(x_{k,j},y_j), \quad \forall j\in {\mathcal {S}}_i\cap {\mathcal {S}}_k. \end{aligned}$$

(19)

Lemma 8

$\mu _i < \mu _k \iff \exists j\in {\mathcal {S}}_k : x_{i,j} < x_{k,j}$.

Proof

Straight part: From Lemma 7, ${\mathcal {S}}_i\cap {\mathcal {S}}_k\ne \emptyset$. If $\mu _i < \mu _k$, then, from (19), $\exists j\in {\mathcal {S}}_k : \psi ^\prime _j(x_{i,j},y_j) < \psi ^\prime _j(x_{k,j},y_j)$ which implies that $\exists j\in {\mathcal {S}}_k : x_{i,j} < x_{k,j}$.

Converse part: $\exists j\in {\mathcal {S}}_k : x_{i,j} < x_{k,j}$. Either $j\in {\mathcal {S}}_i$ or $j\notin {\mathcal {S}}_i$. If $j\in {\mathcal {S}}_i$ then, from (19), $\psi ^\prime _j(x_{i,j},y_j) < \psi ^\prime _j(x_{k,j},y_j)$ implies $\mu _i < \mu _k$. If $j\notin {\mathcal {S}}_i$, then, from (4), $\mu _i\le \psi ^\prime _j(0,y_j) < \psi ^\prime _j(x_{k,j},y_j) < \mu _k$.

Lemma 9

If $\mu _i < \mu _k$, then ${\mathcal {S}}_i\subset {\mathcal {S}}_k$.

Proof

If $j\in {\mathcal {S}}_i$, then, from (4), $\psi ^\prime _j(0,y_j) < \mu _i$. If $\mu _i < \mu _k$ then $\psi ^\prime _j(0,y_j) < \mu _k$. Hence, from (4) we can conclude that $j\in {\mathcal {S}}_k$. Therefore, ${\mathcal {S}}_i\subset {\mathcal {S}}_k$. $\square$

Lemma 10

$\exists m\in {\mathcal {S}}_k : x_{i,m} < x_{k,m} \iff x_{i,j} < x_{k,j},\quad \forall j\in {\mathcal {S}}_k$.

Proof

Straight part: If $\exists m\in {\mathcal {S}}_k : x_{i,m} < x_{k,m}$, then, from Lemmata 8 and 9, we have $\mu _i < \mu _k$ and ${\mathcal {S}}_i\subset {\mathcal {S}}_k$. For $j\in {\mathcal {S}}_i$, from (19), we have $\psi ^\prime _j(x_{i,j},y_j) < \psi ^\prime _j(x_{k,j},y_j)$, which implies $x_{i,j} < x_{k,j}$. For $j\in {\mathcal {S}}_k\setminus {\mathcal {S}}_i$, $x_{i,j} = 0$ and $0 < x_{k,j}$. Hence, $x_{i,j} < x_{k,j}$, $\forall j\in {\mathcal {S}}_k$.

Converse part: It is true from the statement. $\square$

We are now in position to prove Proposition 1.

Proof (Proof of Proposition 1)

1 $\iff$ 2 $\iff$ 3 follows from Lemmata 8 and 10. Now, we show 3 $\iff$ 4.

Straight part: If $x_{i,j} < x_{k,j}, \;\forall j\in {\mathcal {S}}_k$, then, from the fact that 3 $\iff$ 1 and Lemma 9, we can conclude that $\lambda _i = \sum _{j\in {\mathcal {S}}_i}x_{i,j} = \sum _{j\in {\mathcal {S}}_k} x_{i,j} < \sum _{j\in {\mathcal {S}}_k}x_{k,j} = \lambda _k$.

Converse part: Since $\lambda _k = \sum _{j\in {\mathcal {S}}_k}x_{k,j}$, if $\lambda _i < \lambda _k$, then $\exists j\in {\mathcal {S}}_k : x_{i,j} < x_{k,j}$. Since 2 $\iff$ 3, if $\lambda _i < \lambda _k$, then $x_{i,j} < x_{k,j}, \;\forall j\in {\mathcal {S}}_k$. $\square$

1.2 Proof of Proposition 2

As before, we first prove a series of technical lemmata.

Lemma 11

$$\begin{aligned} \forall a,b \in {\mathcal {D}}-\{0\}, \quad b \, \frac{\phi ^\prime (b)}{\phi (b)} \ge a \, \frac{\phi ^\prime (a)}{\phi (a)} \iff b\ge a \end{aligned}$$

Proof

See the proof of Lemma 2 in Brun and Prabhu (2012). $\square$

Lemma 12

${\mathcal {C}}_m\cap {\mathcal {C}}_n\ne \emptyset$.

Proof

Assume the contrary, i.e., if $i\in {\mathcal {C}}_m$, then $i\notin {\mathcal {C}}_n$, and if $k\in {\mathcal {C}}_n$, then $k\notin {\mathcal {C}}_m$. For one such pair $i$ and $k$, from (4), we can conclude that $\psi ^\prime _m(0,y_m)<\mu _i \le \psi ^\prime _n(0,y_n)$ and $\psi ^\prime _n(0,y_n) < \mu _k \le \psi ^\prime _m(0,y_m)$, which is a contradiction. $\square$

Lemma 13

If $\psi ^\prime _n(0,y_n) \le \psi ^\prime _m(0,y_m)$, then ${\mathcal {C}}_m\subseteq {\mathcal {C}}_n$.

