Queueing Theory And Replacement Model
Queueing theory and Replacement model1. Trucks at a single platform weigh-bridge arrive according to Poisson probability distribution. The timerequired to weigh the truck follows an exponential probability distribution. The mean arrival rate is 12trucks per day and mean service rate is 18 trucks per day. Determine(i) What is the probability that no trucks are in the system?(ii) What is the average number of trucks waiting for service?(iii) What is the average time a truck waits for weighing service to begin?(iv) What is the probability that an arriving truck will have to wait for service?Solution:Solution:πΏ 12 π‘ππ’πππ πππ πππ¦, π 18 π‘ππ’πππ πππ οΏ½οΏ½π ππππ‘ππ π πΏ 12 0.67π 18The probability that no trucks are in the system is P0 1 Ο 1 0.67 0.33Average number of trucks in the systemπΏπ π0.67 2.031 π 1 0.67The average number of trucks waiting for serviceLq Ls Ξ΄ 2.03 0.67 1.36ΞΌThe average time a truck waits for weighing service to beginππ πΏπ 1.36 0.113 πππ¦π πΏ12The probability that an arriving truck will have to wait for οΏ½ ππππ‘ππ π πΏ 12 0.67π 182. The city council of a small town has decided to build a tennis court in the central park. Players areexpected to arrive on the average of 10 sets of players per 12 hour a day. Playing time is exponentiallydistributed with a mean of 1 hour. Arrivals are Poisson. What are the expected queueing statisticsassuming the basic single server model?
Solution:πΏ 10πππ ππ’π, π 1 πππ ππ ππππ‘ππ π πΏ 10 0.83π 12Average number of players in the park10π0.8312πΏπ 4.88101 π 1 1 0.8312Average number of players in the queueLq Ls Ξ΄ 4.88 0.83 4.05ΞΌAverage waiting time of players in the parkππ πΏπ 4.88 0.49 ππ’ππ πΏ10Average waiting time of players in the queueππ πΏπ 4.05 0.41 ππ’ππ πΏ103. Customers arrive at a one window drive at a bank according to Poisson distribution with mean 10 perhour. Service time per customer is exponential with mean 5 minutes. The space in front of the windowincluding that for the serviced car can accommodate a maximum of 3 cars. Others can wait outside thisspace. Determine(i) The probability that an arriving customer can drive directly to the space in front of the window.(ii) The probability that an arriving customer will have to wait outside the indicated space.(iii) The length of time an arriving customer is exposed to wait before starting service.Solution:Ξ΄ 10/hour, π 605 12/ ππ’πThe probability that an arriving customer can drive directly to the space in front of the window isπ π₯ 3 π π₯ 0 π π₯ 1 π π₯ 2 π€ πππ ππ πππΏ1 π ππ1 πΏπ
πΏ ππΏ π00πΏπΏ1 ππ10 1 12πΏπΏ1 ππ11πΏπΏ1 ππ10101 12122 πΏπΏ1 ππ221 πΏππΏπΏ1 1 πππΏπΏ1 ππ2210 10021 2.53 0.421212 14412The probability that an arriving customer will have to wait outside the indicated space isπΏπ π 3 π3πΏπ€ πππ π π π ππ π 3 1012π3 0.58The length of time an arriving customer is exposed to wait before starting service isππ πΏππΏ10π0.83πΏπ 12 4.881 π 1 10 1 0.8312Lq Ls ππ Ξ΄ 4.88 0.83 4.05ΞΌπΏπ 4.05 0.41 ππ’ππ πΏ104. Arrivals at telephone booth are considered to be Poisson with an average time of 10 minutes betweenone arrival and the next. The length of the phone call is assumed to be distributed exponentially, withmean 3 minutes.(i) What is the probability that a person arriving at the booth will have to wait?(ii) The telephone department will install a second booth when convinced that an arrival would expectwaiting for at least 3 minutes for a phone call. By how much should the flow of arrivals increase in orderto justify a second booth?(iii) What is the average length of the square that forms from time to time?Solution:πΏ 11/ππππ’π‘ππ , π /ππππ’π‘ππ 103
