Queuing Theory Basics ( PDFDrive ).pdf

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Cuantitativos M. En C. Eduardo Bustos Farias 1
Waiting Line Models:
Queuing Theory Basics

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When you are in queue?
“ In the bank, restaurant, supermarket…
“ In front of restroom during the break of football
game
How much is your patience?
“ Waiting costs your patience and your temper and it
also costs the business.
Time =
For the business, they have to find the optimal
service level that keeps customers happy and
makes them profitable.

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INTRODUCTION
INTRODUCTION
Queuing models are everywhere. For example,
Queuing models are everywhere. For example,
airplanes
airplanes “
“queue up
queue up”
” in holding patterns, waiting for
in holding patterns, waiting for
a runway so they can land. Then, they line up again
a runway so they can land. Then, they line up again
to take off.
to take off.
People line up for tickets, to buy groceries, etc.
People line up for tickets, to buy groceries, etc.
Jobs line up for machines, orders line up to be filled,
Jobs line up for machines, orders line up to be filled,
and so on.
and so on.

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INTRODUCTION
INTRODUCTION
A. K. Erlang (a Danish engineer) is credited with
A. K. Erlang (a Danish engineer) is credited with
founding queuing theory by studying telephone
founding queuing theory by studying telephone
switchboards in Copenhagen for the Danish
switchboards in Copenhagen for the Danish
Telephone Company.
Telephone Company.
Many of the queuing results used today were
Many of the queuing results used today were
developed by Erlang.
developed by Erlang.

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A
A queuing model
queuing model is one in which you have a
is one in which you have a
sequence of times (such as people) arriving at a
sequence of times (such as people) arriving at a
facility for service, as shown below:
facility for service, as shown below:
Arrivals
Arrivals
00000
00000 Service Facility
Service Facility

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Consider St. Luke
Consider St. Luke’
’s Hospital in Philadelphia and the
s Hospital in Philadelphia and the
following three queuing models.
following three queuing models.
Model 1: St. Luke
Model 1: St. Luke’
’s Hematology Lab
s Hematology Lab St. Luke
St. Luke’
’s
s
treats a large number of patients on an outpatient
treats a large number of patients on an outpatient
basis (i.e., not admitted to the hospital).
basis (i.e., not admitted to the hospital).
Outpatients plus those admitted to the 600
Outpatients plus those admitted to the 600-
-bed
bed
hospital produce a large flow of new patients each
hospital produce a large flow of new patients each
day.
day.

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Most new patients must visit the hematology
Most new patients must visit the hematology
laboratory as part of the diagnostic process. Each
laboratory as part of the diagnostic process. Each
such patient has to be seen by a technician.
such patient has to be seen by a technician.
After seeing a doctor, the patient arrives at the
After seeing a doctor, the patient arrives at the
laboratory and checks in with a clerk.
laboratory and checks in with a clerk.
Patients are assigned on a first
Patients are assigned on a first-
-come, first
come, first-
-served
served
basis to test rooms as they become available.
basis to test rooms as they become available.
The technician assigned to that room performs the
The technician assigned to that room performs the
tests ordered by the doctor.
tests ordered by the doctor.
When the testing is complete, the patient goes on to
When the testing is complete, the patient goes on to
the next step in the process and the technician sees
the next step in the process and the technician sees
a new patient.
a new patient.
We must decide how many technicians to hire.
We must decide how many technicians to hire.

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WATS (Wide Area Telephone Service) is an acronym
WATS (Wide Area Telephone Service) is an acronym
for a special flat
for a special flat-
-rate, long distance service offered
rate, long distance service offered
by some phone companies.
by some phone companies.
Model 2: Buying WATS Lines
Model 2: Buying WATS Lines As part of its
As part of its
remodeling process, St. Luke
remodeling process, St. Luke’
’s is designing a new
s is designing a new
communications system which will include WATS
communications system which will include WATS
lines.
lines.

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We must decide how many WATS lines the hospital
We must decide how many WATS lines the hospital
should buy so that a minimum of busy signals will
should buy so that a minimum of busy signals will
be encountered.
be encountered.
When all the phone lines allocated to WATS are in
When all the phone lines allocated to WATS are in
use, the person dialing out will get a busy signal,
use, the person dialing out will get a busy signal,
indicating that the call can
indicating that the call can’
’t be completed.
t be completed.

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The equipment includes measuring devices such as
The equipment includes measuring devices such as
Model 3: Hiring Repairpeople
Model 3: Hiring Repairpeople St. Luke
St. Luke’
’s hires
s hires
repairpeople to maintain
repairpeople to maintain 20
20 individual pieces of
individual pieces of
electronic equipment.
electronic equipment.
electrocardiogram machines
electrocardiogram machines
small dedicated computers
small dedicated computers
CAT scanner
CAT scanner
other equipment
other equipment

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If a piece of equipment fails and all the repairpeople
If a piece of equipment fails and all the repairpeople
are occupied, it must wait to be repaired.
are occupied, it must wait to be repaired.
We must decide how many repairpeople to hire and
We must decide how many repairpeople to hire and
balance their cost against the cost of having broken
balance their cost against the cost of having broken
equipment.
equipment.

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All three of these models fit the general description
All three of these models fit the general description
of a queuing model as described below:
of a queuing model as described below:
PROBLEM
PROBLEM ARRIVALS
ARRIVALS SERVICE FACILITY
SERVICE FACILITY
1
1 Patients
Patients Technicians
Technicians
2
2 Telephone Calls
Telephone Calls Switchboard
Switchboard
3
3 Broken Equipment Repairpeople
Broken Equipment Repairpeople
These models will be resolved by using a
These models will be resolved by using a
combination of analytic and simulation models.
combination of analytic and simulation models.
To begin, let
To begin, let’
’s start with the basic queuing model.
s start with the basic queuing model.

