MADE BY
RAHUL SHAMI
Examples
People at a fast food center
Aircraft waiting to take off
Machinists waiting for a
tool crib
Cars waiting to take Petrol
Telephone systems
The tool crib problem
Waiting time cost of machinists
V/S
Idle time of tool crib servers
Design of Queues
Two people in series
v/s
Two people in parallel
(1)
(2)
λ
λ
2μ
2μ
μ
μ
To find a Distribution:
(1) Do stop watch study
(2) Draw a Histogram
(3) Hypothesize a Distribution
(4) Prove or Disprove hypothesize by
Chi-square Goodness of fit test.
POISSION PROCESS
∆t
Probability of an arrival in ∆t
→ Proportional to ∆t = λ ∆t
Probability of no arrival = 1- λ ∆t
No. of arrivals in time t ~ Poisson
P {N (t) =n}= e-λt(λ t)n/n!
Inter-arrival time ~ Exponential
-λ
t
f (t) = λ e
t≥0
Time till the kth arrival ~ Erlang k
fk(t) = λ (λ t)(k-1)e-λt / (k-1)!
SOME MEASURES OF
PERFORMANCE
Length of queue
Waiting time in the system
Idle time of the system
TOTAL COST = WAITING COST
+ SERVICE COST
CHARACTERISATION OF A
QUEUEING SYSTEM
A/B/C/X/Y
A=Inter-arrival time Distribution
It can be:
EXPONENTIAL
M
DETERMINISTIC
D
ERLANG (type k, k=1,2…)
Ek
GEN INDEP DISTR
GI
B = Service time distribution
It can be :
EXPONENTIAL
M
DETERMINISTIC
D
ERLANG (type k, k=1,2…)
Ek
GENERAL
G
C = no. of parallel servers
1,2,…..
X = restriction on system capacity
1,2…….
Y = queue discipline
FIFO
LIFO
SIRO
RRR
GD
QUEUEING SYSTEM
No. of
people
in the
system
Time
Pn(t) = probability of n people in the
system at time t
λ = arrival rate
μ = service rate
ρ = λ/μ = traffic intensity
Analytical Steady State Solution For
The M / M / 1 Model
Pn(t+∆t) = Pn (t) [ 1-λ ∆t] [1- μ ∆t] +
Pn (t)[ λ ∆t] [ μ ∆t] +
Pn+1(t)[ 1-λ ∆t][ μ∆t] +
Pn-1(t) [ λ ∆t] [1- μ ∆t]
Pn(t+∆t) = Pn (t) [ 1-λ ∆t - μ ∆t] +
Pn+1 (t) [ μ ∆t] +
Pn-1(t) [ λ ∆t]
(for n ≥1)
P0 (t+∆t) = P0(t) [ 1-λ ∆t ] +
P1 (t) [ μ ∆t]
Lt Pn(t+∆t) – Pn(t) / ∆t = Pn’(t)
t  0
= - (λ+ μ) Pn(t) + μ Pn+1(t)
+ λ Pn-1(t)
(for n ≥ 1)
& for n = 0
LHS = - λ P0(t) + μ P1(t)
IN STEADY STATE
Pn’(t) = 0
&
Pn(t) = Pn
Therefore:
0= - (λ+ μ) Pn + μ Pn+1 + λ Pn-1
0= - λ P0+ μP1
We get:
&
P1 = (λ/μ) P0 = ρ P0
Pn+1 = (ρ+1)Pn – ρPn-1
(n ≥1)
2
P2 = ρ P0
Assume for n,
Pn = ρn P0
& it can be shown that,
Pn+1 = ρn+1 P0
Hence it is proved by Induction that:
Also
Pn = ρn P0
for all

P
n 0
n
=1
Hence,
&
P0/1-ρ = 1
P0 = 1- ρ
n≥1
Hence,
Pn = (1- ρ)ρn
for n ≥1
GEOMETRIC DISTRIBUTION
(a) Expected no. of people in the system

L= E[N] =  nP = ρ/1-ρ
n 0
n
(b) Expected no. of people in the queue
Lq = E[Nq] = 0P0 +
= ρ2/1-ρ