The probability that a person arriving at the booth will have to wait1πΏ 10ππ‘ππππ§ππ‘πππ ππππ‘ππ π 0.31π3The second booth will be installed, if the new waiting time Wqβ² is greater than 3. Let the correspondingarrival rate be π1 .Wqβ² 3πΏ1πΏ1 3 31 1π π πΏ1 πΏ13 3 πΏ1 1 3πΏ1 4πΏ1 1 πΏ1 πΌππππππ π ππ πππππ£πππ ππ πΏ1 πΏ 141 13 4 10 20Average length of the square that forms from time to time11ΞΌ103Lw 3 1.437ΞΌ Ξ΄ 1 173 10 305. There are two clerks in a university to receive fees from the students. If the service time for eachstudent is exponential with mean 4 minutes and if the boys arrive in a Poisson fashion at the counter atthe ratio of 10 per hour, determine(i) The probability of having to wait for service.(ii) The expected percentage idle time for each clerk.Solution:πΆ 2, πΏ 10 πππ ππ’π, π 160πππ ππππ’π‘ππ 15 πππ ππ ππππ‘ππ π π 1π0 π 0πΏ1010 1 ππΆ 2 1530 3πΏ ππΏ πππππ .π!π! ππ πΏ 1
2 1 π 0 1 0.67 10150!01015π!π10 22 15 15 .2! 2 15 1010 151!0.4530.2! 30 101 110 22 15 15 .2! 2 15 10 1 1.67 20.1 10.45 30.2! 20 1 1 1.67 0.34 1 0.5The probability of having to wait for serviceπΏ ππ. ππ π π ππ 1 ! ππ πΏ 010 215. 15 0.5 0.1672 1 ! 2(15) 10The expected percentage idle time for each clerk is 1 π 100 1 12 100 100 0.67 100 67%33Replacement ProblemsReplacement of items whose maintenance and repair costs increase with time and ignoringchanges in the value of the money during the period1. The cost of machine is Rs. 6100 and its scrap value is 100. The maintenance cost found fromexperience is as follows:Year:12345678Maintenance cost (Rs.): 100 250 400 600 900 1200 1600 2000When should the machine be replaced?Solution:
Capital Scrap MaintenanceAverage annual costTotal costcostvaluecostπ(π)ππͺ πΊ π(π)πͺ πΊ 020007050130501631.25Here the minimum average annual cost is Rs. 1575. Therefore the machine can be replaced at the end ofsixth year.Yearsn2. Machine A cost Rs. 45000 and the operating costs are estimated at Rs. 1000 for the first yearincreasing by Rs. 10000 per year in the second and subsequent years. Machine B cost Rs. 50000 andoperating cost are Rs 2000 for the first year, increasing by Rs. 4000 in the second and subsequent years.If we now have a machine of type A, should we replace it with B? If so when? Assume that bothmachines have no resale value and future costs are not discounted.Solution:Machine A:Capital Scrap MaintenancAverage annual costTotal costcostvaluee costπ(π)ππͺ πΊ π(π)πͺ πΊ 260004450000310006400010900027250Here the minimum average annual cost of Machine A is Rs. 26000. Therefore the Machine A can bereplaced at the end of third year.YearsnMachine 20005000072000Total costπͺ πΊ π(π)Average annual 6.67205002000020333.33πππͺ πΊ π(π)