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Queuing Systems
...
customers
channel
(server)
system
arrival departure
waiting line (queue)
Single Channel
Waiting Line System
system
arrival departure
server 2
server k
...
.
.
.
server 1 Multi-Channel
Waiting Line System

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Bank Customers Teller Deposit etc.
Doctor’s Patient Doctor Treatment
office
Traffic Cars Light Controlled
intersection passage
Assembly line Parts Workers Assembly
Tool crib Workers Clerks Check out/in tools
Situation Arrivals Servers Service Process
Waiting Line Examples

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THE BASIC MODEL
THE BASIC MODEL
Consider the Xerox machine located in the fourth
Consider the Xerox machine located in the fourth-
-
floor secretarial service suite. Assume that users
floor secretarial service suite. Assume that users
arrive at the machine and form a single line.
arrive at the machine and form a single line.
Each arrival in turn uses the machine to perform a
Each arrival in turn uses the machine to perform a
specific task which varies from obtaining a copy of a
specific task which varies from obtaining a copy of a
1
1-
-page letter to producing
page letter to producing 100
100 copies of a
copies of a 25
25-
-page
page
report.
report.
This system is called a single
This system is called a single-
-server (or single
server (or single-
-
channel
channel) queue.
) queue.

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2.
2. The number of people in the queue (waiting
The number of people in the queue (waiting
for service).
for service).
Questions about this or any other queuing system
Questions about this or any other queuing system
center on four quantities:
center on four quantities:
1.
1. The number of people in the system (those
The number of people in the system (those
being served and waiting in line).
being served and waiting in line).

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3.
3. The waiting time in the system (the interval
The waiting time in the system (the interval
between when an individual enters the system
between when an individual enters the system
and when he or she leaves the system).
and when he or she leaves the system).
4.
4. The waiting time in the queue (the time
The waiting time in the queue (the time
between entering the system and the
between entering the system and the
beginning of service).
beginning of service).

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ASSUMPTIONS OF THE BASIC MODEL
ASSUMPTIONS OF THE BASIC MODEL
1.
1. Arrival Process.
Arrival Process. Each arrival will be called a
Each arrival will be called a
“
“job.
job.”
” The
The interarrival time
interarrival time (the time between
(the time between
arrivals) is not known.
arrivals) is not known.

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Therefore, the
Therefore, the exponential probability
exponential probability
distribution
distribution (or
(or negative exponential
negative exponential
distribution
distribution) will be used to describe the
) will be used to describe the
interarrival times for the basic model.
interarrival times for the basic model.
Probabilidad de que la atención sea completada
dentro de “ t “ unidades de tiempo
X = t

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The exponential distribution is
The exponential distribution is
completely specified by one
completely specified by one
parameter, l, the mean arrival rate
parameter, l, the mean arrival rate
(i.e., how many jobs arrive on the
(i.e., how many jobs arrive on the
average during a specified time
average during a specified time
period).
period).

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Mean interarrival time is the average time
Mean interarrival time is the average time
between two arrivals. Thus, for the
between two arrivals. Thus, for the
exponential distribution
exponential distribution
Avg. time between jobs = mean interarrival time =
Avg. time between jobs = mean interarrival time =
1
1
λ
λ
Thus, if
Thus, if λ
λ = 0.05
= 0.05
mean interarrival time = = = 20
mean interarrival time = = = 20
1
1
λ
λ
1
1
0.05
0.05

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2.
2. Service Process.
Service Process. In the basic model, the time
In the basic model, the time
that it takes to complete a job (the
that it takes to complete a job (the service
service
time
time) is also treated with the exponential
) is also treated with the exponential
distribution.
distribution.
The parameter for this exponential distribution
The parameter for this exponential distribution
is called
is called μ
μ (the
(the mean service rate
mean service rate in jobs per
in jobs per
minute).
minute).

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μ
μT
T is the number of jobs that would be served
is the number of jobs that would be served
(on the average) during a period of
(on the average) during a period of T
T minutes
minutes
if the machine were busy during that time.
if the machine were busy during that time.
The
The mean
mean or
or average
average,
, service time
service time (the
(the
average time to complete a job) is
average time to complete a job) is
Avg. service time =
Avg. service time = 1
1
μ
μ
Thus, if
Thus, if μ
μ = 0.10
= 0.10
mean service time = = = 10
mean service time = = = 10
1
1
μ
μ
1
1
0.10
0.10

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3.
3. Queue Size.
Queue Size. There is no limit on the number
There is no limit on the number
of jobs that can wait in the queue (an infinite
of jobs that can wait in the queue (an infinite
queue length).
queue length).

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4.
4. Queue Discipline.
Queue Discipline. Jobs are served on a first
Jobs are served on a first-
-
come, first
come, first-
-serve basis (i.e., in the same order
serve basis (i.e., in the same order
as they arrive at the queue).
as they arrive at the queue).

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5.
5. Time Horizon.
Time Horizon. The system operates as
The system operates as
described continuously over an infinite
described continuously over an infinite
horizon.
horizon.

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6.
6. Source Population.
Source Population. There is an infinite
There is an infinite
population available to arrive.
population available to arrive.

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QUEUE DISCIPLINE
QUEUE DISCIPLINE
In addition to the arrival distribution, service
In addition to the arrival distribution, service
distribution and number of servers, the queue
distribution and number of servers, the queue
discipline must also be specified to define a queuing
discipline must also be specified to define a queuing
system.
system.
So far, we have always assumed that arrivals were
So far, we have always assumed that arrivals were
served on a first
served on a first-
-come, first
come, first-
-serve basis (often called
serve basis (often called
FIFO, for
FIFO, for “
“first
first-
-in, first
in, first-
-out
out”
”).
).

