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Unformatted text preview: ,5; L"l it? To begin we must first decide upon an appropriate state space. It
is clear that the state of the system must include more information than
merely the number of customers in the system. For instance, it would
not he enough to specify that there is one customer in the system as we
would also have to know which chair he was in. Further, if we only
know that there are two customers in the sysrem, then we would not
know if the man in chair 1 is still being served or if he is just waiting
for the person in chair 2 to finish. To account for these points, the
followingr state space, consisting of the five states, {0, 0), (1,0), (U, l],
{I, ll. and (h, .l ), will he used. The states have the following interpretation: State interpretation (0, 0) There are no customers in the system. {1, 0) There is one customer in the system, and he is in chair 1.
(U, 1) There is one customer in the system, and he is in chair 2.
{'l, I} There are two customers in the system, and both are pres— ently being served.
(h, 1) There are two customers in the system, hot the customer in the first chair has completed his work in that chair and is
waiting for the second chair to become free. it should he noted that when the system is in state (b, 1), the person
in chair 1, though not being served, is nevertheless “blocking” potential
arrivals from entering the system. As a prelude to writing down the balance equations, it is usually
worthwhile to make a transition diagram. This is done by first drawing
a circle For each state and then drawing an arrow labeled by the rate
at which the process goes from one state to another. The transition
diagram for this model is shown in Figure 8.1. The explanation For the
diagram is as Follows: The arrow from state (0, U) to state { l, U) which is labeled A means
that when the process is in state (0, 0), that is when the system is empty,
then it goes to state (1, U) at a rate it, that is via an arrival. The arrow
from (0, l) to (l, I) is similarly explained. When the process is in state (1,0), it will go to state (0, I) when
the custmner in chair I is finished and this occurs at a rate. til; hence
the arrow from (1,0) to (U, I) labeled til. The arrow from (1, l) to
(F). ii is similarly explained. . ._...w.wwerF’.'?‘ﬂ’iﬁ3w' 360 8 Queueing Theory Figure 8.r. A transition diagram When in state (b, I) the process will go to state {0, 1) when the
customer in chair 2 completes his service (which occurs at rate ps2);
hence the arrow from ((3,1) to (O, '1) labeled pl. Also when in state
(I, l) the process will go to state (1, 0} when the man in chair 2 finishes
and hence the arrow from {1, 1) to (1, 0) labeled til. Finally, if the pro
cess is in state (0, I}, then it will go to state (0, 0) when the man in
chair 2 completes his service, hence the arrow from (O, 1) to (0, 0) la—
beled pl. As there are no other possible transitions, this completes the tran—
sition diagram. To write the balance equations we equate the sum of the arrows
(multiplied by the probability of the states where they originate) coming
into a State with the sum of the arrows {multiplied by the probability
of the state) going our of that state. This gives State Rate that the process [eaves = rate that it enters (0, O) “"00 2 H2100: (150i HIPID:i\Pnu+M1PH
(011) (A l' Helium : MPH; ‘l' Hips:
([31) (“4+ H23P1I25Pin {b1 1) id‘lPiJl = iLIPI] These along with the equation
Poo "L P10 'l' Pm + PII ‘l' PM :1 may be solved to determine the limiting probabilities. Though it is easy
to solve the above equations, the resultant solutions are quite involved 3. Exponential Models 36: and hence will not be explicitly presented. HOWBVCE, it is easy to answer
nut questions in terms of these limiting probabilities. First, since a po—
tential customer will enter the system when the state is either (0, 0) or
(O, 1), it follows that the proportion of cusromers entering the system
is P00 + Pm. Secondly, since there is one cusromet in the system when
ever the state is (0, l) or {1, OJ and two customers in the system when ever the srale is (1, '1) or (b, 1), it Follows that L, the average number
in the sysretn, is given by L = p0]. ‘l' lute ‘l’ 2U)” ‘l‘ PM} To derive the average amount of time that an entering customer spends
in the system“ we use the relationship W = L/Aa. Since a potential cus— tomer will enter the system when in state (0, 0) or (U, 1), it follows that
A4 = Ml’m + Pm} and hence W_P01+P10+2{P11+Pml
_—————————.._____
Ml’oo‘l‘ Pm) Example 3c (of) if A = 1, IL] = 1, pt: = 2, then calculate the above quantities
' of interest. fbﬂf‘hﬁ'z 13—15:;1.2,;LL;._3A.,4l16nncalCHlaFB—El‘lé—abmw Solution.
