Uncertainty in the Demand for Service: The Case of Call...

Introduction Theoretical Results Time-Varying Queues

Uncertainty in the Demand for Service:The Case of Call Centers and Emergency

Departments

Shimrit Maman

Technion - Israel Institute of Technology

January 28, 2009

Advisors: Prof. Avishai Mandelbaum and Dr. Sergey Zeltyn


Outline

1 IntroductionMotivationModel DefinitionThe Case of a Financial Call CenterThe Case of an Emergency DepartmentPractical Guidelines

2 Theoretical ResultsQueues Performances in a Random EnvironmentQED-c RegimeQED RegimeED Regime

3 Time-Varying Queues


Motivation

A standard assumption in the modeling of a service systempostulates that the arrival process is a Poisson process with knownparameters. For example, the prevalent approach in call centers isto assume known arrival rates for each basic interval (say,half-hour).

However, as a rule, call centers data contradicts this assumptionand shows a larger variability of the arrival process than the oneexpected from the Poisson hypothesis.

We explain this over-dispersion by the natural uncertainty of thearrival rate.


Literature Review

Henderson S.

Input model uncertainty: Why do we care and what should we do aboutit?. 2003.

Whitt W.

Staffing a call center with uncertain arrival rate and absenteeism. 2006.

Halfin S., Whitt W.

Heavy-traffic Limits for Queues with many Exponential Servers. 1981.

Zeltyn S. and Mandelbaum A.

Call centers with impatient customers: Exact analysis and many-serverasymptotics of the M/M/n+G queue. 2005.

Steckley S., Henderson S. and Mehrotra V.

Forecast Errors in Service Systems. 2007.

Koole G. and Jongbloed G.

Managing uncertainty in call centers using poisson mixtures. 2001.


Model Definition

The M?|M|n + G Queue:

λ - Expected arrival rate of a Poisson arrival process.

µ - Exponential service rate.

n service agents.

G - Patience distribution. Assume that the patience densityexist at the origin and its value g0 is strictly positive.

Random Arrival Rate: Let X be a random variable with cdf F ,E [X ] = 0, and finite σ(X ). Assume that the arrival rate variesfrom day to day in an i.i.d. fashion:

M = λ + λcX , c ≤ 1,

c ≤ 1/2: Conventional variability.

1/2 < c < 1: Moderate variability.

c = 1: Extreme variability.


The Case of a Financial Call Center

Our study focuses on the arrival counts to the Retail queue.

We consider 263 regular weekdays ranging between April 2007and April 2008.

Holidays which exhibit different daily patterns are excludedfrom our analysis.

Each day is divided into 48 half-hour intervals.



Coefficient of Variationsampled CV- solid line, Poisson CV - dashed lineCoefficient of Variation Per 30 Minutes, seperated weekdays

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time

Coe

ffici

ent o

f Var

iatio

n

Sundays Mondays Tuesdays Wednesdays Thursdays

Sampled CV’s � CV’s of Poisson variables with fixed arrival rates

⇒Over-Dispersion



Average Number of ArrivalsAverage Arrival Per 30 Minutes, seperated weekdays

0

200

400

600

800

1000

1200

1400

1600

1800

2000

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time

Ave

rage

Arr

ival

Sundays Mondays Tuesdays Wednesdays Thursdays

(1) Sundays; (3) Tuesdays and Wednesdays;(2) Mondays; (4) Thursdays;



Basic definitions and notation:

λid - The expectation of arrival volume in the i th interval forday type d , i = 1, 2, . . . , 48 and d = 1, 2, 3, 4.

σid - The std of arrival volume for the i th interval for day typed .

cd - The uncertainty coefficient for day type d .

λid - The average call volume in the i th interval over all daysof type d .

σid - The sampled standard deviation of call volumes in thei th interval over all days of type d .



1. Relation between λid and σid:

Consider a Poisson mixture variable Y with random rateM = λ + λc · X , where E [X ] = 0, finite σ(X ) > 0 and1/2 < c ≤ 1. Then,

Var(Y ) = λ2c · Var(X ) + λ + λc · E (X ).

Given λ →∞σ(Y ) ∼ λc · σ(X ), 1

andln(σ(Y )) ∼ c · ln(λ) + ln(σ(X )).

1f (λ) ∼ g(λ) denotes that limλ→∞ f (λ)/g(λ) = 1.



