Dynamic scheduling of a GI/GI/1+GI queue with multiple customer classes

Abstract

We consider a dynamic control problem for a GI/GI/1+GI queue with multiclass customers. The customer classes are distinguished by their interarrival time, service time, and abandonment time distributions. There is a cost c _k >0 for every class k∈{1,2,…,N} customer that abandons the queue before receiving service. The objective is to minimize average cost by dynamically choosing which customer class the server should next serve each time the server becomes available (and there are waiting customers from at least two classes).

It is not possible to solve this control problem exactly, and so we formulate an approximating Brownian control problem. The Brownian control problem incorporates the entire abandonment distribution of each customer class. We solve the Brownian control problem under the assumption that the abandonment distribution for each customer class has an increasing failure rate. We then interpret the solution to the Brownian control problem as a control for the original dynamic scheduling problem. Finally, we perform a simulation study to demonstrate the effectiveness of our proposed control.

Appendix

The proofs of all propositions and lemmas in the paper appear in this Appendix, in the order that they appear in the main body of the paper.

1.1 Proof of Proposition 4.1

The first step is to define the reduced Brownian control problem, which has the one-dimensional state-descriptor $\hat{W}$. The reduced Brownian control problem is to

(35)

using a control process $(\hat{Q}_{1},\hat{Q}_{2},\ldots,\hat {Q}_{N},\hat{I} )$ such that

(36)

The control $(\hat{Q}_{1},\hat{Q}_{2},\ldots,\hat{Q}_{N},\hat {I} )$ is admissible if it is $\mathcal{F}_{t}$-adapted and satisfies (36). The control $(\hat {Q}_{1}^{\star},\hat{Q}_{2}^{\star},\ldots,\hat{Q}_{N}^{\star},\hat {I}^{\star} )$ is optimal if it is admissible and for any admissible $(\hat{Q}_{1},\hat{Q}_{2},\ldots,\hat{Q}_{N},\hat {I} )$

where

The following two claims, which are Propositions 2 and 3 in [5] adapted to our setting, complete the proof. The claims are stated without proof because the arguments in [5] do not rely on their assumption of linear drift, and so apply to our nonlinear drift setting, also.

Claim 1

(Proposition 2 in [5])

The Brownian control problem (6)–(8) is equivalent to the reduced Brownian control problem (35)–(36) in the following sense:

• Every admissible control $(\hat{Q}_{1},\hat{Q}_{2},\ldots,\hat {Q}_{N},\hat{I} )$ for the reduced Brownian control problem yields an admissible control $\hat{Y}$ for the Brownian control problem having $\hat{\mathcal{C}}(\hat{Y}) = \hat{\mathcal{C}}_{R} (\hat{Q}_{1},\hat {Q}_{2},\ldots,\hat{Q}_{N},\hat{I} )$.

• Similarly, for any admissible control $\hat{Y}$ for the Brownian control problem, there exists an admissible control $(\hat {Q}_{1},\hat{Q}_{2},\ldots,\hat{Q}_{N},\hat{I} )$ for the reduced Brownian control problem having $\hat{\mathcal{C}}(\hat{Y}) \geq\hat {\mathcal{C}}_{R} (\hat{Q}_{1},\hat{Q}_{2},\ldots,\hat{Q}_{N},\hat {I} )$.

Claim 2

(Proposition 3 in [5]) The reduced Brownian control problem (35)–(36) is equivalent to the workload control problem in the following sense:

• Every admissible control $(q,\hat{I})$ for the workload control problem yields an admissible control $(\hat{Q}_{1},\hat {Q}_{2},\ldots,\hat{Q}_{N},\hat{I} )$ for the reduced Brownian control problem having $\hat{\mathcal{C}}_{R} (\hat{Q}_{1},\hat {Q}_{2},\ldots,\hat{Q}_{N},\hat{I} ) = \hat{\mathcal{C}}_{w}(q,\hat{I})$.