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QUEUE DISCIPLINE
QUEUE DISCIPLINE
However, this may not always be the case. For
However, this may not always be the case. For
example, in an elevator, the last person in is often
example, in an elevator, the last person in is often
the first out (LIFO).
the first out (LIFO).
Adding the possibility of selecting a good queue
Adding the possibility of selecting a good queue
discipline makes the queuing models more
discipline makes the queuing models more
complicated.
complicated.
These models are referred to as scheduling models.
These models are referred to as scheduling models.

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Various Type of Queues
• Single Channel/Single Phase
• Multi-channel/Single Phase
• Single Channel/Multi-phase
• Queuing Network

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Queuing System Structure
Server departure
arrival
• Arrival characteristic
1. Size of units
- Single
- Batch
2. Arrival rate
- Constant
- Probabilistic
3. Degree of patience
- Patient
- Impatient
• Arrival characteristic
1. Size of units
- Single
- Batch
2. Arrival rate
- Constant
- Probabilistic
3. Degree of patience
- Patient
- Impatient
• Features of lines
1. Length
- Infinite capacity
- Limited capacity
2. Number
- Single
- Multiple
3. Queue discipline
- FIFO
- Priorities
• Features of lines
1. Length
- Infinite capacity
- Limited capacity
2. Number
- Single
- Multiple
3. Queue discipline
- FIFO
- Priorities
• Service facility
1. Structure
2. Service rate
- Constant
- Probabilistic - random services
- State-dependent service time
• Service facility
1. Structure
2. Service rate
- Constant
- Probabilistic - random services
- State-dependent service time
• Population Source
- Finite
- Infinite
• Population Source
- Finite
- Infinite
• Exit
1. Return to service population
2. Do not return to service population
• Exit
1. Return to service population
2. Do not return to service population

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Deciding on the
Optimum Level of Service
Total expected cost
Negative Cost of waiting
time to company
Cost
Low level
of service
Optimal
service level
High level
of service
Minimum
total cost
Cost of providing
service

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Performance Measures
P0 = Probability that there are no customers in the system
Pn = Probability that there are n customers in the system
LS = Average number of customers in the system
LQ = Average number of customers in the queue
WS = Average time a customer spends in the system
WQ = Average time a customer spends in the queue
Pw = Probability that an arriving customer must wait for service
ρ = Utilization rate of each server (the percentage of time that
each server is busy)

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X: # of customer arrivals within time interval of length t
Pr(X=n) =
λ = the mean arrival rate per time unit
t = the length of the time interval
e = 2.7182818 (the base of the natural logarithm)
n! = n(n−1)(n−2) (n−3)… (3)(2)(1)
X: # of customer arrivals within time interval of length t
Pr(X=n) =
λ = the mean arrival rate per time unit
t = the length of the time interval
e = 2.7182818 (the base of the natural logarithm)
n! = n(n−1)(n−2) (n−3)… (3)(2)(1)
( )
!
n
e
t t
n λ
λ −
Very large population of potential customers
• behave independently
• in any time instant, at most one arrives
• arrive at intervals of average duration 1/λ
Arrival Process
Very large population of potential customers
• behave independently
• in any time instant, at most one arrives
• arrive at intervals of average duration 1/λ
X follows Poisson Distribution(λt)
Mean = λt
Variance = λt
λ = arrival rate = # of arrivals per unit of time
t should be expressed in the same time unit as λ
X follows Poisson Distribution(λt)
Mean = λt
Variance = λt
λ = arrival rate = # of arrivals per unit of time
t should be expressed in the same time unit as λ
Important
Important

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Examples of Poisson Distribution
0.0
0.2
0.4
p(x)
Poisson distribution
with parameter 1/2
x
0 1 2 3
0.0
0.2
p(x)
with parameter 2
x
0 1 2 3 4 5
with parameter 1
0.0
0.2
0.4
p(x)
0 1 2 3 4 x

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Service Process
Assume service time is exponentially distributed(μ)
P.D.F. f(t) = μe-μt
Pr(Service ≤ t) = 1 − e-μt
Mean = 1/μ
Variance = 1/μ2
μ= service rate = # of customers served per unit time
Assume service time is exponentially distributed(μ)
P.D.F. f(t) = μe-μt
Pr(Service ≤ t) = 1 − e-μt
Mean = 1/μ
Variance = 1/μ2
μ= service rate = # of customers served per unit time
Properties of exponential distribution
1. Memoryless (The conditional probability is the same as the unconditional probability.)
2. Most customers require short services; few require long service
3. If arrival process follows Possion (λ), then inter-arrival time follows exponential(λ)
Properties of exponential distribution
1. Memoryless (The conditional probability is the same as the unconditional probability.)
2. Most customers require short services; few require long service
3. If arrival process follows Possion (λ), then inter-arrival time follows exponential(λ)
Important
Important

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Examples of Exponential
Distribution

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Queuing Theory Notation
“ A standard notation is used in queuing theory to
denote the type of system we are dealing with.
“ Typical examples are:
“ M/M/1 Poisson Input/Poisson Server/1 Server
“ M/G/1 Poisson Input/General Server/1 Server
“ D/G/n Deterministic Input/General Server/n
Servers
“ E/G/∞ Erlangian Input/General Server/Inf.
Servers
“ The first letter indicates the input process, the second
letter is the server process and the number is the
number of servers.
“ (M = Memoryless = Poisson)

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Terminology
λ = Arrival rate = 1/ Mean arrival interval
μ = Service rate = 1/ Mean service time
ρ = λ/ μ
k = # of Servers
λ = Arrival rate = 1/ Mean arrival interval
μ = Service rate = 1/ Mean service time
ρ = λ/ μ
k = # of Servers
ρ = Utilization rate of each server (the percentage of time that each server is busy)
ρ = Utilization rate of each server (the percentage of time that each server is busy)
Performance measures

