Problem with two server Queue












4












$begingroup$


I am practicing for an exam on queuing theory and I found this question somewhat confusing. Appreciate if somebody can shed some light on it.



There are 2 facilities, A and B, which provide the same type of service. Each facility contains a single server with service times distributed as negative exponential random variables. The mean service times are $1/mu_1=1$ minute and $1/mu_2 = 4$ minutes at facilities A and B, respectively.



A combined total of 60 customers per hour wish to avail themselves of the service provided by facilities A and B. Arrivals of customers at the critical point are Poisson. At that point each customer makes a choice, independently of all others, on which facility he is going to use. This choice is made without any knowledge of the status of the queue in front of either facility. Let $p$ denote the probability that a random customer chooses facility A (and $1 - p$ that he or she chooses B).



Each user is only concerned with minimizing his expected time spent in a service facility, (= average waiting time + average service time). It can be reasonably expected that in the long run (system in steady state and in equilibrium-as far as the distribution of customers between the two facilities is concerned) the customers will "distribute" themselves among the two facilities, in the sense that p will stabilize around a specific value. What is that value of p?






queueing-theory







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    I am practicing for an exam on queuing theory and I found this question somewhat confusing. Appreciate if somebody can shed some light on it.



    There are 2 facilities, A and B, which provide the same type of service. Each facility contains a single server with service times distributed as negative exponential random variables. The mean service times are $1/mu_1=1$ minute and $1/mu_2 = 4$ minutes at facilities A and B, respectively.



    A combined total of 60 customers per hour wish to avail themselves of the service provided by facilities A and B. Arrivals of customers at the critical point are Poisson. At that point each customer makes a choice, independently of all others, on which facility he is going to use. This choice is made without any knowledge of the status of the queue in front of either facility. Let $p$ denote the probability that a random customer chooses facility A (and $1 - p$ that he or she chooses B).



    Each user is only concerned with minimizing his expected time spent in a service facility, (= average waiting time + average service time). It can be reasonably expected that in the long run (system in steady state and in equilibrium-as far as the distribution of customers between the two facilities is concerned) the customers will "distribute" themselves among the two facilities, in the sense that p will stabilize around a specific value. What is that value of p?






    queueing-theory







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      2



      $begingroup$


      I am practicing for an exam on queuing theory and I found this question somewhat confusing. Appreciate if somebody can shed some light on it.



      There are 2 facilities, A and B, which provide the same type of service. Each facility contains a single server with service times distributed as negative exponential random variables. The mean service times are $1/mu_1=1$ minute and $1/mu_2 = 4$ minutes at facilities A and B, respectively.



      A combined total of 60 customers per hour wish to avail themselves of the service provided by facilities A and B. Arrivals of customers at the critical point are Poisson. At that point each customer makes a choice, independently of all others, on which facility he is going to use. This choice is made without any knowledge of the status of the queue in front of either facility. Let $p$ denote the probability that a random customer chooses facility A (and $1 - p$ that he or she chooses B).



      Each user is only concerned with minimizing his expected time spent in a service facility, (= average waiting time + average service time). It can be reasonably expected that in the long run (system in steady state and in equilibrium-as far as the distribution of customers between the two facilities is concerned) the customers will "distribute" themselves among the two facilities, in the sense that p will stabilize around a specific value. What is that value of p?






      queueing-theory







      share|cite|improve this question









      $endgroup$




      I am practicing for an exam on queuing theory and I found this question somewhat confusing. Appreciate if somebody can shed some light on it.



      There are 2 facilities, A and B, which provide the same type of service. Each facility contains a single server with service times distributed as negative exponential random variables. The mean service times are $1/mu_1=1$ minute and $1/mu_2 = 4$ minutes at facilities A and B, respectively.



      A combined total of 60 customers per hour wish to avail themselves of the service provided by facilities A and B. Arrivals of customers at the critical point are Poisson. At that point each customer makes a choice, independently of all others, on which facility he is going to use. This choice is made without any knowledge of the status of the queue in front of either facility. Let $p$ denote the probability that a random customer chooses facility A (and $1 - p$ that he or she chooses B).