• Similarly, for any admissible control $(\hat{Q}_{1},\hat {Q}_{2},\ldots,\hat{Q}_{N},\hat{I} )$ for the reduced Brownian control problem, there exists an admissible control for the workload control problem having $\hat{\mathcal{C}}_{w}(q, \hat{I}) \leq\hat {\mathcal{C}}_{R} (\hat{Q}_{1},\hat{Q}_{2},\ldots,\hat{Q}_{N},\hat {I} )$.  □

1.2 Proof of Corollary 4.1

It follows from Proposition 4.1 that given the control $\hat{Y}^{\star}$ that is optimal for the Brownian control problem (6)–(8), then there exists an admissible control $(q,\hat{I})$ for the workload control problem having

(37)

Suppose that $(q,\hat{I})$ is not optimal for the workload control problem. Then there exists an admissible control $(q^{\star}, \hat {I}^{\star})$ where

(38)

It follows from Proposition 4.1 that there exists an admissible control $\hat{Y}$ for the Brownian control problem having

(39)

This is a contradiction, because (37)–(39) implies $\hat{\mathcal{C}}(\hat{Y}) < \hat{\mathcal{C}}(\hat{Y}^{\star})$.  □

1.3 Proof of Proposition 4.2

The following notation is useful in the proof. For any w≥0, let

(40)

Then, since $\hat{W}^{\star}(0) = \hat{w}$, for the second statement of the proposition, we must show that $E_{\hat{w}} [\hat{W}^{\star}(t) ]/t \rightarrow0$ as t→∞. Note that we explicitly subscript the expectation because it is useful in the proof to also consider what happens when the initial state differs from $\hat{w}$.

We observe that the process $\hat{W}^{\star}$ is a delayed regenerative process with regeneration point 0. The initial cycle length is $T_{0} := \inf \{ t \geq0: \hat{W}^{\star}(t) = 0 \}$. The remaining i.i.d. cycles are defined as the process starting from 0, reaching $\overline{w}>0$, and returning to 0, and we let τ represent the length of one of those cycles. Suppose we can show that

• (i) $\hat{W}^{\star}$ is positive recurrent; that is, that $E_{\hat {w}}[\tau] < \infty$;

• (ii) $E_{\hat{w}} [ \int_{T_{0}}^{T_{0} + \tau} \hat{W}^{\star}(s)\,ds ] < \infty$;

• (iii) $E_{\hat{w}} [ \int_{0}^{T_{0}} \hat{W}^{\star}(s)\,ds ] < \infty$.

Then, also noting that $\int_{0}^{T_{0}} \hat{W}^{\star}(s)\,ds < \infty$ almost surely, it follows from the renewal reward theorem for regenerative processes (see, for example, Theorem 2.3 in Sect. 13 on Regenerative Processes of [26]) that

(41)

In the next paragraph, we establish that (41) implies

(42)

To show (42), it is enough to show that

(43)

To show (43), the first step is to recall that $\hat{\xi }(\cdot)$ in (15) has the same distribution as $\theta \cdot+ \sigma\hat{B}(\cdot)$ for $\hat{B}$ a standard Brownian motion, and then apply Ito’s lemma to find

(44)

Since any sample path of $\hat{W}^{\star}(t)^{2}$ is continuous almost surely, for any t>0, $\int_{0}^{t} \hat{W}^{\star}(s)^{2} \,ds < \infty$ almost surely and $\hat{M}(t) := \int_{0}^{t} \hat{W}^{\star}(s) \,d\hat{B}(s)$ is a local martingale. Then there exists a sequence of stopping times {τ _k ,k=1,2,…} such that τ _k <τ _k+1 for all k=1,2,…,τ _k →∞ as k→∞ almost surely, and $\hat{M}(t \wedge\tau_{k})$ is a martingale for each k=1,2,… , so that $E [ \hat{M}(t \wedge\tau_{k}) ] = 0$. Then

(45)

(46)

(47)

(48)

where

• (45) follows from the definition of the sequence {τ _k ,k=1,2,…};

• (46) follows from Fatou’s lemma;

• (47) follows from (44) and the fact that $\hat{M}(t \wedge\tau_{k})$ is a martingale;

• (48) follows from the monotone convergence theorem.

