Technical Appendix for: When Promotions Meet Operations: Cross-Selling and Its Effect on Call-Center Performance

Technical Appendix for: When Promotions Meet Operations: Cross-Selling and Its Effect on Call-Center Performance In this technical appendix we provide proofs for the various results stated in the manuscript titled: When promotions meet operations: Cross-selling and its effect on call-center performance. We start the technical appendix with the construction of the sample paths for all related processes. Specifically, our construction follows a strong approximation approach (see for example [7] and [8]. Let N i (, i = 1,..., 11, be independent unit rate Poisson processes. Then, one can write the system dynamics through the following equations: Q (t + Z 1 (t = Q ( + Z 1 ( + ÑA(t ÑD 1 (t, (A1 Z2 (t = Z2 ( ÑD 2 (t + 1 {K } [N 7 (pµ s Z1 (u1 {<Y (u N K }du ( ] + N 5 pµ s Z1 (u1 {Q (u=}du ( + 1 {K <}N 8 pµ s Z1 (u1 {Y (u N K }du (A2 and ( Z1 (t = Z1 ( + N 1 t 1 {Z1 (u+z 2 (u<n }du ( ( t 1 {K } [N 5 pµ s Z 1 (u1 {Q (u=}du + N 6 ( t + 1 {K }N 7 pµ s Z 1 (u1 {<Y (u N K }du ( ( t 1 {K <} [N 9 pµ s Z 1 (u1 {Q (u=}du + N 1 ( t + N 3 µ cs Z 2 (u1 {Q (u>}du (1 pµ s Z 1 (u1 {Q (u=}du (1 pµ s Z 1 (u1 {Q (u=}du ] ] where one should recall that Y (t = Z 1 (t + Z 2 (t + Q (t, and we define: ( Ñ A (t := N 1 ( 1 {Z 1 (u+z2 (u<n }du + N 2 1 {Z 1 (u+z2 (u N }du. ( Ñ D2 (t := N 3 µ cs Z2 (u1 {Q (u >}du + N 4 (µ cs Z2 (u1 {Q (u=}du. 1

and finally, ( Ñ D1 (t = 1 {K }N 5 pµ s Z1 (u1 {Q (u=}du ( + 1 {K }N 6 (1 pµ s Z1 (u1 {Q (u=}du ( + 1 {K }N 7 pµ s Z1 (u1 {<Y (u N K }du ( + 1 {K <}N 8 pµ s Z1 (u1 {Y (u N K }du ( + 1 {K <}N 9 pµ s Z1 (u1 {Q (u=}du ( + 1 {K <}N 1 (1 pµ s Z1 (u1 {Q (u=}du ( + N 11 ˆ(udu, (A3 where the rate function ˆ(t is set to satisfy that the sum of the instantaneous rates of all the processes in (A3 would equal µ s Z 1 (t. The construction follows by noting that all input and output processes in the system can be modelled by thinning of Poisson processes. By Lemma 9.4 in [7], there exists a probability space (Ω, F, P, a filtration {F t } t and an 11-dimensional Brownian Motion (B 1 (,..., B 11 ( such that the random variable C i := sup t N i (t t B i (t, (A4 log(2 t has a moment generating function in a neighborhood of the origin and in particular, there exist constants c 1, c 2 and Γ, such that P {C i Γ + x} c 1 e c 2x, x >, for all i =... 1,..., 11. Note that all the time changes of the unit rate Poisson processes in equations (A1-(A3 are bounded by c for some positive constant c. Indeed, this is a result of the fact that we examine only cases in which N R + p µ cs. We can hence write: Q (t + Z1 (t = Q ( + Z1 ( + t µ s Z1 (udu Z1 (t + MZ,Q(t + O(log(2 ct, (A5 2

Z2 (t = Z2 ( µ cs Z2 (udu + pµ s Z1 (udu pµ s Z1 (u1 {Y (u N >K }du + M Z 2 (t + O(log(2 ct, (A6 and Z 1 (t = Z 1 ( + 1 {Z 1 (u+z 2 (u<n }du µ s Z 1 (u1 {Q (u=}du pµ s Z 1 (u1 {<Y (u N K }du + µ cs Z 2 (u1 {Q (u>}du + M Z 1 (t + O(log(2 ct Here, M Z,Q (, M Z 1 ( and M Z 2 ( are sums of time changed Brownian motions. For example, if K >, MZ 1 (t = B 1 ( 1 {Z 1 (u+z2 (u<n }du ( B 5 pµ s Z1 (u1 {Q (u=}du + B 6 ((1 pµ s Z1 (u1 {Q (u=}du ( + B 7 pµ s Z1 (u1 {<Y (u N K }du ( + B 3 µ cs Z2 (u1 {Q (u>}du, (A7 where B i (, i = 1,..., 7, are standard Brnownian motions. Using the Brownian Motion strong law of large numbers (see problem 2.9.3. in [6] and the fact that the time arguments are all bounded by ct for some constant, we have that ( M Z,Q (t, M Z 2 (t, M Z 1 (t (,,, as, (A8 where the convergence is uniform on compact sets. We also define C = C 1 + C 2 +... + C 11 with C i, i = 1,..., 11 as defined in equation (A4. Defining the processes T 1 (t := pµ s Z 1 (u1 {Y (u N >K }du, T 2 (t := 1 {Z 1 (u+z 2 (u<n }du, T 3 (t := µ s Z 1 (u1 {Q (u>}du, T 4 (t := µ cs Z 2 (u1 {Q (u>}du, T 5 (t := pµ s Z 1 (u1 {<Y (u N K }du, 3

one can re-write equations (A5-(A7 as follows: Q (t = Q ( + Z1 ( + t µ s Z1 (udu Z1 (t + MZ,Q(t + O(log(2 ct, (A9 Z2 (t = Z2 ( µ cs Z2 (udu + pµ s Z1 (udu T1 (t + MZ 2 (t + O(log(2 ct, (A1 and Z1 (t = Z1 ( + T2 (t µ s Z1 (udu + T3 (t + T4 (t T5 (t + MZ 1 (t + O(2 log(ct. (A11 Notational conventions and organization of the appendix. We let Ξ (t := (Q (t, Z2 (t, Z1 (t, and let X be the state-space in which Ξ (t obtains values. We let ν be the unique stationary distribution of Ξ (t (which exists by Lemma 2.2. We use the notation ξ for a general element in X and for a given ξ we let q(ξ, z 2 (ξ and z 1 (ξ be its corresponding coordinates. Finally, we use E ξ [ ] for the expectation with respect to the initial condition ξ. Accordingly, we let E ν [ ] be the expectation with respect to an initial condition that is distributed according to the stationary distribution ν. The rest of this appendix is organized as follows. Each of the sections A, B, C is dedicated to prove Theorem 4.1 under one of the conditions 1, 2 and 3, respectively. D is dedicated to proving the result in 5 of the main paper. Finally, E provides proofs for some Lemmas that were given in the paper and several auxiliary results that are used in this appendix. A. Condition 1 The main result of this section is the following Theorem: Theorem A.1 Consider a sequence of systems with N = R + βr + γ R + o( for some < β pµ s µ cs and fix a sequence K with K R δ, as, 4

