asymptotic solution of palm's integral equation

Asymptotic Solution of Palm's Integral EquationAuthor(s): Olavi HellmanSource: Operations Research, Vol. 11, No. 4 (Jul. - Aug., 1963), pp. 553-560Published by: INFORMSStable URL: http://www.jstor.org/stable/168001 .

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ASYMPTOTIC SOLUTION OF PALM'S

INTEGRAL EQUATION

Olavi Heilman

University of Turku, Turku, Finland

(Received November, 1962)

The asymptotic solution of Palm's nonlinear integral equation is obtained, valid for channels sufficiently far away from the first channel. It is shown how the asymptotic formula may be applied, in certain cases, for the deter- mination of the number of channels sufficient to serve a given Poisson input.

3ALM l" has investigated the following queuing problem. The infinite system Li, L2, ..., Ln ... of parallel channels is receiving traffic

by a Poisson input. Each call enters at Li, proceeding, if L1 was occu- pied, along the line of channels until it arrives at an unoccupied chamnel. By assuming an exponential service time Palm derives an expression for (Pr(t), the probability that every call entering Lr during time (to, to+t) will be served, given that at moment to a call was forwarded to Lr?i. The application of Palm's expression of sor(t) implies, however, the solution of an algebraic equation of the r+1 degree, a complication that makes the use of the general formula of qPr(t) very difficult. In what follows we shall derive an asymptotic expression for sOr(t), valid for large r.

Every channel Lr will receive a call, with any probability 1-,B, 0 < < 1, given in advance, after a sufficiently long time since the foregoing arrival at Lr. It is now possible to associate the time tr with the channel Lr such that 1-fl will be the probability of no call entering Lr+i in a period (to, to+t()), given that at moment to a call entered Lr+l. By using the asymptotic formula of pOr(t) it will be shown that, in the range of validity of the formula, the t(? form a monotonically increasing sequence in r. By using this sequence one may, in cases where the asymptotic formula may be used, decide how many channels to build in order to exclude those channels that would practically never be entered by a call: The designer of the system of channels is ready to declare a time T such that a channel Lk and the subsequent channels will be superfluous if the probability of no call entering Li in a period (to, to+ T), given that a call entered Lk at moment to, will be at least 1-fl, where d is a small number also given by the designer (e.g., d-0.05). If now

t:) < T< t(g)

553



554 Olavi Hellman

where r1 is such that the asymptotic formula is applicable, no channels should be built after Lr?+l.

PALM'S INTEGRAL EQUATIONS AND THEIR SOLUTION

BY ASSUMING an exponential service time with unit expectation value,* Palm['] obtains the following system of integral equations for the sr(t) (cf. also reference 2 where Palm's calculations are reproduced):

snr()-S?l(t-I ( e X9or(-x)dipr- (x) (r1) (1)

Palm solves equation (1) by using the Laplace transformation:

(pr(t) = C Crkearkt (2)

where Crk = Br-1( 1 -ark) /Br' (- ark).

Br (t) = xr+l + f, (+)t(t+ 1) ...( t+r-1) Xl (3)

Br'(t) =dBrMl) dt,

and where the ark, k =0, 1, , r, are the roots of the equation

Br(t) =0-

The numbers ark have the following properties: Each ark is real and positive, besides, ari>ar, -Fl1 for 1 <i<r and -aro< -ar+o0<0O. Further

Br-1 ( 1-ark) /BrI(-ark) _ 0

for all r and k. Since (Pr() 1 for all r, it follows that

Ek=O Grk=. (4)

Consequently 0 Cr k _ 1 (5)

for 0?k?r.

CERTAIN PROPERTIES OF THE QUANTITIES acro AND OF THE FUNCTION Br(t).

LEMMA 1. Br+](-ar0) <-ar0(r+1)! for r?1. Proof. Since ari>ar,i1+l, i=1, 2, *, r, we obtain ari-aro>i.

Further, since Br+i(t)-tBr(t+1)+XBr(t), r>1, it follows that

Br+l(-aro) =aro (arl+1-aro) ... (arr+1 aro) < -aro (r+1) !