Finally, it follows from (45)–(48) that to establish (43), it is enough to establish (41).

In summary, the proof is complete once we show (i)–(iii). For this, it is helpful to define the operator

where

and u is any twice differentiable function having domain ℜ⁺. In the following, assuming $\overline{w} > \hat{w}$, we show that for

for any k∈{0,1,2,…}

(49)

and

(50)

This is sufficient to complete the proof because when $\hat{w} = 0$, (49) and (50) imply (i) and (ii) by letting k=0 and k=1, and for $\hat{w} >0$,

which implies (iii) by letting k=1.

The argument to show (49)

For any x≥0, let

Then it is straightforward to check that

solves

Moreover, since r(x)<0 for x>0, u(x) is decreasing. This implies that u(x)≥0 for $x\in [0,\overline{w} ]$ because $u (\overline{w} )=0$. Furthermore, there exists a twice differentiable function $\tilde{u}$ having domain ℜ⁺ such that $\tilde{u}(x)=u(x)$ for $x\in [0,\overline{w} ]$ and $\tilde{u}'$ is bounded.

From Ito’s lemma¹, by recalling $\hat{\xi}(\cdot) \stackrel{D}{=} \theta\cdot+ \sigma\hat {B}(\cdot)$ for a standard Brownian motion $\hat{B}(\cdot)$ with $\hat{B}(0)=0$,

Since $\tilde{u}'$ is bounded, the stochastic integral in the above is a martingale and, therefore, so is

Then

is also a martingale, and so

Since $\hat{W}^{\star} (t\wedge T^{\overline{w}} )\in [0,\overline{w}]$, $\tilde{u} (\hat{W}^{\star} (t\wedge T^{\overline{w}} ) )\geq0$, and so

Taking the limit as t→∞ in the above and using the monotone convergence theorem implies

which establishes (49).

The argument to show (50)

For any $\tilde{w} > \overline{w}$, let

Next, for any x≥0, let

where

Then it is straightforward to check that

solves

Note that there can be at most one x _q >0 such that q(x)>0 for all x∈[0,x _q ) and q(x)<0 for all x>x _q . Then $v_{\tilde {w}}(0)=v_{\tilde{w}} (\tilde{w} )=0$ implies that such x _q exists and $x_{q}\in (0,\tilde{w} )$ so that $v_{\tilde{w}}(x)$ is strictly increasing for x∈[0,x _q ] and strictly decreasing otherwise. This implies that $v_{\tilde{w}}(x) \geq0$ for all $x\in [0,\tilde{w} ]$. Furthermore, there exists a twice differentiable function $\tilde{v}_{\tilde{w}}$ having domain ℜ⁺ such that $\tilde{v}_{\tilde{w}}(x)=v_{\tilde{w}}(x)$ for all $x\in [0,\tilde{w} ]$ and $\tilde{v}_{\tilde{w}}'$ is bounded.

From Ito’s lemma, by recalling $\hat{\xi}(\cdot) \stackrel{D}{=} \theta \cdot+ \sigma\hat{B}(\cdot)$ for a standard Brownian motion $\hat {B}(\cdot)$ with $\hat{B}(0)=0$,

Since $\tilde{v}'$ is bounded, the stochastic integral in the above is a martingale and, therefore, so is

as well as

If we renew $\hat{W}^{\star}$ once the stochastic process reaches $\overline{w}>\hat{w}$, we have $\hat{W}^{\star}(s) \in(0,\tilde{w})$ for all $s \in[0, t \wedge T_{0}^{\tilde{w}})$. Then, for all $s \in[0, t \wedge T_{0}^{\tilde{w}})$, $d\hat{I}^{\star}(s) = 0$. Therefore, since $\hat{I}^{\star}$ is continuous, it follows that

Then the same argument as in the previous paragraph shows that

(51)

The monotone convergence theorem then implies

(52)