with δ. Then, under T P [K ], E[I ] N R, as, (A12 and E[(Q K + ] R, as. (A13 The first part of the theorem will be proved in Corollary A.3 and the second part will be proved in A.1. The intuition behind the latter is based on the large extra capacity of the system. Specifically, since T P [K ] dictates that whenever the queue length is greater than K - every service or cross-selling completion is followed by an admission of a customer from the queue into service - we have that whenever the queue is longer than K, the depletion rate of the queue is roughly µ s R + µ cs (N R. In particular, the queue depletion rate is much greater than the input rate to the queue leading to extremely small excess queue above the level of K. Most of this section is dedicated to the proof of Theorem A.1. The main complication arises from the fact that we are interested in convergence of the steady-state variables Q and I rather then mere convergence on finite time intervals. Before proceeding with the proof we state and prove the asymptotic optimality result for this section: Corollary A.1 Assume that N2 R N1 R, in addition to Assumptions 4.1 and 4.2. Then, the following is asymptotically optimal: Staffing: Staff with N2 agents. Control: Use T P [K ] with K = δ R such that δ and K W /2. Proof: By Little s law: E[W ] W = E[Q ] W K + E[(Q K + ] W. (A14 Equation (A13 now implies that lim sup E[W ] W 1 2, (A15 5

and in particular that T P [K ] is asymptotically feasible. Also, since the system is stable, we have by Little s law that E[Z 1 ] = /µ s. But recall that E[Z 2 ] = N E[Z 1 ] E[I ]. Hence, (A12 implies that µ cs E[Z 2 ] µ cs (N 2 R 1, as. (A16 Recall that for each, rµ cs (N 2 R (C(N 2 C (R constitutes an upper bound for the optimal value of the cross-selling problem (1. Equation (A16 implies that the upper bound is asymptotically achieved leading to asymptotic optimality of the sequence (N 2, T P [K ]. We proceed now to prove Theorem A.1. The proof is composed of several components: In Lemma A.1 we show that the process Z1 (t cannot take values that are much smaller than R. Using a Lyapunov function argument, the bound is then extended to the steady state distribution in Corollary A.2. In Proposition A.1 we show that the steady state queue length is negligible with respect to. In Lemmas A.3 and A.4 we prove the convergence to fluid limits that satisfy certain characteristics, and, finally, Corollary A.3 uses all the previous components to complete the proof of Theorem A.1. Lemma A.1 Consider a sequence of systems, where the th system is staffed with N = R+βR+ γ R + o( R for β pµ s µ cs, max(β, γ >, and operated with T P [K ], where K R δ, as, (A17 for some δ (,. Then, for all ɛ >, there exist t (ɛ and (ɛ (independent of the initial conditions, such that for all (ɛ and T t (ɛ, E [ sup t (ɛ t T (Z 1 (t µs ] ɛ, (A18 and P { sup t (ɛ t T ( Z1 (t } > ɛ c 3 e c 4/ log(2 ct, µ s (A19 6

{ } for two positive constants c 3 and c 4, where (Z 1 (t µs = max µ s Z1 (t,. Before proceeding to the proof we have the following corollary which follows directly from Lemma A.1 by initializing the system with its stationary distribution. Corollary A.2 Under the condition of Lemma A.1, there exists T such that P { ( Z 1 µ s > ɛ } c 3 e c 4/ log(2 ct, (A2 for all large enough. Remark A.1 Note that Lemma A.1 and Corollary A.2 are not restricted to any form of staffing sequences. Hence, we will make use of these also for the analysis of the service driven regime. Proof of Lemma A.1: Fix δ > and define the set Ω (δ, = { ω Ω : 11 max sup i=1,...,11 t T } B i (ct + C log(2 ct δ. Assume that at time, Z 1 ( < µ s ɛ and define τ = inf {t : Z 1 (t /µ s ɛ 2 }. Consider equations (A5 and (A7. Then, on every interval [s, t with t τ, with Q (u = for all u [s, t we have by equation (A5 that Z 1 (t Z 1 (s (t s µ s ( µ s ɛ 2 ɛ (t s δ = µ s (t s δ. 2 On the other hand for intervals [s, t with t τ and Q (u > K, for all u [s, t, we have by equation (A7 that Z1 (t Z1 (s µ cs (N Z1 ɛ (udu δ µ cs (t s δ, 2 s 7

and finally, on intervals [s, t with t τ and such that < Q (u K for all u [s, t, we have by equation (A5 that Z 1 (t Z 1 (s (K + (t s µ s ( µ s ɛ 2 δ (K ɛ + µ s (t s δ. 2 Our assumptions on the magnitude of the threshold, K, imply that for large enough there exists k > such that K k for all thereafter. Hence, we have that for large enough and on Ω (δ, Z1 (t τ Z1 ɛ ( + µ s µ cs (t τ δ. (A21 2 Choosing δ = ɛ/8 and recalling that by definition Z 1 (, we have that on Ω (δ,, τ 1 µ s ɛ 4 (µ s µ cs ɛ/2. Define now and τ = sup {t τ : Z 1 (t µs ɛ2 }, τ = inf { t τ : Z1 (t < } ɛ. µ s But on Ω (δ, and for every τ s < t τ, we have by our previous argument that Z 1 (t Z 1 (s + η(ɛ(t s ɛ/4 implying that Z 1 (t µ s ɛ, for all t τ. In particular, choosing t (ɛ = 1 µ s ɛ 4 (µ s µ cs ɛ/2, we have that on Ω (ɛ, ( sup Z1 (t ɛ. t (ɛ t T µ s (A22 8

Now, note that 11 ( P ((Ω (δ, c P (C log(2 ct δ + P i=1 sup t T c Bi (t δ, (A23 implying that (A19 holds by equation (A4 and recalling that for a Brownian Motion B(t and any b > ( P sup t T B(t b T 4 /2T 2π b e b2, whenever B( = (see Problem 2.8.2 in [6]. Moreover, since Z 1 c we also have that E [ sup t (ɛ t T (Z 1 (t µs ] ɛ + cc 3 e c 4ɛ/ log(2 ct, (A24 so that there exists large enough so that the above is smaller than 3ɛ. Repeating the argument 2 with 2 ɛ instead of ɛ we have the result of the lemma. 3 Proposition A.1 Under the assumption of Theorem A.1, E[(Q K + ]. (A25 Proof: We prove the result for K. The proof is similar for arbitrary K > with Q ( replaced everywhere with (Q ( K +. We start by re-defining { } Ω (δ, = ω Ω : 11 max sup i=1,...,11 t T/ B i (ct + C log(2 ct δ, where we assume that T is larger than 2t (where t is defined in (A29. Assume first that at time, [Z 1 ( /µ s ] ɛ. Fix a constant K >, assume that Q ( > 2K and let τ = inf{t : Q (t Q ( 3K/2}. 9

