1 generalized solution
1.1 step 1
1.2 step 2
1.3 step 3
generalized solution
3 steps:
1. find minimum upstream count,
n
u
{\displaystyle n_{u}}
2. find minimum downstream count,
n
d
{\displaystyle n_{d}}
3. choose lower of two,
n
p
=
min
{
n
u
,
n
d
}
{\displaystyle n_{p}=\min\{n_{u}\ ,\ n_{d}\}}
step 1
all possible observer straight lines between upstream boundary , point p have constructed observer speeds smaller free flow speed:
c
q
p
=
q
0
Δ
t
−
k
0
Δ
x
=
q
0
(
t
p
−
t
q
)
−
k
0
(
x
p
−
x
q
)
(
5
)
{\displaystyle c_{qp}=q_{0}\delta {t}-k_{0}\delta {x}=q_{0}(t_{p}-t_{q})-k_{0}(x_{p}-x_{q})\qquad (5)}
where
Δ
t
=
t
p
−
t
q
=
(
x
m
−
x
u
)
v
q
p
{\displaystyle \delta {t}=t_{p}-t_{q}={\frac {(x_{m}-x_{u})}{v_{qp}}}}
v
q
p
∈
[
0
,
v
f
]
{\displaystyle v_{qp}\in [0,v_{f}]}
,
Δ
x
=
x
m
−
x
u
{\displaystyle \delta {x}=x_{m}-x_{u}}
thus need minimize
n
q
+
q
0
(
t
p
−
t
q
)
−
k
0
(
x
m
−
x
u
)
{\displaystyle {n_{q}+q_{0}(t_{p}-t_{q})-k_{0}(x_{m}-x_{u})}}
; i.e.,
n
u
=
min
t
q
{
n
q
+
q
0
(
t
p
−
t
q
)
−
k
0
(
x
m
−
x
u
)
}
(
6
)
{\displaystyle n_{u}=\min _{t_{q}}\{n_{q}+q_{0}(t_{p}-t_{q})-k_{0}(x_{m}-x_{u})\}\qquad (6)}
since
d
n
q
/
d
t
≤
q
0
{\displaystyle dn_{q}/dt\leq q_{0}}
, see objective function non-increasing , therefore
t
q
∗
=
t
p
1
{\displaystyle t_{q}^{*}=t_{p_{1}}}
. q should placed @
p
1
{\displaystyle p_{1}}
, have:
c
q
p
=
c
p
1
p
=
q
0
(
x
m
−
x
u
v
f
)
−
k
0
(
x
m
−
x
u
)
=
0
(
7
)
{\displaystyle c_{qp}=c_{p_{1}p}=q_{0}\left({\frac {x_{m}-x_{u}}{v_{f}}}\right)-k_{0}(x_{m}-x_{u})=0\qquad (7)}
thus,
n
u
=
n
p
1
{\displaystyle n_{u}=n_{p_{1}}}
step 2
we have:
n
d
=
min
q
′
{
n
q
′
+
c
q
′
p
}
=
min
q
′
{
n
q
′
+
q
0
Δ
t
−
k
0
Δ
x
}
{\displaystyle n_{d}=\min _{q^{ }}\{n_{q^{ }}+c_{q^{ }p}\}=\min _{q^{ }}\{n_{q^{ }}+q_{0}\delta {t}-k_{0}\delta {x}\}}
repeat same steps find
n
d
{\displaystyle n_{d}}
minimized when
q
′
=
p
2
{\displaystyle q^{ }=p_{2}}
. , @ point
p
2
{\displaystyle p_{2}}
get:
n
d
=
n
p
2
+
q
0
(
x
d
−
x
m
w
)
−
k
0
(
x
d
−
x
m
)
(
8
)
{\displaystyle n_{d}=n_{p_{2}}+q_{0}({\frac {x_{d}-x_{m}}{w}})-k_{0}(x_{d}-x_{m})\qquad (8)}
since fd triangular,
q
0
w
+
k
0
=
k
j
{\displaystyle {\frac {q_{0}}{w}}+k_{0}=k_{j}}
. therefore, (8) reduces to:
n
d
=
n
p
2
+
(
x
d
−
x
m
)
k
j
(
9
)
{\displaystyle n_{d}=n_{p_{2}}+(x_{d}-x_{m})k_{j}\qquad (9)}
step 3
to solution choose lower of
n
u
{\displaystyle n_{u}}
,
n
d
{\displaystyle n_{d}}
.
n
p
=
min
{
n
u
,
n
d
}
=
min
{
n
p
1
,
n
p
2
+
(
x
d
−
x
m
)
k
j
}
(
10
)
{\displaystyle n_{p}=\min\{n_{u}\ ,\ n_{d}\}=\min\{n_{p_{1}}\ ,\ n_{p_{2}}+(x_{d}-x_{m})k_{j}\}\qquad (10)}
this newell s recipe 3-detector problem.
Comments
Post a Comment