Finally, to complete the proof, it follows from (51) and (52) that it is enough to show there exists M>0 that does not depend on $\tilde{w}$ such that

(53)

To prove (53), observe that, for any $\tilde {w} > \overline{w}$, we have

Since g(x)→−∞ as x→∞, we have

Also, we argue that

(54)

To prove, observe that there exists y ^⋆>0 such that g(x)≤0 for all x≥y ^⋆ because g(x)→−∞ as x→∞. Then for any $\tilde{w} \geq y^{\star}$, we have

implying (54). Then, we use L’Hopital’s rule to conclude

Since it is straightforward to see that

the proof is completed.  □

1.4 Proof of Lemma 4.1

It is enough to show that the function ϕ(x,w) is jointly continuous in (x,w) and Lipschitz continuous in w. This is because the fact that there exists a unique solution w _κ of IVP(κ) on a compact interval of ℜ⁺ then follows from Picard’s existence theorem (see, for example, Proposition 2.4 of [32]). The fact that there exists a unique solution w _κ of IVP(κ) on ℜ⁺ follows an iteration argument, which we omit because it is standard.

To see that ϕ(x,w) is continuous in (x,w), let ϵ>0 be arbitrarily small. We show that, for i∈{1,2}, if x _i ≥0 and w _i ≥0 are such that

for

then

Define

so that

Next, define

where

is such that

so that

Note that such a $p \in\mathcal{P}$ can be found because for any $p \in\mathcal{P}$, $\tilde{q}_{1}^{2} \geq0$, and, for fixed x ₁ and x ₂, $\tilde{q}_{2}^{2}$ is a continuous function of p having $\tilde {q}_{1}^{2}(p) \rightarrow\mu_{2} x_{2} > 0$ as $p \rightarrow q_{1}^{1\star} / (\mu_{1}(x_{1} - x_{2}))$. Without loss of generality, assume (x ₁,w ₁) and (x ₂,w ₂) are such that

Then

Since

and

where the last inequality follows from the definition of $(\tilde {q}_{1}^{2},\tilde{q}_{2}^{2},\ldots,\tilde{q}_{N}^{2} )$, it follows that

(55)

so that

To see that ϕ(x,w) is Lipschitz continuous in w, note that it follows from (55) that when x ₂=x ₁=x≥0,

1.5 Proof of Proposition 4.3

From Theorem 4.1, it is sufficient for the proof to construct v:ℜ⁺→ℜ that is twice continuously differentiable and κ>0 that solves the Bellman equation (20)–(21). Define

for

Then, v′ solves

(56)

The first step in showing that (v,κ ^⋆) solves the Bellman equation is to show that (20) is satisfied. Suppose we can show that v′(x)∈[0,c ₁ μ ₁] for all x≥0. Then, we have

for any x≥0. To see this, observe that

for any $q \in\mathcal{A}(x)$. Next, note that as a function of q ₁, $q_{1}^{2} / (\mu_{1}f_{1} )+ (\mu_{2} (x-q_{1}/\mu_{1} ) )^{3}/ (\mu_{2}f_{2}^{2} )$ is convex because

Therefore, $\mbox{argmin}_{q_{1}\in[0,\mu_{1}x]} \{ q_{1}^{2} / (\mu_{1}f_{1} )+ (\mu_{2} (x-q_{1}/\mu_{1} ) )^{3}/ (\mu_{2}f_{2}^{2} ) \}$, for any given x≥0, is derived by solving

(57)

Since the solution of (57) is $q_{1}^{\star}(x)$ defined in Sect. 4.2.1, we have

which implies by (56) that (20) is satisfied.

The argument that v′(x)∈[0,c ₁ μ ₁] for all x≥0 is by contradiction; suppose not. Then, since v′(0)=0 and v′(x)→c ₁ μ ₁ as x→∞ follows from L’Hopital’s rule, there must exist x ₁<x ₂ such that v′(x ₁)=v′(x ₂)>c ₁ μ ₁ and v″(x ₁)>0, v″(x ₂)<0. From (56),

Subtracting the bottom equation from the top yields

This is a contradiction because v″(x ₁)−v″(x ₂)>0, c−v′(x ₁)<0, and g(x ₁)−g(x ₂)<0 since g is a strictly increasing function.

