Es un taller resuelto de teoría de colasFull description
Una cola se caracteriza por el número máximo admisible de clientes que puede contener; estas pueden ser infinitas o finitas; el proceso básico de colas está referido como “cliente” que requi…Descripción completa
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%illo& %illo& 'roo 'roo a!ona a!onall oper opera a un un ca*er ca*ero o auto autom$! m$!co co en el ue ue los los clien clientes tes reali+ reali+an an transa transacc de d/as $biles las lleadas al autoca*ero ocurren al a+ar a+ar con una tasa de lleadas de 24 0.4 clientes por minuto. a. u$l es la medida o el nmero esperado de clientes ue llear$ en un lapso de cinco minutos b. upona ue puede usarse la distribuci,n de probabilidad probabilidad de Poisson para describir el proceso de lleadas. U!lice la tasa de lleadas de la parte a) para calcular las probabili de ue eactamente 0 1 2 9 3 clientes lleuen durante un lapso de cinco minutos. c. e esperan demoras si m$s de tres clientes llean durante cualuier lapso de cinco minutos. u$l es la probabilidad de ue ocurran demoras
2 personas personas en 5 minutos en promedio
a b
P(0) P(1) P(2) P(3)
0.6703 Us Usando la tabla de distribu 0.2681 0.0536 0.0072
2. -n -n el sis sistem tema a de l/ne l/nea a de esper espera a del %illo& %illo& 'roo 'roo a!on a!onal al 'an 'an ("ea ("ea el el probl problema ema 1) 1) sup sup distribuci,n de probabilidad eponencial eponencial con una tasa de ser"icios de 36 clientes por ora probabilidad eponencial para responder las siuientes preuntas: a. u$l es la probabilidad de ue el !empo de ser"icio sea de un minuto o menos b. u$l es la probabilidad de ue el !empo de ser"icio sea de dos minutos o menos c. u$l es la probabilidad de ue el !empo de ser"icio sea de m$s de dos minutos
a) b) c)
0.45 0.4511 1188 8836 364 4 La prob probab abil ilid idad ad de ue ue e ell !em !empo po de ser ser 0.6# 0.6#88 8805 0578 788 8 la prob probab abili ilida dad d de de ue ue el !empo !empo de ser" ser" 0.3011#4212 m$s de dos minutos minutos
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ciones bancarias sin descender de sus autom,"iles. -n las maanas clientes por ora o
dades
iones
ona ue los !empos de ser"icio del autoca*ero siuen una o 0.6 clientes por minuto. U!lice la distribuci,n de
icio sea de un minuto o menos cio sea de dos minuto o menos
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PROBLEMA #5
-l escrit escritori orio o de re
Minutos lamda Mu Pregunta A P0 Pregunta B Lq
Pregunta
10 12
1 Hora
60
¿Cuáll es la probab ¿Cuá probabilid ilidad ad de que no haya haya peticion peticiones es d
16.67; ¿Cuá ¿C uáll es es el el núm númer ero o pr prom omed edio io de de pet petic icio ione nes s que que es esp p
4.166667
. ¿Cuál es el tiempo de espera promedio promedio en minutos Horas
Minutos
!q
0.416666667
25
Pregunta "
¿Cuá ¿C uáll es es el el tie tiemp mpo o pr prom omed edio io en el es escr crit itor orio io de ref efer er
! Pregunta E P
Horas
Minutos
0.5
30
¿Cuá ¿C uáll es es la la pr prob obab abil ilid idad ad de qu que e una una nu nuev eva a lle llega gada da t
83.33;
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uda. upona ue puede u!li+arse una distribuci,n de probabilidad de Poisson con una ue los !empos de ser"icio sian una distribuci,n de probabilidad eponencial con una tasa
e ayuda en el sistema?
ran ser atendidas?
antes de que comience a ser atendido?
encia en minutos (tiempo de espera más tiempo de servicio)?
nga que esperar a que la atiendan?
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Problema #$ =o"ies =o"ies >oni >onitt es un establecim establecimient iento o de renta renta de pel/culas pel/culas en ?@? 9 "ideo "ideo Apico Apico para clientes clientes clientes llean a =o"ies =o"ies >onit >onit con una tasa de lleadas de 1.25 clientes por minuto. minuto. -l emplead minuto. upona lleadas Poisson 9 !empos de ser"icios eponenciales. a. u$l es la probabilidad probabilidad de ue no a9a clientes en el sistema b. u$l es el nmero promedio de clientes ue esperan ser atendidos c. u$l es el e l !empo promedio ue un cliente espera para ue comiencen a atenderlo d. u$l es la probabilidad de ue un cliente ue llea tena ue esperar a ue lo a!endan e. Bndican las caracter/s!cas de operaci,n ue el sistema de mostrador de salida de un empleado proporciona un ni"el de ser"icio aceptable
tasa de lleada tasa de ser"icio a)
1.25 2 P(0)
b)
mero promedio de clientes ue esperan ser atendidos.
1.0417
c)
>iempo promedio ue un cliente espera para ue comiencen a atenderlo.
