θ+p-1(s) = Bg+,p-1(s+ λ+p-1– λ+p-1θ+p-1(s).Combining the two last equations yields the
important relation:s= σp-1(s) – λ+p-1(1 – Bg+,p-1(σp-1(s))).Compared with the model without fragmentation
we see that the two results are similar in the way
the remaining service time of the lower priority
packets is introduced in the expression, however,
for the model with fragmentation the influence
from the lower priority traffic is limited to the
remaining service time of a single fragment.A second observation we may mention is that
when the fragments are small the difference in
the service times of a packet and the correspond-
ing service times introduced by fragmentation
may be small. This can be seen from the LSTs of
the two variants. If we let ti= ibfwe may writewith dBp(ti) = Bp(ti) – Bp(ti–1).This is a well known approximation of theintegral so when bfissufficiently small we will haveA major difference between the two models is
that the service of a low priority packet may be
interrupted after the completion of a fragment. It
will therefore also be of interest to find the wait-
ing time for the last fragment of a packet of pri-
ority class p. We denote this waiting time Wl,p.
We have Wl,p= Wf,p+ Dpwhere Dpconsists of
the service times of all the fragments from the
‘tagged’ packet of class pplus the delay cycles
generated from packets of the 1, 2, ..., p– 1
higher priority classes. The probability that a
‘tagged’ packet will consist of exactly ifragmentsis given by so the probability that the lastfragment of a packet has exactly i fragments prior
in the queue hip(from that particular packet is:and thecorresponding generating function is:The LST of service time distribution of all these
fragments is then given asAs explained in [Takagi 1991] the corresponding
LST of Dpis obtained from Bpf*(s) by the rela-
tion:Finally we get the LST for the waiting time of
the last fragment in a packet from priority class
pas:.4.3 Waiting Time Distribution for the
Highest Priority Traffic
This corresponds to the case p= 1 and in this
case we have σ 0 (s) = s, which gives the LST of
the waiting time as:whereWM/G/1(s) is the LST of the waiting time distri-
bution in an M/G/1 queue with input rate λ 1 and
LST of the service time given as B 1 (s):.
If we let w 1 (t) denote the density function for the
waiting time we get by inverting the equation
above:where wM/G/1(t) is the density functions for the
M/G/1 queuing model and ~b 1 – (t) the DF of the
remaining service time for an arbitrary low pri-
ority packet.In fact the latter formula may be explained as
follows: In the long run an arriving high priority
packet will find the server either idle or serving ahigh priority packet with probabilityand will ‘see’ the system as an M/G/1 queue, or
will find the server occupied with a low priority1 − ρ^1−
1 −ρ 1w 1 (t)= 1 −
ρ 1 −
1 −ρ 1⎛
⎝⎜
⎞
⎠⎟
wM/G/1(t)+
ρ 1 −
1 −ρ 1b ̃ 1 −(t)*wM/G/1(t)WM/G/1(s)=
s( 1 −ρ 1 )
s−λ 1 +λ 1 B 1 (s)W 1 *(s)=WM/G/1(s)1− ρ^1−
1 −ρ 1+ ρ^1−
1 −ρ 1⎛ B ̃ 1 −(s)
⎝⎜⎞
⎠⎟
Wl*,p(s)=Wf*,p(s)D*p(s)B∗g,p(s)=Gp(e−sbf)=
∑∞
i=1e−stidBp(ti)Bp∗(s)=∫∞
t=0e−stdBp(t)Bg,p∗(s)≈Bp∗(s).1
gp
igiphpi=1
gp
(i+1)gip+1 i=0, 1 ,...Hp(z)=1
gp∑∞
i=0(i+1)gpi+1zi=1
gp
G′p(z).Bf
∗
p(s)=1
gpG′p(Bf(s)) =1
gpG′p(e−sbf).D∗p(s)=Bf
∗
p (σp−^1 (s)) =1
gp
G′p(
e−σp−^1 (s)bf