LEMMA 2. ar+l,o<X ?2/(r+1)!. Proof. Now Br+l(0)=Xr?2 -ar0o<-ar?l,o<0, and, according to

* One obtains the formulas, valid for the exponential channel with parameter , from the formulas derived below by substituting X/g for X.



Palm's Integral Equation 555

Lemma 1, Br+,(-aro) <-aro(r?+) !. If -y denotes the point of inter- section of the t-axis and of the straight line through the points (0, Xr+2) and (-aro,-aro(r+1) 1), it follows, because the arl in

Br+l(t) (t+ar+?,O) (t+ar+1,1) ... (t+ar+i,r+l)

are all positive, that -aro < - y < - ar+1,o for all r. Now

y= aro/{ [aro(r+ 1) !/Xr+2]+ 11 <Xr+2/ (r+ 1) !.

Hence ar+1,0 < Xr-2/( r+1)!

LEMMA 3. Br( -r0) =0{ XexX /(r-1) !},

where

aro=e I(rXl/r!)e/{1+Xe \k- (XM/k!)[1/(r - k)]+e XZ?r??i (Xt/l!)}.

Proof. The function - &4o is the solution of the linear part of Br(t) = 0. We expand Br(t) by Maclaurin series as follows:

,Xr l =rr1

Br(t) r? +tE Z I )(r-1)!X 1=0

+ t2 (+ ()Olt+ 1) (Olt+2) * (Oit+ r- 1)

[l/(lt+ 1) + I/(lt+2) +* + /(Oit+r-l)1X1,

where 0 < 01 < 1, and substitute - UrO for t:

= -xX2r+2 r 1 (r+1)! Br( - r0) = 2( !)2-1

5 ( (+1_Ij (- Oldro)

.. (r -I-ldrO) N2(r!)/ 1=0 (r+

OlarOo r - r--1 r,

where we wrote mnore briefly r-1 Ak 00 oo

N=1+Xe 1 - +e k=0 k;! -r-k k =r+1 -r

Now there exists r0 such that &ao <1 for r> ro. Therefore, for a sufficiently large r,

(r+1)( 0) . . . lrO) ( + +

1=01 (r+-l) (i lir0+ar

(r+ (r+1)1 ( 1)! r-1

l ro 1= l1! r+1-l- * By y (r) O[x(r) ] we shall indicate that Olimroo [y(r)/x(r) I < 1.



556 Olavi Hellman

It follows that all terms in the above expression of Br(-&ro) will have an order of magnitude less than or equal to 2e [X2r/(r-1)!]. REMARK. N-*1 as r-- oo .

Proof.

x 1 1 r-I x llr 1 rl

-571- z- i = rl 1 1-(r-r )/r)or r i=o (r1!r r .o1

r

and 1?i- (Xl/l!) =O(X"+l/(r+1)!).

THE APPROXIMATE VALUES OF arc AND OF THE Cri

THEOREM 1. Let r be such that Xr+l/(r-2) !?1. Then aro= ro-e, where aro 0[ e_ (Xr+l/r !) ] and E= O[e-x(X2r?2/r !(r-1)!].

Proof. That dro = O[eX (Xr?l/r!)] is clear from Lemma 3. Further- more, it follows from Lemma 2 that EI <Xr?l/r!. Indeed,

I| |dro I- 11 _< I - d-rO +E I < Xr+'/r!.

Now Br( -aro+ E) = Br(dro) + EBr( - ro+ 026) = 0,

where O < 02 < 1, which yields

e =-Br( dr)/Bro ( drO+6026).

The substitution of the Maclaurin expansion

(l+t) ... (r-l+t) -(r-l) !

+t(1+03t) . .. (r-1+03t) [l/(1+03t) + *

where 0<03<1, into

Brt() = r+ 1 lElr (X'/l!) [(I lt) ... (r-l+t) ]/(r+1-1) !

* = l+t[l/(l+t1+ 1*/(r-l+t) ]}

leads to the expression

BAlt) = (r +l) ![El=-r (Xl/l!) 1/(r-1+1) +fi(t) +f2(t)], (6) where

f' (t) = tE_ 1(0 W/1! J[(1+ 03t) ... (r -1+03t)]/(r+1-1) !