The second step in showing that (v,κ ^⋆) solves the Bellman equation is to show that (21) is satisfied. It is immediate that v′(0)=0. The fact that v′ is bounded follows from the fact that v′ is continuous and v′(x)→c ₁ μ ₁ as x→∞ by L’Hopital’s rule. To see that v′ is nonnegative, note that it follows from its definition that either v′(x)≥0 for all x, in which case the proof is complete, or there exists x ₀ such that v′(x)>0 for all x<x ₀ and v′(x)<0 for all x>x ₀. This is a contradiction because v′(x)→c ₁ μ ₁>0 as x→∞, and so there cannot exist such an x ₀.  □

1.6 Proof of Lemma 4.2

Note that c ₁ μ ₁ γ ₁>c ₂ μ ₂ γ ₂ and c ₂ μ ₂>c ₁ μ ₁ implies that γ ₁>γ ₂,

and

Then

Since κ ₁(L) and κ ₂(L) are continuous in L, we conclude that there exists L>0 such that κ ₁(L)=κ ₂(L). Since κ ₂(L)>0 for all L≥0, it must be that at L such that κ ₁(L)=κ ₂(L), κ ₁(L)=κ ₂(L)>0.  □It is useful for the proof of Proposition 4.4 to note that

for all x∈ℜ (see, for example, Lemma EC.3 in [2] to see that xΦ(x)+ϕ(x)>0 for all x∈ℜ), so that Ψ is a strictly decreasing function. Also, Ψ(x)→∞ as x→−∞ and Ψ(x)→0 as x→∞.

1.7 Proof of Proposition 4.4

From Theorem 4.1, it is sufficient for the proof to construct v:ℜ⁺→ℜ that is twice continuously differentiable and κ>0 that solves the Bellman equation (20)–(21). We first do this when c ₁ μ ₁ γ ₁>c ₂ μ ₂ γ ₂ and c ₁ μ ₁≥c ₂ μ ₂, so that a static priority control is optimal, and second do this when c ₁ μ ₁ γ ₁>c ₂ μ ₂ γ ₂ and c ₁ μ ₁<c ₂ μ ₂, so that a dynamic priority control is optimal.

The proof when c ₁ μ ₁ γ ₁>c ₂ μ ₂ γ ₂ and c ₁ μ ₁≥c ₂ μ ₂. Define

for

Note that κ ^⋆>0 because

and $\varPsi (-\sqrt{\frac{2}{\sigma^{2} \gamma_{2}}}\theta )>0$. Also,

(58)

so that it is straightforward to see v′(0)=0 and v′(x) increases to c ₂ μ ₂ as x→∞. Furthermore, v′ solves

(59)

To show that (v,κ ^⋆) solves the Bellman equations (20)–(21), first note that it is immediate from the representation of v′ in (58) and the properties of Ψ that (21) is satisfied. To see that (20) is also satisfied, we show that

(60)

Then it follows from (59) and (60) that (20) holds, i.e., that

We show that (60) holds by considering each of the two possibilities: c ₁ μ ₁=c ₂ μ ₂ or c ₁ μ ₁>c ₂ μ ₂. In the case that c ₁ μ ₁=c ₂ μ ₂, it follows from the fact that c ₁ μ ₁ γ ₁>c ₂ μ ₂ γ ₂ that γ ₁>γ ₂, so that

and (60) holds. Otherwise, in the case that c ₁ μ ₁>c ₂ μ ₂, first note that to show (60) holds it is equivalent to show