 (n  1)P
n 1
n
(c) Expected idle time of the server
= 1/λ
WAITING TIME DISTRIBUTION
W(t) = P(T≤t)
∞
∑
n=0
P(n+1 service completions in time ≤ t │arrival found n people in system) Pn
 (x ) n e  x dx  n
=  
 (1  )
n!
n 0  0


t
t
=
=

(
1


)
e

0
t
 (1  )e
0
x
(x )
dx

n!
n
 (1 ) x
n
-μ(1-ρ)t
=
1-e
dx
EXPECTED WAITING TIME
W = 1/(μ-λ)
And VARIANCE [T] = 1/(μ-λ)2
For M/M/c queue
Pn = (λn/n! μn )P0
(1≤n≤c)
=
n
n-c
(λ /c c!
n
)
μ P0
(n≥c)
 c1 1   
1    c 
     

and
 n 0 n!    c!    c   
P0 =
n
c
1
QUEUEING THEORY
M/M/1
Solution of the steady state differential equation for {Pn} by
generalizing function.
Pn+1=(ρ+1)Pn –ρPn-1
P1=ρ P0
(n≥1)
n
Multiplying by Z & sum from n=1 to ∞
We get,
1 

z n 1
p z
n 1
n 1


  (   1) Pn z   Z Pn 1 z
n 1
n
n 1
n 1
or
1/z[P(z)-P1z-P0] = (ρ+1)[P(z)-P0]-ρz [P(z)]
As P1=ρ P0 we get,
or
1/z[P(z)-(ρz+1) P0] = (ρ+1)[P(z)- P0]- ρz[P(z)]
P(z)= P0 /1-zρ
P(1) =1= P0 /1-ρ
as P0 is prob. ρ<1
or
p0=1-ρ

 pn  1
n 0
P(z)=(1-ρ)/(1-ρz )= (1-ρ)[1+zρ+(zρ)2+………..]
Pn=(1-ρ) ρn

P ( z )   (1   ) 
n 0
n
z
n
L  EN 

np 
n
n 0


1 
Lq  0 p0   (n  1) pn
= L-(1-p0)
n 0
=L-ρ

L
q

1 
2

1 

,
L
q

 {Nq / Nq  0}   ( n  1) p   ( n  1) p
n 1
,
,
n
n
p’n=p {n in system/n≥2}
= p {n in system ∩ n≥2}/p{n≥2}
n
pn

 2
1  p 0  p1 
pn

p
n 2
n
pn
1
L q   (n  1) 2 
1 

'
L'q 
1
1 
(n≥2)
Example:
M/M/1
λ=5/hr
1/μ=10 min.
ρ=5/6
L= ρ/1-ρ=5
Lq=25/6
Lq’ =6
p0 = 1-ρ = 1/6 ≈ 16.7%
μ=6/hr
4 seats in waiting room
p {finding no seat}= p{N≥5} =ρ5 = 0.402
Tq= Time spent waiting in queue
Wq(0)= p{Tq≤0} = p{Tq=0} =p0 =1-ρ
Wq(t) for t>0

=[

n 1
p{n completion in ≤ t / arrival found n in
system }pn]+ Wq(0)
(  1) n 1 x
 (1  )  
e dx  (1  )
(n  1)!
n 1
0

t
n
t
 (1  ) e
x
0
(x) n 1
dx  (1  )

n 1 (n  1)!

t
 (1  ) e x (1) dx  (1  )
0
=1-ρe-μ(1-ρ)t
Therefore, Wq(t)=
1-ρ,
1-ρe-μ(1-ρ)t
t =0
t >0
with the probability distribution of Tq we can calculate
expected waiting time

Wq  E[Tq ] tdw q ( t )
0



 0(1  )   t (  )e (  ) t dt


0




Wq 

(   ) t
t
(



)
e
dt

0

(   )
Also of interest would be the total time the customer had to spend in
system including services .Denoting this random variable by T, it’s CDF
by W (t), it’s density by w (t)
& its expected value by W
w (t)=(μ- λ ) e- (μ- λ) t
W=E (t)=1/(μ-λ)
(t>0)
PROOF OF LITTLE’S FORMULA
L= λ W

Pn   Pr
{n arrivals during a waiting time of T/T=t} dW (t)
0
Where W(t) is the cumulative dist. function of the waiting time.
Thus,