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Single Server Case
Poisson arrivals, exponential service rate, no priorities, no balking,
steady state)
STATE ENTRY RATE LEAVING RATE
0 λP0 µP1
1 λP0 + µP2 (λ + µ)P1
2 λP1 + µP3 (λ + µ)P2
3 λP2 + µP4 (λ + µ)P3
: : :
: : :
Pn = (λ/μ)nP0 = ρnP0 for n = 1,2,3,...
1 =
1
if
1
0
0
0
0
<
=
= −
∞
=
∞
=
∑
∑ ρ
ρ ρ
P
n
n
n
n P
P
0 1 2 3 4 ....
λ
μ
λ
μ
λ
μ
λ
μ
ONLY IF
λ< μ or λ/μ = ρ <1
Steady state exists!
ONLY IF
λ< μ or λ/μ = ρ <1

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Single Server Case
⇒ P0 = 1 − ρ
⇒ Pn = ρn (1 − ρ )
LS = E[N] where N = no. of customers in system (denote S)
=
= ρ /(1 − ρ )
LQ = E[Nq] where Nq = no. of customers in queue (denote Q)
=
= ρ2/(1− ρ )
WS = LS/ λ
WQ = LQ/ λ
∑
∞
= 0
n
n
nP
( )
∑
∞
=
−
0
1
n
n
P
n
Little’s Law
LS = λ WS
LQ= λ WQ
Little’s Law
LS = λ WS
LQ= λ WQ

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Single Server Queue
Performance (M/M/1)
P0 = 1 – λ/μ
Pn = (λ/μ)nP0
LQ =
LS = LQ + λ/μ = LQ + ρ = ρ/(1− ρ)
WQ = LQ / λ
WS = WQ + 1/μ
Pw = 1 – P0 = ρ
P0 = 1 – λ/μ
Pn = (λ/μ)nP0
LQ =
LS = LQ + λ/μ = LQ + ρ = ρ/(1− ρ)
WQ = LQ / λ
WS = WQ + 1/μ
Pw = 1 – P0 = ρ
( ) ρ
ρ
λ
μ
μ
λ
−
=
− 1
2
2

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The Schips, Inc. Truck Dock
Problem
Schips, Inc. is a large department store chain that has six branch
stores located throughout the city. The company’s Western Hills
store, which was built some years ago, has recently been
experiencing some problems in its receiving and shipping
department because of the substantial growth in the branch’s
sales volume. Unfortunately, the store’s truck dock was
designed to handle only one truck at a time, and the branch’s
increased business volume has led to a bottleneck in the truck
dock area. At times, the branch manager has observed as
many as five Schips trucks waiting to be loaded or unloaded.
As a result, the manager would like to consider various
alternatives for improving the operation of the truck dock and
reducing the truck waiting times.

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Problem
One alternative the manager is considering is to speed up the
loading/unloading operation by installing a conveyor system at
the truck dock. As another alternative, the manager is
considering adding a second truck dock so that two trucks could
be loaded and/or unloaded simultaneously.
What should the manager do to improve the operation of the truck
dock? While the alternatives being considered should reduce
the truck waiting times, they may also increase the cost of
operating the dock. Thus the manager will want to know how
each alternative will affect both the waiting times and the cost of
operating the dock before making a final decision
Truck arrival information: truck arrivals occur at an average rate of
three trucks per hour.
Service information: the truck dock can service an average of four
trucks per hour.

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Problem
Options:
1. Using conveyor to speed up service rate
2. Add another dock server
Assumptions:
“ The waiting cost is linear
“ Poisson Arrivals
“ Exponential service time

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Schips, Inc. - Current
Situation
Pw = Probability that an arriving customer must wait for
service
Pw = Probability that an arriving customer must wait for
service

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Schips, Inc. - Alternative I
“ Alternative I: Speed up the loading/unloading
operations by installing a conveyor system (costs
of different conveyer system are not provided
here, but you should consider it when you
evaluate the total cost)

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M/M/k Queue
Server 1
Departure
kμ
Arrival
λ
Server 2
Server k
Multiple server, single queue (Poisson arrivals, I.I.D. exponential service rate,
no priority, no balking, steady state)
ONLY IF
λ< kμ or λ / kμ = ρ <1
ONLY IF
λ< kμ or λ / kμ = ρ <1

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M/M/k Queue Performance Measures

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The Schips, Inc. Problem
(continued)
Alternative I
Alternative II:
k = 2
P0 = 0.4545
LQ = 0.123
LS = 0.873
WQ = 0.041
WS = 0.291
Pw = 0.2045

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Economic Analysis of Queuing
System
Cost of waiting vs. Cost of
capacity
Cost of waiting vs. Cost of
capacity
COST
CAPACITY
CAPACITY
WAITING
TOTAL

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The Schips, Inc. Problem
(Cost analysis, FYI)
cW = Hourly waiting cost for each customer
cServer = Hourly cost for each server
LS = Average number of customers in system
k = Number of servers
cW = Hourly waiting cost for each customer
cServer = Hourly cost for each server
LS = Average number of customers in system
k = Number of servers
Total waiting cost/hour = cWL
Total server cost/hour = cServerk
Total cost per hour = cWL+ cServerk
Total waiting cost/hour = cWL
Total server cost/hour = cServerk
Total cost per hour = cWL+ cServerk
Total Hourly Cost Summary for The Schips Truck Dock Problem
cW = $25/hour, cServer = $30/hour
Incremental cost of using conveyor: $20/hour for every Δμ = 2
Total Hourly Cost Summary for The Schips Truck Dock Problem
cW = $25/hour, cServer = $30/hour
Incremental cost of using conveyor: $20/hour for every Δμ = 2
System μ Avg. # of Trucks in
system (L)
Total Cost/Hour cWL + cSk
1-server 4 3 (25)(3)+(30)(1)=$105
1-server
+conveyor
6 1 (25)(1) +(30+20)(1) = $75
1-server
+conveyor
8 0.6 (25)(0.6) + (30+40)(1) = $85
1-server
+conveyor
10 0.43 (25)(0.43) + (30+60)(1) =
$100.71
2-server 4 0.873 (25)(0.873) + (30)(2) = $81.83