Here the minimum average annual cost of Machine B is Rs. 20000. Therefore the Machine B can bereplaced at the end of third year.Total cost of Machine A for the first year π π . 46000Total cost of Machine A for the second year π π . 57000 β 46000 π π . 11000 20000Total cost of Machine A for the third year π π . 78000 β 57000 π π . 21000 20000 The Total cost of Machine A for the third year is greater than the minimum average annual cost ofMachine B. Machine A is replaced by Machine B at the end of second year.Replacement of items whose maintenance costs increases with time and value of the moneyalso changes with Time1. A manufacturer is offered two machines A and B. A has cost price of Rs. 2,500, its running cost is Rs.400 for each of first years and increased by Rs. 100 every subsequent year, Machine B having the samecapacity as A, costs Rs. 1250 and has running cost of Rs. 600 for 6 years, increasing by Rs. 100 per yearthereafter. Taking moneyβs value as 10% per year, which machine should be purchased? Scrap value ofboth the machines is negligibly small.Solution:π 10% 0.1π£ 11 0.90911 π 1 0.1Machine ACapital cost C Rs. 2500, Scrap Value S factorππ πDiscountedrunningcostπΉπ ππ .2310.5338.7359.24373.2381.69πͺ πΉπ ππ 5176.45549.65931.29ππ 3516.7592πͺ πΉπ ππ πππ 05876.01877.51
Hence machine A should be replaced after 9th year.Machine B:Capital cost C Rs. 1250, Scrap Value S 501250Running DiscountfactorcostπΉπππ οΏ½οΏ½οΏ½π ππ 85πͺ πΉπππ πππ 5.86866.3351πͺ πΉπ ππ πππ 11844.52Hence machine A should be replaced after 8th year.Since the weighted average cost in 9 years of machine A is Rs. 876 and weighted average cost in 8 yearsof machine B is Rs. 840, it is advisable to purchase machine B.Replacement of items that fail suddenly1. A plant has 6 numbers of cells and a study has been made on the nature of deterioration with time.The following is the probability of failure after replacement.Months after replacement 12345Probability of failure0.2 0.15 0.1 0.3 0.25The cost of replacement as and when it fails has been found to be Rs. 200 per unit. Now it has beenproposed to replace all the items at a specified frequency and in such a case the cost of such areplacement comes to Rs. 60 per unit. Can we accept this proposal? If so, at what frequency we have toundertake replacement of all the cells at a time?Solution:Let ππ be the probability that a cell fails during the ith month of its life.Let ππ represent the number of replacements made at the end of ith month when all 6 cells are newinitially. Then we have
Month012345Expected number of failuresπ0π1 π0 π1 6 0.2π2 π0 π2 π1 π1 6 0.15 1.2 0.2π3 π0 π3 π1 π2 π2 π1 6 0.1 1.2 0.15 1.14 0.2π4 π0 π4 π1 π3 π2 π2 π3 π1 6 0.3 1.2 0.1 1.14 0.15 1.008 0.2π4 π0 π5 π1 π4 π2 π3 π3 π2 π4 π1 6 0.25 1.2 0.3 1.14 0.1 1.008 0.15 2.2926 0.2 6 1.2 1.14 1.008 2.2926 2.5837Individual replacement policy:π΄π£πππππ ππππ ππ πππππ πππ 1 0.2 2 0.15 3 0.1 4 0.3 5 0.25 3.25ππ΄π£πππππ ππ’ππππ ππ πππππ’πππ πππ ππππ‘ πππ‘ππ ππ’ππππ ππ πππππ 6 1.85 πππ ππππ‘ ππππππ ππππ ππ πππππ 3.25πΆππ π‘ ππ πππππ£πππ’ππ πππππππππππ‘ π΄π£πππππ ππ’ππππ ππ πππππ’πππ πππ ππππ‘ πππππ£πππ’ππ πππππππππππ‘ πππ π‘ πππ ππππ‘ 1.85 200 π π . 370 πππ ππππ‘ Group replacement policy:End ofmonth(n)12345Total cost of replacing cells(T)6 60 1.2 200 6006 60 1.2 1.14 200 8126 60 1.2 1.14 1.008 200 1029.66 60 1.2 1.14 1.008 2.2926 200 1488.126 60 1.2 1.14 1.008 2.2926 2.5837 200 2004.86Averagecost permonthT/n600406343.2372401Minimum average cost per month is 343. So group replacement can be made at the end of third month.Group replacement period 3 monthsIndividual replacement cost/month Rs. 370Minimum group replacement cost/month Rs. 343