1
n
 t
Pn 
(

t
)
e
dW( t )
n≥0

n! 0
&
L  EN 

 nP
n 0

n

n
   (t ) n e t dW( t )
n 1 n! 0

n 1
(

t
)
L   (t )e t dW( t )
n 1 ( n  1)!
0


L   (t )dW( t )e t (e t )
0
L=λ (ET)=λ W
For M/M/1
Wq=L/μ
M/M/1/K
Pn = (1-ρ)ρn/1- ρk+1
= 1/k+1
(ρ≠1)
(ρ=1)
[1  (k  1) k  k k 1 ]
L
(1   k 1 )(1  )
TRANSIENT BEHAVIOUR OF M/M/1/1 SYSTEM
dP1(t)/dt = -μP1(t) + λP0(t) ………(1)
dP0(t)/dt =+-λP0(t) + μP1(t) ………(2)
P0(t) +P1(t)=1
Hence (1) is equivalent to
P’1(t)= -μP1(t) + λ[1-P1(t)]
So,
P’1(t) + (λ+μ)P1(t)=λ
Ordinary first order linear differential equation with constant coefficients
P1(t)=c e- ( λ+ μ) t + λ /(λ + μ)
P1(t) at t=0
i.e.
P1(0) is used to determine c
c=P1(0)-λ/λ+μ
&
P1(t) = λ/(μ+λ)[1-e-(λ+μ)t]+P1(0)e-(λ+μ)t
&
P0(t) = μ/(μ+λ)[1-e-(λ+μ)t]+P0(0)e-(λ+μ)t
Lim t→∞
P1=1/ρ+1
&
P0=ρ/ρ+1
P0(t) = a+ b e-ct
Where
a= μ /(μ + λ)= ρ/ρ+1
b= -λ /(μ + λ)+P0(0)
c= μ + λ
If b>0
P0(t) is asymptotic to a= ρ/ρ+1
If P0(t) =P0
a+b
b=0
& P0(t) becomes μ / (μ +λ)
P0 (t)
a
t
M/M/C
Pn=
λn= λ
μn= n μ
=cμ
n≤c
n>c
λnP0/n!μn
1≤n≤c
λnP0/cn-c c! μn
n ≥c

1
1     c 
P0        

n
!

c
!

c




 n 0  
  
c 1
n
c
1
BIRTH AND DEATH PROCESSES
Pn (t+∆t) = Pn (t) [ 1-λn ∆t] [1- μn ∆t] + Pn+1 (t) [ 1-λn+1∆t] [ μn+1 ∆t] +
Pn-1(t) [ λn-1 ∆t] [1- μn-1 ∆t]+0[∆t]
(for n ≥1)
P0 (t+∆t) = P0 (t) [ 1-λ ∆t ] + P1 (t) [ μ ∆t] +0[∆t]
dPn/dt=- Pn(t) (λn+ μn) + Pn+1(t) μn+1 + Pn-1(t) λn-1
dP0/dt=- P0(t) λ0 + P1(t) μ1
0 =- Pn + Pn+1 μn+1 + Pn-1 λn-1
0 =- P0λ0+ P1μ1
Pn+1=(λn+ μn) Pn/μn+1 - (λn-1)Pn-1/μn+1
P1=(λ0/μ0)P0
(for n ≥1)
(for n ≥1)
By iteration we get,
P2=(λ1λ0/μ2μ1)P0
Pn 
 n 1 n  2 ............ 0
P0
 n  n 1 ...............1
n
 P0 
i 1
i 1
i

P
n 0
n
1
n


 
1  P0 1   i 1 
i 
n 1 i 1

(for n ≥1)
For M/M/1 system
λi = λ
,
μi = μ
M/M/∞
If
λ n= λ
,
μn= nμ
We get,
Pn=P0λn/n!μn
P0=e-λ/μ
Pn= e-λ/μ (λ / μ)n / n!
POISSION


 i 1
 e  1

i
n 1 i 1

n
M/G/1 SYSTEM
Service time
B (t), b (t)
Arrival process Poisson λ
X (ti) embedded Markov process where denotes number in system and t1,t2,…….are
successive service completion time.
X (ti) → No. of customer the ith customer leave behind as he departs
Xn+1=
Xn+An+1 -1
An+1
(Xn≥1)
(Xn=0)
Xn =No. in system at nth departure point
An+1=No. of customers who arrived during the service time S(n+1) of the (n+1)th
customer
S(n+1) is independent of n=S
An+1 only depend on S
P{A=a/S=t}=e-λt(λt)a/a!

