1. Model Antrian M/M/s

2. M/M/s

3. M/M/s (s > 1)
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4. 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 smPs + lPs-2 = {l + (s-1)m} Ps-1
s (s+1)mPs+1 + lPs-1 = (l + sm) Ps
s (s+2)mPs+2 + lPs = (l + (s+1)m) Ps+1
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5. 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 =
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6. M/M/s (cont.)
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7. M/M/s (cont.) Therefore, if we denote Pn = Cn× P0 , then (l / m)n
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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,...
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8. M/M/s (cont.) if 0 £ n £ s if s £ n So, if l < sm =>
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if 0 £ n £ s
if s £ n 30 29

9. 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
= P r S rj
s! j=0 dr
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10. M/M/s (cont.) (l / m)s d ¥ Lq = P0 ------------ r ------ S rj
s! dr j=0
(l / m)s d
= P r
s! dr (1 - r)
(l / m)s r
= P
s! (1 - r)2
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11. M/M/s (cont.)
(l / m)s r Lq = P , r = l / sm s! (1 - r) (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
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12. 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)}
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13. 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
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14. 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

15. 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

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

17. M/M/s Example I
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18. M/M/s Example I (cont.) = 0.429 (@ 43% of time, system is empty)
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= % of time, system is empty)
as compared to s = 1: P0 = 0.20
(l / m)s r
Lq = P
s! (1 - r)2
= × {0.82 ´ 0.4} / { 2! ´ ( )2 }
= 0.152
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Dalam M/M/1: P0=1-ρ=1-0.8=0.2 34 30

19. M/M/s Example I (cont.) Wq = Lq / l = 0.152 / (1/10) = 1.52 (min)
W = Wq + 1 / m = / (1/8) = 9.52 (min)
What proportion of time is both repairman busy? (long run)
P(N ³ 2) = 1 - P0 - P = = 0.228
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20. 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)?
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21. 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
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22. 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
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23. 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 = /3 = minute
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24. 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