MODELING WEBCHAT SERVICE CENTER WITH MANY LPS SERVERS

Transcription

1 MODELING WEBCHAT SERVICE CENTER WITH MANY LPS SERVERS Jiheng Zhang Oct 26, 211

2 Model and Motivation Server Pool with multiple LPS servers LPS Server K Arrival Buffer.

3 Model and Motivation Server Pool with multiple LPS servers LPS Server K Arrival Buffer. Customer contact centers via web-chat.

4 Modeling Servers work at different speed depending how many customer in service. Service Rate /.5 hour service rate number of customers in service

5 Modeling Servers work at different speed depending how many customer in service. Service Rate /.5 hour service rate number of customers in service Classify the pool of N homogeneous servers into groups. Group k: all servers serving k customers.

6 Modeling The classical -model: K K µ1 µ2 µk µ1 µ2 µk

7 Modeling The classical -model: K K K K µ1 µ1 µ2 µ2 µk µk µ1 µ1 µ2 µ2 µk µk

8 Literature Review Many-server Queues Puhalskii 27, Mandelbaum, Massey & Reiman 1998,... Perry & Whitt 21 now... LPS Queues Zhang, Dai and Zwart 21, 211 Zhang & Zwart 28, Gupta & Zhang 211,... Averaging Principle Coffman, Puhalskii & Reiman 1995, 1998 Kurtz 1992, Hunt & Kurtz 1994,...

9 Notations Server pool Z(t) = (Z (t), Z 1 (t),..., Z K (t)) N K+1

10 Notations Server pool Z(t) = (Z (t), Z 1 (t),..., Z K (t)) N K+1 Queue Q(t) N

11 Notations Server pool Z(t) = (Z (t), Z 1 (t),..., Z K (t)) N K+1 Queue Q(t) N K Z k (t) = N, and Q(t)(N Z K (t)) =. k=

12 Notations Server pool Z(t) = (Z (t), Z 1 (t),..., Z K (t)) N K+1 Queue Q(t) N Service K Z k (t) = N, and Q(t)(N Z K (t)) =. k= ( t S k µ k ) Z k (s)ds

13 Notations Server pool Z(t) = (Z (t), Z 1 (t),..., Z K (t)) N K+1 Queue Q(t) N Service K Z k (t) = N, and Q(t)(N Z K (t)) =. k= ( t S k µ k ) Z k (s)ds Key assumption: exponential service time

14 Notations Server pool Z(t) = (Z (t), Z 1 (t),..., Z K (t)) N K+1 Queue Q(t) N Service K Z k (t) = N, and Q(t)(N Z K (t)) =. k= ( t S k µ k ) Z k (s)ds Key assumption: exponential service time The index (routing) i (t) = min{ k K : Z k (t) > }.

15 Z Z 3 Background Stochastic Model Fluid Model Approximations Optimality System Dynamics some simulations 2 One Sample Path of Stochastic Process Z(t) 15 Z 1 Z 2 State 1 5 Z 4 Z 5 Z Time One Sample Path of Index Process i * (t) Index Time FIGURE: λ = 2, N = 2, K = 6 and µ = (1, 1.6, 1.8, 2.2, 2.3, 2.4)

16 System Dynamics When will z k (s) jump by 1

17 System Dynamics When will z k (s) jump by 1 i (s ) = k 1, and an arrival happens at s

18 System Dynamics When will z k (s) jump by 1 i (s ) = k 1, and an arrival happens at s a service completion from group k + 1 at s

19 System Dynamics When will z k (s) jump by 1 i (s ) = k 1, and an arrival happens at s a service completion from group k + 1 at s When will z k (s) jump by -1

20 System Dynamics When will z k (s) jump by 1 i (s ) = k 1, and an arrival happens at s a service completion from group k + 1 at s When will z k (s) jump by -1 i (s ) = k, and an arrival happens at s

21 System Dynamics When will z k (s) jump by 1 i (s ) = k 1, and an arrival happens at s a service completion from group k + 1 at s When will z k (s) jump by -1 i (s ) = k, and an arrival happens at s a service completion from group k at s