Then, plugging equation (A7 into equation (A5, and using the fact that Z 2 (t = N Z 1 (t whenever Q (t >, we have that on Ω (δ, τ τ Q (t τ Q ( + (t τ µ s Z1 (udu µ cs N Z1 (udu + δ. (A26 Recall from the proof of Lemma A.1, that there exists δ small enough so that on Ω (δ,, Z 1 (t µ s ɛ for all t. In particular, on Ω (δ, and for all t τ T, ( ( + ( µ s Z1 (t µ cs Z2 (t = µ s Z1 (t 1 µ s µs Z1 (t 1 µ s 1+β µcs µ s Z1 (t ( ( + µ s ɛ µ Z1 (t 1 µ s + 1+β µ s Z1 (t, where µ = µ s µ cs. Hence, µ s ɛ µ β µ s, (A27 Q (t τ Q ( + ( µ s ɛ µ β (t τ + δ. (A28 µ s Choosing ɛ and δ small enough and letting η := (µ s ɛ µ β µ s and t = 3K/2 + δ, (A29 η we must have that τ t / on Ω (δ,. By similar considerations as in the proof of Lemma A.1, we now have that for all t / t T/, Q (t Q ( K. In particular, on Ω (δ,, Q (2t / 2 q(ξ 2 sup ξ A:q(ξ>2K q(ξ K, (A3 where A := {ξ X : (z 1 (ξ µ s ɛ}. As in the proof of Lemma A.1, one can prove that P ((Ω (δ, c c 5 e c 6δ, for some positive constants c 5 and c 6. Since Q (t Q ( + A (t we then have that E ξ [Q (2t / 2 ] q(ξ 2 sup ξ A:q(ξ>2K q(ξ K + E [ A (2t /1 (Ω (δ, c ]. (A31 1

Using the Cauchy-Schwartz inequality and noting that E[2A (2t ] c 7 for some constant c 7 and for all, we can choose large enough so that E ξ [Q (2t / 2 ] q(ξ 2 sup ξ A:q(ξ>2K q(ξ K 2. (A32 Also, using again the fact that Q (t Q ( + A (t, we readily have that, sup E[Q (2t / 2 q(ξ 2 ] c 8, ξ A:q(ξ 2K (A33 for some constant c 8 (that depends on K but is independent of. After some simple manipulations we have that q(ξ 2 E ξ [Q (2t / 2 ] K 2 q(ξ c 8 + ( K 2 q(ξ c 8 E ξ [Q (2t / 2 q(ξ 2 ] 1{ξ / A} (A34 We now follow an argument very similar to the proof of Theorem 5 in [3]. Specifically, let ν ( be the stationary distribution of the process Ξ (t. Then, E ν [Q ( 2 ] = E ν [Q (2t / 2 ], (A35 and in particular, ( = q(ξ 2 E ξ [Q (2t / 2 ] ν (dξ, ξ Ξ where q(ξ is the queue component of the state ξ. By equation (A34 we now have that E ν [Q (] 2c 8 K 2 ( [( ] K E K ν 2 Q ( c 8 E[Q (2t / 2 Q ( 2 Q (] 1{Ξ ( / A} (A36 We now have the following Lemma whose proof we postpone to E. Lemma A.2 Under the assumption of Theorem A.1, and for any integer m 1, lim sup E ν [( Q m ] ( <. 11

that Using this Lemma together with Corollary A.2, we have by the Cauchy-Schwartz inequality lim sup E ν [Q ( m 1{ξ / A}] =. Applying this with some minor manipulation to (A36 we get that E ν [Q (] c 9, for some constant c 9 and all large enough. Consequently, lim sup E[Q ] =. (A37 Lemma A.3 Fluid Limits Consider a finite interval [, T ] and suppose that ( Q (, Z 1 (, Z 2 ( ( Q(, Z1 (, Z 2 (. ( Then, under the assumptions of Theorem A.1, the sequence Q (t ; Z 1 (t ; Z 2 (t ; T i (t, i = 1,..., 5 is tight in D[, T ] and every subsequence { k } k 1 contains a further subsequence that converges to some limit almost surely uniformly on compact sets. Moreover, any such limit process ( Q(t; Z1 (t; Z 2 (t; T i (t, i = 1,..., 5, satisfies the following equations: Z 1 (t + Q(t = Q( + Z 1 ( + t Z 2 (t = Z 2 ( µ cs Z 1 (t = Z 1 ( + T 2 (t µ s µ s Z1 (udu, (A38 Z 2 (udu + pµ s Z1 (udu T 1 (t, Z 1 (udu + T 3 (t + T 4 (t T 5 (t, (A39 (A4 T 1 (t1 { Q(t>} = pµ s Z1 (t, (A41 T 2 (t1 { Z1 + Z 2 < 1+β µs } = 1, (A42 12

T 3 (t1 { Q(t>} = µ s Z1 (t, (A43 T 4 (t1 { Q(t>} = µ cs Z2 (t, (A44 T 5 (t1 { Q(t>} =. (A45 Proof: We start by establishing the existence of the fluid limits. To this end, note that T i ( are increasing continuous functions with T i ( = and for t > s 5 i=1 T i (t T i (s c(t s. (A46 This follows directly from the fact that Z 1 (t + Z 2 (t N c for some c >. Invoking the Arzelà-Ascoli Theorem (see for example [1] together with (A8 we have that the sequence ( T 1,..., T 5 (t, M Z,Q (t, M Z 1 (t, M Z 2 (t is relatively compact. In particular, from equations (A9-(A11 it follows that the sequence ( T i (t, i = 1,.., 5; Z 1 (t ; Z 2 (t ; Q (t, M Z,Q (t, M Z 1 (t, M Z 2 (t is relatively compact so that every subsequence contains a further subsequence that converges to some limit. It is trivial that every limit must satisfy the equations (A38-(A4. To see why equations (A41-(A45 must hold consider for example equation (A41: Choose t with Q(t >. It is then possible to choose large enough along the subsequence such that for all > on the subsequence Q (t/ > ɛ for some ɛ > (and this can be shown to also hold in some small neighborhood of t. In particular for large enough and for any s in some neighborhood of t, Q (s > K (by the assumption on K, so that T1 (t = pµ s Z1 (t. Lemma A.4 Fix ɛ > and assume < β pµs µ cs. For any process ( Q(t; Z1 (t; Z 2 (t; T i (t, i = 1,..., 5, 13