Proof

If $i\in {\mathcal {C}}_m$, then, from (4), $\mu _i > \psi ^\prime _m(0,y_m)$. If $\psi ^\prime _n(0,y_n) \le \psi ^\prime _m(0,y_m)$, then $\mu _i>\psi ^\prime _n(0,y_n)$. Hence, from (4) we can conclude that $i\in {\mathcal {C}}_n$. Therefore, ${\mathcal {C}}_m\subseteq {\mathcal {C}}_n$. $\square$

Since ${\mathcal {C}}_m\cap {\mathcal {C}}_n\ne \emptyset$, from (2), we have

$$\begin{aligned} \mu _i = \psi ^\prime _m(x_{i,m},y_m) = \psi ^\prime _n(x_{i,n},y_n) , \quad \forall i\in {\mathcal {C}}_m\cap {\mathcal {C}}_n. \end{aligned}$$

(20)

Lemma 14

We have $\psi ^\prime _n(0,y_n) < \psi ^\prime _m(0,y_m)$ if and only if

$$\begin{aligned} \exists i\in {\mathcal {C}}_n : \frac{u_m \, x_{i,m}}{r_m} \, \phi ^\prime (\rho _m) < \frac{u_n \, x_{i,n}}{r_n} \, \phi ^\prime (\rho _n) . \end{aligned}$$

Proof

Straight part: From Lemma 12, ${\mathcal {C}}_m\cap {\mathcal {C}}_n\ne \emptyset$. If $\psi ^\prime _n(0,y_n) < \psi ^\prime _m(0,y_m)$, then, from (20), $\exists i \in {\mathcal {C}}_n: \psi ^\prime _m(x_{i,m},y_m) - \psi ^\prime _m(0,y_m) < \psi ^\prime _n(x_{i,n},y_n)-\psi ^\prime _n(0,y_n)$, i.e. $\frac{u_m \, x_{i,m}}{r_m} \, \phi ^\prime (\rho _m) < \frac{u_n \, x_{i,n}}{r_n} \, \phi ^\prime (\rho _n)$.

Converse part: Assume $\exists i\in {\mathcal {C}}_n : \frac{u_m \, x_{i,m}}{r_m} \, \phi ^\prime (\rho _m) < \frac{u_n \, x_{i,n}}{r_n} \, \phi ^\prime (\rho _n)$. Then $\psi ^\prime _m(x_{i,m},y_m) - \psi ^\prime _m(0,y_m) < \psi ^\prime _n(x_{i,n},y_n)-\psi ^\prime _n(0,y_n)$. Either $i\in {\mathcal {C}}_m$ or $i\notin {\mathcal {C}}_m$. If $i\in {\mathcal {C}}_m$, then, from (20), $\psi ^\prime _n(0,y_n) < \psi ^\prime _m(0,y_m)$. If $i\notin {\mathcal {C}}_m$, then, from (3), $\psi ^\prime _m(0,y_m) \ge \mu _i = \psi ^\prime _n(x_{i,n},y_n) > \psi ^\prime _n(0,y_n)$. Hence, $\psi ^\prime _n(0,y_n) < \psi ^\prime _m(0,y_m)$. $\square$

Lemma 15

We have $\psi ^\prime _n(0,y_n) < \psi ^\prime _m(0,y_m)$ if and only if

$$\begin{aligned} \frac{u_m \, x_{i,m}}{r_m} \, \phi ^\prime (\rho _m) < \frac{u_n \, x_{i,n}}{r_n} \, \phi ^\prime (\rho _n), \ \forall i\in {\mathcal {C}}_n. \end{aligned}$$

Proof

Straight part: If $\psi ^\prime _n(0,y_n) < \psi ^\prime _m(0,y_m)$, then from Lemma 13, ${\mathcal {C}}_m\subset {\mathcal {C}}_n$. For $i\in {\mathcal {C}}_m$, from (20), $\psi ^\prime _m(x_{i,m},y_m) -\psi ^\prime _m(0,y_m) < \psi ^\prime _n(x_{i,n},y_n)-\psi ^\prime _n(0,y_n)$, i.e., $\frac{u_m \, x_{i,m}}{r_m} \, \phi ^\prime (\rho _m)<\frac{u_n \, x_{i,n}}{r_n} \, \phi ^\prime (\rho _n)$. For $i\in {\mathcal {C}}_n\setminus {\mathcal {C}}_m$, $x_{i,m} = 0$ and $0 < x_{i,n}$. Hence, since $\phi$ is strictly increasing, $\frac{u_m \, x_{i,m}}{r_m} \, \phi ^\prime (\rho _m) =0 <\frac{u_n \, x_{i,n}}{r_n} \, \phi ^\prime (\rho _n)$.

Converse part: the proof is a direct consequence of Lemma 14. $\square$

We are now in position to prove Proposition 2.

