Side_1_360

From the last expression we also find the

integral

where

If we assume that all the lower priority classes

have negative exponentially distributed service

times (with mean service times bk= μp-1for pri-

ority class p) we find by applying the formula

above:

If some of the lower classes have constant ser-

vice times we only have to replace the term

F(t/b 1 ; μkb 1 , ρ 1 ) with the corresponding term

for those classes.

The result when fragmentation is performed is

found to be:

4.3.3 Convolution of the Waiting

Time in a M/D/1 Queue with an

Exponentially Distributed Time

We shall use the expression

for the normalised waiting time for the M/D/1

queue to express the convolution

(in terms of a sum of terms q(t– k; ρ) that are

weighted with a geometrical factor as given

above). To show this expansion we write

F(t; μ; ρ) = μe–μtG(t;μ;ρ) where

By applying the Lemma 1 (below) we get:

.

Integrating (and collecting) we get:

The corresponding result for F(t; μ, ρ) is then:

Introducing the new summing variable j= k– i

in the first (double) sum we have

and the corresponding

expression may be written as:

( 1 −ρ)

k= 0

⎣⎦t

∑

i= 0

k

∑

ρ

μ+ρ

⎛

⎝⎜

⎞

⎠⎟

k((ρ+μ)(k−t))i

i!

e−ρ()k−t

=( 1 −ρ)

j= 0

⎣⎦t

∑

i= 0

⎣⎦t−j

∑

ρ

μ+ρ

⎛

⎝⎜

⎞

⎠⎟

i+j

((ρ+μ)(i−(t−j)))

i

i!

e−ρ()i−()t−j

=( 1 −ρ)

j= 0

⎣⎦t

∑

ρ

μ+ρ

⎛

⎝⎜

⎞

⎠⎟

j

i= 0

⎣⎦t−j

∑

(ρ(i−(t−j)))

i

i!

e−ρ()i−()t−j

=

j= 0

⎣⎦t

∑

ρ

μ+ρ

⎛

⎝⎜

⎞

⎠⎟

j

qt(−j;ρ)

k= 0

⎣⎦t

∑

i= 0

k

∑ =

j= 0

⎣⎦t

∑

i= 0

⎣⎦t−j

∑

Ft(;μ,ρ)=

μ( 1 −ρ)

μ+ρ

ρ

μ+ρ

⎛

⎝⎜

⎞

⎠⎟

k

k= 0

⎣⎦t

∑

⎛

⎝

⎜⎜

((ρ+μ)(k−t))

i

i= 0 i!

k

∑ e

−ρ()k−t

−

ρ

μ+ρ

⎛

⎝⎜

⎞

⎠⎟

k

eμ()k−t

k= 0

⎣⎦t

∑

⎞

⎠

⎟⎟.

G(t;μ,ρ)=−^1 −ρ

μ+ρ

ρ

μ+ρ

⎛

⎝⎜

⎞

⎠⎟

k

k= 0

⎣⎦t

∑

⎛

⎝

⎜⎜ eμk

− ρ

μ+ρ

⎛

⎝⎜

⎞

⎠⎟

k

eμk

k= 0

⎣⎦t

∑

((ρ+μ)(k−t))

i

i!

e−()ρ+μ()k−t

i= 0

k

∑

⎞

⎠

⎟

⎟

G(t;μ,φ)=

(1−ρ) eμk

x=k

t

∫

k= 0

⎣⎦t

∑

[]ρ(k−x)k

k!

e−()ρ+μ()k−xdx=

−

1 −ρ

μ+ρ

ρ

μ+ρ

⎛

⎝⎜

⎞

⎠⎟

k

k= 0

⎣⎦t

∑ eμk

ζ= 0

()μ+ρ()k−t

∫

ζk

k!

e−ζdζ

G(t;μ,φ)= eμx

x= 0

t

∫ q(x;ρ)dx=

(1−ρ) eμk

k= 0

⎣⎦x

∑

x= 0

t

∫

[]ρ(k−x)

k

k!

e−()ρ+μ()k−xdx.

∫t

x=0

WM/D/ 1 (x)dx=b 1 G(t/b 1 ;ρ)

G(t;ρ)= lim

μ→ 0

F(t;μ, ρ)

=

1

ρ

∑
t

k=0

(q(t−k;ρ)−(1−ρ)).

W 1 c(t)=1−

1 −ρ

1 −ρ 1

q(t/b 1 ;ρ 1 )

−

1

1 −ρ 1

(P

∑

k=2

ρkF(t/b 1 ;μkb 1 ,ρ 1 )

)

b 1

bk

(

G(t/b 1 ;ρ 1 )−H(t−bk)G

(

t−bk

b 1

;ρ 1

))

W 1 c,f(t)=1−
1 −ρ
1 −ρ 1 q(t/b^1 ;ρ^1 )−
ρ− 1
1 −ρ 1

b 1

bf

(

G(t/b 1 ;ρ 1 )−H(t−bf)G

(

t−bf

b 1

;ρ 1

))

q(x;ρ)=(1−ρ)

∑
x

k=0

[ρ(k−x)]k

k!

e−ρ(k−x)

F(t;μ;ρ)=μ

∫t

x=0

e−μ(t−x)q(x;ρ)dx