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Unformatted text preview: 2.15
a} Let D‘HL, divided by gm), have the quotient ditto) and remainder swat) so that Di+L = g(D)zﬁl{D) + dﬂtD} H
Mulﬁrtlrms by 5i and sums over i. stﬂlDL = E; 512mm} + It sthD) Since 2. sicﬁlm) has degree less than L, this must be the remainder (and Bi Stame) the
quotient) on dividing 5(D)DL by ng). Thus (:03) = E sg_c{i}(D) b] Two polynomials are equal if and only if theireoefﬁeients are equal, so the above
poin actuality nnplies Cj = 31:53“) 1.16 a} Consider the two scenarios below and note that mese scenarios are indistinguishable to
the reeeiver. If the reoeiver releases the packet as a; in the queenoned recqation, then on error occurs on
scenario 2. If the receiver returns an ask but doesn't release apacket (Le. the appropriate
action for scenario 2). men under scenario 1., the transmitter erroneously goes on to packet.
3. anllglr1 ifthe receiver returns a trait, the problem is only poslpmed since the transmitter
would then transmit (2.52) in scenario 1 and {2,Jt1) in scenario 2. As explained on page 66.
packets at] and it: might be idtZIIL'lulJ bit sittings, so the receiver can not resolve its ambing
by are hit values of the packets. b) The soenarios below dem0nsttate incorrect operation for the modiﬁed oonditions. packet l I)
accepted ’ 2 .1 1' a) T = T1+ Tr + Hg ‘3) q =(1PJUPr‘} A packet is transmitted once with probability.r q, mice with probability {1—q}q. thrcc times twith probability (Iq}2q, etc. Thus the expected numbarof transmissions of a given yacht
15 .. i r 1
E{ transmissions perpackct} = 2 ,q{1_q}t—t = 3 i=1 To verify the above Summation. not: that far any x, {I E Jr. : l1  d—x'
' ﬁx
i=1 :i=l Using It for (lq} abutc gives the: desired result. ll
Efln
ME "—I 
E: a.
r—w.
_ I an
a
H._.a’

Ft.
H
PL .—
WM :2) Efﬁmc pct packet} = (T. +Tf+2Tdeq
= (1.3]f0.993 =1.303 a Note that p; and Pf have very little cﬁ'w on Eitjme pcrpacket] in snap and wait systems
unless may an: unusually large. 2.20 The simplest example is for node A to send packets U timing}: 111 in the ﬁrst :1 frames. In
Gate of delayed aelotowledgements (in. no return packets in the interim), node A goes back
and retransrnits packet CI. If the other node has received all the packets. it is waiting for
packet n, and if the modulus m equals 11, this repeat of packet 0 is interpreted as packet 11. The right hand side of Eq. (2.24) is satisﬁed with equality if SN = SNm(t1)+nl. This
occurs if node A sends packets I] through ml in dte ﬁrst It Ernmes withnoretum packets
from node B. The last such frame has SN = nl. whereas Slim at that time (say [1) 13 0. Continuing dds scenario, we find an. example where the right hand side of Eq. (1.25) is
satisﬁed with equality. If all the frames above are t‘:t:trtet:tl3,r received, then after the last
frame, RN becomes equal to It lfannther frame is sentiment A {now call this time 11} and 
if Shimm is still D, then when it is received at B (say at 12), we have Rth) = SNmm(t1)+n. 2.21. Let KNEE) be the value of RN at node B at an arbitrary time 't; Slim“ is the smallest packet
number not yet acknowledged at A at time t {which is regarded as ﬁtted) and Shim _1 is the largest packet number sentfmm A at time t Since RNE't} is non decreasing in t, it is
Sufﬁcient to Show that RN{I+Tm+Td_) S Slim and to show that RNttTm—Td) 2 SNmn. For the first inequality. note that the packet numbered SNM (by deﬁnitimt of SNmﬂ but not entered the DLC unit at node A by time t, and thus can not have started transmission by.
time I. Since then: is a delay of at least Tm+Td from the time a. packet transmission starts
until the completion of its reception. packet S'N'm;IL can not have been received by time
I+Tm+Td. Because of the correctness of the protocol. RNIII+Tm+Td} can be no greater that tlte number of a packet not yet received. i.e. Shim. for the second inequality, note that for the transmitter to have a given value of SNmin. at cm: L that value must have been msmitted earlier as the request number in ii. ﬂame
coming back from node E. The latest time that such a frame could have been formed is t Tm~Td, so by this time RN mus: have been at least SNmm. 2.23 After a given packet is tranSmitted ftbm node A, the second subsequent fratth mansmission
tetmhtaﬁen from B entries the acknowledgement (recall that the ﬂame transmission in
progress from B when A ﬁnishes its maritime]: cannot carry the act: for that
transmission; recall also that propagation and processing delays are negligible. Thus q is
the probability of 11] frame terminations from A before the second. frame termination from
B. This can be rephrased as theprobnbilitythat outofthe nextn ﬂame terminations from
either node, either nl or II come from node: A. Since successive frame terminations are
equally liker and independently from A or B, this pmbability is II 1 an
q: 2 MIL)! 2‘” = (n+1}2 i=lle‘l 2.24 If an isolated error in the feedback direction more, then the so]: for a given packet is held
up by one frame in the feedback diJ'ElCliOI'I (i.e., the number RN in the feedback frame
following the feedback frame in error reoelenowledges the old packet as well as any new
packet that might have been received in the interim). Thus q is now the proleab‘ilitj.r of nl.