Proof (Proof of Proposition 2)

1 $\iff$ 2 $\iff$ 3 follows from Lemmata 14 and 15. Next, we show 3 $\iff$ 4.

Straight part: If $\frac{u_m \, x_{i,m}}{r_m} \, \phi ^\prime (\rho _m) < \frac{u_n \, x_{i,n}}{r_n} \, \phi ^\prime (\rho _n), \;\forall i\in {\mathcal {C}}_n$, then from the fact that 3 $\iff$ 1 and Lemma 13, we can conclude that

$$\begin{aligned} u_m \, \rho _m \, \phi ^\prime (\rho _m)&= \sum _{i \in {\mathcal {C}}_m}{\frac{u_m \, x_{i,m}}{r_m} \, \phi ^\prime (\rho _m)} \nonumber \\&< \sum _{i \in {\mathcal {C}}_n}{\frac{u_n \, x_{i,n}}{r_n} \, \phi ^\prime (\rho _n)} \nonumber \\&= u_n \, \rho _n \, \phi ^\prime (\rho _n) \end{aligned}$$

(21)

Observe that it implies $\rho _n>0$. Assume first that $\rho _m>0$. Note that $\psi ^\prime _n(0,y_n) < \psi ^\prime _m(0,y_m)$ can also be written as $u_n \phi (\rho _n) < u_m \phi (\rho _m)$. Together with inequality 21, it implies that $\rho _n \, \frac{\phi ^\prime (\rho _n)}{\phi (\rho _n)} > \rho _m \, \frac{\phi ^\prime (\rho _m)}{\phi (\rho _m)}$, which is equivalent to $\rho _n>\rho _m$ according to Lemma 11. Clearly, $\rho _n>\rho _m$ also holds if $\rho _m=0$. We thus get $u_n \, \phi (\rho _m) < u_n \, \phi (\rho _n) < u_m \, \phi (\rho _m)$ in both cases, which implies that $u_n < u_m$, as claimed.

Converse part: To prove $4 \Longrightarrow 1$, we prove that $\lnot 1 \Longrightarrow \lnot 4$. Assume that $u_n \phi (\rho _n) \ge u_m \phi (\rho _m)$. Since $\lnot 1 \iff \lnot 2$, we get $\frac{u_m \, x_{i,m}}{r_m} \, \phi ^\prime (\rho _m) \ge \frac{u_n \, x_{i,n}}{r_n} \, \phi ^\prime (\rho _n)$, $\forall i\in {\mathcal {C}}_n$. According to Lemma 13, $u_n \phi (\rho _n) \ge u_m \phi (\rho _m)$ implies ${\mathcal {C}}_n \subseteq {\mathcal {C}}_m$, and thus

$$\begin{aligned} u_m \rho _m \phi ^\prime (\rho _m)&\ge \sum _{i \in {\mathcal {C}}_m}{\frac{u_m \, x_{i,m}}{r_m} \, \phi ^\prime (\rho _m)} \\&\ge \sum _{i \in {\mathcal {C}}_n}{\frac{u_n \, x_{i,n}}{r_n} \, \phi ^\prime (\rho _n)} \\&= u_n \rho _n \phi ^\prime (\rho _n). \end{aligned}$$

Let us first assume that $\rho _n>0$ and $\rho _m>0$. Together with $u_n \phi (\rho _n) \ge u_m \phi (\rho _m)$, it implies that $\rho _n \,\frac{\phi ^\prime (\rho _n)}{\phi (\rho _n)} \le \rho _m \, \frac{\phi ^\prime (\rho _m)}{\phi (\rho _m)}$, which is equivalent to $\rho _n \le \rho _m$ according to Lemma 11. Clearly, $\rho _n \le \rho _m$ still holds if $\rho _n=0$. The case $\rho _m=0<\rho _n$ is impossible because ${\mathcal {C}}_n \subseteq {\mathcal {C}}_m$. We thus obtain that $\rho _n \le \rho _m$ in all cases. But from $u_n \phi (\rho _n) \ge u_m \phi (\rho _m)$ and $\phi (\rho _m)\ge \phi (\rho _n)$, we deduce that $u_n \ge u_m$. We thus conclude that if $u_n < u_m$, then $u_n\phi (\rho _n) < u_m \phi (\rho _m)$, i.e., $\psi ^\prime _n(0,y_n) < \psi ^\prime _m(0,y_m)$. $\square$

1.3 Proof of Lemma 1

We give below the proof of lemmata 1.