0.8333333333 =inutos
d)
Probabilidad de ue un cliente ue llea tena ue esperar a ue lo a!endan.
e)
-l ser"icio es aceptable aceptable por ue la tasa de ser"icio es ma9or a la tasa de
0.375 -s la probabilidad de ue n
50 seundos
0.625 -s la la pr probabilidad de de u ue t
Problema #%
Para ara la l/ne l/nea a de espe esperra con con un solo solo canal anal de 'ur 'urer ?ome ?ome de la se
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DE1clien ienteF(1 minuto) (60 min)F1ora E 60 cli P0)
IE(1 IE(1.2 .25 5 cli clien entte) e)F( F(1 1 min minut uto) o) (60 (60 mi min) n)F1 F1o orra E 75 clie clie
L)
L E (60G2)F(75(75H60))E 3.2 tres personas en la cola
L)
L E 60F(75H60)E 4 clientes en el sistema
%)
% E60F E60F(( 75(7 75(75H 5H60 60)) ))E0 E0.0 .05 5 prom promed edio io clie clien nte tess en la co coll
%)
%E1F(75H60)E0.07 !empo en el sistema
P&)
P&E60F75E0.8
-l sis sistem ema a de dell ser ser"i "ici cio o es es m$s m$s de deCc Ccie ient nte e por por ue ue la la pr probab obabil il
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e las "en en casa. ?urante las noces entre semana los o del mostrador de salida a!ende a dos clientes por
o a9a clientes en el sistema.
na ue esperar a ue lo a!endan.
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.
ntes por ora ntes Po E 1H60F75 E 0.2
a
idad de espera es m$s alta ue la persona ue no
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Problema # 10
>rosper rosper >ire ompan ompan99 decidi, contr contrat atar ar a un nue"o mec$ni mec$nicc ue ue orde ordene nen n un *ue *ueo nue nue"o de llan llanttas. ?os me mecc$nic $nicos os soli solicc contra contratad tado o por K 14 por ora ora 9 puede puede atender atender a un promedi promedi epe eperi rien enci cia a 9 pue puede de ate atend nder er a un un prome promedi dio o de cua cuatr tro o clie clien n clientes llean al taller >rosper a la tasa de dos clientes por a. u u$ $les les son son las las caracter cter/s /s! !cas de ope operaci,n ci,n con cada ada mec eponenciales b. i la empresa asina un costo de cliente en espera de K30
a)
Mecanico !periencia "imitada# DE llean 2 clientes por ora IE a!ende 3 clientes por ora PoE 1HDFJ E1H2F3E0.3333 LEDG2F(I(IHD))E ((2))G2F(3(3H2))E1.3 personas en la cola LEDF(IHD)E2F(3H2)E2 clientes clientes en el sistema
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LE DF(IHD)E2F(4H2)E1 cliente esta en el sistema % E DF(I(IHD))E2F(4(4H2))E0.25 DF(I(IHD))E2F(4(4H2))E0.25 promedio de clientes en la c %E 1F(JHD)E1F(4H2)E 0.5 !empo en el sistema P&EDFIE2F4E0.5
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o para ue se encarue de todos los cambios para los clientes itaron el traba*o. Uno !ene eperiencia limitada puede ser o de tres clientes por ora. -l otro !ene "arios aos de es por ora pero deben paarle K20 por ora. upona ue los ora. $nico suponiendo lleadas de Poisson 9 !empos de ser"icio por ora cu$l mec$nico proporciona el costo opera!"o menor
b) osto >otal por ora: K30 >E 30(2)14(1) >E K74
>E30(1) >EK50
-l me mecanico e eperimentado !
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la
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PROBLEMA # 11
$.%
Lamda
10 (minutos (minutos)) 6 (clientes por ora) ora)
&iem'o de ser( Mu Pregunta A
Calcule Calc ule las las caracter caracter&sti &sticas cas de operac operaci' i'
Lq
0.2#762
L
0.7142# Horas
!q
0.11#05 Horas
!
0.28571
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Problema 1 PeteN PeteNs =aret =aret es una peuea peuea !enda !enda de abarrot abarrotes es local local con s, upona upona ue los los comprador compradores es acen cola cola en la ca*a con base base e de Poisson Poisson con una una tasa tasa de lleada lleadass de 15 clien clientes tes por por ora. ora. L en la ca*a siuen una distribuci,n distribuci,n de probabilidad eponencial 20 clientes por ora. a. alcule las caracter/s!cas de operaci,n de esta l/nea de esp b. i la meta meta de de ser"ici ser"icio o del erent erente e es limit limitar ar el !emp !empo o de esp esp proceso de cobro en la ca*a a no m$s de cinco minutos uO r al sistema de cobro en la ca*a actual
tasa de lleada >asa de ser"icio
a)
0.25 Pe Personas por minuto. 0.33 Pe Personas por minuto
Po L L %
0.25 2.25 3 #
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60
de la l&nea de espera de clientes suponiendo llegadas oisson y tiempos de servicio e!ponenciales.
Minutos
7.14 Minutos
17.14
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lo una ca*a reistradora. reistradora. n la distribuci,n de probabilidad s !empos de ser"icio con una tasa de ser"icio de ra. era pre"io al inicio del comendar/a con respecto
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Un est estudi udio o de la oper operaci aci,n ,n de ser"ic ser"icio io de co comid mida a de ml!pl ml!pl mues muestr tra a ue ue el !emp !empo o prom promed edio io ent entre re la lle llea ada da de un cli clie e su pedido completo es de 10 minutos. ?urante el *ueo los de ser"iH cio de comida reuiere un promedio de 2 minutos u$l es la tasa de ser"icios por canal en
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es canales en el parue de beisbol de los ed 'irds te en el mostrador de ser"icio de comida 9 su par!da con clientes llean a ra+,n de cuatro por minuto. La operaci,n por pedido. por minuto acer un pedido io de comida
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Problema % Q una planta planta de ensamble ensamble par!cular par!cular llean llean traba*o traba*oss al a+arR supon supona a ue las tasa tasa de lleadas lleadas es de cin !empos !empos de ser"icio (en minutos minutos por por traba*o) traba*o) no siuen la distribu distribuci,n ci,n de probabi probabilida lidad d eponencial eponencial.. ? operaci,n de la planta de ensamble son los siuientes.
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co traba*os por ora. Los os diseos propuestos para la