Since the minimum group replacement cost/month is lesser than the individual replacementcost/month, the group replacement policy is best and hence all the cells are to be replaced once in threemonths and the cells which fail during this three month period are to be replaced individually.2. The probability ππ of failure just before age n (in months) of 1000 light bulbs is given. If individualreplacements costs Rs. 12.5 and group replacement costs Rs. 3.0 per bulb, find the optimal replacementpolicy.nππ123450.1 0.2 0.25 0.3 0.15Solution:Let ππ be the probability that a light bulb fails during the ith month of its life.Let ππ represent the number of replacements made at the end of ith month when all 1000 cells are newinitially. Then we haveMonth012345Expected number of failuresπ0π1 π0 π1 1000 0.1π2 π0 π2 π1 π1 1000 0.2 100 0.1π3 π0 π3 π1 π2 π2 π1 1000 0.25 100 0.2 210 0.1π4 π0 π4 π1 π3 π2 π2 π3 π1 1000 0.3 100 0.25 210 0.2 291 0.1π4 π0 π5 π1 π4 π2 π3 π3 π2 π4 π1 1000 0.15 100 0.3 210 0.25 291 0.2 396 0.1 1000 100 210 291 396 330Individual replacement policy:π΄π£πππππ ππππ ππ πππ π‘ ππ’πππ πππ 1 0.1 2 0.2 3 0.25 4 0.3 5 0.15 3.2ππ΄π£πππππ ππ’ππππ ππ πππππ’πππ πππ ππππ‘ 1000 3133.2πΆππ π‘ ππ πππππ£πππ’ππ πππππππππππ‘ π΄π£πππππ ππ’ππππ ππ πππππ’πππ πππ ππππ‘ πππππ£πππ’ππ πππππππππππ‘ πππ π‘ πππ ππππ‘ 313 12.5 π π . 3913 πππ ππππ‘ Group replacement policy:
End ofmonth(n)12345Averagecost permonthT/n42503438350438663918Total cost of replacing cells(T)1000 3 100 12.5 42501000 3 100 210 12.5 68751000 3 100 210 291 12.5 105131000 3 100 210 291 396 12.5 154631000 3 100 210 291 396 330 12.5 19588Minimum average cost per month is 3438. So group replacement can be made at the end of secondmonth.Group replacement period 2 monthsIndividual replacement cost/month Rs. 3913Minimum group replacement cost/month Rs. 3438Since the minimum group replacement cost/month is lesser than the individual replacementcost/month, the group replacement policy is best and hence all the tube lights are to be replaced oncein two months and the cells which fail during this two month period are to be replaced individually.3. The probability ππ of failure just before age n shown below for 1000 bulbs. If the individualreplacements costs Re. 1 and the group replacement costs Rs. 0.3 per item, find the optimalreplacement policy.nππ123450.3 0.1 0.1 0.2 0.3Solution:Let ππ be the probability that a bulbs fails during the ith month of its life.Let ππ represent the number of replacements made at the end of ith month when all 1000 cells are newinitially. Then we haveMonth012345Expected number of failuresπ0π1 π0 π1 1000 0.3π2 π0 π2 π1 π1 1000 0.1 300 0.3π3 π0 π3 π1 π2 π2 π1 1000 0.1 300 0.1 190 0.3π4 π0 π4 π1 π3 π2 π2 π3 π1 1000 0.2 300 0.1 190 0.1 187 0.3π4 π0 π5 π1 π4 π2 π3 π3 π2 π4 π1 1000 0.3 300 0.2 190 0.1 187 0.1 305 0.3 1000 300 190 187 305 489