Mondaysln(sd) vs ln(average) per 30 minutes. Sundays

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8

ln(Average Arrival)

ln(S

tand

ard

Dev

iatio

n)

Tue-Wedln(sd) vs ln(average) per 30 minutes. Sundays

0

1

2

3

4

5

6

0 1 2 3 4 5 6 7 8

ln(Average Arrival)

ln(S

tand

ard

Dev

iatio

n)

Results:

Two clusters exists. Denote j(i) = 1 for i = 1, 2, . . . , 21, andj(i) = 1 for i = 22, 23, . . . , 48.Very good fit (R2 > 0.97).Significant linear relations:

ln(σid) = cdj(i) · ln(λid) + ln(σ(Xdj(i))) ∀ d , i



Mondaysln(sd) vs ln(average) per 30 minutes. Sundays

y = 0.8292x - 0.2882R2 = 0.992

y = 0.8356x - 0.6561R2 = 0.9766

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8

ln(Average Arrival)

ln(S

tand

ard

Dev

iatio

n)

00:00-10:30 10:30-00:00

Tue-Wedln(sd) vs ln(average) per 30 minutes. Sundays

y = 0.8027x - 0.1235R2 = 0.9899

y = 0.8752x - 0.8589R2 = 0.9882

0

1

2

3

4

5

6

0 1 2 3 4 5 6 7 8

ln(Average Arrival)

ln(S

tand

ard

Dev

iatio

n)

00:00-10:30 10:30-00:00

Results:

Two clusters exists. Denote j(i) = 1 for i = 1, 2, . . . , 21, andj(i) = 1 for i = 22, 23, . . . , 48.Very good fit (R2 > 0.97).Significant linear relations:

ln(σid) = cdj(i) · ln(λid) + ln(σ(Xdj(i))) ∀ d , i



2. Fitting a Gamma Poisson mixture model to the data:(Jongbloed and Koole [’01])

Assume a prior Gamma distribution for the arrival rate

λ + λcXd= Gamma(a, b). Then, the distribution of Y is Negative

Binomial.

1 Maximum likelihood estimators of a and b.

2 Goodness of fit test including FDR control method to correctthe multiple comparisons.



H0,id : Midd= Gamma(aid , bid)

Sundays 10:00-10:30

1500 2000 2500

0.0

0.2

0.4

0.6

0.8

1.0

Mondays 10:00−10:30

CD

F

EmpiricalGPM(29.51,58.5),l=1726.43p.value= 0.674

Monday 14:30-15:00

700 800 900 1000 1100 1200

0.0

0.2

0.4

0.6

0.8

1.0

Mondays 14:30−15:00

CD

F

EmpiricalGPM(47.78,19.01),l=908.22p.value= 0.982

Results:

Very good fit.

Only 13 hypotheses are rejected (out of 192).



3. Relation between our main model and Gamma Poissonmixture model:

Let M = λ + λcXd= Gamma(a, b). Then,

E [M] = ab = λ; Var(M) = λb and Var(X ) = λ1−2c · b.

We derive the following relations

b = σ2(X ) · λ2c−1,

a = σ−2(X ) · λ2−2c .

and conclude that

ln(b) = (2c − 1) · ln(λ) + ln(σ2(X )).



Mondaysln(sd) vs ln(average) per 30 minutes. mondays

0

1

2

3

4

5

0 1 2 3 4 5 6 7 8

ln(Average Arrival)

ln(b

)

Tue-Wedln(sd) vs ln(average) per 30 minutes. Tuesdays-wednesdays

0

1

2

3

4

5

0 1 2 3 4 5 6 7 8

ln(Average Arrival)

ln(b

)

Results:

Two clusters.

Very good fit.

Significant linear relations:

ln(bid) = cdj(i) · ln(λid) + ln(σ2(Xdj(i))) ∀ d , i



Mondaysln(sd) vs ln(average) per 30 minutes. mondays

y = 0.6975x - 0.894R2 = 0.9453

y = 0.7699x - 2.0427R2 = 0.9076

0

1

2

3

4

5

0 1 2 3 4 5 6 7 8

ln(Average Arrival)

ln(b

)

00:00-10:30 10:30-00:00

Tue-Wedln(sd) vs ln(average) per 30 minutes. Tuesdays-wednesdays

y = 0.6146x - 0.5349R2 = 0.9257

y = 0.7825x - 2.0107R2 = 0.9457

0

1

2

3

4

5

0 1 2 3 4 5 6 7 8

ln(Average Arrival)

ln(b

)

00:00-10:30 10:30-00:00

Results:

Two clusters.

Very good fit.

Significant linear relations:

ln(bid) = cdj(i) · ln(λid) + ln(σ2(Xdj(i))) ∀ d , i



4. Asymptotic distribution of X:

X =M − λ

λc=

M − ab

(ab)c=

M − E [M]

σ(M)/σ(X )

Let Wad=

Gamma(a, b)− ab

b√

a=

Gamma(a, 1)− a√a

.

