Model Antrian M/M/s

M/M/s

M/M/s (s > 1) 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 58 24

M/M/s (cont.) State Rate In = Rate Out 0 mP1 = lP0 1 2mP2 + lP0 = (l + m) P1 2 3mP3 + lP1 = (l + 2m) P2 .... ................... s-1 smPs + lPs-2 = {l + (s-1)m} Ps-1 s (s+1)mPs+1 + lPs-1 = (l + sm) Ps s+1 (s+2)mPs+2 + lPs = (l + (s+1)m) Ps+1 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 25

M/M/s (cont.) Now, solve for P1 , P2, P3... in terms of P0 P1 = (l / m) P0 P2 = (l / 2m) P1 = (1/2!) × (l / m)2 P0 P3 = (l / 3m) P2 = (1/3!) × (l / m)3 P0 ......... Ps = (1/s!) × (l / m)s P0 Ps+1 = (1/s) × (l / m) Ps = 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 26

M/M/s (cont.) 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 27

M/M/s (cont.) Therefore, if we denote Pn = Cn× P0 , then (l / m)n 15 April 2019 Therefore, if we denote Pn = Cn× P0 , then (l / m)n Cn = ---------- , for n = 1, 2, ...., s. n! and , for n = s+1, s+2,... R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 28

M/M/s (cont.) if 0 £ n £ s if s £ n So, if l < sm => 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology if 0 £ n £ s if s £ n 30 29

M/M/s (cont.) Now solve for Lq: Note, r = l / sm Lq = S¥n=s (n - s) Pn = S¥j=0 j Ps+j ; Note, n = s + j ¥ (l / m)s = S j ---------- rj P0 j=0 s! (l / m)s ¥ d = P0 ------------ r S ------ rj s! j=0 dr 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 30

M/M/s (cont.) (l / m)s d ¥ Lq = P0 ------------ r ------ S rj s! dr j=0 (l / m)s d 1 = P0 ------------ r ------ --------- s! dr (1 - r) (l / m)s r = P0 ------------ --------- s! (1 - r)2 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 31

M/M/s (cont.) (l / m)s r Lq = P0 ----------- --------- , r = l / sm s! (1 - r)2 (Lq : avg # in queue) Wq = Lq / l (Wq: avg waiting time in Q) W = Wq + 1 / m (W: avg waiting time in sys.) L = l (Wq + 1/m) (L: avg # in the system) = Lq + l / m 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 32

Steady-State Parameters of M/M/s Queue r = l / sm P(L(¥) ³ s) = {(l/m)s P0} / {s!(1- l/sm)} = {(sr)s P0} / {s! (1 - r)} 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 16

Steady-State Parameters of M/M/s Queue (cont.) L = sr + {(sr)s+1 P0} / {s (s!) (1 - r)2} = sr + {r P (L(¥) ³ s) } / {1 - r} W = L / l Wq = W - 1/m Lq = l Wq = {(sr)s+1 P0} / {s (s!) (1 - r)2} = {r P (L(¥) ³ s) } / {1 - r} L - Lq = l / m = sr 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 16

The M/M/s Model Assumptions Interarrival times have an exponential distribution with a mean of 1/l. Service times have an exponential distribution with a mean of 1/m. Any number of servers (denoted by s). With multiple servers, the formula for the utilization factor becomes r = l / sm but still represents that average fraction of time that individual servers are busy. © The McGraw-Hill Companies, Inc., 2003

Comparison of Multi-Server Systems Measure M/M/11 M/M/12 M/M/13 Lq 6.821 2.247 0.951 Wq 0.682 0.225 0.0.95 E 0.909 0.833 0.767 Pr{ Tq = 0 } 0.318 0.551 0.715 Pr{ Tq > 1 } 0.251 0.061 0.014

Values of L for the M/M/s Model for Various Values of s © The McGraw-Hill Companies, Inc., 2003 Figure 14.8 Values of L for the M/M/s model for various values of s, the number of servers.

M/M/s Example I 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 33

M/M/s Example I (cont.) = 0.429 (@ 43% of time, system is empty) 15 April 2019 = 0.429 (@ 43% of time, system is empty) as compared to s = 1: P0 = 0.20 (l / m)s r Lq = P0 ----------- --------- s! (1 - r)2 = 0.429 × {0.82 ´ 0.4} / { 2! ´ (1 - 0.4)2 } = 0.152 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology Dalam M/M/1: P0=1-ρ=1-0.8=0.2 34 30

M/M/s Example I (cont.) Wq = Lq / l = 0.152 / (1/10) = 1.52 (min) W = Wq + 1 / m = 1.52 + 1 / (1/8) = 9.52 (min) What proportion of time is both repairman busy? (long run) P(N ³ 2) = 1 - P0 - P1 = 1 - 0.429 - 0.343 = 0.228 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 35

M/M/s Example 2 Many early examples of queuing theory applied to practical problems concerning tools for service. Attendants manage the tools, while mechanics, assumed to be from an infinite calling population, arrive for service. Assume Poisson arrivals at rate 2 mechanics per minute and exponentially distributed service times for attendance with mean 40 seconds. How many attendants are required for statistical equilibrum (Steady state)? 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 45

M/M/s Example 2 (cont.) l = 2 per minute, and m = 60/40 = 3/2 per minute. Since, the offered load is greater than 1, that is, since, l / m = 2 / (3/2) = 4/3 > 1, more than one server is needed if the system is to have a statistical equilibrium. The requirement for steady state is that s > l / m = 4/3. Thus, at least s = 2 attendants are needed. The quantity 4/3 is the expected number of busy server, and for s ³ 2, r = 4 / (3s) is the long-run proportion of time each server is busy. (What would happen if there were only s = 1 server?) s = 2 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 45

M/M/s Example 2 (cont.) Let there be s = 2 attendants. First, P0 is calculated as = {1 + 4/3 + (16/9)(1/2)(3)} -1 = {15 / 3}-1 = 1/5 = 0.2 The probability that all servers are busy is given by P(L(¥) ³ 2) = {(4/3)2 (1/5)} / {2!(1- 2/3)} = (8/3) (1/5) = 0.533 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 16

M/M/s Example 2 (cont.) Thus, the time-average length of the waiting line of mechanics is Lq = {(2/3)(8/15)} / (1 - 2/3) = 1.07 mechanics and the time-average number in system is given by L = Lq+ l/m = 16/15 + 4/3 = 12/5 = 2.4 mechanics Using Little’s relationships, the average time a mechanic spends at the tool crib is W = L / l = 2.4 / 2 = 1.2 minutes while the avg time spent waiting for an attendant is Wq = W - 1/m = 1.2 - 2/3 = 0.533 minute 15 April 2019 R. Rumani M / rrm@ittelkom.ac.id Telkom Institute of Technology 30 16

Example 3  QUIZ until 10:00 A grocery store has three registers open. Customers arrive to check out at an average of 1 per minute (Poisson). The service time averages 2 minutes (exponential). Please calculate these parameters : (Po, L, Lq, W, Wq) © The McGraw-Hill Companies, Inc., 2003 l = 1 customer per minute m = (1/2) customers per minute s = 3