(61)

because when (61) holds

Then, in the case that c ₁ μ ₁>c ₂ μ ₂ and γ ₁<γ ₂, (61) holds because (61) is equivalent to

which holds because (c ₁ μ ₁ γ ₁−c ₂ μ ₂ γ ₂)/(γ ₁−γ ₂)<0 and v′(x)≥0 for all x≥0 from its explicit expression in (58). Finally, in the case that c ₁ μ ₁>c ₂ μ ₂ and γ ₁≥γ ₂, (61) holds because (61) is equivalent to

which holds because v′(x)≤c ₂ μ ₂ for all x≥0 by (58), c ₂ μ ₂<c ₁ μ ₁ by assumption, and

The proof when c ₁ μ ₁ γ ₁>c ₂ μ ₂ γ ₂ and c ₁ μ ₁<c ₂ μ ₂ is true. Define

(62)

and

(63)

Then, it is straightforward to verify that $v_{1}'$ and $v_{2}'$ in (62) and (63) solve

(64)

with initial conditions

and that

(65)

It follows from (64) and (65) that $v_{1}'' ( L^{\star}) = v_{2}'' ( L^{\star})$. Therefore,

is twice-continuously differentiable.

The first step in showing (v,κ ^⋆) solves the Bellman equation is to show that (20) is satisfied. For this, observe that

Therefore, if we can show that

(66)

and

(67)

it follows from (64) that (20) holds, i.e., that

Furthermore, it also follows that

To see (67) holds, replace κ ^⋆ in (63) by κ ₂(L ^⋆) in Lemma 4.2 so that

(68)

Then $v_{2}'(x)$ for all x≥L ^⋆ is strictly increasing function in x because −Ψ(x) is strictly increasing in x for all x∈ℜ. So (67) holds since

The argument that (66) holds is by contradiction. Suppose that there exists x ₀<L ^⋆ for which $v_{1}'(x_{0})> (c_{1}\mu_{1}\gamma_{1}-c_{2}\mu_{2}\gamma_{2} )/ (\gamma_{1}-\gamma_{2} )$. Then there exists x ₁∈(x ₀,L ^⋆) such that $v''_{1}(x_{0})<0$ and $v'_{1}(x_{1})= (c_{1}\mu_{1}\gamma_{1}-c_{2}\mu_{2}\gamma_{2} )/ (\gamma_{1}-\gamma_{2} )$. From (64),

so that, subtracting the bottom equation from the top shows that

This is a contradiction because $v_{1}'' ( L^{\star}) = v_{2}'' ( L^{\star})\geq0$ (since $v_{2}'$ is a strictly increasing function), $v_{1}''(x_{1}) < 0$, and x ₁<L ^⋆ implies that

The second step in showing (v,κ ^⋆) solves the Bellman equation is to show that (21) is satisfied, i.e., that

• (i) v′(0)=0;

• (ii) v′ is bounded;

• (iii) v′ is nonnegative.

For (i), recall that $v'_{1}(0)=0$. For (ii), we can show that $v_{2}'(x) \rightarrow c_{1} \mu_{1}$ as x→∞ by noting that Ψ(x)→0 as x→∞ in (68). For (iii), by calculus and algebra, it can be shown that $v_{1}'$ in (62) has the equivalent representation

Then, depending on how large κ ^⋆ is, $v'_{1}$ is either nonnegative for all x≥0 or there exists a unique x ₀>0 such that $v_{1}'(x) > 0$ for all x<x ₀ and $v_{1}'(x) < 0$ for all x>x ₀. Since $v'_{1}(L^{\star}) = v'_{2}(L^{\star})= \frac{c_{1} \mu_{1} \gamma_{1} - c_{2} \mu_{2} \gamma_{2}}{\gamma_{1} - \gamma_{2}}>0$, x ₀>L ^⋆ if such x ₀ exists. Therefore, $v'_{1}$ is nonnegative for all x<L ^⋆. Note that $v'_{2}(x) >0$ for all x>L ^⋆ because $v_{2}'(L^{\star}) >0$ and $v_{2}'$ is strictly increasing. Therefore, v′ is nonnegative.  □

1.8 Proof of Lemma 6.1

We already proved that ϕ(x,w) is jointly continuous in (x,w) in the proof of Lemma 4.1. Therefore, we can apply Theorem 2.1 in [16] to conclude that w _κ (x) is jointly continuous in (x,κ).  □

