Why Wait?!?Bryan Gorney

Joe WalkerDave Mertz

Josh StaidlMatt Boche

Purpose: To minimize costs of businesses with waiting lines, taking into consideration the cost of servers and the long-run loss of revenue by making customers wait too long.

Outline:

I. Rudiments of Queuing Theorya. 1 Customer, 1 Serverb. Many Customer, No Serverc. Many Customers, 1 Server

II. Concept of a Stationary DistributionIII. Traffic Intensity

a. Average Queue Lengthb. Little’s Principlec. Average Waiting Time

IV. Store Profit Maximization

1 Customer, 1 Server

• N(t) = number of individuals at checkout counter at time t.

• N(t) has only two possible values of 0 or 1.

• “being served” corresponds to N(t) = 1.

• “finished being served” corresponds to N(t) = 0.

To obtain the time-dependent transition probabilities for this Markov chain, let

Xt =

be the time-dependent distribution vector for the states “being served” and “finished being served”.

That is,

p(t) = P[N(t) = 1]

and

q(t) = P[N(t) = 0]

p(t)q(t))(

• q(t) = 1 – p(t)

• lim p(t) = 0

lim Xt =

State diagram

P[service is completed in t] = t

P[service is not completed in t] = 1 – t

t t

1 0

t

t

(0

1)

1

The transition matrix for each t time step is given by

A =

Thus, Xt+t = AXt

which in matrix form gives:

(1 – t 0

t 1

)

p(t + t)

q(t + t)( ) = (1 – t 0

t 1

) p(t)q(t))(

Performing matrix multiplication yields:

p(t + t) = (1 – t)p(t)

Rearranging so difference quotient is on the left-hand side:

By letting t go to zero, the difference quotient becomes a derivative, and

p(t + t) – p(t)

t= -p(t)

p(t + t) – p(t)

tt 0limp’(t) = = -p(t)

This is the exponential differential equation with solution:

p(t) = P[N(t) = 1] = pe-t

Since we assumed that there is an individual initially being served, N(0) = 1. This implies that p(0) = 1 which gives

p0 = 1. Finally we have:

p(t) = P[N(t) = 1] = e-t

Since q(t) = 1 – p(t), the probability of completing the service is:

q(t) = 1 – e-t

• Next let T denote the time at which service is completed. It is considered to be the time until transition from one state to another.

• T is a continuous random variable with range 0 < T < .

• P[T > t] = P[N(t) = 1].

• P[T < t] = 1 – e-t (complement).

• Left-hand side = cumulative distribution function of the continuous random variable T.

• Right-hand side = exponential distribution.

P[T > s + t given T > s] = P[T > t] (Memoryless property)

for all s,t > 0.

Density function is the derivative of the cumulative distribution which is:

f(t) = P[T < t] = e-t, 0 < t < .

E(T) = tf(t)dt = te-tdt =

Thus the average value of T is . We interpret the service rate as:

ddt

0

0

1

1

1Mean Time Until Transition=

Many Customers, No Server

1. The probability that an arrival occurs in a short interval [t, t+t] is proportional to the length of the interval t.In symbols:

P[ N(t+t) - N(t) = 1] = t

for some constant >0.

Let N(t) be a time dependent function that denotes the number of customer arrivals in a given interval [0,t], and assume the following:

2. The Memoryless Property:The probability that an arrival occurs in [t, t+t] doesnot depend on the time of previous arrivals:

P[ N(t+t) – N(t) = n | N(s) = m ] = P[ N(t+t) – N(t) = n]

for all 0 s t.

3. Occurrences of arrivals in non-overlapping intervals areindependent.

4. The Probability of two or more arrivals in [t, t+t] is negligible.

Assume Pn(t) = P[N(t) = n]

N(t) has possible values 0,1,2,… since it is possibleto have any number of customers in line.

Considering a time step of size t, N(t) forms a MarkovChain, with a transition matrix of:

A = (1-t 0 0 0 . . . t 1-t 0 0 . . . 0 t 1- t 0 . . .