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Discrete distribution
“ Suppose the bank has only one server, the
interarrival and service rate are both discrete
distribution.
“ This bank wants to
simulate for 150
customers arrival.
“ This bank wants to know the queuing length
and waiting time of their current service.
Interarrival distribution Service time distribution
Value Prob Value Prob
1 0.05 1 0.15
2 0.15 2 0.15
3 0.35 3 0.25
4 0.35 3 0.20
5 0.10 4 0.10
Cum. Prob. 1 5 0.05
6 0.05
7 0.03
8 0.02
Cum. Prob. 1

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Waiting Times in Queue
0
2
4
6
8
10
12
14
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 141 146
Customer
Queue Length Versus Time
(Shown only at times just after arrivals)
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 50 100 150 200 250 300 350 400 450 500
Customer Arrival Times
Single server queueing simulation (starting empty and idle)
Customer IA_Time Arrival_Time Service_Time Queue_Time Start_Time Depart_Time Before entry After entry Before entry After entry
1 1 1 3 0 1 4 0 3 0 0
2 3 4 3 0 4 7 0 3 0 0
3 3 7 3 0 7 10 0 3 0 0
4 3 10 2 0 10 12 0 2 0 0
5 3 13 2 0 13 15 0 2 0 0
6 4 17 3 0 17 20 0 3 0 0
7 4 21 6 0 21 27 0 6 0 0
8 4 25 3 2 27 30 2 5 0 1
9 1 26 3 4 30 33 4 7 1 2
10 4 30 7 3 33 40 3 10 0 1
11 3 33 3 7 40 43 7 10 0 1
12 2 35 3 8 43 46 8 11 1 2
13 4 39 3 7 46 49 7 10 2 3
14 4 43 3 6 49 52 6 9 1 2
15 3 46 1 6 52 53 6 7 1 2
Number in queue
Server work

Metodos
“ If the arrival rate
keeps the same, but
the service rate is
faster…
Service time distribution
Value Prob
1 0.25
2 0.25
3 0.20
3 0.20
4 0.10
5 0.00
6 0.00
7 0.00
8 0.00
Cum. Prob. 1
Waiting Times in Queue-Faster Service Rate
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 141 146
Customer
Queue Length Versus Time- Faster Service
0
0.5
1
1.5
2
2.5
0 50 100 150 200 250 300 350 400 450 500

Metodos
Discrete distribution :
L5-QSim2-3servers
Interarrival distribution Service time distribution
Value Prob Value Prob
1 0.80 1 0.15
2 0.15 2 0.15
3 0.03 3 0.25
4 0.01 3 0.2
5 0.01 4 0.1
1 5 0.05
6 0.05
7 0.03
8 0.02
1
“ If the bank has more frequent
arrival, they definitely need more
servers. Now they have 3 servers.

Metodos
Discrete distribution : L5-QSim2-3servers (con’t)
Waiting Times in Queue-3 servers
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 141 146
Customer
Queue Length Versus Time- 3 servers
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 50 100 150 200 250
Queue Length Versus Time- 2 servers
0
5
10
15
20
25
30
0 50 100 150 200 250
“ If you change to 2 servers, then….
“ The waiting time and queuing length with 3 servers…
Waiting Times in Queue-2 servers
0
5
10
15
20
25
30
35
40
45
50
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101 106 111 116 121 126 131 136 141 146
Customer

Metodos
Waiting Line Models
Extensions
Notation for Classifying Waiting Line Models
M = Designates a Poisson probability distribution for the
arrivals or an exponential probability distribution for
service time
D = Designates that the arrivals or the service time is
deterministic or constant
G = Designates that the arrivals or the service time has
a general probability distribution with a known mean
and variance
Code indicating
arrival
distribution
Code indicating
service time
distribution
Number of
parallel
servers
others

Metodos
Waiting Line Models
Extensions
Notation for Classifying Waiting Line Models
M/M/1
M/M/k
M/G/1 (M/D/1 is a special case, D for deterministic service
time)
G/M/1 And more….
G/G/1
G/G/k
Code indicating
arrival
distribution
Code indicating
service time
distribution
Number of
parallel
servers
others

Metodos
M/G/1 Queue Performance
Measures
M/G/1 System: Steady state results (λ<μ)
P0 = 1−ρ (ρ = λ/μ)
LQ =
LS = LQ + λ/μ = LQ + ρ
WQ = LQ / λ
WS = WQ + 1/μ
Pw = 1 – P0 = ρ
)
1
(
2
2
2
2
ρ
ρ
σ
λ
−
+
μ = service rate
1/ μ = mean service time
σ2 = variance of service time distribution
M/D/1 Queue: σ2 = 0
LQ =
μ = service rate
1/ μ = mean service time
σ2 = variance of service time distribution
M/D/1 Queue: σ2 = 0
LQ = )
1
(
2
2
ρ
ρ
−

Metodos
An Example: Secretary Hiring
Suppose you must hire a secretary and you have to select one
of two candidates.
Secretary 1 is very consistent, typing any document in exactly
15 minutes.
Secretary 2 is somewhat faster, with an average of 14 minutes
per document, but with times varying according to the
exponential distribution.
The workload in the office is 3 documents per hour, with
interarrival times varying according to the exponential
distribution. Which secretary will give you shorter turnaround
times on documents?

