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TProblem Problem:
A ramp has an arrival rate of 200 cars an hour and the ramp meter only permits 250 cars per hour, while the ramp can store 40 cars before spilling over.
(A) What is the probability that it is half-full, empty, full?
(B) How many cars do we expect on the ramp?
Example Solution:
ρ
=
200
/
250
=
0.8
{\displaystyle \rho =200/250=0.8\,\!}
Part (A)
What is the probability that it is half-full, empty, full?
Half-full
P
(
n
)
=
(
1
−
ρ
)
1
−
ρ
N
+
1
(
ρ
)
n
→
P
(
20
)
=
(
1
−
0.8
)
1
−
0.8
40
+
1
(
0.8
)
20
=
0.0023
{\displaystyle P(n)={\frac {\left({1-\rho }\right)}{1-\rho ^{N+1}}}\left(\rho \right)^{n}\to P(20)={\frac {\left({1-0.8}\right)}{1-0.8^{40+1}}}\left({0.8}\right)^{20}=0.0023\,\!}
Empty
P
(
n
)
=
(
1
−
ρ
)
1
−
ρ
N
+
1
(
ρ
)
n
→
P
(
0
)
=
(
1
−
0.8
)
1
−
0.8
40
+
1
(
0.8
)
0
=
0.20
{\displaystyle P(n)={\frac {\left({1-\rho }\right)}{1-\rho ^{N+1}}}\left(\rho \right)^{n}\to P(0)={\frac {\left({1-0.8}\right)}{1-0.8^{40+1}}}\left({0.8}\right)^{0}=0.20\,\!}
Full
P
(
n
)
=
(
1
−
ρ
)
1
−
ρ
N
+
1
(
ρ
)
n
→
P
(
40
)
=
(
1
−
0.8
)
1
−
0.8
40
+
1
(
0.8
)
40
=
2.65
∗
10
−
5
{\displaystyle P(n)={\frac {\left({1-\rho }\right)}{1-\rho ^{N+1}}}\left(\rho \right)^{n}\to P(40)={\frac {\left({1-0.8}\right)}{1-0.8^{40+1}}}\left({0.8}\right)^{40}=2.65*10^{-5}\,\!}
Part (B)
How many cars do we expect on the ramp?
E
(
n
)
=
(
ρ
)
(
1
−
ρ
)
1
−
(
N
+
1
)
(
ρ
)
N
+
N
ρ
N
+
1
1
−
ρ
N
+
1
=
(
0.8
)
(
1
−
0.8
)
1
−
(
40
+
1
)
(
0.8
)
40
+
40
(
0.8
)
40
+
1
1
−
(
0.8
)
40
+
1
=
4
{\displaystyle E(n)={\frac {\left(\rho \right)}{\left({1-\rho }\right)^{}}}{\frac {1-\left({N+1}\right)\left(\rho \right)^{N}+N\rho ^{N+1}}{1-\rho ^{N+1}}}={\frac {\left({0.8}\right)}{\left({1-0.8}\right)^{}}}{\frac {1-\left({40+1}\right)\left({0.8}\right)^{40}+40\left({0.8}\right)^{40+1}}{1-\left({0.8}\right)^{40+1}}}=4\,\!}