)

t t t

1-t 1-t 1-t

State Diagram of Transition Matrix

Xt+t = AXt is the state equation for each time step size t, where:

(P0(t)P1(t) :Pn(t) :

)Xt =

After Matrix Multiplication: we have the followingsystem of equations:

P0(t+t) = (1-t)P0(t) P1(t+t) = tP0(t)+(1-t)P1(t)

Pn(t+t) = tPn-1(t)+(1-t)Pn(t)

These equations can then be rearranged to make difference quotients.

Letting t go to zero we obtain a system of differentialequations. For n 1,

Pn’(t) = limt0

= [Pn-1(t) – Pn(t)],

and for n = 0,

P0’(t) = -P0(t)

This equation is the exponential differential equation with initial condition that there are no arrivals (P0(0) = 1). This gives:

P0(t) = e-t

Pn(t+ t) – Pn(t) t

Similarly, letting n=1,

P1’(t) = [P0(t)- P1(t)]= [e-t - P1(t)]

This is a first-order linear differential equationwhich can be solved by multiplying by theintegrating factor et.

ddt

[etP1(t)] = so that etP1(t) = t+c

The initial condition P0(0) =1 implies Pn(0) = 0for all n 1, thus:

P1(t) = te-t

It can be shown by induction that the system of differential equations has solution:

Pn(t) = e-t(t)n for t0, n = 0,1,2,… n!

•Known as a homogeneous Poisson process with rate . ( does not depend on time variable t)•Mean is t (t customers during time interval of length t)

Ex. = 2 customers/minute, t =5 minutes t = 5*2 = 10 customers over 5 minutes

Many Customers, 1 Server

• Let N(t) = number of individuals at checkout counter at time t.

• N(t) = any of the integer values 0, 1, 2, … at time t.

• Probability of arrival in t is t.

• Probability of departure in t is t.

• Three mutually exclusive events:

I. Exactly one arrival in (t,t + t)

II. Exactly one departure in (t,t + t)

III. No arrivals or departures in (t,t + t).

00 11 22

tttt tt

tttt tt

1 – 1 – t - t - tt1 – 1 – tt 1 – 1 – t - t - tt

• The time step for this Markov chain is t and the transition matrix is:

1-1-t t t 0 0 0 …t 0 0 0 …

t 1 – t 1 – t – t – t t t 0 0 …t 0 0 …

0 0 t 1 – t 1 – t – t – t t t 0 …t 0 ….. ........ ..

...... ......

....

..( )A =A =

• Let Pn(t) = P[N(t) = n] be the probability that there are n customers in the queue at time t. Thus letting

PP00(t)(t)

PP11(t)(t)

PPnn(t)(t)

......

......

( )XXtt = =

• We obtain the equation Xt + t = AXt for the single-server queue.

• The state equation gives the following (infinite) system of equations for the single-server queue:

P0(t + t) = (1 – t)P0(t) + tP2(t)

P1(t + t) = (1 – t - t)P1(t) + tP0(t) + tP2(t)

Pn(t + t) = (1 – t - t)Pn(t) + tP0 - 1(t) + tPn + 1(t)

......

......

System of differential equations by letting t go to zero; more specifically, for n > 1,

limlimt 0t 0

PPnn(t + (t + t) – Pt) – Pnn(t)(t)

ttP’P’nn(t) =(t) =

= -(= -(PPnn(t) (t) PPn – 1n – 1(t) + (t) + PPn – n –

11(t)(t)And, PAnd, P00’(t) = -’(t) = -PP00(t) + (t) + PP11(t).(t).

Many classical methods are available for the solving thesystem:

PPnn’(t) =’(t) = -(-(PPnn(t) (t) PPn – 1n – 1(t) + (t) + PPn – 1n – 1(t)(t)

PP00’(t) = -’(t) = -PP00(t) + (t) + PP11(t)(t)

However, they involve ideas beyond the scope of our analysis -instead, we will be using the system of equations to obtainsteady-state (time-independent) behavior.

Stationary Distribution

Looking for whether the system of time-dependent probabilities settles down and displays no more “transient” behavior - analogous to finding the fixed points for deterministic systems. In this case, the fixed points will be “stationary” distributions.