Metodos
Secretary Hiring - Queuing Model

Metodos
M/M/s with Finite Population
The number of customers in the system
is not permitted to exceed some
specified number
Example: Machine maintenance problem

Metodos
M/M/s with Limited Waiting
Room
Arrivals are turned away when the
number waiting in the queue reaches a
maximum level
Example: Walk-in Dr.s office with limited
waiting space

Metodos
“ λ = Mean number of
arrivals per time period
“ e.g., 3 units/hour
“ μ = Mean number of
people or items served
per time period
“ e.g., 4 units/hour
“ 1/μ = 15 minutes/unit
Remember: λ & μ Are
Rates
© 1984-1994 T/Maker Co.
If average service time is 15
minutes, then μ is 4
customers/hour

Metodos
Summary
“ Queuing system design has an important impact on
the service provided by an enterprise
“ Steady state performance measures can provide
useful information in assessing service and
developing optimal queuing systems
“ The general procedure of solving a queuing problem:
“ Many queuing systems do not have closed-form
solutions. Simulation is a powerful tool of analyzing
those systems.
Identify
Queue Type
Identify
Queue Type
Estimate Arrival
& Service Processes
Estimate Arrival
& Service Processes
Calculate
Performance
Measures
Calculate
Performance
Measures
Conduct
Economic Analysis
Conduct
Economic Analysis

Metodos
Zapatería Mary’s
Los clientes que llegan a la zapatería Mary’s son en
promedio 12 por minuto, de acuerdo a la distribución
Poisson.
El tiempo de atención se distribuye
exponencialmente con un promedio de 8 minutos por
cliente.
La gerencia esta interesada en determinar las
medidas de performance para este servicio.

Metodos
SOLUCION
“ Datos de entrada
λ = 1/ 12 clientes por minuto = 60/ 12 = 5 por
hora.
μ = 1/ 8 clientes por minuto = 60/ 8 = 7.5 por
hora.
“ Calculo del performance
P0 = 1- (λ / μ) = 1 - (5 / 7.5) = 0.3333
Pn = [1 - (λ / μ)] (λ/ μ) = (0.3333)(0.6667)n
L = λ / (μ - λ) = 2
Lq = λ2/ [μ(μ - λ)] = 1.3333
W = 1 / (μ - λ) = 0.4 horas = 24 minutos
Wq = λ / [μ(μ - λ)] = 0.26667 horas = 16 minutos
P0 = 1- (λ / μ) = 1 - (5 / 7.5) = 0.3333
Pn = [1 - (λ / μ)] (λ/ μ) = (0.3333)(0.6667)n
L = λ / (μ - λ) = 2
Lq = λ2/ [μ(μ - λ)] = 1.3333
W = 1 / (μ - λ) = 0.4 horas = 24 minutos
Wq = λ / [μ(μ - λ)] = 0.26667 horas = 16 minutos
Pw = λ / μ = 0.6667
ρ = λ / μ = 0.6667

Metodos
USANDO WINQSB

Metodos

Metodos
Datos de entrada para WINQSB
μ
λ

Metodos
Medidas de desempeño

Metodos

Metodos
OFICINA POSTAL TOWN
La oficina postal Town atiende público los Sábados
entre las 9:00 a.m. y la 1:00 p.m.
Datos
- En promedio, 100 clientes por hora visitan la oficina postal
durante este período. La oficina tiene tres dependientes.
- Cada atención dura 1.5 minutos en promedio.
- La distribución Poisson y exponencial describen la llegada de
los clientes y el proceso de atención de estos respectivamente.
La gerencia desea conocer las medidas relevantes al servicio en
orden a:
– La evaluación del nivel de servicio prestado.
– El efecto de reducir el personal en un dependiente.
La gerencia desea conocer las medidas relevantes al servicio en
orden a:
– La evaluación del nivel de servicio prestado.
– El efecto de reducir el personal en un dependiente.

Metodos
SOLUCION
Se trata de un sistema de colas M / M / 3 .
Datos de entrada
λ = 100 clientes por hora.
μ = 40 clientes por hora (60 / 1.5).
¿Existe un período estacionario (λ kμ )?
λ = 100 kμ = 3(40) = 120.

Metodos
USANDO EL WINQSB

Metodos

Metodos
Sistemas de colas M/G/1
Supuestos
- Los clientes llegan de acuerdo a un proceso Poisson con
esperanza λ.
− El tiempo de atención tiene una distribución general con
esperanza μ.
− Existe un solo servidor.
- Se cuenta con una población infinita y la posibilidad de
infinitas filas.

Metodos
TALLER DE REPARACIONES TED
Ted repara televisores y videograbadoras.
Datos
- El tiempo promedio para reparar uno de estos artefactos es de
2.25 horas.
- La desviación estándar del tiempo de reparación es de 45
minutos.
- Los clientes llegan a la tienda en promedio cada 2.5 horas, de
acuerdo a una distribución Poisson.
- Ted trabaja 9 horas diarias y no tiene ayudantes.
- El compra todos los repuestos necesarios.
+ En promedio, el tiempo de reparación esperado debería
ser de 2 horas.
+ La desviación estándar esperada debería ser de 40
minutos.

Metodos
Ted desea conocer los efectos de usar nuevos
equipos para:
1. Mejorar el tiempo promedio de reparación
de los artefactos;
2. Mejorar el tiempo promedio que debe esperar
un cliente hasta que su artefacto sea reparado.
Ted desea conocer los efectos de usar nuevos
equipos para:
1. Mejorar el tiempo promedio de reparación
de los artefactos;
2. Mejorar el tiempo promedio que debe esperar
un cliente hasta que su artefacto sea reparado.

Metodos
SOLUCION
Se trata de un sistema M/G/1 (el tiempo de atención
no es exponencial pues σ = 1/μ).
Datos
“ Con el sistema antiguo (sin los nuevos equipos)
λ = 1/ 2.5 = 0.4 clientes por hora.
μ = 1/ 2.25 = 0.4444 clientes por hora.
σ = 45/ 60 = 0.75 horas.
“ Con el nuevo sistema (con los nuevos equipos)
μ = 1/2 = 0.5 clientes por hora.
σ = 40/ 60 = 0.6667 horas.