Proof (Proof of Lemma 1)

From (2), if $x_{i,j} > 0$, then

$$\begin{aligned} \mu _i = \psi ^\prime _j(x_{i,j},y_j)= u_j \left[ \phi (\rho _j)+ \frac{x_{i,j}}{r_j} \, \phi ^\prime (\rho _j) \right] , \end{aligned}$$

from which we conclude that

$$\begin{aligned} \sum _{i \in {\mathcal {C}}_j}\mu _i = u_j \, \left[ N_j \, \phi (\rho _j) + \rho _j \, \phi ^\prime (\rho _j) \right] = (N_j-1) \, u_j \, \phi (\rho _j) + \frac{\partial D_K}{\partial y_j}(\mathbf{x}) \end{aligned}$$

Similarly, we have

$$\begin{aligned} \sum _{i \in {\mathcal {C}}_{j+1}}\mu _i = (N_{j+1}-1) \, u_{j+1} \, \phi (\rho _{j+1}) + \frac{\partial D_K}{\partial y_{j+1}}(\mathbf{x}) \end{aligned}$$

Now,

$$\begin{aligned} \sum _{i \in {\mathcal {C}}_{j}}\mu _i - \sum _{i \in {\mathcal {C}}_{j+1}}\mu _i&= \sum _{i\in {\mathcal {C}}_j\setminus {\mathcal {C}}_{j+1}}\mu _i \\&= (N_j \! - \! N_{j+1}) \, u_j \phi (\rho _j) + u_j \phi ^\prime (\rho _j) \! \sum _{i\in {\mathcal {C}}_j\setminus {\mathcal {C}}_{j+1}} \! \frac{x_{i,j}}{r_j}. \end{aligned}$$

Thus,

$$\begin{aligned} \frac{\partial D_K}{\partial y_j}(\mathbf{x}) - \frac{\partial D_K}{\partial y_{j+1}}(\mathbf{x})&= \left( u_{j+1} \phi (\rho _{j+1}) - u_j \phi (\rho _j) \right) (N_{j+1}-1) \\&+\,u_j \phi ^\prime (\rho _j) \!\! \sum _{i\in {\mathcal {C}}_j\setminus {\mathcal {C}}_{j+1}}\frac{x_{i,j}}{r_j}. \end{aligned}$$

From Proposition 2, we have $u_{j+1} \phi (\rho _{j+1}) \ge u_j \phi (\rho _j)$. Moreover $\phi ^\prime (\rho _j)>0$ because $\phi$ is strictly increasing. Since the second term on the RHS is strictly positive if ${\mathcal {C}}_j\setminus {\mathcal {C}}_{j+1}\ne \emptyset$, we can conclude that $\frac{\partial D_K}{\partial y_j}(\mathbf{x}) \ge \frac{\partial D_K}{\partial y_{j+1}}(\mathbf{x})$, with strict inequality if ${\mathcal {C}}_j\setminus {\mathcal {C}}_{j+1}\ne \emptyset$. $\square$

Proof of results in Sect. 3.2

1.1 Proof of the results in Sect. 3.2.1

We give below the proofs of lemmata 2–5.

Proof (Proof of Lemma 2)

Proof of part $1$ : Since $\psi ^\prime _j(x, y)$ is strictly increasing in each of its two arguments, $\hat{y}_j \le y_j$ and $\hat{x}_{i,j} \le x_{i,j}$ implies that $\hat{\mu }_i = \psi ^\prime _j(\hat{x}_{i,j},\hat{y}_j) \le \psi ^\prime _j(x_{i,j},y_j) =\mu _i$. The proofs of parts $2$, $3$ and $4$ follow similarly. $\square$

Proof (Proof of Lemma 3)

Assume the contrary. From Lemma 2.$4$, $\hat{y}_m > y_m$ and $\hat{x}_{i,m} \ge x_{i,m}$ implies $\hat{\mu }_i > \mu _i$. However, from Lemma 2.$1$, $\hat{y}_n \le y_n$ and $\hat{x}_{i,n} \le x_{i,n}$ implies $\hat{\mu }_i \le \mu _i$, which is a contradiction. $\square$

Proof (Proof of Lemma 4)

From (2), if $i \in {\mathcal {C}}_j$, then

$$\begin{aligned} \mu _i = \psi ^\prime _j(x_{i,j},y_j)=u_j \, \left[ \phi (\rho _j)+\frac{x_{i,j}}{r_j} \, \phi ^\prime (\rho _j) \right] \end{aligned}$$

Thus,

$$\begin{aligned} \sum _{i\in {\mathcal {C}}_j} \mu _i = \left( N_j-1 \right) \, u_j \, \phi (\rho _j) + \psi ^\prime _j(y_j,y_j) \end{aligned}$$

Since $N_j = \hat{N}_j$ (from Assumption 1) and $\psi ^\prime _j(x, y)$ is strictly increasing in each of its two arguments, we can conclude that $\sum _{i\in {\mathcal {C}}_j}\mu _i$ is a strictly increasing function of $y_j$. $\square$

Proof (Proof of Lemma 5)

From Lemma 4, $\hat{y}_m > y_m$ is equivalent to $\sum _{i\in {\mathcal {C}}_m}\hat{\mu }_i > \sum _{i\in {\mathcal {C}}_m}\mu _i$, which, since ${\mathcal {C}}_m = {\mathcal {C}}_n$, is equivalent to $\sum _{i\in {\mathcal {C}}_n}\hat{\mu }_i > \sum _{i\in {\mathcal {C}}_n}\mu _i$. Again, from Lemma 4, we can conclude that $\hat{y}_n > y_n$.

1.2 Proof of the results in Sect. 3.2.2

We give below the proofs of propositions 3, 4 and 5.