(3) Solving the balance equations, yields that _ I: ._ n: _ 2 l
Pun — 5%: P10 “ is Pl! — 35
h r _ :
Pm — 347, Pb: — f:
Hence,
_ e3 ' _ is
L "‘ i?! W, _ 15
(I \ f" ‘ ' ‘ \N E .X'lror 1 i: n "this med/El . we COJIGJ'LCJC’V {a Silkya , Jerri/(Ly f/meu‘ ﬁll—of fHQUOm
I r'
.« 0 system in which the server is able to serve two customers at the same
time. Whenever the server completes a service, he then serves the next
two customers at the same time. However, if there is only one customer
in line, then he serves that customer by himself. We shall assume that
his service time is exponential at rate p. whether he is serving one or
two customers. As usual, we suppose that customers arrive at an ex
ponential rate Pt. One example of such a system might be an elevator
or a cable car which can take at most two passengers at any time. it would seem that the state of the system would have to tell us not
only how many customers there are in the system, but also whether one
or two are presently being served. However, it turns out that we can
solve the problem easier not by concentrating on the number of cus~
tomers in the system, but rather on the number in qtietre. So let us define
the state as the number of customers waiting in queue, with two states
when there is no one in queue. That is, let us have as a state space 0’ 3 U, 1, 2, . . . , . . . with the interpretation
Stare Interpretation
0' No one in service
0 Server busy; no one waiting
n, n > O 1'? customers waiting la) The transition diagram is shown in Figure 8.2 and the balance equations are
State Rate at which the process leaves = rate at which it enters
0! API)‘ 2 lLPo
0 (A + ulPU = APU + up, + up:
“I n 2‘ 1 + “Jinn = h‘Prl—I + l‘l‘IJnrl Figure 8.2.
l Now the set of equations
(A l tall)” = kink, + plans; 3'? = l, 2,. . . (3.9]
have a solution of the form
ll)» : “HIDE: To see this, substitute the above in Equation (3.9) to obtain (h + pJain—PU : hull—i139 + pJointljljo Or
(A '1' Ma = h ‘1' pot" Solving this for (1 yields the three roots:
‘1‘\!1+47\/.L J—1+Vl+4)\/p.
____#____#._.—— LIZ], or.==_.—————"—'—'_"1 and a:
2. 2 tot possible, it foiiows that \/1+ ink/p. — 1 As the firsL two are clearly 1 L1 = .—————'#'—'—'
2
Hence,
PM ; an?!)
it
Pi] _ X Pan where the bottom equation foiiows from the first balance equation. (‘39::
ond balance equation as one of these equation ‘ To obtain PU, we use 7: m+m+2m=1 can ignore the see
aiways redundant.) 01' Ol‘ 01' and thus (a) PW: pull ea)
hluﬂ — 0:) where V1 + 4h/u — 1 2 Ct , It should be noted that for the above to be valid we need or a 1 or
equivalently ,Vu < 2 which is intuitive SlnCC the maximum service rate
is Eu which must be larger than the arrival rate A to avoid overloading
the system. All the relevant quantities of interest now can be determined. For
instance, to determine the proportion of customers that are served alone,
We first note that the rate at which customers are served alone is
AP“. + p.13}, since when the system is empty 3 cusromer will be served
aloneupon the next arrival and when there is one customer in queue
he will be served alone upon a departure. the rare ativ_l1_icli_cu§torners
are served..is.)r,_it_follows—that——« l_ a '" . ... ... .._, ... .. _,.....\..,_.‘_,.,_' ‘ ‘ Al’ni + “.101 ll
h i = p”. + E P. l Proportion of customers that are served alone = dw,Lam twa 5MVM9 0le m Capm’fg "5 4 , M/M/2/4/c/ / 47L 211 A A:( /)m‘:\
MGM W ’U‘AL/ME‘"
r  1%! a“ an
104A = "114/4 3} 7C: :10
3
3A z,+ ﬁx; : 4km+ 2/4 7,; Z} 7L1: '3‘ Lo
3?‘?;'+'3(L‘7{.23 %/\R:+D}AZ§ 3? X5: 73% 2'9
{X +2/JJ71/3 5 1X12+ 2px;, :) ML: 3H3 K”
1/515; : A W}
4
Z 7% 7' ’L
1150
=:> 7w  0405! 7c: : 0.403;
Zr}. :— O‘(§f2 Zr; 7; 0‘0578 lq:0.00¢72
(a) [F Zvr 3,, .—= 04733
4  ..
(b) )9: 2 no)“ : o 83/78
F0
* 4 A h 516*!
“ft “5550 J71 * ~‘ ‘7
w a 7' 0 2’6 60 ﬂuffy CL) : f. 113 + lzq 7.04347}?
(6!) 2.14, + {79. : {.2061};
2'0: (a) 10 4 3
H) mm: In : 0DV03875’ (a) ‘We approximate 1E and MS as independent M/ M / 1 queues.
0 1E
: '2 letters/day
,u. : 25 lettersfday
w = mg + 1/11, : 4/25 +1/25 21/5 day
0 MS
A : 15 letters/clay
u; = m” + 1/11 2 3/50 + 1/25 :1/10day
We approximate 1E + MS as an M/M/Q queue.
A = 20 + 15 = 35 letters/day
w = new + 1/11. : 419/1275 +1/25 2 4/513 0.08 day
B y forming a. typing pool= performance improves for both departments. I 0 ...
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 Fall '07
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