satisfying equations (A38-(A45, there exists t (ɛ (independent of Z 1 ( and Z 2 (, such that for all t t (ɛ Z 1 (t 1 µ s ɛ, (A47 for every fluid limit ( Q(t, Z 1 (t, Z 2 (t. Moreover, there exists t t (ɛ, such that for all t t : Ī(t ɛ, (A48 where Ī(t = 1 + β µ s Z 1 (t Z 2 (t. Proof: The argument is very simple. Assume that the statement [ Z 1 (t 1 µ s ] ɛ is violated at time, that is Z 1 ( < 1/µ s ɛ. By equation (A38, for every interval [s, t], on which Q(u = and Z 1 (u < 1/µ s ɛ for u [s, t], we have that d ( Q(t + Z1 (t dt 1 µ s (1/µ s ɛ, (A49 or equivalently d ( Q(t + Z1 (t dt µ s ɛ. (A5 Also, by equation (A4, on intervals [s, t] such that Q(u > and Z 1 (u < 1/µ s ɛ for u [s, t], we have that d Z 1 (u du µ cs Z2 (u, and since we assumed that Z 1 (u < 1/µ s ɛ, u [s, t], we have that Z 2 (u β µ s interval and d Z 1 (u du Combining equations (A5 and (A52 we have that for all t (A51 + ɛ on this ( β µ cs + ɛ. (A52 µ s d Z 1 (t dt [ ( ] β µ s ɛ µ cs + ɛ, (A53 µ s for each u with Z 1 (u < 1/µ s ɛ. In particular, if Z 1 ( < 1/µ s ɛ, we have that t (ɛ Z 1 (/(µ s ( β µ s + ɛ µ s ɛ, with Z 1 ( t (ɛ 1/µ s ɛ. Note that by this argument Z 1 is increasing 14

as long as it is below 1/µ s ɛ, implying that Z 1 (t 1/µ s ɛ, t t (ɛ. (A54 Now, we claim that there exists a time t t (ɛ, such that t t, Q(t =. Indeed, assume that at time t (ɛ, Q(t > and let t = inf { t t (ɛ : Q(t = }. Then, for all t (ɛ t t, d Q(t ( ( + = 1 µ dt s Z1 (t µ cs Z2 (t = µ s Z1 (t 1 µ s µs Z1 (t 1 µ s ( 1+β µ cs µ s Z ( ( + 1 (t µ s ɛ µ Z1 (t 1 µ s + 1+β µ s Z 1 (t (A55 µ s ɛ µ β µ s, where µ = µ s µ cs. Choosing ɛ small enough, we have that d Q(t η for some η >. In particular, t Q( t (ɛ/η. Moreover, since d Q(t for all t t (ɛ, we also have that Q(t = for all t t. We can now set t = t. Now, since for all t t, Q(t =, we have by equation (A38 that d Z 1 (t = 1 µ dt s Z1 (t for all t t and it is straightforward to show the existence of a time t (ɛ t, such that for all t t (ɛ, Z 1 (t 1 µ s ɛ. To prove the second part of the lemma, assume that at some time t t (ɛ, Ī(t >. Then, letting t = inf{t t : Ī(t = }, we have that on [t, t], dī(t = µ s Z 1 (t µ cs Z2 (t + pµ s Z1 (t. But since Z 1 (t 1 µ s ɛ for all t t (ɛ, we also have that ( ( β 1 dī(t µ sɛ µ cs + ɛ + pµ s ɛ. (A56 µ s µ s Hence, choosing ɛ small enough, we have the existence of η >, such that dī(t η >, for all t t. In particular, there exists a time t at which Ī(t =. Moreover, by repeating a similar argument starting at the first time after t in which Ī(t ɛ/2, we have that Ī(t ɛ for all 15

t t. The following Theorem shows that the number of idle server doesn t exceed the negative part of the threshold. It applies to both cases β = and β > and will be used also in the proofs in B and C. Theorem A.2 Consider a sequence of systems, where the th system is staffed with N = R + βr + γ R + o( R for β pµ s µ cs, max(β, γ >, and operated with T P [K ], where K R δ, as, (A57 for some δ (,. Then, E[ ( (N Z [K ] + ] N R, as, (A58 where for a real number x, [x] = max( x,, and [x] + = max(x,. Proof: We prove here the theorem only for the case β >. The case β = which is more involved is proved in E. To that end, initialize the th system with (Q (, Z 1 (, Z 2 ( distributed according to the stationary distribution. Then, the process (Q (t, Z 1 (t, Z 2 (t is a stationary process implying that E[I (t] = E[I ] for all t. Now, by Proposition A.1 and since Z 1 + Z 2 N, we have that the sequence ( Q, Z 1, Z 2 is tight and every limit point is of the form (, Z 1 (, Z 2 (. In particular, the sequence of processes is tight and every limit point ( Q(t, Z 1 (t, Z 2 (t satisfies the fluid limit ( Q (t, Z 1 (t, Z 2 (t equations (A38-(A45. We can thus apply Lemma A.4 to conclude the existence of t such that for all t t, Ī(t ɛ and this holds for any limit point. In particular, we have that for all t t, lim sup I (t ɛ, a.s., (A59 16

where the bound also holds when applying expectation since I (t N c, for some constant c >. That is, lim sup E[I (t] ɛ. Recall now that I (t has the distribution of I. Thus, we can conclude that (A6 lim sup E[I ] ɛ. (A61 Since this is true for any ɛ we have the assertion of the Lemma. The following corollary follows directly from Theorem A.2 by noting that under the assumptions of Theorem A.2 we have that [K ] =. Corollary A.3 Under the assumptions of Theorem A.1, E[I ] N R. (A62 B. Condition 2 The following proposition establishes the asymptotic feasibility of the proposed solution under Condition 2. Proposition B.1 Assume µ cs µ s and N agents are used for the th system with lim inf N R N 1 R 1. Then, using T P [] for all, we have that lim sup E[W ] W 1. (A63 17

This is a rather straightforward result. Recall that by the definition of N 1, E[W,µ F CF S s (N W,µ F CF S s (N > ] W. Now, T P [] dictates that in the presence of a positive queue length, every service completion will be followed by an admission of a customer from the queue. Thus, whenever all agents are busy the system will deplete customers at a faster rate than the associated M/M/N (using here the fact that µ cs µ s and the result will hold recalling that, by Condition 2, N R N 1 R for large enough. The asymptotic optimality result is then the following corollary: Corollary B.1 Assume that µ cs µ s, and lim inf N 2 R N 1 R 1, as well as that Assumptions 4.1 and 4.2 hold. Then, the following is asymptotically optimal: Staffing: Staff with N 2 agents. Control: Use T P []. Proof: Proposition B.1 guarantees the asymptotic feasibility of pairs (N 2, T P []. The asymptotic optimality argument is exactly the same as in the proof of Corollary A.1 using Theorem A.2. Proof of Proposition B.1: Recall that W,µ F CF S s (N is the steady state waiting time in an M/M/N system with service rate and service rate µ s. Also, let W be the steady state waiting time under T P [] for a cross-selling system with N agents, arrival rate, service rate µ s and cross-selling rate µ cs. Then, we have the following intuitive result. Lemma B.1 Fix. Assume N R. If µ cs µ s and T P [] is used then, E[W W > ] E[W,µ F CF S s (N W,µ F CF S s (N > ] = 1 Nµ s (A64 18