      Each user is only concerned with minimizing his expected time spent in a service facility, (= average waiting time + average service time). It can be reasonably expected that in the long run (system in steady state and in equilibrium-as far as the distribution of customers between the two facilities is concerned) the customers will "distribute" themselves among the two facilities, in the sense that p will stabilize around a specific value. What is that value of p?






      queueing-theory




      queueing-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 13 '12 at 16:24









      KaminiKamini

      374




      374






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          Let $lambda=60$, then the arrival rate to station $A$ is $60p$ and the arrival rate to station $B$ is $60(1-p)$, so the utilization at station $A$ is $rho_A=p$ and the utilization at station $B$ is $rho_B=4(1-p)$ (so we must have $p>frac34$ for the system to be stable). The expected size in system is
          begin{align}
          mathbb E[L_A] &= sum_{n=1}^infty n(1-rho_A)rho_A^n= frac{rho_A}{1-rho_A} = frac p{1-p}\
          mathbb E[L_B] &= sum_{n=1}^infty n(1-rho_B)rho_B^n= frac{rho_B}{1-rho_B} = frac{4(1-p)}{4p-3}.
          end{align}
          By Little's law, the mean sojourn times are
          begin{align}
          mathbb E[W_A] &= frac{mathbb E[L_A]}{lambda_A} = left(frac p{1-p}right)frac1p = frac1{1-p}\
          mathbb E[W_B] &= frac{mathbb E[L_B]}{lambda_B} = left(frac{4(1-p)}{4p-3}right)frac1{4(1-p)} = frac1{4p-3}.
          end{align}
          Over time we can reasonably expect the distribution of customers to tend so that the expected waiting time is the same at each station. Therefore
          $$mathbb E[W_A]=mathbb E[W_B]implies frac1{1-p}=frac1{4p-3}implies p=frac45. $$