We first prove a lemma that shows that ${\mathcal {S}}^+$ is empty if and only if the load of each and every server is constant under the transformation.

Lemma 16

$y_j=\hat{y}_j, \forall j\in {\mathcal {S}}\iff {\mathcal {S}}^+=\emptyset$.

Proof

If ${\mathcal {S}}^+=\emptyset$ then ${\mathcal {S}}^-={\mathcal {S}}$. That is, $\hat{y}_j \le y_j, \forall j\in {\mathcal {S}}$. We also have $\sum _{j\in {\mathcal {S}}}\hat{y}_j = \sum _{j\in {\mathcal {S}}}y_j$. This is possible only if $\hat{y}_j=y_j,\forall j\in {\mathcal {S}}$.

The converse is true by definition of ${\mathcal {S}}^+$. $\square$

Proof

[Proof of Proposition 3] Assume by contradiction that we can find a server $s \in {\mathcal {S}}_1$ such that $s \in {\mathcal {S}}^-$. Then, according to Corollary 1, ${\mathcal {S}}_1 \subset {\mathcal {S}}^-$. Since ${\mathcal {S}}^+ \ne \emptyset$ and $\hat{y}_j > y_j$ for all $j \in {\mathcal {S}}^+$, we have $\sum _{j \in {\mathcal {S}}^+}{\hat{y}_j} > \sum _{j \in {\mathcal {S}}^+}{y_j}$, i.e.,

$$\begin{aligned} \sum _{i\in {\mathcal {C}}}{\left( \sum _{j \in {\mathcal {S}}^+}{\hat{x}_{i,j}} \right) } > \sum _{i\in {\mathcal {C}}}{\left( \sum _{j \in {\mathcal {S}}^+}{x_{i,j}} \right) }, \end{aligned}$$

from which we conclude that there exists $i$ such that $\sum _{j \in {\mathcal {S}}^+}{\hat{x}_{i,j}} > \sum _{j \in \mathcal{S}^+}{x_{i,j}}$. Since ${\mathcal {S}}_k = {\mathcal {S}}_1 \subset {\mathcal {S}}^-$ for all $k \in {\mathcal {C}}_{min}$, we necessarily have $i \not \in {\mathcal {C}}_{min}$ and thus $\hat{\lambda }_i \le \lambda _i$. Therefore,

$$\begin{aligned} \hat{\lambda }_i = \sum _{j \in {\mathcal {S}}^-}{\hat{x}_{i,j}} + \sum _{j \in {\mathcal {S}}^+}{\hat{x}_{i,j}} \le \sum _{j \in {\mathcal {S}}^-}{x_{i,j}} + \sum _{j \in {\mathcal {S}}^+}{x_{i,j}} = \lambda _i. \end{aligned}$$

Thus,

$$\begin{aligned} \sum _{j \in {\mathcal {S}}^-}{\hat{x}_{i,j}} \le \sum _{j \in {\mathcal {S}}^-}{x_{i,j}} + \left( \sum _{j \in {\mathcal {S}}^+}{x_{i,j}} - \sum _{j \in {\mathcal {S}}^+}{\hat{x}_{i,j}} \right) < \sum _{j \in {\mathcal {S}}^-}{x_{i,j}}. \end{aligned}$$

We therefore conclude that class $i$ is such that $\sum _{j \in {\mathcal {S}}^+}{\hat{x}_{i,j}} > \sum _{j \in {\mathcal {S}}^+}{x_{i,j}}$ and $\sum _{j \in {\mathcal {S}}^-}{\hat{x}_{i,j}} <\sum _{j \in {\mathcal {S}}^-}{x_{i,j}}$. Therefore, we can find a server $m \in {\mathcal {S}}^+$ and a server $n \in {\mathcal {S}}^-$ such that $\hat{x}_{i,m}> x_{i,m}$ and $\hat{x}_{i,n} < x_{i,n}$. But according to Lemma 3, this is impossible. We therefore conclude that ${\mathcal {S}}_1 \subset {\mathcal {S}}^+$. $\square$

Proof

[Proof of Proposition 4] We first prove that if $S^+=\emptyset$ then ${\mathcal {S}}_1={\mathcal {S}}_K$. From Lemma 16, this is equivalent to proving that if $y_j=\hat{y}_j$, $\forall j \in {\mathcal {S}}$ then ${\mathcal {S}}_1 = {\mathcal {S}}_K$. Assume the contrary, that is ${\mathcal {S}}_1\subsetneq {\mathcal {S}}_K$. Then, $\exists m: m\in S_K, m\notin S_1$.

Since $y_m=\hat{y}_m$, from Lemma 4, we get $\sum _{i\in {\mathcal {C}}_m} \mu _i = \sum _{i\in {\mathcal {C}}_m} \hat{\mu }_i$, which we can rewrite as

$$\begin{aligned} \sum _{i\in {\mathcal {C}}_{max}} \mu _i + \sum _{i\in {\mathcal {C}}_m\setminus {\mathcal {C}}_{max}} \mu _i = \sum _{i\in {\mathcal {C}}_{max}} \hat{\mu }_i + \sum _{i\in {\mathcal {C}}_m\setminus {\mathcal {C}}_{max}} \hat{\mu }_i. \end{aligned}$$

(22)

We shall show that the above equality is not possible, which then proves the claim.