Having Lemma B.1 the proposition readily follows. Specifically, fix a sequence of staffing levels N with lim inf N R N 1 R 1. Then, by Lemma B.1, E[W W > ] W 1 W (N µ s. (A65 But, from the definition of N 1 we have that lim sup 1 W ( N 1 µ s 1, (A66 so that, recalling that R = µ s, we have lim sup E[W W > ] W lim sup 1 W ( N 1 µ s W ( N 1 µ s W (N µ s 1, (A67 and the proof is completed by noting that E[W ] = E[W W > ]P {W > } E[W W > ]. C. Condition 3 Our optimality results under Condition 3 are closely related to the results of Gans and Zhou [4]. We start then with a description of the system analyzed in [4] as well as with some results comparing this system with the cross-selling system. While [4] considers a system that is essentially different from the cross-selling system, we prove that in this asymptotic regime the two problems are, in some sense, equivalent. Specifically, we prove that the model in [4] constitutes an upper bound on the expected profit for the cross-selling model and that this upper bound is asymptotically achieved under the appropriate staffing and control. To simplify the presentation of the results in which we use this asymptotic equivalence, we give here a brief description of the model considered in [4]: Consider a call center with two types of jobs: Type-H and Type-L. Type-H jobs arrive at rate H, are processed at rate µ H and served FCFS 19

within their class. A constraint of the form E[W ] W limits the expected delay that these jobs may face. An infinite backlog of type-l jobs awaits processing at rate µ L. A pool of homogeneous servers process all jobs, and a system controller must maximize the rate at which type-l jobs are processed, subject to the service-level constraint placed on the type-h work. Given a fixed number of agents, the problem of finding the optimal control is formulated as a constrained, average-cost Markov Decision Process (MDP and the structure of effective routing policies is determined. When µ H = µ L, the suggested policies are globally optimal and have a very simple threshold structure. We refer to this model as the G&Z model. To create a basis for comparison of the two models (Cross-Selling vs. G&Z one may consider cross-selling transactions against processing of type-l jobs and service transactions against processing of type-h jobs. Clearly, the dynamics of the two models are different. In the cross-selling system, rather than having an infinite backlog of cross-selling jobs, these become available only upon a completion of a service job, and if they are not processed right away they disappear. Intuitively then, the processing rate of type-l jobs in the G&Z model constitutes an upper bound on the cross-selling rate in the cross-selling model. We prove this formally in Lemma C.1. The above differences also illustrate the relative technical complexity of the cross-selling model. While in the G&Z model there is an infinite backlog of type-l jobs, the availability of cross-selling jobs is strongly dependent on the number of customers in the service phase in our model. The technical implication of this difference, is that any description of the system dynamics of the crossselling system must be at least two-dimensional, regardless of whether µ s = µ cs or not. Our asymptotic analysis, however, allows us to reduce the dimensionality of the problem whenever µ s = µ cs and prove that using T P the upper bound, as given by the G&Z model, is asymptotically achieved. The following is an adaptation of Definition 7 from [4]. Definition C.1 Fix. A randomized threshold reservation policy with threshold K and probability p acts as follows at each event epoch in which there are no type-h calls waiting to be served: 1. A type-h customer will enter service immediately upon arrival if there are any idle agents. 2. Upon service completion (of either a type-l or a type-h job: If there are K or fewer idle agents, the policy does nothing. If there are K + 1 or more idle agents, then with probability 1 p the policy puts enough type-l jobs into service so that exactly K agents are idle, and with 2

probability p the policy puts enough type-l jobs into service so that exactly K 1 agents are idle. Note that without randomization the threshold reservation policy defined in Definition C.1 can be thought of as the TP control adapted to the G&Z model. Denote by T P (N, p the randomized threshold policy of G&Z with threshold K determined through (A68 and with a randomization probability p. The following is a version of the optimality result of [4] for the case µ s = µ cs. We only cite the parts of the Theorem that are relevant for our results. Theorem C.1 (Theorem 1 - Gans and Zhou: Consider a G&Z model with arrival rate, service rates µ H = µ L = µ s = µ cs, N agents and average delay bound W. Then one of two cases holds: Either 1. The problem is infeasible, or 2. A randomized threshold reservation policy with a threshold K and probability p is optimal, for some p [, 1]. Moreover, the threshold K is chosen so that { K (N = max k [ N, ] } ξ k (N N µ s W. (A68 Here ξ k (N = P {Z,µ F CF S s (N = N Z,µ F CF S s (N N + k} and Z,µ F CF S s (N is the steadystate number of busy servers in an M/M/N system with arrival rate and service rate µ s. Remark C.1 Note that under T P (N, (i.e. when setting p =, the steady state number of busy agents in the G&Z system, denoted by E[ Z ], satisfies E[ Z ] = E[Z F CF S,µ s N + K ]. (N Z,µ F CF S s (N Given two random variables X and Y, we use the notation X st Y to denote that a random variable X is stochastically greater than Y. Let CS π (t be the cumulative cross-selling completions up to time t when the control π is used. Also, let T H π (t be the cumulative completion of type-l jobs up to time t in the G&Z model when the control π is used. Note that, by the same argument as in Lemma 2.1, letting Z,π be the steady state number of busy agents in the G&Z model under the control π, we have that the steady state throughput rate of type-l jobs equals µ cs (E[ Z,π ] R. The following lemma does not assume µ s = µ cs. 21

Lemma C.1 Fix, µ s, µ cs, N and W. Let πg&z be the optimal control in the G&Z system with µ H = µ s and µ L = µ cs. Then, for any policy π Π(N we have that T H π g&z (t st CS π (t, t. (A69 In particular, µ cs (E[Z,π ] R µ cs (E[ Z,π g&z ] R. (A7 Proof: We use a sample path construction and a coupling argument. We will show that under our sample path construction the inequality (A69 holds a.s. This, in turn, implies the stochastic ordering in (A69. We construct the coupled sample paths as follows: fix a common sample path of arrivals, service times and cross-selling times for both systems. Specifically, let {t n } n=1, {s n } n=1, and {c n } n=1 be, respectively, the sequence of arrival times, service times and potential cross-selling times (that is, if cross-selling is exercised on customer n, his cross-selling time will be c n. Then, our sample path construction uses the same sequences, {t n }, {s n } and {c n } for both systems. For simplicity of notation label the cross-selling system by 1 and the G&Z system by 2. Fix a scheduling policy π 1 for system 1 and use the same scheduling policy for system 2. This is clearly possible because whenever system 1 can schedule a customer to cross-sell system 2 can schedule a type-l job to service. It is now straightforward to show by induction on the event epochs (arrival, or service completion of any type that both systems will have exactly the same sample paths, and we would have that pathwise T H π1 (t = CS π1 (t, t, (A71 and µ cs (E[ Z,π 1 ] R = µ cs (E[Z,π 1 ] R. (A72 Since we fixed the scheduling policy for system 1 we can continue by saying that max µ cs (E[ Z,π ] R µ cs (E[Z,π 1 ] R, (A73 π and in particular, µ cs (E[ Z,π g&z ] R sup µ cs (E[Z,π 1 ] R. (A74 π 1 22