          share|cite|improve this answer











          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f236503%2fproblem-with-two-server-queue%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            Let $lambda=60$, then the arrival rate to station $A$ is $60p$ and the arrival rate to station $B$ is $60(1-p)$, so the utilization at station $A$ is $rho_A=p$ and the utilization at station $B$ is $rho_B=4(1-p)$ (so we must have $p>frac34$ for the system to be stable). The expected size in system is
            begin{align}
            mathbb E[L_A] &= sum_{n=1}^infty n(1-rho_A)rho_A^n= frac{rho_A}{1-rho_A} = frac p{1-p}\
            mathbb E[L_B] &= sum_{n=1}^infty n(1-rho_B)rho_B^n= frac{rho_B}{1-rho_B} = frac{4(1-p)}{4p-3}.
            end{align}
            By Little's law, the mean sojourn times are
            begin{align}
            mathbb E[W_A] &= frac{mathbb E[L_A]}{lambda_A} = left(frac p{1-p}right)frac1p = frac1{1-p}\
            mathbb E[W_B] &= frac{mathbb E[L_B]}{lambda_B} = left(frac{4(1-p)}{4p-3}right)frac1{4(1-p)} = frac1{4p-3}.
            end{align}
            Over time we can reasonably expect the distribution of customers to tend so that the expected waiting time is the same at each station. Therefore
            $$mathbb E[W_A]=mathbb E[W_B]implies frac1{1-p}=frac1{4p-3}implies p=frac45. $$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Let $lambda=60$, then the arrival rate to station $A$ is $60p$ and the arrival rate to station $B$ is $60(1-p)$, so the utilization at station $A$ is $rho_A=p$ and the utilization at station $B$ is $rho_B=4(1-p)$ (so we must have $p>frac34$ for the system to be stable). The expected size in system is
              begin{align}
              mathbb E[L_A] &= sum_{n=1}^infty n(1-rho_A)rho_A^n= frac{rho_A}{1-rho_A} = frac p{1-p}\
              mathbb E[L_B] &= sum_{n=1}^infty n(1-rho_B)rho_B^n= frac{rho_B}{1-rho_B} = frac{4(1-p)}{4p-3}.
              end{align}
              By Little's law, the mean sojourn times are
              begin{align}
              mathbb E[W_A] &= frac{mathbb E[L_A]}{lambda_A} = left(frac p{1-p}right)frac1p = frac1{1-p}\
              mathbb E[W_B] &= frac{mathbb E[L_B]}{lambda_B} = left(frac{4(1-p)}{4p-3}right)frac1{4(1-p)} = frac1{4p-3}.
              end{align}
              Over time we can reasonably expect the distribution of customers to tend so that the expected waiting time is the same at each station. Therefore
              $$mathbb E[W_A]=mathbb E[W_B]implies frac1{1-p}=frac1{4p-3}implies p=frac45. $$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Let $lambda=60$, then the arrival rate to station $A$ is $60p$ and the arrival rate to station $B$ is $60(1-p)$, so the utilization at station $A$ is $rho_A=p$ and the utilization at station $B$ is $rho_B=4(1-p)$ (so we must have $p>frac34$ for the system to be stable). The expected size in system is
                begin{align}
                mathbb E[L_A] &= sum_{n=1}^infty n(1-rho_A)rho_A^n= frac{rho_A}{1-rho_A} = frac p{1-p}\
                mathbb E[L_B] &= sum_{n=1}^infty n(1-rho_B)rho_B^n= frac{rho_B}{1-rho_B} = frac{4(1-p)}{4p-3}.
                end{align}
                By Little's law, the mean sojourn times are
                begin{align}
                mathbb E[W_A] &= frac{mathbb E[L_A]}{lambda_A} = left(frac p{1-p}right)frac1p = frac1{1-p}\
                mathbb E[W_B] &= frac{mathbb E[L_B]}{lambda_B} = left(frac{4(1-p)}{4p-3}right)frac1{4(1-p)} = frac1{4p-3}.
                end{align}
                Over time we can reasonably expect the distribution of customers to tend so that the expected waiting time is the same at each station. Therefore
                $$mathbb E[W_A]=mathbb E[W_B]implies frac1{1-p}=frac1{4p-3}implies p=frac45. $$






                share|cite|improve this answer











                $endgroup$



                Let $lambda=60$, then the arrival rate to station $A$ is $60p$ and the arrival rate to station $B$ is $60(1-p)$, so the utilization at station $A$ is $rho_A=p$ and the utilization at station $B$ is $rho_B=4(1-p)$ (so we must have $p>frac34$ for the system to be stable). The expected size in system is
                begin{align}
                mathbb E[L_A] &= sum_{n=1}^infty n(1-rho_A)rho_A^n= frac{rho_A}{1-rho_A} = frac p{1-p}\
                mathbb E[L_B] &= sum_{n=1}^infty n(1-rho_B)rho_B^n= frac{rho_B}{1-rho_B} = frac{4(1-p)}{4p-3}.
                end{align}
                By Little's law, the mean sojourn times are
                begin{align}
                mathbb E[W_A] &= frac{mathbb E[L_A]}{lambda_A} = left(frac p{1-p}right)frac1p = frac1{1-p}\
                mathbb E[W_B] &= frac{mathbb E[L_B]}{lambda_B} = left(frac{4(1-p)}{4p-3}right)frac1{4(1-p)} = frac1{4p-3}.
                end{align}
                Over time we can reasonably expect the distribution of customers to tend so that the expected waiting time is the same at each station. Therefore
                $$mathbb E[W_A]=mathbb E[W_B]implies frac1{1-p}=frac1{4p-3}implies p=frac45. $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Apr 27 '16 at 4:48

























                answered Mar 1 '16 at 13:39









                Math1000Math1000

                19.1k31745




                19.1k31745






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f236503%2fproblem-with-two-server-queue%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Ellipse (mathématiques)

                    Image
                    Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                    Read more

                    Quarter-circle Tiles

                    Image
                    up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                    Read more

                    Mont Emei

                    Image
                    Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                    Read more