The system of differential equations:

PPnn’(t) =’(t) = -(-(PPnn(t) (t) PPn – 1n – 1(t) + (t) + PPn – 1n – 1(t)(t)

PP00’(t) = -’(t) = -PP00(t) + (t) + PP11(t)(t)

has a fixed point or stationary distribution provided its rates of change are zero, that is:

Pnn’’(t) = 0 for n 0.

lim Pn’(t) = 0 for n 0, is equivalent to assuming that

lim Pn(t) = Pn exists, where Pn does not depend upon t.

Applying Pn’(t) = 0 to the system of equations, we obtain:

1.) 0 = -P0 + P1

and

2.) 0 = -(+ )Pn + Pn-1 + Pn+l, for n > 1.

Solving for P1, using equation 1:

P1 = (P0

Solving for P2, using P1 and equation 2:

P2 = (2 P0

Continuing on, using induction, shows:

Pn = (n P0

t

t

Queue size must be a non-negative integer, hence,P0 + P1 + P2 + … = 1

So,P0 + ()P0 + (2P0+ … = 1

Which is,

P0 Σ (n = 1.

This sums to 1/(1-()) provided that () < 1, which is the condition for the stationary distribution.

Using this sum, P0 = 1 – (so the stationary distribution for the single-server queue is:

Pn = ()n(1-())

n = 0

Traffic Intensity• Average Queue Length• Little’s Principle• Average Waiting Time

Average Queue Length

Let ρ = (λ / μ). (called the traffic intensity of the queue)

By assumption, ρ < 1, as λ < μ.

Using ρ in: Pn = (λ / μ)n (1 – (λ / μ)),

Pn = ρn (1 – ρ).

Obtaining the average number Na of customers in the queue:

Na = E(N) = Σ nPn = Σ n(ρn(1 – ρ))

= (1 – ρ) Σ nρn = (1 – ρ) Σ ρnρn-1

= (1 – ρ)ρ Σ = (1 – ρ)ρ Σ

= (1 – ρ)ρ = (1 – ρ)ρ(1 – ρ)-2

=

n = 0

n = 0

ddρ

ρn

n = 0

n = 0

n = 0

ddρ

n = 0

ddρ

ρn

(1 / (1 – ρ))

ρρ - 1

Average Number of Customers in Queue

0

1

2

3

4

5

6

7

8

9

10

0 0.2 0.4 0.6 0.8 0.9 1

Traffic Intensity (p)

Nu

mb

er o

f C

ust

om

ers

ρρ - 1

Na =

Average Waiting Time

Little’s Principle – the average number of customers in a queueing system is equal to the average arrival rate of customers to that system times the average time spent in that system.

Naa = = λTa

Ta = Na a / / λ = (ρ / ((1 – ρ)λ) = ((λ / μ) / ((1 – (λ / μ))λ))

= (1 / μ) / (1 – (λ / μ)) = (1 / μ) / (1 – ρ) = 1/(μ - λ)

Average Waiting Time

0

2

4

6

8

10

12

0 0.2 0.4 0.6 0.8 0.9 1

Traffic Intensity

Un

its

of

Tim

eTa = 1/ (μ – λ )

Model Formulation of Store Profit Maximization

C(ŧ), which is the Customer Attrition Function, models the loss of profit per day and is given below:

C(ŧ) ={0 for ŧ Ta0

ATa0 for ŧ Ta0

Ta0 represents the threshold where customers stop returning to the store due to long waiting times.

Ŧ0

ŧ

C(ŧ)

Graph of Customer Attrition Function

C(ŧ) = C(1/(μ-λ)) for one server C(ŧ) = C(1/(μ-λ)) for servers

This models the loss of profit due to customers having to wait in lines

Customer waiting costs plus employee costs

J() = C(1/(μ-λ)) +K 1 <

= # number of employees (checkers)K = cost associated with retaining those employees

Conclusion

The object of the manager is to minimize J() byfinding the number of employees that eliminates the attrition factor by decreasing costs.

References

A Course in Mathematical Modeling, by Douglas Mooneyand Randall Swift

Dr. Steve Deckelman