For future reference let V (N = rµ cs (E[ Z,π g&z ] R (C (N C (R, so that V (N is the optimal throughput rate in the G&Z model with N agents. Now, let N be a sequence with N N 1 for all and N R R ˆγ, as, (A75 for some ˆγ >. The existence of such a sequence is guaranteed since by 9 of [2] we have that N 1 R R γ, (A76 for some γ >. Also, let Ȳ,p be the steady state overall number of customers in a G&Z system with N agents and using the control T P (N, p. Also, let Y be the steady state overall number of customers in a cross-selling system with N agents and using T P [K ] with K determined through (A68. Accordingly, we let Z and Z be the number of busy agents in the above two systems. Define the scaled variables X,p = Ȳ,p N N R, and X = Y N N R. Lemma C.2 Assume that µ s = µ cs, N satisfies (A75 and K is defined through (A68. Then, there exists a function δ(, such that K δ(ˆγ, Ŵ, as, (A77 R where δ(, is finite for all positive and finite arguments. Lemma C.2 is proved in E. It is a key component in the proof of the following convergence result where we let D := D[, be the space right continuous processes with left limits endowed with the J 1 Skorohod topology. We say that a sequence of processes x ( x( in D if the convergence holds in D[s, T ] for each < s < T <. We let Ȳ,p (t, t be the process representing the overall number of customers in a G&Z system with N agents and using the control T P (N, p. Also, let Y (t, t be the process representing the overall number of customers in a cross-selling system with N agents and using T P [K ] with K determined 23

through (A68. Define the scaled processes X,p (t = Ȳ,p (t N N R, and X (t = Y (t N N R. Proposition C.1 Diffusion Limits: With N satisfying (A75, and assuming that X,p ( X(, and X ( X(, as, (A78 we have that and X,p ( X( in D as, X ( X( in D as (A79 (A8 where X( is a diffusion process with infinitesimal drift function βµ s, x m(x = (β + xµ s, δ x (A81 and infinitesimal variance term σ 2 = 2µ s, where δ := δ(ˆγ, Ŵ is given in (A115. Proof: We start by proving the result for the sequence X (. We use coupling with a Birth and Death (B&D process corresponding to a state dependent M/M/1 system for which the limits are known. In particular, consider the B&D process with rates: ˆ i = for all i, where i is the number of customers in the system, and ˆµ i = (N + K + iµ s 1 i K 1 N µ s i K otherwise, (A82 where one should recall that K is negative in this setting. We denote this process by Ŷ ( and define its scaled version by ˆX ( = Ŷ ( +K. Initializing the N R th B&D process with ˆX ( K and recalling the convergence of K / R, the sequence ˆX (, converges weakly to X( with the diffusion parameters given in equation (A81 (see for example [8]. In order to complete our proof 24

we have to show that for any T > d T ( ˆX, X P, as (A83 where d T (, = sup <t T ˆX (t X (t. By the convergence together theorem (see for example Theorem 11.4.7 of [1] we will have the desired result. In order to evaluate d T ( ˆX, X, we can use a coupling argument and deduce that d T ( ˆX, X sup <t T [Z (t (N + K ], (A84 N R which converges to zero by Remark E.1. To complete the proof, then, we present the coupling argument, in which we omit the superscript for simplicity of notation. We initialize the crossselling system with all agents busy and no customer in queue and we initialize the B&D process with K customers in system. We generate arrivals from the same Poisson process. We generate the departures from the same Poisson process with thinning. Let Ŷ (t be the value of the state dependent M/M/1 process at time t. Y (t, as before, is the number of customers in the crossselling system at time t. We prove by induction that Ŷ (t Y (t (N + K, for all t. Ŷ (t (Y (t (N + K sup s t[z(t (N + K], for all t By our initial conditions the assumption holds at the first departure from the system. Assume that it holds for the first n 1 departures and consider the n th, let the time of this departure be t n. By our inductive assumption the n th departure will be a departure in both systems if Ŷ (t n = Y (t n (N + K while preserving the ordering. It will be a departure in the M/M/1 system, and not in the cross-selling system, only if Ŷ (t n > and Ŷ (t n > Y (t n (N + K thus preserving the ordering. It will be a departure in the cross-selling system and not in the M/M/1 queue only if = Ŷ (t n > Y (t n (N + K again preserving the ordering. Also, whenever Ŷ (t n > Y (t n (N + K > the difference between the two processes cannot increase, since every departure will necessarily be a departure in the M/M/1 system. The difference can only increase when = Ŷ (t n > Y (t n (N +K, in which case Ŷ (t n (Y (t n (N +K = [Y (t n N +K] = [Z(t n N +K], where the last equality follows from the fact Y (t = Z(t whenever Y (t N. Thus, the second part of the inductive assumption is preserved. Note that the result would still hold as long as Ŷ ( = (Y ( (N K+. 25

The proof for the sequence X,p ( is much simpler. It is trivial to show through a coupling argument that for any p [, 1] one can construct the sample path of the processes X, (, X,p (, X,1 (, so that X, (t X,p (t X,1 (t, t. Note that for p = the overall number of customers in the G&Z system has exactly the same law as the state dependent M/M/1 defined through equation (A82 above. For p = 1 the same holds with K replaced with K + 1. But, by [8] the scaled versions of these two M/M/1 systems will have the same limit X(. The proof is completed by applying the convergence together theorem. We now show that the process-wise convergence also implies in this case convergence of the steady-state variables. Corollary C.1 With µ s = µ cs and N satisfying (A75, there exists a random variable X such that for any p [, 1] X,p X, as, (A85 and X X, as, (A86 where X has the steady-state distribution of the diffusion process X( in Proposition C.1 and the convergence also holds in expectation. The proof of Corollary C.1 is given in E. Note that since the result is independent of p we have by Remark C.1 that whenever N satisfies (A75 E[Z,µ F CF S s (N R Z,µ F CF S s (N N + K ] (E[ Z ] R N R, as. (A87 This readily follows by noting that one can write Z = N (Ȳ,p N. Corollary C.1 leads immediately to the following result, which establishes asymptotic optimality for a fixed sequence of staffing levels, N. Corollary C.2 Assume µ s = µ cs and consider a sequence of cross-selling systems such that N satisfies equation (A75, and for each, T P [K ] is used with K determined through equation (A68. Then, lim sup E[W ] W 1, (A88 26

and V (N, T P [K ] V (N 1, as. (A89 The following lemma will help us to translate the result of Corollary C.2 to the more general asymptotic optimality result that we need. Lemma C.3 Assume µ s = µ cs in addition to Assumptions 4.1 and 4.2. Let N and K be N2 determined through (18 and (19 and assume that lim sup R < 1. Then, N 1 R lim inf N R R >, (A9 and lim sup N R R <. (A91 Corollary C.3 Assume that µ s = µ cs in addition to Assumptions 4.1 and 4.2. Also, assume that N2 lim sup R < 1. Then, the following is asymptotically optimal for the cross-selling system: N 1 R Staffing: Staff with N agents where N is given by equations (18 and (19. Control: Use T P [K (N ] where K (N is given by equation (19. D. Proofs for 5 D.1 Proof of Lemma 5.1 The argument is straightforward and we only briefly outline it. For any policy π Π(N we construct the corresponding sample paths as follows: We generate the arrivals from a Poisson stream. In addition, we generate an infinite sequence of service times {s i } i 1 and cross-selling times {c i } i 1. When constructing the actual sample paths the service s i will be assigned to the i th customer to begin service and the cross-selling time c i will be assigned to the i th customer to begin cross-selling. Clearly, under this construction the process (Z 2 (t, Y (t is invariant to the order in 27