P{A  a} 
P{Xn+1=j / Xn=i} =
A a
P
0  S  t dB(t )
P{A=j-i+1}

=
e t (t ) ji 1
0 ( j  i  1)! dB(t)
0
(j ≥ i-1, i ≥1)
(j < i-1, i ≥1)
Imbedded Markov process as it is only dependent on ( i , j)
POLLACZEK-KHINTCHINE FORMULA
Xn+1=Xn-U(Xn)+A
Where
U (Xn) =
1
0
(Xn>0)
(Xn=0)
E{Xn+1}= E {Xn}=L(D)
L(D)=L(D)-E [U(Xn)]+ E(A)
E [U (Xn)]=E(A)

 A 
E[U(X n )]  E(A)   E 
dB( t )

S  T 
0

  tdB( t )  E(S) 
0


S
Squaring, we get
X2n+1=X2n+U2(Xn)+A2-2XnU(Xn) -2AU(Xn) +2AXn
Taking expected values we get,
0=E[U2(Xn)]+E[A2]-2E[XnU(Xn)]-2E[AU(Xn)]+2E[AXn]
But
U2(Xn)= U(Xn) & XnU(Xn)=Xn
We therefore get
0 = ρ+E(A2)-2L(D)-2ρ2+2L(D)ρ
L(D) = ρ-2ρ2+E[A2] / 2(1-ρ)
E(A2) = Var(A)+[E(A)]2
= Var(A)+ρ2
Where
Var (A)=E {Var (A/S)} + Var {E(A/S)}
We get,
=E (λs)+ Var (λs)
=ρ+λ2σs2
Therefore
L(D)=ρ+(ρ2+λ2σ2s)/2(1-ρ)
It can be shown that
L(D) = L
in this case
L = ρ + (ρ2+λ2σs2) /2(1-ρ)
EXAMPLE:
λ=10/hr
1/μ=5min
M/M/1
Training will in an improvement in the variance of services time
but at a slight increase in mean.
1/μ=5.5min
Std. deviation decreases from 5 to 4 min.
Present
After training
M/M/1
M/G/1
Mean ↑ 10%
Std. deviation ↓ 20%
► Variance ↓ ≈ 36%
L
W
5
30 min
8.625
51.750
Hence the mean effects result much more.
Calculate reduction in variance required to makeup an increase of 0.5 in the mean.
L=5= ρ+(ρ2+λ2+σ2s)/2(1-ρ)
where ρ=λ/μ=10(10/2)(1/60)
This yields σ2s<0 →
even with σ2s= 0 (deterministic service time)
L is greater than S.
L= ρ + (ρ2)/2(1-ρ) = 6.0
BUSY PERIOD M/M/1
G(X)=
{Given first service time =t busy period generated
by all arrivals occurring during t ≤ x-t }dB(t)
x
P
r
0
Where Gn (x) is the n fold conv. of G (x)
e t (t ) n n
G( x )   
G ( x  t )dB( t )
n!
0 n 0
x 

G * (x) 
sx
e
G ( x )dx

0
Taking transformation on both sides

G (x) 
*
0
e  t (t ) n
n
G
( x  t )dB( t )dk

0 n 0 n!




x

 t
n
e
(

t
)
G * (x)   
dB( t )  e  xs G n ( x  t )dx
n!
0 n 0
t
 
 t
n
e
(

t
)
G * (x)   
e st [G * (s)] n dB( t )
n!
0 n 0

G * (x) 
 t  tG*(s ) st
e
e dB( t )
 e
0
= B*[s + λ – λ G*(s)]
E (x)= [-d G*(s) / ds ] s=0
= -G*’(0)
= -B*’(0)/(1+λB*’(0)) = (1/μ)/(1-λ/μ)=1/(μ-λ)
Same result as M/M/1
P0=1-ρ
1-P0=ρ
Hence ρ/1-ρ= E [x] /E [Length of idle period]=E[x]/(1/λ)
Therefore
E (x)=1/μ-λ
GI/M/1 System
(Bn ≤ Xn+1 ; Xn ≥ 0)
Xn+1=Xn-Bn +1
Pij = Pr{Xn+1= j /Xn=i} =
Pr {B= i+1-j}
0
i+1 ≥ j
i+1<j
e  t ( t ) i 1 j
dA(t )

0 (i  1  j )!

