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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