1.9 Proof of Lemma 6.2

Let w ₁>w ₂>0. Then, recalling that θ≤0,

1.10 Proof of Lemma 6.3

The proof for Lemma 6.3 is identical to the proof of Lemma 4 in the Appendix of [5], and so is omitted.  □

1.11 Proof of Lemma 6.4

Since ϕ(0,0)=0, it follows from the definition of IVP(κ) that

Hence, the fact that $w'_{\kappa}$ is continuous on ℜ⁺ and the existence of x _κ such that $w'_{\kappa} (x_{\kappa } )<0$ imply that the set

is not empty and

To complete the proof, we have to show two things; (i) $\overline {\mathcal{E}}:= \{ x>x_{\kappa}:w'_{\kappa}(x)=0 \}$ is empty and (ii) $w'_{\kappa}(x)\geq0$ for all x∈[0,x ₀].

Proof of (i)

The argument that $\overline{\mathcal{E}}$ is empty is by contradiction; suppose not that $\overline{\mathcal{E}}$ is not empty. We can define

Then, by the definitions of x ₀ and x ^κ, we have

(69)

and

which from the definition of IVP(κ) implies that

(70)

Assume we can show that

(71)

Also, (69) and Lemma 6.2 show that

(72)

Then (71) and (72) imply that

which contradicts (70).

To establish (71), assume we can show that

(73)

Let

so that

Since x ^κ>x ₀, there exists $q^{0}\in\mathcal{A} (x_{0} )$ such that $q_{i}^{0}\leq q_{i}^{\kappa\star}$ for all i∈{1,2,…,N} with at least one of the inequalities being strict. This implies that $m_{i} (q_{i}^{0} )\leq m_{i} (q_{i}^{\kappa\star} )$ for all i∈{1,2,…,N}, because h _i is increasing. Since (73) implies that (c _i −w _κ (x ₀)/μ _i )>0 for all i∈{1,2,…,N}, it follows that

(74)

The definition of ϕ implies that

(75)

Together, (74) and (75) show (71).

We now establish (73). First suppose that w _κ (x ₀)≥max(c ₁ μ ₁,c ₂ μ ₂,…,c _N μ _N ). Then c _i −w _κ (x ₀)/μ _i ≤0 for all i∈{1,2,…,N}, so that for any $q\in\mathcal{A} (x_{0} )$

which implies ϕ(x ₀,w _κ (x ₀))≤0. This is a contradiction because it follows from IVP(κ) that

but we have assumed κ>0. Next, let c _i μ _i for i∈{1,2,…,N} be ordered so that $c_{i_{1}}\mu_{i_{1}}\leq c_{i_{2}}\mu_{i_{2}}\leq \cdots \leq c_{i_{N}}\mu_{i_{N}}$. We may assume that at least one of inequalities is strict because otherwise we have min(c ₁ μ ₁,c ₂ μ ₂,…,c _N μ _N )=max(c ₁ μ ₁,c ₂ μ ₂,…,c _N μ _N ) and the proof is complete. Let s∈{1,2,…,N−1} be the largest value with $c_{i_{s}}\mu_{i_{s}}<c_{i_{s+1}}\mu_{i_{s+1}}$ and suppose $c_{i_{s}}\mu_{i_{s}}\leq w_{\kappa} (x_{0} )<c_{i_{s+1}}\mu_{i_{s+1}}$. Then, $c_{i_{j}}\mu_{i_{j}}-w_{\kappa} (x_{0} )>0$ for all j∈{s+1,s+2,…,N} and $c_{i_{j}}\mu_{i_{j}}-w_{\kappa} (x_{0} )\leq0$ for all j∈{1,2,…,s} so that

where $q^{s}\in\mathcal{A} (x_{0} )$ such that $q_{i_{j}}^{s}=0$ for all j∈{s+1,s+2,…,N}. This is again a contradiction for the same reason as earlier in this paragraph. If we repeat this argument, we can prove that w _κ (x ₀)<min(c ₁ μ ₁,c ₂ μ ₂,…,c _N μ _N ) and this completes the proof of (i).