=
0
j1
Pio  1   Pi
i 1
i+1≥ j ≥ 1
i+1<j
bn= Pr {B=n}
Pij≡
1-b0
b0
0………………
1   bk
b1
b0…………….
1   bk
b2
b1……………
1
k 0
2
k 0
steady state vector q={qn}
qP =q

q i   q i k 1b k
(i≥1)
k 0

i
i 0
k 0
q0   [qi (1   bk )]
GI/M/1
qP =q

q i   q i k 1b k
(i≥1)
k 0
q0 


l 0
l
[q i (1   b k )]
k 0
solving by method of operators
Dqi=qi+1
Hence for i≥1
Hence
We get
qi-[qi-1b0+qib1+qi+1b2+………………]=0
qi-1[D-b0-Db1-D2b2-D3b3………]=0

 D   bn Dn  0
for trivial sol
n 0


bn Dn 
n 0
Therefore z=B(z)
Generating fn of b
Call it B( )
……………………….(i)
Hence if rj denotes the jth root (0<rj<1) of equation
Therefore the sol. is
qi 
 c j rj
j
i
(i≥1)
Where cj are constant
However for (i) there is atmost one admissible root between 0 & 1 and
that root is r0
qi=c r0i
(i≥1)
Constants c and q0 are to be determined from boundary conditions.


i 0
qi  1
One root
y=B(z)
y=z
0 < β (0) = b0<1
C(1)=∑bn=1
' ( z) 


n 1

nb n z n 1
>0
 ' ' ( z )   n(n  1)bn z
n 2
>0
n2
Therefore β(z) is monotone , non decreasing &convex in fact as bn>0
β(z) is strictly convex
y
y
y=β(z)
y=z
z
1
ro
No root between 0&1
λ/μ>1
Numerical procedure for obtaining the root r0
z
Ex.
Newton-Raphson method guarantees a solution
It converge because of the convexity of β(z)
Using B(z) =z
& qi=cr0i
Yields c=q0=1-r0
so that
Wq (t)=
qn=(1-r0)r0n
1-r0
1-r0e-μ(1-r0)t
n≥0 ρ<1
t=0
t>0
Example:
Inter-arrival time probability distribution
t (min)
α (t)
2
3
4
0.2
0.7
0.1
1/λ=2.9min
A (t)
0.2
0.9
1.0
1/μ=2min
z
β(z)
0
0.2
0.4
0.6
0.8
1.0
0.24
0.32
0.42
0.56
0.75
1.00
qn=0.54(0.46)n
L(A) = r0/1-r0 = 0.85
Lq(A) = r02/1/r0 = 0.39
Wq=r0/μ(1-r0)=1.70min
W= 3.70min
n ≥0
MORDIFICATION OF GI/M/1
Finite waiting room of size N
Pij≡
1-b0
1-∑bi
.
.
.
.
1-∑bi
1-∑bi
(0,k) Policy
b0
b1
0………………
b0……………..
bN-1 ………….bb b1 b0
bN-1………….b2 b1 b0
(0,0)
0
0
Pij ≡
0,0
1,0
.
.
.
k,0 1-∑bi
(1,1) 1-∑bi
(2,1) 1-∑bi
.
.
.
(1,0) ……….(k,0) (1,1) (2,1)……
1
0
0
0 ……..
0
1…....0
0
0 ………
0 ……………..0
bk bk-1………
……………………… b1 b0
…………………………. b2 b1 b0
GI/G/1
LINDEY’S EQUATION:- Integral equation of Wiener-Hopf type
Wq(n+1) =
Wq(n)+S(n)-T(n)
0
for (Wq(n)+Sq(n)-T(n)>0)
for (Wq(n)+Sq(n)-T(n)<0)
Wq(n+1) = max (0,Wq(n)+S(n)-T(n))
Wq(n) is a discrete Markov Process
Wq(n+1)(t) = Pr{ Wq(n) + S(n)- T(n) ≤t}
Let
U(n)= S(n)-T(n) with C.D.F U(n)(x)
Then by convolution formula
t

Wq( n 1) ( t ) 
Wq
(n)
( t  x )dU ( n ) x

In steady state :
t
Wq ( t ) 

Wq ( t  x )dU ( n ) x
(0≤t<∞)

= 0
The limit Wq(t) =
iff
(t<0)
(n)
q
Lim W
n 
t
exists
ρ<1
U is the conv. of S & (-T)
SOLUTION OF WIENER-HOPF INTEGRATION
Define
Wq-(t)=
0
t>0
t

Wq ( t  x )dU ( x )
t<0

t

q
W ( t )  Wq ( t ) 

Wq(t - x)dU( x )
(-∞<t<∞)
-

W q (s) 



W q (s) 



e st Wq ( t )dt 


e st Wq ( t )dt
0

e st Wq ( t )dt   e st Wq ( t )dt
0
(Two sided ) LST
RHS:
 t
t

Wq ( t  x )dU( x ) 