22 System Dynamics When will z k (s) jump by 1 i (s ) = k 1, and an arrival happens at s a service completion from group k + 1 at s When will z k (s) jump by -1 i (s ) = k, and an arrival happens at s a service completion from group k at s Dynamic Equation for Z k, < k < K t ( t Z k (t) = Z k () + 1 {i (s )=k 1}dΛ(s) + S k+1 µ k+1 t ( t ) 1 {i (s )=k}dλ(s) S k µ k Z k (s)ds ) Z k+1 (s)ds

23 System Dynamics Dynamic Equation for Z Z (t) = Z () t ( t 1 {i (s )=}dλ(s) + S 1 µ 1 ) Z 1 (s)ds

24 System Dynamics Dynamic Equation for Z Z (t) = Z () t Dynamic Equation for Z K Z K (t) = Z K () + t ( t 1 {i (s )=}dλ(s) + S 1 µ 1 ( t 1 {i (s )=K 1}dΛ(s) S K µ K ) Z 1 (s)ds ) Z K (s)1 {Q(s )=} ds

25 System Dynamics Dynamic Equation for Z Z (t) = Z () t Dynamic Equation for Z K Z K (t) = Z K () + t Dynamic Equation for Q Q(t) = Q() + t ( t 1 {i (s )=}dλ(s) + S 1 µ 1 ( t 1 {i (s )=K 1}dΛ(s) S K µ K ( t 1 {i (s )=K}dΛ(s) S K µ K ) Z 1 (s)ds ) Z K (s)1 {Q(s )=} ds ) Z K (s)1 {Q(s)>} ds

26 System Dynamics Arrival process, Λ(t), is assumed to be non-homogeneous Poisson process with rate λ(t). Arrival Rate/.5 Hour on Monday Arrival Rate/.5 Hour on Tuesday Arrival Rate/.5 Hour on Wednesday arrival rate arrival rate arrival rate Time of the Day Time of the Day Time of the Day Arrival Rate/.5 Hour on Thursday Arrival Rate/.5 Hour on Friday Arrival Rate/.5 Hour on Weekdays arrival rate arrival rate arrival rate Time of the Day Time of the Day Time of the Day

27 A Heavy Traffic Regime Large number of servers to accomodate large demand.

28 A Heavy Traffic Regime Large number of servers to accomodate large demand. Consider a sequence of system indexed by n 1 n Nn N, and 1 n λn (t) λ(t), as n.

29 A Heavy Traffic Regime Large number of servers to accomodate large demand. Consider a sequence of system indexed by n 1 n Nn N, and 1 n λn (t) λ(t), as n. But each server s service rate µ = (µ 1,..., µ K ) is fixed.

30 A Heavy Traffic Regime Large number of servers to accomodate large demand. Consider a sequence of system indexed by n 1 n Nn N, and 1 n λn (t) λ(t), as n. But each server s service rate µ = (µ 1,..., µ K ) is fixed. Fluid Scaling: Z n (t) = Zn (t) n, Q n (t) = Qn (t) n.

31 Fluid Model Let (z, q) be the fluid counterpart of ( Z n, Q n ).

32 Fluid Model Let (z, q) be the fluid counterpart of ( Z n, Q n ). K S = {(z, q) [, N] K+1 R + : z k = N and q(n z K ) = } k=

33 Fluid Model Let (z, q) be the fluid counterpart of ( Z n, Q n ). K S = {(z, q) [, N] K+1 R + : z k = N and q(n z K ) = } How do we deal with t k= 1 {i (s )=k}d Λ n (s)?

34 Fluid Model Let (z, q) be the fluid counterpart of ( Z n, Q n ). S = {(z, q) [, N] K+1 R + : How do we deal with t K z k = N and q(n z K ) = } k= 1 {i (s )=k}d Λ n (s)? Introduce the mapping f : R K+2 [, 1] K+1, µ k+1 z k+1 1, if k = I(z) 1, ( λ f k (z, λ) = 1 µ ) kz k +, if k = I(z), λ, otherwise,

35 Fluid Model Let (z, q) be the fluid counterpart of ( Z n, Q n ). S = {(z, q) [, N] K+1 R + : How do we deal with t K z k = N and q(n z K ) = } k= 1 {i (s )=k}d Λ n (s)? Introduce the mapping f : R K+2 [, 1] K+1, µ k+1 z k+1 1, if k = I(z) 1, ( λ f k (z, λ) = 1 µ ) kz k +, if k = I(z), λ, otherwise, where I(z) = min{ k K : z k > }.