For $i \in {\mathcal {C}}_{max}$, since $\lambda _i > {\hat{\lambda }}_i$, $\sum _{j\in {\mathcal {S}}_i}x_{i,j} > \sum _{j\in {\mathcal {S}}_i}\hat{x}_{i,j}$. Thus, there exists an $n\in {\mathcal {S}}_i$ such that $x_{i,n}>\hat{x}_{i,n}$. Since $y_n=\hat{y}_n$, from Lemma 2.$3$, we can conclude that $\mu _i > \hat{\mu }_i$, and that $\sum _{i \in {\mathcal {C}}_{max}}{\mu _i}>\sum _{i \in {\mathcal {C}}_{max}}{\hat{\mu }_i}$, which, upon substitution in (22), leads to

$$\begin{aligned} \sum _{i\in {\mathcal {C}}_m\setminus {\mathcal {C}}_{max}} \mu _i < \sum _{i\in {\mathcal {C}}_m\setminus {\mathcal {C}}_{max}} \hat{\mu }_i. \end{aligned}$$

If ${\mathcal {C}}_m\setminus {\mathcal {C}}_{max}=\emptyset$, then the above inequality cannot be possible which then proves the claim. So, assume ${\mathcal {C}}_m\setminus {\mathcal {C}}_{max}\ne \emptyset$. Then the above inequality implies that $\exists i\notin C_{min}\cup C_{max} : \mu _i < \hat{\mu }_i$. Since $y_j=\hat{y}_j, \forall j\in {\mathcal {S}}_i$, application of Lemma 2.4 leads to $x_{i,j} < \hat{x}_{i,j}, \forall j\in {\mathcal {S}}_i$, and consequently to $\lambda _i = \sum _{j\in {\mathcal {S}}_i}x_{i,j} < \sum _{j\in {\mathcal {S}}_i}\hat{x}_{i,j} = {\hat{\lambda }}_i$. However, for $i\notin {\mathcal {C}}_{min}\cup {\mathcal {C}}_{max}$, $\lambda _i={\hat{\lambda }}_i$. Hence, there is a contradiction, and we can conclude that ${\mathcal {S}}_1={\mathcal {S}}_K$.

We now show that if ${\mathcal {S}}^+ \ne \emptyset$ then ${\mathcal {S}}_1\ne {\mathcal {S}}_K$. Assume the contrary, that is ${\mathcal {S}}_1 ={\mathcal {S}}_K$. From Proposition 3, if ${\mathcal {S}}^+ \ne \emptyset$ then ${\mathcal {S}}_1 = {\mathcal {S}}_K \subset {\mathcal {S}}^+$ which implies that ${\mathcal {S}}^-= \emptyset$, i.e., a contradiction. $\square$

Proof

[Proof of Proposition 5] If ${\mathcal {S}}^+ =\emptyset$ then the proposition is true. So, assume ${\mathcal {S}}^+\ne \emptyset$. Then, from Proposition 3, ${\mathcal {S}}_1\subset {\mathcal {S}}^+$. In order to prove the proposition, assume by contradiction that there exists a server $j \in \{ S_1+1, \ldots , S_K-1 \}$ such that $j \in {\mathcal {S}}^-$ and $j+1 \in {\mathcal {S}}^+$. Again, if $S_1+1=S_K$ then the proposition is true. So, assume that $S_1+1<S_K$.

Since $j\in {\mathcal {S}}^-$ and $j+1\in {\mathcal {S}}^+$, from Lemma 5,

$$\begin{aligned} \sum _{i\in {\mathcal {C}}_j}\hat{\mu }_i \le \sum _{i\in {\mathcal {C}}_j}{\mu }_i, \end{aligned}$$

(23)

and

$$\begin{aligned} \sum _{i\in {\mathcal {C}}_{j+1}}\hat{\mu }_i > \sum _{i\in {\mathcal {C}}_{j+1}}{\mu }_i, \end{aligned}$$

(24)

Moreover, from the contrapositive of Lemma 5, we can conclude that ${\mathcal {C}}_j{\setminus }{\mathcal {C}}_{j+1}\ne \emptyset$. Note that since $j<S_K$, classes $i \in {\mathcal {C}}_{max}$ do not belong to ${\mathcal {C}}_j{\setminus }{\mathcal {C}}_{j+1}$. Similarly, since $j>S_1$, classes $i \in {\mathcal {C}}_{min}$ do not belong to ${\mathcal {C}}_j{\setminus }{\mathcal {C}}_{j+1}$.

