and with LTS.
We denote Wpthe waiting time for a packet of
priority class pand we denote the corresponding
Distribution Functions (DF) by Wp(t) = P(Wp t)
and with LST
4.1 The Unsaturated Case
We consider the unsaturated case ρ< 1, and we
define the higher priority intensity and load
(from the ρhighest priority classes) by
andFurther we also define the service time distribu-
tion of an arbitrary packet in one of the higher
priority classes denoted Bp+for 1, ..., p. The cor-
responding LST is given as the weighted sum
with mean and ithmoment given as
for i= 2, 3, ...
Similarly it is also efficient to define the service
time distribution of an arbitrary packet in one of
the lower priority classes p+ 1, ..., P, which we
denote Bp–. The corresponding LST is given as
the weighted sum
where the rate and correspond-
ing load Further the mean and
ithmoment are given as
andfor i = 2, 3, ...We also define the remaining service time B~p–
(of the corresponding service time Bp–) and the
LST is given byBased on the definitions above and by using the
results found in [Takagi 1991] we may write the
LST of the waiting time Wpon the following
compact form:Wp*(s) = Wp+(σp-1(s)) whereand where WM/G/1(s) is the LST of the waiting
time distribution in an M/G/1 queue with inputrate and LST of the servicetime given asFurther the function σp–1(s) is defined throughthe LST of the busy period distribution, θ+p-1(s),
generated by packets of class 1, 2, ..., p– 1:where θ+p-1(s) is the unique solution of the equa-
tionCombining the two last equations yields the
important relation:≤
̃bp(t)=^1
bp
P(Bp>t)=
1
bp∫∞
τ=tbp(τ)dτB ̃∗p(s)=^1 −B∗p(s)
sbpWp∗(s)=∫∞
t=0e−stdWp(t).λ+p=
∑pk=1λk ρ+p=∑pk=1ρk.Bp+(s)=1
λ+p∑pk=1λkB∗k(s)b+p=^1
λ+p∑p
k=1λkbk=
ρ+p
λ+pandb+p(i)=^1
λ+p∑p
k=1λkb(ki)B−p(s)=
1
λ−p∑P
k=p+1λkBk∗(s)λ−p=∑P
k=p+1λkρ−p=∑P
k=p+1ρk.b−p=1
λ−p∑P
k=p+1λkbk=ρ−p
λ−pb−(i)
p =1
λ−p∑P
k=p+1λkb(ki)B ̃p−(s)=^1
ρ−p∑P
k=p+1ρkB ̃∗k(s).Wp+(s)=WM/G/ 1 (s)(
1 −
ρ−p
1 −ρ+p
+
ρ−p
1 −ρ+p
B ̃−p(s))λ+p(
=
∑pk=1λk)
B+p(s)=(
1
λ+p∑pk=1λkBk∗(s))
:
WM/G/ 1 (s)=s(
1 −ρ+p)
s−λ+p+λ+pB+p(s).
σp− 1 (s)=s+λ+p− 1 −λ+p− 1 θ+p− 1 (s)θ+p− 1 (s)=Bp+− 1(
s+λ+p− 1 −λ+p− 1 θ+p− 1 (s))
.
s=σp− 1 (s)−λ+p− 1 (1−B+p− 1 (σp− 1 (s))).