Then lima→∞

MGFa(t) = et2/2, t ∈ R, and this limit is the moment

generating function of the standard normal distribution Norm(0, 1).

Conclusion: As λ →∞ (equivalent to a →∞), the randomvariable X/σ(X ) converges in distribution to a standard normaldistributed variable.

X

σ(X )D→ Norm(0, 1).


The Case of an Emergency Department

Consider 194 weeks between from January 2004 till October2007 (five war weeks are excluded from data).

The analysis is performed using two resolutions: hourly arrivalrates (168 intervals in a week) and three-hour arrival rates (56intervals in a week).

Meanwhile, we do not clean our data (Jewish holidays are notexcluded as they should be).



Coefficient of Variation

1 Arrival-Rate Variability: The Case of Emergency

Department

Consider 194 weeks between from January 2004 till October 2007 (five war weeks are excludedfrom data). The analysis is performed using two resolutions: hourly arrival rates (168intervals in a week) and three-hour arrival rates (56 intervals in a week). Meanwhile, we donot clean our data (Jewish holidays are not excluded as they should be).

First, we redraw Figure 1.2 from Section 1.6 on Financial call center, comparing betweencoefficient of variation and inverted square root of mean arrival rates. Figure 1 shows that,in contrast to the call center study, inverted square root of mean is relatively close to CV’s.Peaks at graphs correspond to night periods with small arrival rates.

Therefore, variability of arrival rates at this resolution is somewhat larger (but not muchlarger) the the variability of iid Poisson random variables.

Figure 1: Coefficient of variation

One-hour resolution Three-hour resolution

20 40 60 80 100 120 140 1600

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

interval

coefficient of variationinverted sq. root of mean

5 10 15 20 25 30 35 40 45 50 550

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

interval

coefficient of variationinverted sq. root of mean

If we fit regression model of ln(σ) versus ln(λ), Figure 2 demonstrates a linear patternwith the slope that is very close to 0.5.

Note. Derivation of asymptotic linear relation (1.35) in Section 1.6 is based on c > 1/2.It is unclear how it works for c that is close to 1/2 and relatively small λ.

Finally, Figure 3 shows that if we increase time resolution, variability of arrival rates issignificantly larger than the variability of iid Poisson random variables.

1

In contrast to the call center study, inverted square root ofmean is relatively close to CV’s.

Peaks at graphs correspond to night periods with small arrivalrates.



ln(σ) versus ln(λ) plotsFigure 2: ln(σ) versus ln(λ) plots

One-hour resolution Three-hour resolutiony = 0.497x+ 0.102, R2 = 0.968 y = 0.527x+ 0.087, R2 = 0.947

0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

ln(Mean Arrival Rate)

ln(S

tand

ard

Dev

iatio

n)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.50

0.5

1

1.5

2

2.5

ln(Mean Arrival Rate)ln

(Sta

ndar

d D

evia

tion)

Figure 3: Coefficient of variation

Eight-hour resolution One-day resolution

2

A linear pattern with the slope that is very close to 0.5.

Derivation of the asymptotic linear relation is based onc > 1/2. It is unclear how it works for c that is close to 1/2and relatively small λ.


Practical Guidelines

Determine ”uncertainty coefficient c” via regression analysis.

Check if Gamma model is reasonable.

Calculate X distribution (asymptotic analysis).

Apply our QED-c results in order to determine appropriatestaffing.


Queues Performances in a Random Environment

Assume a set of D days. At each day an arrival rate, m, isindependently generated. Let Y be a system performance measure,and denote by YM the corresponding random performance measurein the random environment.

Denote:

arrj - Number of arrivals on day j

yj - The value of the performance measure YM , on day j .

If the performance measure is related to system performance, e.g.offered load and queue length, by SLLN

Y =1

D

D∑j=1

yjD↑∞−→ E [YM ] .



We classify performance measures that relate to the customers,e.g. delay probability and waiting time, into two classes:

Short-Term Performance Measure: What will happentomorrow?

Y =1

D

D∑j=1

yjD↑∞−→ E [YM ] .

Long-Term Performance Measure: What would be theperformances in the long-run?

Y =

∑Dj=1 arrj · yj∑D

j=1 arrj

D↑∞−→ E [M · YM ]

E [M]> Y .



Lemma

Assume that the system is in steady state. The long-term valueand the short-term value of YM are asymptotically equivalent for allc < 1.

limλ→∞

E [M · YM ]

E [M]= lim

λ→∞E [YM ] .