Metodos
Sistemas de colas M/M/k/F
Se deben asignar muchas colas, cada una de un
cierto tamaño límite.
Cuando una cola es demasiado larga, un modelo de
cola infinito entrega un resultado exacto, aunque de
todas formas la cola debe ser limitada.
Cuando una cola es demasiado pequeña, se debe
estimar un límite para la fila en el modelo.

Metodos
Características del sistema M/M/k/F
- La llegada de los clientes obedece a una distribución Poisson
con una esperanza λ.
- Existen k servidores, para cada uno el tiempo de atención se
distribuye exponencialmente, con esperanza μ.
− El número máximo de clientes que puede estar presente en
el sistema en un tiempo dado es “F”.
- Los clientes son rechazados si el sistema se encuentra
completo.

Metodos
Tasa de llegada efectiva.
- Un cliente es rechazado si el sistema se encuentra completo.
- La probabilidad de que el sistema se complete es PF.
- La tasa efectiva de llegada = la tasa de abandono de clientes
en el sistema (λe).
λe = λ(1 - PF)

Metodos
COMPAÑÍA DE TECHADOS RYAN
Ryan atiende a sus clientes, los cuales llaman y
ordenan su servicio.
Datos
- Una secretaria recibe las llamadas desde 3 líneas telefónicas.
- Cada llamada telefónica toma tres minutos en promedio
- En promedio, diez clientes llaman a la compañía cada hora.

Metodos
Cuando una línea telefónica esta disponible, pero la
secretaria esta ocupada atendiendo otra llamada, el
cliente debe esperar en línea hasta que la secretaria
este disponible.
Cuando todas las líneas están ocupadas los clientes
optan por llamar a la competencia.
El proceso de llegada de clientes tiene una
distribución Poisson, y el proceso de atención se
distribuye exponencialmente.

Metodos
La gerencia desea diseñar el siguiente sistema con:
- La menor cantidad de líneas necesarias.
- A lo más el 2% de las llamadas encuentren las líneas ocupadas.
La gerencia esta interesada en la siguiente información:
El porcentaje de tiempo en que la secretaria esta ocupada.
EL número promedio de clientes que están es espera.
El tiempo promedio que los clientes permanecen en línea esperando ser
atendidos.
El porcentaje actual de llamadas que encuentran las líneas ocupadas.

Metodos
SOLUCION
“Se trata de un sistema M / M / 1 / 3
“Datos de entrada
λ = 10 por hora.
μ = 20 por hora (1/ 3 por minuto).
“ WINQSB entrega:
P0 = 0.533, P1 = 0.133, P3 = 0.06
6.7% de los clientes encuentran las líneas
ocupadas.
Esto es alrededor de la meta del 2%.
sistema M / M / 1 / 4
P0 = 0.516, P1 = 0.258, P2 = 0.129, P3 = 0.065, P4 = 0.032
3.2% de los clntes. encuentran las líneas ocupadas
Aún se puede alcanzar la meta del 2%
sistema M / M / 1 / 5
P0 = 0.508, P1 = 0.254, P2 = 0.127, P3 = 0.063, P4 = 0.032
P5 = 0.016
1.6% de los cltes. encuentran las linea ocupadas
La meta del 2% puede ser alcanzada.

Metodos
“Otros resultados de WINQSB
Con 5 líneas telefónicas
4 clientes pueden esperar
en línea

Metodos

Metodos
Sistemas de colas M/M/1//m
En este sistema el número de clientes potenciales es
finito y relativamente pequeño.
Como resultado, el número de clientes que se
encuentran en el sistema corresponde a la tasa de
llegada de clientes.
Características
- Un solo servidor
- Tiempo de atención exponencial y proceso de llegada
Poisson.
- El tamaño de la población es de m clientes (m finito).

Metodos
CASAS PACESETTER
Casas Pacesetter se encuentra desarrollando cuatro
proyectos.
Datos
- Una obstrucción en las obras ocurre en promedio cada 20
días de trabajo en cada sitio.
- Esto toma 2 días en promedio para resolver el problema.
- Cada problema es resuelto por el V.P. para construcción
¿Cuanto tiempo en promedio un sitio no se
encuentra operativo?
-Con 2 días para resolver el problema (situación actual)
-Con 1.875 días para resolver el problema (situación nueva).

Metodos
SOLUCION
Se trata de un sistema M/M/1//4
Los cuatro sitios son los cuatro clientes
El V.P. para construcción puede ser considerado
como el servidor.
Datos de entrada
λ = 0.05 (1/ 20)
μ = 0.5 (1/ 2 usando el actual V.P).
μ = 0.533 (1/1.875 usando el nuevo V.P).

Metodos
Medidas del V.P V.P
Performance Actual Nuevo
Tasa efectiva del factor de utilización del sistema ρ 0,353 0,334
Número promedio de clientes en el sistema L 0,467 0,435
Número promedio de clientes en la cola Lq 0,113 0,100
Número promedio de dias que un cliente esta en el sistema W 2,641 2,437
Número promedio de días que un cliente esta en la cola Wq 0,641 0,562
Probabilidad que todos los servidores se encuentren ociosos Po 0,647 0,666
Probabilidad que un cliente que llega deba esperar en el sist. Pw 0,353 0,334
Medidas del V.P V.P
Performance Actual Nuevo
Tasa efectiva del factor de utilización del sistema ρ 0,353 0,334
Número promedio de clientes en el sistema L 0,467 0,435
Número promedio de clientes en la cola Lq 0,113 0,100
Número promedio de dias que un cliente esta en el sistema W 2,641 2,437
Número promedio de días que un cliente esta en la cola Wq 0,641 0,562
Probabilidad que todos los servidores se encuentren ociosos Po 0,647 0,666
Probabilidad que un cliente que llega deba esperar en el sist. Pw 0,353 0,334
Resultados obtenidos por WINQSB
Resultados obtenidos por WINQSB

Metodos
Análisis económico de los sistemas
de colas
Las medidas de desempeño anteriores son usadas
para determinar los costos mínimos del sistema de
colas.
El procedimiento requiere estimar los costos tales
como:
- Costo de horas de trabajo por servidor
- Costo del grado de satisfacción del cliente que espera en la
cola.
-Costo del grado de satisfacción de un cliente que es atendido.