36 Fluid Model Divide the space S into two regions S + = {(z,..., z K, q) S : q > }, S = {(z,..., z K, q) S : q = }. The fluid model can be defined by ODE s in the two regions.

37 Fluid Model Divide the space S into two regions S + = {(z,..., z K, q) S : q > }, S = {(z,..., z K, q) S : q = }. The fluid model can be defined by ODE s in the two regions. on S + q (t) = λ(t) µ K N.

38 Fluid Model on S z (t) = f (z(t), λ(t))λ(t) + µ 1 z 1 (t), z k(t) = f k 1 (z(t), λ(t))λ(t) + µ k+1 z k+1 (t) f k (z(t), λ(t))λ(t) µ k z k (t), < k < K, z K(t) = f K 1 (z(t), λ(t))λ(t) µ K z K (t), q (t) = f K (z(t), λ(t))λ(t).

39 Fluid Model on S z (t) = f (z(t), λ(t))λ(t) + µ 1 z 1 (t), z k(t) = f k 1 (z(t), λ(t))λ(t) + µ k+1 z k+1 (t) f k (z(t), λ(t))λ(t) µ k z k (t), < k < K, z K(t) = f K 1 (z(t), λ(t))λ(t) µ K z K (t), q (t) = f K (z(t), λ(t))λ(t). The ODEs can be written into a vector form, (z, q ) = Ψ(t, z, q).

40 Existence and Uniqueness of Fluid Model Solution THEOREM (EXISTENCE AND UNIQUENESS) Assume that λ(t) is a continuous function of t. Then, there exists a unique solution to the fluid model, i.e., the ODEs, with initial condition (z(), q()) S.

41 Existence and Uniqueness of Fluid Model Solution THEOREM (EXISTENCE AND UNIQUENESS) Assume that λ(t) is a continuous function of t. Then, there exists a unique solution to the fluid model, i.e., the ODEs, with initial condition (z(), q()) S. We can also extend the result to piecewise continuous function λ(t).

42 Invariant State of Fluid Model When the arrival rate is constant, we can study the stead state of the fluid model.

43 Invariant State of Fluid Model When the arrival rate is constant, we can study the stead state of the fluid model. Assumption: µ k is increasing in k.

44 Invariant State of Fluid Model When the arrival rate is constant, we can study the stead state of the fluid model. Assumption: µ k is increasing in k. Then there exists some k such that { λ = µ k +1z k +1 + µ k z k, N = z k +1 + z k.

45 Invariant State of Fluid Model When the arrival rate is constant, we can study the stead state of the fluid model. Assumption: µ k is increasing in k. Then there exists some k such that { λ = µ k +1z k +1 + µ k z k, N = z k +1 + z k. This implies that z k = µ k +1N λ, z k µ k +1 µ +1 = λ µ k N. k µ k +1 µ k

46 Invariant State of Fluid Model When the arrival rate is constant, we can study the stead state of the fluid model. Assumption: µ k is increasing in k. Then there exists some k such that { λ = µ k +1z k +1 + µ k z k, N = z k +1 + z k. This implies that z k = µ k +1N λ, z k µ k +1 µ +1 = λ µ k N. k µ k +1 µ k Define it to be the invariant state I.

47 Invariant State of Fluid Model THEOREM Given any initial value (z(), q()) S the fluid model converges, fast, to the invariant state I.

48 Z 1 Background Stochastic Model Fluid Model Approximations Optimality Invariant State of Fluid Model 2 Stochastic Model: N=2,λ=4,µ=(1,1.6,1.8,2.2,2.3,2.4) Z 15 THEOREM 1 Given any initial value (z(), q()) S the fluid model 5 converges, fast, to the invariant state I Time 2 Fluid Model: N=2,λ=4,µ=(1,1.6,1.8,2.2,2.3,2.4) Z 2 Z 3 Z 4 Z 5 Z Time

49 Functional Law of Large Numbers THEOREM (FWLLN) If ( Z n (), Q n ()) (z, q ), then ( Z n, Q n ) converges weakly to the fluid model solution (z, q) with (z(), q()) = (z, q ).