Since ${\mathcal {C}}_{j+1} \subset {\mathcal {C}}_j$, we have, for all $i \in {\mathcal {C}}_{j+1}$,

$$\begin{aligned} \mu _i = \psi ^\prime _j(x_{i,j},y_j)=u_j \left[ \phi (\rho _j)+\frac{x_{i,j}}{r_j} \, \phi ^\prime (\rho _j) \right] . \end{aligned}$$

Therefore, $\sum _{i\in {\mathcal {C}}_{j+1}}\hat{\mu }_i > \sum _{i\in {\mathcal {C}}_{j+1}}\mu _i$ is equivalent to

$$\begin{aligned} \left( N_{j+1}-1 \right) \! \phi (\hat{\rho }_j) + \psi ^\prime _j \left( \sum _{i \in {\mathcal {C}}_{j+1}}{\hat{x}_{i,j}} , \hat{y}_j \right) > \left( N_{j+1}-1 \right) \, \phi (\rho _j) + \psi ^\prime _j \left( \sum _{i \in {\mathcal {C}}_{j+1}}{x_{i,j}}, y_j \right) \end{aligned}$$

and since $\hat{y}_j \le y_j$, this implies that $\sum _{i \in {\mathcal {C}}_{j+1}}{\hat{x}_{i,j}} > \sum _{i \in {\mathcal {C}}_{j+1}}{x_{i,j}}$. Since $\hat{y}_j \le y_j$, necessarily $\sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1}}{\hat{x}_{i,j}} < \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1}}{x_{i,j}}$. However, since all classes $k \in {\mathcal {C}}_{min} \cup {\mathcal {C}}_{max}$ do not belong to $\mathcal{C}_j\setminus {\mathcal {C}}_{j+1}$, we know that $\hat{\lambda }_i = \lambda _i$ for all $i \in \mathcal{C}_j{\setminus }{\mathcal {C}}_{j+1}$, and thus

$$\begin{aligned} \sum _{l=1}^j{\sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1}}{x_{i,l}}} = \sum _{l=1}^j{\sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1}}{\hat{x}_{i,l}}}, \end{aligned}$$

from which we obtain

$$\begin{aligned} \sum _{l<j}{\sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1}}{x_{i,l}}} = \sum _{l<j}{\sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1}}{\hat{x}_{i,l}}} + \left( \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1}}{\hat{x}_{i,j}} - \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1}}{x_{i,j}}\right) \!, \end{aligned}$$

and therefore

$$\begin{aligned} \sum _{l<j}{\sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1}}{x_{i,l}}} < \sum _{l<j}{\sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1}}{\hat{x}_{i,l}}}. \end{aligned}$$

(25)

Subtracting (24) from (23), we obtain

$$\begin{aligned} \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1}}{\hat{\mu }_i} < \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1}}{\mu _i}. \end{aligned}$$

Hence, for each server $l<j$,

$$\begin{aligned}&\left( N_j-N_{j+1} \right) \, \phi (\hat{\rho }_l) + \psi ^\prime _l \left( \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1} }{\hat{x}_{i,l}} , \hat{y}_l \right) < \left( N_j-N_{j+1} \right) \, \phi (\rho _l)\\&\quad +\,\psi ^\prime _l \left( \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1} }{x_{i,l}}, y_l \right) . \end{aligned}$$

But, for $l<j$ and $l \in {\mathcal {S}}^+$, it implies that

$$\begin{aligned} \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1} }{\hat{x}_{i,l}} < \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1} }{x_{i,l}}, \end{aligned}$$

and thus

$$\begin{aligned} \sum _{l<j,l \in {\mathcal {S}}^+}{ \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1} }{\hat{x}_{i,l}} } < \sum _{l<j,l \in {\mathcal {S}}^+}{ \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1} }{x_{i,l}} }. \end{aligned}$$

(26)

From (25), we have

$$\begin{aligned}&\sum _{l<j,l \in {\mathcal {S}}^-}{ \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1} }{\hat{x}_{i,l}} > \sum _{l<j,l \in {\mathcal {S}}^-}{ \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1} }{x_{i,l}} }}\\&\quad + \left( \sum _{l<j,l \in {\mathcal {S}}^+}{ \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1} }{\! x_{i,l}} } \! - \! \sum _{l<j,l \in {\mathcal {S}}^+}{ \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1} }{\! \hat{x}_{i,l}} } \right) , \end{aligned}$$

and using (26) it leads to

$$\begin{aligned} \sum _{l<j,l \in {\mathcal {S}}^-}{ \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1} }{\hat{x}_{i,l}} } > \sum _{l<j,l \in {\mathcal {S}}^-}{ \sum _{i \in {\mathcal {C}}_j \setminus {\mathcal {C}}_{j+1} }{x_{i,l}} }. \end{aligned}$$

(27)

According to (24), for each server $l<j$,

$$\begin{aligned} \left( N_{j+1}-1 \right) \, \phi (\hat{\rho }_l) + \psi ^\prime _l \left( \sum _{i \in {\mathcal {C}}_{j+1} }{\hat{x}_{i,l}} , \hat{y}_l \right) > \left( N_{j+1}-1 \right) \, \phi (\rho _l) + \psi ^\prime _l \left( \sum _{i \in {\mathcal {C}}_{j+1} }{x_{i,l}}, y_l \right) . \end{aligned}$$