Metodos
SERVICIO TELEFONICO DE WILSON
FOODS
Wilson Foods tiene un línea 800 para responder las
consultas de sus clientes
Datos
- En promedio se reciben 225 llamadas por hora.
- Una llamada toma aproximadamente 1.5 minutos.
- Un cliente debe esperar en línea a lo más 3 minutos.
-A un representante que atiende a un cliente se le paga $16
por hora.
-Wilson paga a la compañía telefónica $0.18 por minuto cuando
el cliente espera en línea o esta siendo atendido.
- El costo del grado de satisfacción de un cliente que espera en
línea es de $20 por minuto.
-El costo del grado de satisfacción de un cliente que es
atendido es de $0.05.
¿Qué cantidad de representantes
para la atención de los clientes
deben ser usados para minimizar
el costo de las horas de operación?
¿Qué cantidad de representantes
para la atención de los clientes
deben ser usados para minimizar
el costo de las horas de operación?

Metodos
SOLUCION
“Costo total del modelo
Costo total por horas de
trabajo de “k”
representantes para la
atención de clientes
CT(K) = Cwk + CtL + gwLq + gs(L - Lq)
Total horas para sueldo
Costo total de las
llamadas telefónicas
Costo total del grado de satisfacción
de los clientes que permanecen en línea
Costo total del grado de satisfacción
de los clientes que son atendidos
CT(K) = Cwk + (Ct + gs)L + (gw - gs)Lq

Metodos
Datos de entrada
Cw= $16
Ct = $10.80 por hora [0.18(60)]
gw= $12 por hora [0.20(60)]
gs = $0.05 por hora [0.05(60)]
Costo total del promedio de horas
TC(K) = 16K + (10.8+3)L + (12 - 3)Lq
= 16K + 13.8L + 9Lq

Metodos
Asumiendo una distribución de llegada de los
clientes Poisson y una distribución exponencial del
tiempo de atención, se tiene un sistema M/M/K
λ = 225 llamadas por hora.
μ = 40 por hora (60/ 1.5).
El valor mínimo posible para k es 6 de forma de
asegurar que exista un período estacionario (λKμ).
WINQSB puede ser usado para generar los
resultados de L, Lq, y Wq.

Metodos
En resumen los resultados para K= 6,7,8,9,10.
K L Lq Wq CT(K)
6 18,1249 12,5 0,05556 458,62
7 7,6437 2,0187 0,00897 235,62
8 6,2777 0,6527 0,0029 220,50
9 5,8661 0,2411 0,00107 227,12
10 5,7166 0,916 0,00041 239,70
K L Lq Wq CT(K)
6 18,1249 12,5 0,05556 458,62
7 7,6437 2,0187 0,00897 235,62
8 6,2777 0,6527 0,0029 220,50
9 5,8661 0,2411 0,00107 227,12
10 5,7166 0,916 0,00041 239,70
Conclusión: se deben emplear 8 representantes
para la atención de clientes
Conclusión: se deben emplear 8 representantes
para la atención de clientes

Metodos
Sistemas de colas Tándem
En un sistema de colas Tándem un cliente debe
visitar diversos servidores antes de completar el
servicio requerido
Se utiliza para casos en los cuales el cliente llega de
acuerdo al proceso Poisson y el tiempo de atención
se distribuye exponencialmente en cada estación.
Tiempo promedio total en el sistema =
suma de todos los tiempo promedios en las estaciones
individuales
Tiempo promedio total en el sistema =
suma de todos los tiempo promedios en las estaciones
individuales

Metodos
COMPAÑÍA DE SONIDO BIG BOYS
Big Boys vende productos de audio.
El proceso de venta es el siguiente:
- Un cliente realiza su orden con el vendedor.
- El cliente se dirige a la caja para pagar su pedido.
- Después de pagar, el cliente debe dirigirse al empaque para
obtener su producto.

Metodos
Datos de la venta de un Sábado normal
- Personal
+ 8 vendedores contando el jefe
+ 3 cajeras
+ 2 trabajadores de empaque.
- Tiempo promedio de atención
+ El tiempo promedio que un vendedor esta con un
cliente es de 10 minutos.
+ El tiempo promedio requerido para el proceso de pago es de
3 minutos.
+ El tiempo promedio en el área de empaque es de 2
minutos.
-Distribución
+ El tiempo de atención en cada estación se distribuye
exponencialmente.
+ La tasa de llegada tiene una distribución Poisson de 40
clientes por hora.
Solamente 75% de
los clientes que llegan
hacen una compra
¿Cuál es la cantidad promedio de tiempo ,
que un cliente que viene a comprar
demora en el local?
¿
¿Cu
Cuá
ál es la cantidad promedio de tiempo ,
l es la cantidad promedio de tiempo ,
demora en el local?
demora en el local?

Metodos
SOLUCION
“Estas son las tres estaciones del sistema
de colas Tándem
M / M / 8
M / M / 3
λ
=
4
0
λ
=
3
0
λ
=
3
0
M / M / 2
W1 = 14 minutos
W2 = 3.47 minutos
W3=2.67 minutos
Total = 20.14 minutos.

Metodos
Balance de líneas de ensamble
Una línea de ensamble puede ser vista como una
cola Tándem, porque los productos deben visitar
diversas estaciones de trabajo de una secuencia
dada.
En una línea de ensamble balanceada el tiempo
ocupado en cada una de las diferentes estaciones de
trabajo es el mismo.
El objetivo es maximizar la producción

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