which customers are admitted from the queue. In particular, we will have exactly the same sample paths under π which is obtained from π by admitting customers to service in a FCFS manner. This invariance guarantees (through Little s Law that if π is feasible so will be π. Moreover, both controls will admit the same cross-selling rate since Z 2 has the same probability law under both π and π. D.2 Proof of Lemma 5.2 We use a coupling argument to prove this assertion. Consider two cross-selling systems with the same number of agents, N, in both systems. Let system 1 be the system that uses the policy π and system 2 be the system that uses π (the work conserving system. We assume that both systems are initiated empty and we let {t i } i 1 and {s i } i 1 and {c i } i 1 be, respectively, the sequences of arrival times, service times and cross-selling times in system 1. Specifically, customer i arrives at time t i and requires a service time of s i. If cross-selling is exercised at customer i the cross-selling will require c i units of time. If cross-selling is not exercised on customer i we will have c i =. We construct the sample path of system 2 from system 1 as follows: We use the same stream of arrivals, services and cross-selling times. We cross-sell to customer i in system 2 if and only if we cross-sell to him in system 1. To differ from system 1, upon service completion of customer i, if cross-selling is not exercised, a customer from the queue will be admitted to service (unless the queue is empty. Let b j i, j = 1, 2, be the time at which customer i begins service in system j (In particular, the waiting time of customer i in system j is given by t i b j i. Let Q j (t, j = 1, 2, be the queue length at time t in system j. Also, let CS j (t be the number of customers that left system j up to time t after cross-selling was exercised on them. In order to prove that the assertion of the lemma holds it suffices to show the following: 1. Q 1 (t Q 2 (t, for all t. 2. CS 1 (t CS 2 (t. Indeed, if Q 1 (t Q 2 (t, the assumed feasibility of system 1 will imply the feasibility of system 2. Moreover, if CS 1 (t CS 2 (t, then system 2 performs at least as well as system 1 in terms of cross-selling rate. Since we exercise cross-selling on customer i in system 2 only if we exercise cross-selling on this customer in system 1, it suffices to prove that for all i 1, b 2 i b 1 i. That is, in system 2 all the customers begin service earlier. 28

We will now proceed by induction on the customer number to prove that indeed i 1, b 2 i b 1 i. The conditions clearly holds for the first customer since both systems are initiated empty. Assume the condition holds up to customer n 1 and consider customer n. Specifically consider the following cases: If at time t n there are idle agents in system 2 the customer will be admitted to service immediately upon arrival (in system 2 and the inductive assumption will be kept. Otherwise, let us consider the time b 2 n 1 at which customer n 1 will begin service in system 2 (while he might still be waiting for service in system 1. Let r j i time of customer i n 1 in system j at time b 2 n 1. That is, be the remaining handling r j i = [ s i + c i [b 2 n 1 b j i ]+] +. (A92 In particular, by our inductive assumption ri 2 ri 1, i n 1, and by work conservation and the fact that customer n had to wait in queue, we have that t n < b 2 n 1+min i n 1 {ri 2 ri 2 > } and b 2 n = b 2 n 1 + min i n 1 {ri 2 ri 2 > }. If we can show that for system 1 b 1 n b 2 n 1 + min i n 1 {ri 1 ri 1 > }, then we are done. To see that this indeed the case note that since b 2 i b 1 i for all i n 1 and since the handling times are common for both system, we have that at time b 2 n 1 the overall number of customers in system 1 is at least as large as the overall number of customers in system 2. Now, recall that we assumed that all agents are busy in system 2 at time t n, this implies that on the interval [b 2 n 1, t n all agents are busy (otherwise customer n would not have to wait by work conservation. Hence, at time b 2 n 1 the number of customers in system 2 (and then in both systems will be at least N. For system 1 this implies that the number of idle agents at time b 2 n 1 is smaller than the queue length. Formally, if Z 1 (t is the number of busy agents in system 1 at time t, then we just argued that Z 1 (b 2 n 1 + Q 1 (b 2 n 1 N, and in particular I 1 (b 2 n 1 Q 1 (b 2 n 1, where I 1 (t is the number of idle agents in system 1 at time t. Hence, even if at time b 2 n 1 system 1 admits all waiting customers to service, by the assumption that t n < b 2 n 1 + min i n 1 {ri 2 ri 2 > } b 2 n 1 + min i n 1 {ri 1 ri 1 > }, customer n must find either a non-empty queue or an empty queue but with all agents busy. If he finds an empty queue with all agents busy he will enter at time b 2 n 1 + min i n 1 {ri 1 ri 1 > }, otherwise he will have to wait more. In any case we have that b 2 n 1 + min i n 1 {ri 1 ri 1 > }. 29

D.3 Proof of Proposition 5.1 Since we fix we omit the superscript from the all the notation. Let π be any feasible policy for the original cross-selling problem which exists by our assumption that N N 1. Define π L to be an adaptation of π to a system where there is a limited number of trunk lines, L. The adaptation of π to the system with finite buffer is straightforward. Note that π(i, j defines what action to take in an event epoch when the system is in state i, j. Then, we take π L (i, j = π(i, j, i, j : j L, z 2 N. Also, from any feasible policy, π L, in the finite buffer the system we can construct a corresponding policy, π L for the infinite buffer system by setting π L (i, j =, i, j : j > L. Our aim is now to show that the problems have an asymptotically equal optimal value. For this purpose it suffices to show that starting from a work conserving policy π, the sequence that we construct π L (which is also work conserving and hence within the set of possible solutions for the LP, achieves asymptotically the same value, as L. And vice versa, i.e. that starting from a sequence of policies {π L }, the sequence of adapted policies for the infinite buffer system, {π L }, achieves asymptotically the same value for the infinite buffer system. Hence, it suffices to show that for any ɛ >, for all L large enough Ṽ (N, π ˆV (N, L, π L ɛ, (A93 and Ṽ (N, π L ˆV (N, L, π L ɛ. (A94 Here, Ṽ (N, π and ˆV (N, L, π L are, respectively, the cross-selling rates in the infinite and finite buffer systems, equipped with π and π L, N agents and L trunk lines (in the finite buffer system. Then, by definition V LP (N, L = sup π L ˆV (N, L, π L and V (N = sup π Ṽ (N, π, where the supremum is taken over feasible policies for each system. Recalling that the cross-selling rate under any policy π equals µ cs E[Z π 2 ], in order to prove (A93 it suffices to show that [ ] E[Z2 π ] = lim E Z L,πL 2 L B, where Z L,πL 2 B is the steady state number of agents busy cross-selling in the finite buffer system with L trunk lines and using a control π L. We start, by fixing π, a feasible policy for the infinite buffer system and proving (A93. We will consider only this direction since the proof of (A94 is analogous. Consider the truncation of 3