*1/(1 +03t) + * + 1/(r-1+03t) ]

+1/(r+1-1)[1/(1+t)+ * * * (r -.+t)]




and

f2(t) = t2Zr (Xr/1..[(1+03t) . (r-1+03t)]/(r+ 1-1) ![1/(1+t)

+ . *** +1/(r-3+t)][1/(1+3t)+ ** +l/(r-t+6t)1.

As is easily seen,

El = 1/ ! 1 (r- 1+ 1) _ 1/ (r+ 1) 1tr z- // !) f2 (rO+ 02E) ?0,

and, because Xr+l/(r-2) <1,

fl(-&rO+02E) ?(2&rO- 026 )/(1- IrO-0261)ZE.r (/i) .

It follows then that

Br'(- rO+02E) >r![1-2(arO-02E) (r+1)/(1 --o+02e)]Et- (1/i!).

Consequently

1El I IBr(-aro) l/{r![1-2(aro-021E)(r+1)/(1-aro+02E)]Z=o (1l/,1}

Since 2(4o- 02E) (?+1)/(1 -arO+02E) =0[2e-X/(r-1)]

for r such that Xr?l/(r-2) !I 1 and since, by Lemma 3,

1Br( - aro) j = 0{X2e [X2r/ (r- 1) !]} it is clear that

JE =Ote-x[X2r+2/r!(r-1) !]}.

THEOREM 2. Let Xr+l/(r -2) !<1. -Then

CrO 1- Ofmax[X( I+X) ,2ex]/ (r+1)}

and max,?<zr Cr=-O{max[X(I+X) ,2e-X]/(r+1) } for 1I1, 2, . r. Proof. We shall first estimate Cr0=Bri1(1-aro)/Br'(-aro). Ac-

cording to the mean value theorem we have

Br-1 ( 1 - arO) = Xr+ E r(I) (r -I) !.+ (arO) ( 1-.04arO) .. *r (e- -04arO)

*[1/(1-04arO)+ +1 /(r--04aro)]}X,

where 0 < 04 < 1. We now easily obtain that

Br_1(laro) =!{ =r (Xl/l!) +O[e-IlXr+l/(r_ 1) !]I)

where the inequality Xr?l/(r- 2) ! 1 was used. According to formula (6)

B' (-arO) = (r+ 1) ![I 1 zr ( X1/1!) l /(r-1+1 ) ASf (-arn) +ft2 (-ar).



558 Olavi Hellman

It would be easy to show that, for Xr+/(r- 2) ! 1,

Ifi(-arO) I = 0[2X ( Xr/r !)]

and I f2(-arO) = Of X [X2r/r! (r1) !]} .

Hence, for Xr+l/(r -2) !<1,

C _Br_ 1-arO)_ 1 {Z r (Xl/l!)+O[e Xr+l/(r-1) !]}

rOBr'(- arO) (r +l-) {l_ (Xl/l!)l/(r+1-1)+0[2 (Xr+'/r!)]

1 Eji (XZ?l/l!)1/(r-1)+0{2[Xr+l/(r-l)!]}

(r+l) 1l-r (X1/1!)1/ (r+ I- 1) +0[2 (Xr/r!)

In connection with Lemma 3 it was shown that

and, evidently,

Et-o 1=0 l!)1 (r + 1-)> 1/ (r + 1)=0 X/l!

=e/I(r + ) +O e"[Xr+l/ (r+1) (r+1) !]} . Hence

rl-1= (Xit+1/1!)l/(r-l)+Of2 Arl(r-1) !]}I

1=ro ( /1!)l1/(r+l1-1) +O[2 (Xr+ /r!)]

<X(1+X) +0{2eN[Xr+l/(r-2)!]}

+0 {2e [Xr+l/(r- 1)!]}

which, for Xr+l/ (r -2) !< 1, permits us to express Cro as follows:

CrO= 1-O{max[X(1+X),2e ]/(r+l)}. (7)

Since Z1_o Crl=I

now implies, for Xr+l/(r-2) !; 1, that

EZi= Crl=O{max[X(1+X),2ei]/(r+1)},

and because all the Cr1 are positive, it is obvious that

maxlj<1zr Crl=O{max[X(1+X),2ei]/(r+1)}.