Individual replacement policy:π΄π£πππππ ππππ ππ ππ’πππ πππ 1 0.3 2 0.1 3 0.1 4 0.2 5 0.3 3.1ππ΄π£πππππ ππ’ππππ ππ πππππ’πππ πππ ππππ‘ 1000 3233.1πΆππ π‘ ππ πππππ£πππ’ππ πππππππππππ‘ π΄π£πππππ ππ’ππππ ππ πππππ’πππ πππ ππππ‘ πππππ£πππ’ππ πππππππππππ‘ πππ π‘ πππ ππππ‘ 323 1 π π . 323 πππ ππππ‘ Group replacement policy:End ofmonth(n)12345Total cost of replacing cells(T)1000 0.3 300 1 6001000 0.3 300 190 1 7901000 0.3 300 190 187 1 9771000 0.3 300 190 187 305 1 12821000 0.3 300 190 187 305 489 1 1771Averagecost permonthT/n600395326321354Minimum average cost per month is 321. So group replacement can be made at the end of fourthmonth.Group replacement period 4 monthsIndividual replacement cost/month Rs. 323Minimum group replacement cost/month Rs. 321Since the minimum group replacement cost/month is lesser than the individual replacementcost/month, the group replacement policy is best and hence all the bulbs are to be replaced once in fourmonths and the cells which fail during this four month period are to be replaced individually.
Queueing theory and Replacement model 1. Trucks at a single platform weigh-bridge arrive according to Poisson probability distribution. The time required to weigh the truck follows an exponential proba
Purpose Simulation is often used in the analysis of queueing models. A simple but typical queueing model Waiting line Server Calling population Queueing models provide the analyst with a powerful tool for designing and evaluating the performance of queueing systems. Typical measures of system performance Server util
802.16 networks was conducted in [7], [8] by combining link-layer queueing with physical-layer transmission. A vacation queueing model was adopted in [9] to analyze the link-layer queueing performance of OFDM-TDMA systems with round-robin scheduling. A queueing model for OFDMA systems was used in [10] to design a scheduling scheme that balances
theory to construct the dynamic system of screening system in this paper. Queueing theory is the mathematical study of waiting lines [23]. In queueing theory, a model is constructed so that queue length
the understanding of teletra c, queueing theory fundamentals and related queueing behavior of telecommunications networks and systems. These concepts and ideas form a strong base for the more mathematically inclined students who can follow up with the extensive literature on
probability, operational research, telecommunication, i ndustrial engineering, com-puter science, management science publish articles on queu eing extensively. The ow of new theories and methodologies in queueing has become very hard to keep . Queueing Theory and its Applications, A Personal View 13 distrib
of Queueing Theory Applied to Emergency Care. Here is a picture of the participants at our meeting on October 25, 2012. Figure 1. Emergency Care/Queueing Seminar: (Left to Right) Jed Keesling, Trent Register, Joshua Hurwitz, Jean Larson, James Maissen, Hayriye Gulbu-dak, Evan Milliken,
have faded. But our lack of completeness is also explained by the time constraints of this survey. Queueing applications are abundant in Canada. The diverse areas where queueing analysis has been applied include: ship-ping and canals, grocery store checkout line count estimation, vehic
LΓ³pez Austin, Alfredo, βEl nΓΊcleo duro, la cosmovisiΓ³n y la tradiciΓ³n mesoamericanaβ, en . CosmovisiΓ³n, ritual e identidad de los pueblos indΓgenas de MΓ©xico, Johanna Broda y FΓ©liz BΓ‘ez-Jorge (coords.), MΓ©xico, Consejo Nacional para la Cultura y las Artes y Fondo de Cultura EconΓ³mica, 2001, p. 47-65. LΓ³pez Austin, Alfredo, Breve historia de la tradiciΓ³n religiosa mesoamericana .