Proof of (ii)

Finally, to complete the proof of the lemma, it remains to prove that $w'_{\kappa}(x)\geq0$ for all x∈[0,x ₀]. Suppose there exists x ₁∈(0,x ₀) such that $w'_{\kappa} (x_{1} )<0$. Then, by applying the same argument in the proof of (i) to x ₁, we conclude that $w'_{\kappa}(x)<0$ for all x>x ₁. This is a contradiction because $w'_{\kappa} (x_{0} )=0$ and x ₀>x ₁.  □

1.12 Proof of Lemma 6.5

If $\kappa\in\mathcal{D}$, then by definition of $\mathcal{D}$, there exists x _κ >0 such that w _κ (x) is decreasing if x>x _κ . There are two possibilities: (i) w _κ (x) is bounded for all x≥0, or (ii) w _κ (x)→−∞ as x→∞. We show that case (i) leads to a contradiction, which is enough to complete the proof.

Suppose w _κ (x) is bounded for x≥0. Then lim _x→∞ w _κ (x) exists and is finite, which implies $w_{\kappa}'(x) \rightarrow0$ as x→∞, and so, from the definition of IVP(κ),

(76)

Next, since $\kappa\in\mathcal{D}$, there exists $\overline{w}$ such that $w_{\kappa}(x) < \overline{w}$ for all x>0. Then, by Lemma 6.2,

(77)

Let x>x _κ and define

so that

(78)

It follows from (73) in the proof of Lemma 6.4 that for x ₀ such that $w_{\kappa}'(x_{0}) = 0$, w _κ (x ₀)<min(c ₁ μ ₁,c ₂ μ ₂,…,c _N μ _N ), so that we may assume $\overline{w} < \min(c_{1},\mu_{1}, c_{2} \mu_{2},\ldots,c_{N}\mu_{N})$. Then $c_{k}\mu_{k} - \overline{w} >0$ for all k∈{1,2,…,N}, and it follows from (78) that

(79)

because $\max (q_{1}^{\star} (x ),q_{2}^{\star} (x ),\ldots,q_{N}^{\star} (x ) )\rightarrow\infty$ as x→∞ from the fact that $\sum_{k=1}^{N}q_{k}^{\star } (x )/\mu_{k}=x$ for every x>0. Finally, (79) and (77) imply that

which contradicts (76).  □

1.13 Proof of Lemma 6.6

Fix κ>0. First note that if there exists x ₀>0 such that w _κ (x ₀)≥min(c ₁ μ ₁,c ₂ μ ₂,…,c _N μ _N ), then by the same argument as the one in the second to the last paragraph of the proof of Lemma 6.4,

(80)

It follows from (80) that

Since w _κ is continuous, and the same argument as above shows that $w_{\kappa}'(x) >0$ whenever w _κ (x)≥min(c ₁ μ ₁,c ₂ μ ₂,…,c _N μ _N ), it follows that $w_{\kappa}'(x) >0$ for all x≥x ₀.

Next, we claim that there can exist at most one $\underline{x} >0$ such that $w_{\kappa}(\underline{x}) = \min (c_{1}\mu_{1},c_{2}\mu_{2},\ldots,c_{N}\mu_{N} )$. To see this, note that the argument in the previous paragraph shows that $w_{\kappa}'(x) >0$ for all $x \geq \underline{x}$.

Finally, to complete the proof, we must show that w _κ (x) is increasing on $[0,\underline{x})$. This argument is by contradiction. Assume there exists x ₁ such that $w_{\kappa}'(x_{1}) < 0$. Then, by Lemma 6.4, $\kappa\in\mathcal{D}$, and there exists x _κ such that w _κ is decreasing on [x _κ ,∞). This is a contradiction, because the previous two paragraphs show that $w_{\kappa}'(x) >0$ for all $x \geq\underline{x}$.  □