But, for $l<j$, $l \in {\mathcal {S}}^-$, it implies that

$$\begin{aligned} \sum _{i \in {\mathcal {C}}_{j+1} }{\hat{x}_{i,l}} > \sum _{i \in {\mathcal {C}}_{j+1} }{x_{i,l}}, \end{aligned}$$

and thus

$$\begin{aligned} \sum _{l<j, l \in {\mathcal {S}}^-}{ \sum _{i \in {\mathcal {C}}_{j+1} }{\hat{x}_{i,l}} } > \sum _{l<j, l \in {\mathcal {S}}^-}{ \sum _{i \in {\mathcal {C}}_{j+1} }{x_{i,l}} } \end{aligned}$$

(28)

Now, summing (28) and (27) gives

$$\begin{aligned} \sum _{l<j, l \in {\mathcal {S}}^-}{ \sum _{i \in {\mathcal {C}}_j }{\hat{x}_{i,l}} } > \sum _{l<j, l \in {\mathcal {S}}^-}{ \sum _{i \in {\mathcal {C}}_j} }{x_{i,l}}. \end{aligned}$$

(29)

However, for each server $l \in {\mathcal {S}}^-$, we have $\hat{y}_l \le y_l$ and thus $\sum _{l<j, l \in {\mathcal {S}}^-}{\hat{y}_l} \le \sum _{l<j, l \in {\mathcal {S}}^-}{y_l}$. Since, for $l < j$, $y_l$ can also be written as $y_l=\sum _{i \in {\mathcal {C}}_j}{x_{i,l}} + \sum _{i \not \in {\mathcal {C}}_j}{x_{i,l}}$, it yields

$$\begin{aligned} \sum _{l<j, l \in {\mathcal {S}}^-}{ \sum _{i \notin {\mathcal {C}}_j }{\hat{x}_{i,l}} } \le \sum _{l<j, l \in {\mathcal {S}}^-}{ \sum _{i \notin {\mathcal {C}}_j }{x_{i,l}} } + \left( \sum _{l<j, l \in {\mathcal {S}}^-}{ \sum _{i \in {\mathcal {C}}_j} }{x_{i,l}} - \sum _{l<j, l \in {\mathcal {S}}^-}{ \sum _{i \in {\mathcal {C}}_j }{\hat{x}_{i,l}} } \right) , \end{aligned}$$

and using (29),

$$\begin{aligned} \sum _{l<j, l \in {\mathcal {S}}^-}{ \sum _{i \notin {\mathcal {C}}_j }{\hat{x}_{i,l}} } < \sum _{l<j, l \in {\mathcal {S}}^-}{ \sum _{i \notin {\mathcal {C}}_j }{x_{i,l}} }. \end{aligned}$$

(30)

Therefore, there exists a class $i \notin {\mathcal {C}}_j$ such that

$$\begin{aligned} \sum _{l<j, l \in {\mathcal {S}}^-}{ \hat{x}_{i,l} } < \sum _{l<j, l \in {\mathcal {S}}^-}{ x_{i,l} }. \end{aligned}$$

(31)

It implies that, for this class $i$, we can find a server $n in {\mathcal {S}}_i$ and $n \in {\mathcal {S}}^-$ such that $\hat{x}_{i,n} < x_{i,n}$. Since ${\mathcal {C}}_{max} \subsetneq {\mathcal {C}}_j$, we know that $i \not \in {\mathcal {C}}_{max}$. Moreover, since ${\mathcal {S}}_k ={\mathcal {S}}_1 \subset {\mathcal {S}}^+$ for all $k \in {\mathcal {C}}_{min}$, $i\not \in {\mathcal {C}}_{min}$. We therefore have $\hat{\lambda }_i =\lambda _i$. Thus,

$$\begin{aligned} \sum _{l \in {\mathcal {S}}^-}{\hat{x}_{i,l}} + \sum _{l \in \mathcal{S}^+}{\hat{x}_{i,l}} = \sum _{l \in {\mathcal {S}}^-}{x_{i,l}} + \sum _{l \in {\mathcal {S}}^+}{x_{i,l}}, \end{aligned}$$

which implies

$$\begin{aligned} \sum _{l \in \mathcal{S}^+}{\hat{x}_{i,l}} = \sum _{l \in {\mathcal {S}}^+}{x_{i,l}} + \left( \sum _{l \in {\mathcal {S}}^-}{x_{i,l}} - \sum _{l \in {\mathcal {S}}^-}{\hat{x}_{i,l}} \right) , \end{aligned}$$

and with (31), it yields

$$\begin{aligned} \sum _{l \in \mathcal{S}^+}{\hat{x}_{i,l}} > \sum _{l \in {\mathcal {S}}^+}{x_{i,l}}. \end{aligned}$$

This implies that there exists a server $m<j$, $m \in {\mathcal {S}}^+$ such that $\hat{x}_{i,m} > x_{i,m}$. But, according to Lemma 3, there cannot be two servers $m,n \in {\mathcal {S}}$ such that $m \in {\mathcal {S}}^+$, $n \in {\mathcal {S}}^-$, $\hat{x}_{i,m} > x_{i,m}$ and $\hat{x}_{i,n} < x_{i,n}$. This is a contradiction. Therefore, if $j \in {\mathcal {S}^-}$, then $j+1 \in {\mathcal {S}}^-$ for all servers $j \in {\mathcal {S}}$. $\square$