Remark: Note that the above does not hold for c = 1. In thiswork we focus on long-term performance measures. Thetechniques of solutions, when considering the short-termperformance measures, are similar.


QED-c Regime

QED-c staffing rule:

n =λ

µ+ β

(λ

µ

)c

+ o(√

λ), β ∈ R, c ∈ (1/2, 1).

Assume an M|M|n + G queue with fixed arrival rate λ.

Take λ to ∞.

β > 0: Over-staffing.

β < 0: Under-staffing.

For both cases we provided asymptotically equivalent expressions(or bounds) for P{Wq > 0}, P{Ab|V > 0} and E [V |V > 0].Calculations, based on building blocks from Zeltyn[’05], are carriedout via the Laplace Method.


Square-Root Staffing versus QED-c staffing

QED-c StaffingR = λ/µ β SRS2

c = 0.6 c = 0.75 c = 0.9

1000.5 105 108 (+3%) 116 (+10%) 132 (+25%)1 110 116 (+5%) 132 (+20%) 163 (+48%)

1.5 115 124 (+8%) 147 (+28%) 195 (+69%)

5000.5 511 521 (+2%) 553 (+8%) 634 (+24%)1 522 542 (+4%) 606 (+16%) 769 (+47%)

1.5 534 562 (+5%) 659 (+23%) 903 (+69%)

10000.5 1016 1032 (+2%) 1089 (+7%) 1251 (+23%)1 1032 1063 (+3%) 1178 (+14%) 1501 (+46%)

1.5 1047 1095 (+5%) 1267 (+21%) 1752 (+67%)

20000.5 2022 2048 (+1%) 2150 (+6%) 2468 (+22%)1 2045 2096 (+2%) 2300 (+12%) 2936 (+43%)

1.5 2067 2143 (+4%) 2449 (+18%) 3403 (+65%)

2Square-Root-Staffing


QED-c Regime

Theorem

Assume random arrival rate M = λ + λcµ1−cX , c ∈ (1/2, 1),E [X ] = 0, finite σ(X ) > 0, and staffing according to the QED-cstaffing rule with the corresponding c . Then, as λ →∞,

a. Delay probability: PM,n{Wq > 0} ∼ 1− F (β).

b. Abandonment probability: PM,n{Ab} ∼ E [X − β]+n1−c

.

c. Average waiting time: EM,n[Wq] ∼E [X − β]+n1−c · g0

Remark: For simplifying the formulae we takeM = λ + µ1−c · λcX instead of M = λ + λcX .


QED-c Regime

Examples: Consider two distributions of X

Uniform distribution on [-1,1],

Standard Normal distribution.

(1) β = −0.5, c = 0.7

Delay Probability

0 1000 2000 3000 4000 50000.6

0.65

0.7

0.75

0.8

0.85

0.9

P{W

>0}

λ

P{W>0} vs. λ, c=0.7, β=-0.5

short-term U(-1,1)long-term U(-1,1)approx U(-1,1)short-term N(0,1)long-term N(0,1)approx N(0,1)

