TProblem Problem:
Application of Single-Channel Undersaturated Infinite Queue Theory to Tollbooth Operation. Poisson Arrival, Negative Exponential Service Time
Arrival Rate = 500 vph,
Service Rate = 700 vph
Determine
Percent of Time operator will be free
Average queue size in system
Average wait time for vehicles that wait
Note: For operator to be free, vehicles must be 0
Example Solution:
ρ
=
λ
μ
=
(
500
700
)
=
0.714
{\displaystyle \rho ={\frac {\lambda }{\mu }}=\left({\frac {500}{700}}\right)=0.714\,\!}
P
(
n
)
=
ρ
n
(
1
−
ρ
)
=
(
0.714
)
0
(
1
−
0.714
)
=
28.6
%
{\displaystyle P\left(n\right)=\rho ^{n}\left({1-\rho }\right)=\left({0.714}\right)^{0}\left({1-0.714}\right)=28.6\%\,\!}
Q
=
ρ
1
−
ρ
=
0.714
0.286
=
2.5
{\displaystyle Q={\frac {\rho }{1-\rho }}={\frac {0.714}{0.286}}=2.5\,\!}
w
=
λ
μ
(
μ
−
λ
)
=
500
700
(
700
−
500
)
=
0.003571
h
o
u
r
s
=
12.8571
sec
{\displaystyle w={\frac {\lambda }{\mu \left({\mu -\lambda }\right)}}={\frac {500}{700\left({700-500}\right)}}=0.003571hours=12.8571\sec \,\!}
Q
=
λ
w
=
500
∗
0.00357
=
1.785
{\displaystyle Q=\lambda w=500*0.00357=1.785\,\!}
S
e
r
v
i
c
e
T
i
m
e
=
1
μ
=
1
700
=
0.001429
h
o
u
r
s
=
5.142
sec
{\displaystyle ServiceTime={\frac {1}{\mu }}={\frac {1}{700}}=0.001429\ hours=5.142\sec \,\!}
t
=
1
(
μ
−
λ
)
=
1
(
700
−
500
)
=
0.005
h
o
u
r
s
=
18
s
e
c
o
n
d
s
{\displaystyle t={\frac {1}{\left({\mu -\lambda }\right)}}={\frac {1}{\left({700-500}\right)}}=0.005\ hours=18{\rm {seconds}}\,\!}