50 Functional Law of Large Numbers THEOREM (FWLLN) If ( Z n (), Q n ()) (z, q ), then ( Z n, Q n ) converges weakly to the fluid model solution (z, q) with (z(), q()) = (z, q ) Comparison with Fluid Approximation for Z 2 (t) N=5 N=1 N=2 Fluid Time 1 Comparison with Fluid Approximation for Z 3 (t) N=5

51 Z 1 Z 5 Background Stochastic Model Fluid Model Approximations Optimality Functional Law of Large Numbers 2 Stochastic Model: N=2,λ=4,µ=(1,1.6,1.8,2.2,2.3,2.4) Z 15 Z 2 Z 3 Z 4 1 Z Time 2 Fluid Model: N=2,λ=4,µ=(1,1.6,1.8,2.2,2.3,2.4) Time

52 The Averaging Principle The key is to show that t f k (z(s), λ(s))λ(s)ds. The averaging principle t 1 {i n (s )=k}d Λ n (s) is close to lim lim 1 t+δ 1 δ n δ {i n (s )=k}d Λ n (s) = f k (z(t), λ(t))λ(t). t

53 Random Measure and Martingale Representation We define the random measures by where ν n ([, t] A) = t 1 {m n (t) A}ds, { m n k (t) = Zk n(t), if k = I(Zn (t)),, otherwise.

54 Random Measure and Martingale Representation We define the random measures by where ν n ([, t] A) = t 1 {m n (t) A}ds, { m n k (t) = Zk n(t), if k = I(Zn (t)),, otherwise. Martingale Representation M a(u) n = 1 ( Λ n (u) n u ) λ n (τ)dτ, M n k (u) = 1 n S k(nu) u, k = 1,..., K.

55 Random Measure and Martingale Representation t Z k n (t) = 1 n Zn k () n {m n (s) A k 1 }dma(s) n 1 1 n {m n (s) A k }dma(s) n t ) t ) M k (µ n k Zk n (s)ds + M k+1 (µ n k+1 Zk+1 n (s)ds + λn (s)ν n (ds dy) + λn (s)ν n (ds dy) [,t] A k 1 [,t] A k t µ k Z k n (s)ds + µ k+1 t t Z k+1 n (s)ds, < k < K,

56 A Missing Gap Interchange of Steady State and Heavy Traffic Limits ( Z n (t), Q n t (t)) ( Z, n Z ) n

57 A Missing Gap Interchange of Steady State and Heavy Traffic Limits ( Z n (t), Q n t (t)) ( Z, n Z ) n n (z(t), q(t))

58 A Missing Gap Interchange of Steady State and Heavy Traffic Limits ( Z n (t), Q n t (t)) ( Z, n Z ) n n (z(t), q(t)) t (z, q )

59 A Missing Gap Interchange of Steady State and Heavy Traffic Limits ( Z n (t), Q n t (t)) ( Z, n Z ) n n n (z(t), q(t)) t (z, q )

60 Simulation with general service time distributions Exponential Erlang-2 LN(1, 4) Approximation Group 3.911e e e-4 ±8.1e-4 ±6.9e-4 ±8.9e-4 Group ±.3 ±.22 ±.55 Group ±.92 ±.84 ±.165 Group ±.42 ±.3276 ±.4524 Group ±.447 ±.3336 ±.461 Group ±.54 ±.28 ±.68 Group 6 Sojourn Time ±.7 ±.5 ±.11

61 Optimal Staffing and Control Consider the case with constant arrival rate λ. Utility function [ 1 T ] V T(N, n K) = c(n) + E g( Z n (s), Q n (s))ds, T

62 Optimal Staffing and Control Consider the case with constant arrival rate λ. Utility function [ 1 T ] V T(N, n K) = c(n) + E g( Z n (s), Q n (s))ds, T Think about g(z, q) = 1 λ ( q + k kz k ).

63 Optimal Staffing and Control Consider the case with constant arrival rate λ. Utility function [ 1 T ] V T(N, n K) = c(n) + E g( Z n (s), Q n (s))ds, T Think about g(z, q) = 1 λ ( q + k kz k ). Choose N and K to asymptotically minimize the total cost.

64 Optimal Staffing and Control PROPOSITION Assume λ(t) λ and µ k is increasing. lim lim V n (N, K) V(N, K), in probability as n, T n where V(N, K) = c(n) + g(i).

65 Optimal Staffing and Control 2.8 λ=4,µ=(1,1.6,1.8,2.2,2.3,2.4) 2.6 cost C number of servers N

66 Future Research Dynamic control on finite horizon with time-varying arrival Diffusion approximation

67 Questions?

68 Thank you!