Since

e ( t x )s Wq ( t  x )e sx dU( x )dt
 
Wq(t-x)=0
α2= W(s) . U*(s)
U*(s)=A*(-s) . B*(s)
Hence
Or
W-q(s)+ W q(s)= Wq(s) A*(-s) B*(s)
W q(s) =
W-q(s)/[A*(-s).B*(s)-1]
for x ≥t
THEORY OF TOLL TRAFFIC ENGINEERING IN USA
BY Rojer I. Wilkinson
Pg. 421 system Tech. J
March 1956
TOLL TRAFFIC ENGINEERING
Poisson input , s-servers, Blocked customers cleared (BCC)
The Erlang Loss Formula
Customers finding the system busy leave (BCC)
λj = λ
0
μj=jμ
for j=0,1,2……(s-1)
when j=s
j=0,1,2……s
By Birth and Death process formula
j


j!


 
Pj 
k
s


k!



k 0 

(j=0,1,2…..s)
&
Pj=0
for j>s
This is called the truncated Poisson distribution λ/μ is defined as
a = the offered load . which is dimensionless and is numerically equal to the
mean number of arrivals that occur during a service time .Measured in Erlang
a/s is called the traffic intensity (ρ) .
prob. that the s-serven are busy is given by putting j = s & we get the Erlang
loss formula.
a s s!
B(s, a )  s
…………….(2)
k
 (a / k!)
k 0
This is also called the Erlang’s first formula and is denoted by Ei,s (a) .
The Erlang loss formula gives both the proportion of time all servers are busy
and the proportion of arriving customers who find the system full .
Another useful quantity is the ‘carried load a’.
s 1

j1
js
a '   jPj  s Pj
…………………..(3)
Gives mean number of busy servers.
In Bcc Pj= 0
for j > s & hence
a’ = a[1-B(s,a)]
…………….. (4)
If the no. of servers s becomes infinite carried load a’ becomes = offered load a.
In an existing system, a’ lends itself to direct measurement & the formula (3)
permits estimation of offered load & probability of blocking.
Service occupancy ρ =a’/s
From (2) ,(4) & (5) we can see that as s↑ & a↑ keeping B(s, a)
Constant, ρ the service occupancy increases.
This is expressed by the statement large service groups are more
efficient than smaller ones.
The above results are true for general service times.
Above analysis ignores retrials which meet NC condition.
Study by C.clos in 1944 by observing action of customers
who received busy signal on 1100 calls in NC→.
Exponential distribution with mean 250 sec
MATHEMATICAL MODEL:
No. of trunks = s
No. of waiting positions = y
Where y is so large that few calls are rejected i.e. those calls abandoned
after the first attempt.
Offered load = a erlangs
Conversation holding times = 1
Therefore,
Arrival rate = a
Return time for a call = 1/x
x → Rate of return
.
Define f (m, n) ≡ m calls in progress on the s trunks. &
n calls are waiting on the y storage position.
f (m ,n) = a f(m -1,n) dt + (m + 1) f(m + 1 , n) dt + x (n+1) f(m-1, n+1) dt
+a f(s , n – 1) dt* + [ 1- (a*** + xn** +m) dt ] f(m,n) ………..(6)
Where,
0 ≤ m ≤ s, 0 ≤ n ≤ y
* include this term only when m = s
** omit xn when m = s
*** omit a when m = s & n = y
equation (6) reduces to
( a*** + xn** + m) f (m , n) = a f(m -1 , n) + x (n + 1) f(m – 1 , n +1) +
(m +1) f (m +1 ,n) + a f(s ,n -1) …………(7)
These can be solved in terms of f (s, y) & then using the normalizing equation.
s
y
  f (m, n )  1
We get
m 0 n 0
the entire f (m, n) array
The proportion of time ‘NC exists’

y
 f ( m, n )
n 0
& the “carried load ” = expected no. of servers busy.

s
y
  mf (m, n )
m 0 n 0
Proportion of call attempt meeting NC including all re-trial will be
W(s, a, x) = (Expected over flow calls per unit time)/( Expected calls
offered per unit time)
in which
n
s
y
  nf (m, n )
m 0 n 0
When y is chosen so large that f(s, y) is negligible, as will be now used.
L=a
because B(s, a) =0
no call blocked
W(s, a, x) = x n/ a +x n
0.6
For circuit use length = 150 sec
Avg. return time = 250 sec
No. of trunks = 6
Proportional
Of calls
Meeting NC
0.5
0.4
0.3
0.2
0.1
2
4
6
L= Load carried in erlangs
If a= 4.15 erlangs = L
Proportional of call meeting NC= 27.5%
This gives an idea of the grade of service being given to customers.
8