THEOREM 3. Let Xr+l/(r-2) !<1 and let O<t<t., where drOtl=O(1). Then

?r( t) - earot +O(3max{max[e,X(1+X) ,2e ] max[X(1+X),2e-)

Proof. We first rewrite the expression of vr (t) as follows:

(Pr(t) = e-arO[CrO+MI J(r,t) ],

where M (r,t) eaTot = L e I CrieaTIt




Since ar,>aro+l (cf. the proof of Lemma 1) and since Cr IO for 1<l<r it follows that

Mk(r,t) _e 1~ t?t rl Cle-(a0

< max On Zl-D et 1<!! I _ r

< max CrI/ (et-1). l1 < I < r

Now, according to Theorem 2,

max Crl= O{max[X(1 +X),2eX]/(r+ 1)}.

Furthermore, the inequality Xr+1/(r-2) !<1, Lemma 3, and the fact that =rd 0(1) together imply that tl = O( eXr2). Consequently

MVI(r,t) =O{max[X(1+X),2e-"]/(r+l) (eo(eXr2)_1)} (8)

Since Xr+l/ (r -2) ! < 1, it follows from Theorem 1 and from the relation arotl=0(1) that

arotl= rOti+O[1/(r- 1)].

We obtain now from (7) and (8) for Xr?l/ (r-2) !<1 that

~P(ti )-e-ar0t1+O[h(r t)], (9) with

h (r,t) = 3m {feot-' Imax[X( 1 +X) ,2e ] max[X( 1+X) ,2ef]} hrt)= 3ra> r-1 ' (r+ 1)(e - 1) ' r+1

where we wrote t instead of O(eXr2). Since r> 2,

= Fmaxfmax[e,X(1 +X) ,2ei] max[X( 1 +X),2e-x]l O[h(r,t)-OL3maxX r-1 (r+l)(et- 1) J

It is obvious that (9) holds also for O<t<t<.

AN APPLICATION OF THE ASYMPTOTIC FORMULA

THE RESULTS of the previous section hold for r ? ro where ro is an integer defined through the inequality

Xro+l/(r-2) !<1.

(Clearly, ro = 2 for X< 1. For X> 3, as is easily verified by using Stirling's formula, ro-[R]*, where

R = 2+2 [Xe+V/(Xe)2+4eln(X'3/V\2ir)].

* Here [R] denotes the integer nearest to R.



560 Olavi Hellman

Now let r1 _ r0 and let rot= ir be such that the use of the formula

- or(t) =erOt

(10)

is justified (cf. Theorem 3) and let the sequence {tri+k} be defined as follows:

tr +k = T/ ( arl +k,O) , ( 1 1 )

where k = 0, 1, 2, c. Then e T will be the probability that no call will enter Lrl+k+i in a period (to,to+trl+k), given that a call entered Lr1+k+l at moment to. According to (11) we have

t(O) .+,=| (rl+k+ 1) /X]t(o)* ( 12)

It is easy to show that (ri+k+1)/X>I for all k=0, 1, 2, * and for all X> 0. Consequently, the trl+ky k=0, 1, 2, .., form, for a fixed ,B, a monotonically increasing sequence in r.

By using the sequence { tr$+k} one may decide, if the requirements associated with the formula (10) are satisfied, how many channels to build in order to exclude those channels that would practically never be entered by a call: One chooses a small ,B (e.g., ,B=0.05) and then a time T such that the absence of the channels Lk, Lk?l, - -will not affect significantly the function of the system: the probability of no call entering Lk in a period (to,to+ T), given that a call entered Lk at moment to, will be at least 1-, B. One now first solves -r from the equation

1 - = e-T

and then calculates the sequence t) k= -0, 1, 2, *, from the equation

trl+k=e lnI/(I-B) (r1+k) !2(lk If now

t(O) <T<t(o) ~(k1?>0) trl+kl = - rl+kl?k +(>

no channel will be built after Lr, +k, +1.

REFERENCES

1. C. PALM, Ericsson Technics 1, 1-189 (1943). 2. A. Y. KHINTCIIINE, Mathematical Methods in the Theory of Queueing, Griffin,

1962.



asymptotic solution of palm's integral equation

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