0 1000 2000 3000 4000 50000

0.05

0.1

0.15

0.2

0.25

0.3

0.35

P{A

band

on}

λ

P{Abandon} vs. λ, c=0.7, β=-0.5


0 1000 2000 3000 4000 50000.6

0.65

0.7

0.75

0.8

0.85

0.9

P{W

>0}

λ

P{W>0} vs. λ, c=0.7, β=-0.5


0 1000 2000 3000 4000 50000

0.05

0.1

0.15

0.2

0.25

0.3

0.35

P{A

band

on}

λ



Abandonment Probability

0 1000 2000 3000 4000 50000.6

0.65

0.7

0.75

0.8

0.85

0.9

P{W

>0}

λ

P{W>0} vs. λ, c=0.7, β=-0.5


0 1000 2000 3000 4000 50000

0.05

0.1

0.15

0.2

0.25

0.3

0.35

P{A

band

on}

λ



0 1000 2000 3000 4000 50000.6

0.65

0.7

0.75

0.8

0.85

0.9

P{W

>0}

λ

P{W>0} vs. λ, c=0.7, β=-0.5


0 1000 2000 3000 4000 50000

0.05

0.1

0.15

0.2

0.25

0.3

0.35

P{A

band

on}

λ




QED-c Regime

(2) β = 0.5, c = 0.6

Delay Probability

0 1000 2000 3000 4000 50000.2

0.25

0.3

0.35

0.4

0.45

0.5

P{W

>0}

λ

P{W>0} vs. λ, c=0.6, β=0.5


0 1000 2000 3000 4000 50000

0.02

0.04

0.06

0.08

0.1

P{A

band

on}

λ

P{Abandon} vs. λ, c=0.6, β=0.5


0 1000 2000 3000 4000 50000.2

0.25

0.3

0.35

0.4

0.45

0.5

P{W

>0}

λ

P{W>0} vs. λ, c=0.6, β=0.5


0 1000 2000 3000 4000 50000

0.02

0.04

0.06

0.08

0.1

P{A

band

on}

λ




0 1000 2000 3000 4000 50000.2

0.25

0.3

0.35

0.4

0.45

P{W

>0}

λ

P{W>0} vs. λ, c=0.6, β=0.5


0 1000 2000 3000 4000 50000

0.02

0.04

0.06

0.08

0.1

P{A

band

on}

λ



0 1000 2000 3000 4000 50000.2

0.25

0.3

0.35

0.4

0.45

P{W

>0}

λ

P{W>0} vs. λ, c=0.6, β=0.5


0 1000 2000 3000 4000 50000

0.02

0.04

0.06

0.08

0.1

P{A

band

on}

λ




Constraint Satisfaction

Formulation of the Problem:

Define the optimal staffing level by

n∗λ = argminn

{PM,n{Wq > 0} ≤ α

}.

The staffing level nλ is called asymptotically feasible if

lim supλ→∞

PM,nλ{Wq > 0} ≤ α.

In addition, nλ is asymptotically optimal if

|n∗λ − nλ| = o(f (λ)).


QED Regime

c = 1/2Assume M = λ +

√µλ · X , E [X ] = 0 and finite σ(X ) > 0. Let

λ →∞.

Theorem (Staffing)

a. The optimal staffing level satisfies

n∗ =λ

µ+ β∗

sλ

µ+ o(

√λ),

where β∗ is the unique solution of the equation

α = E [α(β − X )]

with respect to the unknown β.α(·) is the Garnett function

α(β) =

"1 +

rg0

µ·h(βp

µ/g0)

h(−β)

#−1

.


QED Regime

Theorem (Staffing). Continued.

b. Introduce the staffing level

n∗β =

&λ

µ+ β∗

sλ

µ

’.

Then the staffing level n∗β is both asymptotically feasible and

asymptotically optimal (f (λ) =√

λ)

Proof Outline

Given X=x,

nβ =λ

µ+ β ·

sλ

µ=

λ +√

µλ · xµ

+ b(β, x) ·

sλ +

√µλ · x

µ

⇒ as λ →∞, b(β, x) ∼ β − x .


QED Regime

Theorem (Performance Measures)

Under the square-root staffing level, as λ →∞,

a. PM,n∗β{Ab} ∼ 1√

n· E (γ∗X ). γ∗X

b. ρn∗β= 1− β∗√

n+ o

(1√n

).

c.PM,n∗β

{Ab}EM,n∗β

[Wq]∼ g0.


QED Regime

c < 1/2

Assume Λ = λ + µ1−cλc · X , c < 1/2, E [X ] = 0 and finiteσ(X ) > 0. Let λ →∞.

Lemma

Assume the square-root staffing level

n =λ

µ+ β

sλ

µ+ o(

√λ), −∞ < β < ∞

Then, asymptotically, the random part of the arrival rate does notaffect the system’s performances. Namely, the queue is asymptoti-cally equivalent to the M|M|n + G queue with deterministic arrivalrate λ, for all c < 1/2.The M|M|n+G queue was analyzed comprehensively by Zeltyn[’05].


ED Regime

c = 1, Discrete Random Arrival Rate

Assume M = λX , where X is a discrete random variable whichtakes values x1 > x2 > . . . > xI > 0, with probabilitiesp1, p2, . . . , pI , respectively. In addition, let E [X ] = 1, σ(X ) < ∞and λ →∞.

Let

κ = argmins

{s∑

i=1

xipi ≥ α

}.

Assume that the inequality is strict.

Define

α∗4=

α−∑κ−1

i=1 xipi

xκpκ.


ED Regime

Theorem (Staffing)


n∗ =λxκ

µ+ β∗

sλxκ

µ+ o(

√λ);

here β∗ is the unique solution of the equation

α∗ =

"1 +

rg0

µ·h(βp

µ/g0)

h(−β)

#−1

.

b. Introduce the staffing level

n∗β =

&λxκ

µ+ β∗

sλxκ

µ

’.