ﬁ‘ame terminations from A before 3 frame tet'nﬁnaﬁons fromB (one for the frame in
progress, one for the frame in error. and one for the frame acnnlly carrying the eelc see the
solution to problem 2.23). This is the probability that rtl or more of the next n+1 frame
terminations come born A; since each is from A or B independently and with
equal probability. mm”)! 2"H = [n+2+(n+i]nJ2]2'“'1 i!(l1+ii)i 1.25 As in the solution to problem 2.23, q is the probability of n—1 frame terminations coming
from node A before two frame terminations come from node B. Frame terminations from
A {and similarly from B) can be regarded as alternate points in a Poisson point process
from A {or from B}. There are two cases to Consider. 1n the first. the initial frame is
received from A after an even numbered point in the Poisson process at B. and in the
second. the initial frame is received after an odd numbered point. In the ﬁrsr case. :1 is the
probability that Err2 Poisson events from A occur before 4 Poisson events occur from B.
This is the probability, in a combined Poisson point process of Poisson events from A and
B. that 2112 or more Poisson events some from A out of the next 2n+1 events in the
combined process. In the second case. (1 is the probability mat ZnZ Poisson evean ﬂoor A
occur before 3 events occur from B. Since these cases are equally likely, 2n+l
s=12L 2 i=2n2 2n
(2n+1}! ) 2nt 1
7—:— 2 + _ E
11(2n+1lll Zeno on)! 2"
i!{2rli)i l 2 2.26 We view the sysrcm from the receiver and as}: for the erpecmd number of frames, 1',
arriving at the receiver starting immediately after a frame containing a packet that is accepIEd and running until the next frame containing a packet that is accepted. By the assumptio
of the problem, if the packet iii a frame is accepted, then the next frame must contain the
next packet in order (if not, the nansmitter must have gone hack to some eariier packet,
which is in'tpossibie since that earlier packet was accepted earlier and by assumption was
asked in time to avoid the go back). Since the next frame aftera packer acceptance must contain the awaited packet, that p
is accepted with probability 1p. With probability p. on the other hand, that next frame
contains an error. In this case, some number of frames, sat],Ir j. follow this next frame before the awaited packet is again contained in a ﬂame. This new frame might again
contain an error, but the expected number cffmmes until the awaited packet is accep starting with this new frame, is again 1'. Thus, given an error in die frame after a packet
acceptance, and given j further frames before the awaited packet is repeated, the e number of frames ﬁ'otn one acceptance to the next is l+j+"f. Note thatj is the number offrames that the transmitter sends, after the above frame in
up to and including the frame in transmission when feedback arrives concerning the in error. Thus the expected value of j is 13. Combining the events of error and no enor
the next frame after a packet acceptance, we have 'Y= (111) + P(1+3'FT}=1+ NWT}
Solving for Yand for v = 1H. 't' = {HISPMIp} v = [l—PHUeﬂp} 2.27“ Note that the sending DLC could save only one packet if it waited for animowledgenre rather than continuing to transmit. Similarly the sending DLC could save an arbitrarily
large number of packets by taking packets from the network layer at a rate faster than can be transmitted. Thus what is desired is to Show that at most [3+1 packets need be stored without ever forcing the transmitter to wait Thus we assume that a new packet '
admitted from the network layer only when there are no previously transmitted packets
are known to have been unsuccessful on the last transmission (Le. the system repeats
naked packers before accepting and msmining new packets; the system. accepts and
transmits new packets while waiting for feedback information on old packets). When the system is ﬁrst initiated, one packet is admitted to the sending DLC from the
network layer. We use this as the basis of an inducrive argument on successive times ' which a new frame is generated. By the inductive hypothesis, at most [3+] packets
stored at the end of the previous frame generation instant. At the time of generating new frame, there are at most I] outstanding frames [including the one just being com for which feedback has :10! been received. A new packet be accepted from the
layer only if all packets stored are also in frames for which no feedback has yet been
received. Thus if a new packet is anceptcn. the total number sawed is increased to at ﬁ+L and if no new packet is accepted. the totai number saved remants (by the indu ‘
hypothesis) no more than [311. ...
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 Spring '09
 ChoiSunghyun
 Probability theory, Poisson process, Point process, Bi Stame

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