the resulting Markov chain to the subspace of the domain in which {j L}. Then, the restricted Markov chain has the same law as the finite buffer system with π L. Hence, E[Z π 2 ] = E [ Z L,πL 2 B ] P {Y π L} + E[Z π 2 1 {Y π >L}]. (A95 By feasibility of π, E[Q π ] W. Using Markov s inequality we have P {Y π > L} = P {Q π > L N} W L N (A96 Moreover, by the Cauchy-Schwartz inequality E[Z π 2 1 {Y π >L}] E[(Z π 2 2 ]P {Y π > L}. (A97 Since, by definition, Z π 2 N, we then have that E[Z π 2 1 {Y π >L}], as L. (A98 Plugging (A96 and (A98 back into equation (A95 we have that [ ] E[Z2 π ] = lim E L Z L,πL 2 B. (A99 E. Proofs of auxiliary results Proof of Lemma 2.2. It is immediate to see that the chain is irreducible. Because the rates are bounded we can use uniformization and define a related Discrete Time Markov Chain (DTMC. Define the set C = {(i, max{n N + K} : i N}. Let τ C be the first hitting time in the set C. Accordingly, E x [τ C ] is the expected hitting time given that we start at state x. C is a compact set and it is easy to prove that sup x C E x [τ C ] < M C < (an elaborate derivation of the bound, M C, would similar to the proof of Lemma 8 in [4] and we omit the detailed argument. Stability is now established by applying theorem 1.4.1 from [9]. 31

Proof of Lemma A.2: The proof uses a Lyapunov function argument and applies Theorem 5 of [3]. We give only the outline of the proof. Towards that end, using the equations for the evolution of the queue length (as in the proof of Theorem A.1 as well as Lemma A.1, we can easily show that there exists a time t 1 and strictly positive constants δ and γ such that [ ] E ξ e Q (t 1 sup ξ:q(ξ>δ e Q ( e γ, for some constant γ >. Fix Φ (ξ = e q(ξ/ for all ξ X. Let φ (t := sup ξ X (Φ 1 (ξe ξ (Φ (Ξ (t 1. Then, using the fact that Q (t Q (+A (t for all t, it is straightforward that φ (t 1 < for all and moreover that Applying Theorem 5 of [3] we have for all that lim sup φ (t 1 <. E ν [Φ (Ξ (] eδ φ (t 1 1 e γ, and by our definition of Φ ( we have consequently that [ ] lim sup E e Q ( <. The result of the Lemma now follows. Proof of Lemma B.1: Since we fix we omit the superscript throughout the proof of the Lemma. Recall the state descriptor S(t = {Z 2 (t, Y (t}. Consider the set A where all agents are busy, that is A = {(i, j : j N, i }. Let S A (t be the process one gets when restricting the Markov chain to the set A (and in particular Q A (t = (Y A (t N +. Since the new state space is clearly irreducible S A (t is a Markov chain. In particular, we will have that E[Q Y N] = E[Q A ]. For the restricted Markov chain we can couple the queue length with an M/M/1 queue with service rate Nµ s as follows. Let Q B (t be the queue length in this M/M/1 queue. Initiate Q A ( = Q B ( = 1. Generate arrivals from the same Poisson process and departures from the same Poisson process with rate Nµ s + Nµ cs with thinning. Since we assumed that µ cs µ s, it is straightforward to show by induction on the event epochs (arrivals and departures, that for all 32

t, Q B (t Q A (t. But we know that E[Q B ] = E[Q A ] Nµ s, which implies that Nµ s, and the assertion of the lemma is now obtained by applying Little s law. Completing the proof of Theorem E.1. The theorem was proved for the case β > within the proof of Theorem A.1. Indeed, in Theorem A.1 a non-negative value is assumed for K, but the proof remains practically the same for any sequence of thresholds K with K = O(. It remains, then, to prove Theorem A.2 for the case β =. Recall from Remark A.1 that Lemma A.1 holds irrespective of whether β > or not. Moreover, since N = R + γ R + o(, we have that for all, Lemma A.1 holds with (Z1 µ s replaced with Z1 µ s. The proof of Theorem A.2 for the case β = is now similar to the proof of Proposition A.1 but we give it for completeness. We assume for the proof that K but the more general case follows similarly by replacing I ( with (I ( [K ] + and I with (I [K ] + throughout. Recall the definition { } Ω (δ, = ω Ω : 11 max sup i=1,...,11 t T/ B i (ct + C log(2 ct δ. Fix ɛ > and assume that t (ɛ in Lemma A.1 is. Fix K >, assume that I ( > 2K and let τ = inf{t : I (t I ( K}. Using the identity I (t = N Z 1 (t Z 2 (t as well as equations (A1 and (A11, we have that on Ω (δ,, τ τ τ I (t τ I ( (t τ +µ s Z1 (udu+µ cs Z2 (udu pµ s Z1 (udu+δ. (A1 Recall from the proof of Lemma A.1, that there exists δ small enough so that on Ω (δ,, Z 1 (t µ s ɛ for all t. From now on we assume that δ does indeed satisfy this requirement. By definition Z2 (t N Z1 (t for all t. In particular, on Ω (δ,, we have from equation (A1 that I (t τ I ( + ((1 + pµ s ɛ + µ cs ɛ(t τ p(t τ + δ. (A11 33

Let η := (1 + pµ s ɛ + µ cs ɛ p and choose ɛ and δ small enough so that η >. Also, let then, we must have that τ t = K + δ, (A12 η t / on Ω (δ,. By similar considerations as in the proof of Lemma A.1, we now have that for all t / t T/, I (t I ( K. From here the proof follows almost verbally the proof of Proposition A.1 with the appropriate replacements of Q with I. We point out, however, that an analogue of Lemma A.2 is not required here as I N and consequently I / trivially has finite moments of all orders. Remark E.1 The argument in the Proof of Theorem A.2 can be easily modified to show that if I ( Î(, as and the assumptions of Theorem A.2 hold, then (I ( [K ] +, as, (A13 where the convergence is uniform on compact subsets of (,. Indeed, fix ɛ > and consider the set Ω (δ, = { ω Ω : 11 max sup i=1,...,11 t T B i (ct + C log(2 ct δt ɛ 2 }. Then, it is straightforward to show that there exists such that for all, P ((Ω (δ, c ɛ/2. Define the stopping time τ = inf{t : I (t ɛ/2 }. Following the arguments in the beginning of the proof of Theorem A.2, paralleling (A11, one can show that on Ω (δ,, we now write I (t τ I ( + ((1 + pµ s ɛ + µ cs ɛ(t τ p(t τ + δt + ɛ 2. (A14 By the convergence of I (/ we may choose η(ɛ and possibly re-define so that for all, P {I ( > η(ɛ } ɛ/2. By choosing δ appropriately, it is now straightforward to modify the argument in the proof of Theorem A.2 to show that whenever I ( η(ɛ, there 34