4.7 Finite Population Source Model

Transcription

1 Characteristics 1. Arrival Process R independent Source All sources are identical Interarrival time is exponential with rate for each source No arrivals if all sources are in the system. OR372-Dr.Khalid Al-Nowibet 1

2 Characteristics 2. Interarrival time is exponential with rate µ Number of services is Poisson Process with rate µ 3. Multiple Server: Number of servers = s Identical, Independent and Parallel servers Random choice of idle servers 4. System size is infinite 5. Queue Discipline : FCFS Notation M / M / s / / FCFS Finite Population OR372-Dr.Khalid Al-Nowibet 2

3 State of the system Finite Population = R system is in state n if there are n customers in the system (waiting or serviced) If system in state n R n are expected to arrive If system in state R 0 are expected to arrive Let P n be probability that there are n customers in the system in the steady-state. n = 0, 1, 2, 3,,R OR372-Dr.Khalid Al-Nowibet 3

4 Consider: Facility with 6 identical machines Failures on any machine occur at rate If machine fail is taken to the maintenance Each machine needs one worker for maintenance. Maintenance workshop has 3 identical workers. Each worker works at rate µ OR372-Dr.Khalid Al-Nowibet 4

5 Operating Machines Workshop n = 0 Arrival Rate = 6 OR372-Dr.Khalid Al-Nowibet 5

6 Operating Machines Workshop n = 1 Arrival Rate = 5 OR372-Dr.Khalid Al-Nowibet 6

7 Workshop Operating Machines n = 2 Arrival Rate = 4 OR372-Dr.Khalid Al-Nowibet 7

8 n = 3 Arrival Rate = 3 Workshop Operating Machines OR372-Dr.Khalid Al-Nowibet 8

9 n = 4 Arrival Rate = 2 Workshop Operating Machines OR372-Dr.Khalid Al-Nowibet 9

10 n = 5 Arrival Rate = Workshop Operating Machines OR372-Dr.Khalid Al-Nowibet 10

11 n = 6 Arrival Rate = 0 Workshop States of System n = 0, 1, 2, 3, 4, 5, 6 OR372-Dr.Khalid Al-Nowibet 11

12 Rate Diagram: 1. Arrival Rate: if system in stat n Arrival rate = (6 n) 0 n 6 for number of Machines= R Arrival rate = (R n) 0 n R 2. Service Rate : (Finite Number of servers s = 3) If system in state 0 < n 3 Service rate = nµ, 0 n 3 If system in state n > 3 Service rate = 3µ, 3 < n 6 for number of servers = s Service rate = nµ, 0 n s Service rate = sµ, s < n R OR372-Dr.Khalid Al-Nowibet 12

13 Balance Equations: Average Rate out of State n 2 = Average Rate in to State n µ 2µ 3µ 3µ 3µ 3µ cut-1 6P 0 = µp 1 cut-2 5P 1 = 2µP 2 cut-3 4P 2 = 3µP 3 cut-4 3P 3 = 3µP 4 cut-5 2P 4 = 3µP 5 cut-6 P 5 = 3µP 6 For R-Machines and s-servers 0 n s (R n)p n = nµp n+1 s<n<r (R n)p n = sµp n+1 OR372-Dr.Khalid Al-Nowibet 13

14 Solution of Balance Equations: let ρ=/µ 6P 0 = µp 1 P 1 = (6/µ)P 0 = (6)ρ P 0 5P 1 = 2µP 2 P 2 = (5/2µ)P 1 P 2 = (6.5 2 /2µ 2 )P 0 = (30/2)ρ 2 P 0 4P 2 = 3µP 3 P 3 = (4/3µ)P 2 P 3 = ( /3.2µ 3 )P 0 = (120/6)ρ 3 P 0 3P 3 = 3µP 4 P 4 = (3/3µ)P 3 P 4 = ( /3.3.2µ 4 )P 0 = (360/18)ρ 4 P 0 2P 4 = 3µP 5 P 5 = (2/3µ)P 4 P 5 = ( / µ 5 )P 0 = (720/54)ρ 5 P 0 1P 5 = 3µP 6 P 6 = (/3µ)P 5 P 6 = ( / µ 6 )P 0 = (720/162)ρ 6 P 0 OR372-Dr.Khalid Al-Nowibet 14

15 Solution of Balance Equations: Computing P 0 : P 0 + P 1 + P 2 +P 3 +P 4 + P 5 + P 6 = 1 P n =1 n P 0 +(6)ρP 0 +(15)ρ 2 P 0 +(20)ρ 3 P 0 +(20)ρ 4 P 0 + (13.33)ρ 5 P 0 + (4.44)ρ 6 P 0 = 1 P 0 [1+(6)ρ + (15)ρ 2 + (20)ρ 3 +(20)ρ 4 + (13.33)ρ 5 + (4.44)ρ 6 ] = 1 P 0 = [1+(6)ρ + (15)ρ 2 + (20)ρ 3 +(20)ρ 4 + (13.33)ρ 5 + (4.44)ρ 6 ] 1 OR372-Dr.Khalid Al-Nowibet 15