Then the staffing level n∗β is both asymptotically feasible, as

well as asymptotically optimal with f (λ) =√

λ.Proof


ED Regime

What if α =κ∑

i=1

xipi ?

n =λxκ+1

µ+ β · h(λ),

1h(λ)√

λ→ constant ⇒ PM,n{Wq > 0} > α. (β ∈ R)

2h(λ)√

λ→ ∞ ⇒ PM,n{Wq > 0} = α. (β > 0)


ED Regime

What if α =κ∑

i=1

xipi ?

n =λxκ+1

µ+ β · h(λ),

1h(λ)√

λ→ constant

⇒ PM,n{Wq > 0} > α. (β ∈ R)

2h(λ)√

λ→ ∞ ⇒ PM,n{Wq > 0} = α. (β > 0)


ED Regime

What if α =κ∑

i=1

xipi ?

n =λxκ+1

µ+ β · h(λ),

1h(λ)√


2h(λ)√

λ→ ∞ ⇒ PM,n{Wq > 0} = α. (β > 0)


ED Regime

What if α =κ∑

i=1

xipi ?

n =λxκ+1

µ+ β · h(λ),

1h(λ)√


2h(λ)√

λ→ ∞

⇒ PM,n{Wq > 0} = α. (β > 0)


ED Regime

What if α =κ∑

i=1

xipi ?

n =λxκ+1

µ+ β · h(λ),

1h(λ)√


2h(λ)√

λ→ ∞ ⇒ PM,n{Wq > 0} = α. (β > 0)


ED Regime


Under the staffing level n∗β, as λ →∞,

a. The long-term abandonment probability:

ePM,n∗β{Ab} ∼

κ−1Xi=1

pi

`xi − xκ).

b. Mean server’s utilization:

EM,n∗β[U] ∼

κXi=1

pi +IX

i=κ+1

pi ·xi

xκ.

c. Assume that for each i < κ the equation G (y) = 1− xκxi

has

a unique solution y∗i , and g(y∗i ) > 0. Then, the long-termaverage waiting time

eEM,n∗β[Wq] ∼

κ−1Xi=1

"xipi ·

Z y∗i

0

G(u)du

#.


ED Regime

Example: X takes the values 1.5, 1 and 0.5, with probability 1/3for each.

Delay Probability

-3 -2 -1 0 1 2 30

0.2

0.4

0.6

0.8

1

long

-term

P{W

>0}

β

Long-Term P{W>0} vs. β

XK=1.5

XK=1

XK=0.5


-3 -2 -1 0 1 2 30

0.2

0.4

0.6

0.8

1

long

-term

P{W

>0}

β

Long-Term P{W>0} vs. β

0.5 1 1.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

long

-term

P{A

band

on}

XK

Long-Term P{Abandon} vs. XK

XK=1.5

XK=1

XK=0.5


ED Regime

β = −1

Delay Probability

0 500 1000 1500 20000.4

0.5

0.6

0.7

0.8

0.9

1

long

-term

P{W

>0}

λ

Long-Term P{W>0} vs. λ, c=1, β=-1

exact XK=1.5

approx XK=1.5

exact XK=1

approx XK=1

exact XK=0.5

approx XK=0.5

0 500 1000 1500 20000

0.2

0.4

0.6

0.8

1

long

-term

P{A

band

on}

λ

Long-Term P{Abandon} vs. λ, c=1, β=-1

exact XK=1.5

approx XK=1.5

exact XK=1

approx XK=1

exact XK=0.5

approx XK=0.5


0 500 1000 1500 20000.4

0.5

0.6

0.7

0.8

0.9

1

long

-term

P{W

>0}

λ

Long-Term P{W>0} vs. λ, c=1, β=-1

exact XK=1.5

approx XK=1.5

exact XK=1

approx XK=1

exact XK=0.5

approx XK=0.5

0 500 1000 1500 20000

0.2

0.4

0.6

0.8

1

long

-term

P{A

band

on}

λ

Long-Term P{Abandon} vs. λ, c=1, β=-1

exact XK=1.5

approx XK=1.5

exact XK=1

approx XK=1

exact XK=0.5

approx XK=0.5


ED Regime

c = 1, Continuous Random Arrival Rate

Assume M = λX , where X is a continuous random variable withstrictly continuous cdf F over the distribution support. LetE [X ] = 1 and σ(X ) < ∞. Assume λ →∞.

Theorem (Staffing)


n∗ =λ

µ· δ∗ + o(λ), δ∗ ∈ supp(f ),

where δ∗ is the unique solution of the equation

α =

Z ∞δ

xdF (x).

b. Introduce the staffing level n∗δ =

‰λδ∗

µ

ı.

Then the staffing level n∗δ is both asymptotically feasible, aswell as asymptotically optimal (f (λ) = λ).


