12.Sol.08.Extra.Spring

# 12.Sol.08.Extra.Spring - ,5 L"l it To begin we must...

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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 = i-LIPI] 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.,4l16nncal-CH-la-FB—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'l-ror 1 i: n "this med/El . we COJIGJ'LCJC’V {a Silk-ya , Jerri/(Ly f/meu‘ ﬁll—of fHQUO-m 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 l-LPo 0 (A -+- ulPU = APU- + up, + up: “I n 2‘ 1 + “Jinn = h‘Prl—I + l‘l‘IJn-rl 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) h-l-uﬂ — 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 alone-upon 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_follow-s—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 2-11 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 : 0-DV03875’ (a) ‘We approximate 1E and MS as independent M/ M / 1 queues. 0 1E : '2 letters/day ,u. : 25 let-tersfday 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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12.Sol.08.Extra.Spring - ,5 L"l it To begin we must...

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