16 Solution of Balance Equations: Computing P 0 : P 0 = R n= 0 P n P n is a function of P 0 if P 0 = 0 system is infinite system is finite if an only if P 0 > 0 ρ s = sµ NO< 1 finite sum finite value P 0 > 0 and finite always for any, µ and s No Steady-State Condition on, µ, and s OR372-Dr.Khalid Al-Nowibet 16

17 Performance Measures In steady state e, µ, P 0 L B = E[busy servers] = E[#Cust. in service] L s = L q + L B W s = W q + (1/µ) L s = e W s L q = e W q System is in Steady Stead L B = e W B Know 4 measures all measures are known OR372-Dr.Khalid Al-Nowibet 17

18 Performance Measures 1. Effective Arrival Rate e : e = E[arrival rate] R = = n= 0 ( R n) P n n= 0 R R = RP n n Pn = (R L s ) n= 0 n= 0 R ( R n) P n OR372-Dr.Khalid Al-Nowibet 18

19 Performance Measures 2. Average Customers in System L s : L = n s P n n= 0 3. Average Busy servers L B : L B = E[busy servers] = E[#Cust. in service] L B = 0.P 0 +1.P P s ( Pr{ n s }) L B = 0.P 0 +1.P P s (1 Pr{ n < s }) R OR372-Dr.Khalid Al-Nowibet 19

20 Performance Measures 4. Average Customers in Queue L q : L q = L s L B L q = 0.(P 0 +P 1 +P 2 + +P s )+ 1.P s P s P R = R n= s (n s) P n 5. Utilization of the System U: U = Pr{ n > 0 } = P 1 + P 2 + P P R = 1 P 0 6. Utilization of the Service SU: SU = Pr{ all servers busy } = Pr{ n s} SU = 1 (P 0 + P 1 + P P s 1 ) OR372-Dr.Khalid Al-Nowibet 20

21 Performance Measures 7. Average Time Spent in System W s : L s = e.w s L s W s = e 8. Average Waiting time in Queue W q : L q = e.w q L q W q = e OR372-Dr.Khalid Al-Nowibet 21

22 Example: A factory has 6 identical machines for production. Failures occur on each machine at a rate of 2 failures per day according to a Poisson process. The maintenance department has 3 workers all have the same experience. Once a machine failed, one of the workers is called to repair it. If all workers are busy the machine is put in awaiting list for repair when workers available. The repair time is exponentially distributed with mean 3 hours. Assume factory works 9 hrs a day. 1. What is the probability that all worker are idle? 2. On average how many machine repaired in one day? 3. On average how many machine waiting for repaired? 4. What is the probability that all servers are busy? 5.What is the expected time until a failed machine to restart operation? OR372-Dr.Khalid Al-Nowibet 22

23 Example: Arrivals: = 2 failures/day Poisson = failures/hr Service: E[S] = 3hr Exponential µ= 1/E[S] = 1/3 failures/hr ρ = /µ = 0.222/0.333 = Number of Servers: s = 3 Population size = R = 6 n = 0,1,2,3,4,5,6 M/M/3 Finite Population R=6 OR372-Dr.Khalid Al-Nowibet 23

24 Example: Use rate diagram: P 1 = (6/µ)P 0 = (6)ρ P 0 P 2 = (6.5 2 /2µ 2 )P 0 = (30/2)ρ 2 P 0 P 3 = ( /3.2µ 3 )P 0 = (120/6)ρ 3 P 0 P 4 = ( /3.3.2µ 4 )P 0 = (360/18)ρ 4 P 0 P 5 = ( / µ 5 )P 0 = (720/54)ρ 5 P 0 P 6 = ( / µ 6 )P 0 = (720/162)ρ 6 P 0 n Sum. T n P n np n OR372-Dr.Khalid Al-Nowibet 24

25 Example: 1. Pr{worker are idle} = P{all machines working} = P 0 = Average machines repaired in one day = e = (R L s ) = 2 (6 2.22) = 7.56 machines/day 3. Average machine waiting for repaired = L q L q = R n= s (n s) P n = 1.P P 5 +3.P 6 = machine OR372-Dr.Khalid Al-Nowibet 25

26 Example: 4. Pr{all servers are busy} = P 3 + P 4 + P 5 + P 6 = Expected time until a failed machine return to operation = W s = L s / e = 2.22/7.56 = hrs OR372-Dr.Khalid Al-Nowibet 26