ED Regime


Under the staffing level n∗δ , as λ →∞,

a. The short-term delay probability:

PM,n∗δ{Wq > 0} ∼ F (δ).

b. The long-term abandonment probability:

ePM,n∗δ{Ab} ∼ (E [X |X > δ]− δ) · F (δ).

c. The short-term abandonment probability:

PM,n∗δ{Ab} ∼ E [1− δ/X |X > δ] · F (δ).


ED Regime

Theorem (Performance Measures). Continued.

d. Mean server’s utilization:

EM,n∗δ[U] ∼ 1

δ· E [X |X < δ] · F (δ) + F (δ).

e. Assume that for each x > δ the equation G (yx) = 1− δx has

a unique solution y∗x , and g(y∗x ) > 0. Then, the long-termaverage waiting time

eEM,n∗δ[Wq] ∼

Z ∞δ

x ·Z y∗x

0

G(u)du

!dF (x).


ED Regime

Examples: Consider three distributions of X

Uniform distribution on [0,2].

Uniform distribution on [0.5,1.5].

Normal distribution with mean=1 and std=0.25.

δ = 0.75

Long-Term P{Wq > 0}

0 500 1000 1500 20000.6

0.7

0.8

0.9

1

long

-term

P{W

>0}

λ

Long-Term P{W>0} vs. λ, c=1, δ=0.75

exact U(0,2)approx U(0,2)exact U(0.5,1.5)approx U(0.5,1.5)exact Norm(1,0.252)approx Norm(1,0.252)

0 500 1000 1500 20000.6

0.7

0.8

0.9

1

shor

t-ter

m P

{W>0

}

λ

Short-Term P{W>0} vs. λ, c=1, δ=0.75


0 500 1000 1500 20000.2

0.25

0.3

0.35

0.4

long

-term

P{A

band

on}

λ

Long-Term P{Abandon} vs. λ, c=1, δ=0.75


0 500 1000 1500 20000.2

0.25

0.3

0.35

0.4

shor

t-ter

m P

{Aba

ndon

}

λ

Short-Term P{Abandon} vs. λ, c=1, δ=0.75


Short-Term P{Wq > 0}

0 500 1000 1500 20000.6

0.7

0.8

0.9

1

long

-term

P{W

>0}

λ



0 500 1000 1500 20000.6

0.7

0.8

0.9

1

shor

t-ter

m P

{W>0

}

λ



0 500 1000 1500 20000.2

0.25

0.3

0.35

0.4

long

-term

P{A

band

on}

λ



0 500 1000 1500 20000.2

0.25

0.3

0.35

0.4

shor

t-ter

m P

{Aba

ndon

}

λ




ED Regime

δ = 0.75

Long-Term P{Ab}

0 500 1000 1500 20000.6

0.7

0.8

0.9

1

long

-term

P{W

>0}

λ



0 500 1000 1500 20000.6

0.7

0.8

0.9

1

shor

t-ter

m P

{W>0

}

λ



0 500 1000 1500 20000.2

0.25

0.3

0.35

0.4

long

-term

P{A

band

on}

λ



0 500 1000 1500 20000.2

0.25

0.3

0.35

0.4

shor

t-ter

m P

{Aba

ndon

}

λ



Short-Term P{Ab}

0 500 1000 1500 20000.6

0.7

0.8

0.9

1

long

-term

P{W

>0}

λ



0 500 1000 1500 20000.6

0.7

0.8

0.9

1

shor

t-ter

m P

{W>0

}

λ



0 500 1000 1500 20000.2

0.25

0.3

0.35

0.4

long

-term

P{A

band

on}

λ



0 500 1000 1500 20000.2

0.25

0.3

0.35

0.4

shor

t-ter

m P

{Aba

ndon

}

λ




Time - Varying Queues

Based on the ISA (Iterative Staffing Algorithm), a simulationcode developed by Feldman[’04] with the features of randomarrival rate in the time varying M/M/n + G queue.

The goal is to determine time-dependence staffing levelsaiming to achieve a given constant-over-time long- term delaydelay probability, α.

Example 1: c = 1, Discrete Random Arrival Rate1 Mt = λt · X , where

λt = 100− 20 · cos(t), and X =

1.5 w .p. 1/31 w .p. 1/30.5 w .p. 1/3

2 Service time and patience are distributed exponentially withmean 1 (µ = θ = 1).



Arrivals, Offered Load andStaffing Level

Target α=0.1Target alpha=0.1: Arrivals, Offered Load and Staffing Level

0

20

40

60

80

100

120

140

160

180

200

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time

Arrived Staffing Offered Load


0

20

40

60

80

100

120

140

160

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time



0

20

40

60

80

100

120

140

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time




Theoretical staffing level (homogenous arrival rate) dependsboth on X and β:

n =λxκ

µ+ β

√λxκ

µ.

β for xκ = 1.5Beta for X=1.5

-10

-8

-6

-4

-2

0

2

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time

Target alpha=0.1 Target alpha=0.2 Target alpha=0.3



⇒ β is stabilized for targetα = 0.1, 0.2, 0.3 and 0.4.



β for xκ = 1Beta for X=1

-6

-4

-2

0

2

4

6

8

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time




⇒ β is stabilized for targetα = 0.6, 0.7 and 0.8.

What about target α = 0.5 ?

Extreme values (-2.3, 2.1 &10) and noisy

β for xκ = 0.5Beta for X=0.5

0

2

4

6

8

10

12

14

16

18

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time




⇒ β is stabilized for targetα = 0.9



β for xκ = 1Beta for X=1

-6

-4

-2

0

2

4

6

8

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time




⇒ β is stabilized for targetα = 0.6, 0.7 and 0.8.

What about target α = 0.5 ?

Extreme values (-2.3, 2.1 &10) and noisyβ for xκ = 0.5Beta for X=0.5

0

2

4

6

8

10

12

14

16

18

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time




⇒ β is stabilized for targetα = 0.9



Delay Probability (Target)Delay Probability

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time




Theoretical and Empirical Delay

Probability vs. xκ and βTheoretical and Empirical Delay Probability vs. Beta

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-3 -2 -1 0 1 2 3

Beta

Del

ay P

roba

bilit

y

Theoretical Empirical

Xk=0.5

Xk=1

Xk=1.5



Abandonment ProbabilityAbandonment Probability

0

0.1

0.2

0.3

0.4

0.5

0.6

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time




Abandonment Probability - Low α’sAbandonment Probability

0

0.01

0.02

0.03

0.04

0.05

0.06

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time

Target alpha=0.1 Target alpha=0.2

Target alpha=0.3 Target alpha=0.4

Servers’ UtilizationServers' Utilization

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time






Example 2: c = 1, Continuous Random Arrival Rate

1 Mt = λt · X , where

λt = 100− 20 · cos(t), and X ∼ Uni[0.5, 1.5].

2 Service time and patience are distributed exponentially withmean 1 (µ = θ = 1).

Theoretical staffing level (homogenous arrival rate) dependson the QOS parameter δ:

n =λ

µ· δ



Delay Probability (Target)Delay Probability

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time




QOS δdelta

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Time




Delay Probability vs. δTheoretical and Empirical Delay Probability vs. delta

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5

delta

Del

ay P

roba

bilit

y

Empirical Theoretical


Thank You


Appendix 1

c = 1/2, γ∗X

Denote

β∗x = β∗ − x ,

β∗x = β∗x

rµ

g0= (β∗ − x) ·

rµ

g0.

Then,

γ∗x = α · β∗x ·

"h(β∗x )

β∗x− 1

#.

Back


Appendix 2

Proof Outline (c = 1, Discrete Random Arrival Rate)

1. n =λxκ

µ+ β

sλxκ

µ+ o(

√λ) =

λxi

µ· xκ

xi+ o(λ)

⇒ for i < κ (xi > xκ) we are in the ED regime, for i = κ in the QEDregime, and for i > κ (xi < xκ) in the QD regime.

2. limλ→∞

ePM,n{Wq > 0} = limλ→∞

E [XPM,n{Wq > 0}]

= E limλ→∞

{XPM,n{Wq > 0}} =κ−1Xi=1

xipi + xκpκ · α∗.

3. For a certain value xs of X , the asymptotic long-term delay probability isbounded in the open interval As =

`Ps−1i=1 xipi ,

Psi=1 xipi

´. ⇒ There is a

unique s which for α ∈ As .


Appendix 2

Proof Outline (c = 1, Discrete Random Arrival Rate)

4. For some ε > 0 define

n14=

&λxκ

µ+ (β∗ − ε)

sλxκ

µ

’; n2

4=

$λxκ

µ+ (β∗ + ε)

sλxκ

µ

%.

Garnett function is strictly decreasing in β

⇒ PM,n1{Wq > 0} → α + τ1 ; PM,n2{Wq > 0} → α− τ2.

⇒ For large λ

PM,n1{Wq > 0} > α + τ1/2 ; PM,n2{Wq > 0} < α− τ2/2.

Delay probability is monotonically decreasing in number ofservers

⇒ n1 ≤ n∗ ≤ n2.

ε > 0 is arbitrary ⇒ q.e.d. Back

Uncertainty in the Demand for Service: The Case of Call...

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