Massachusetts
Institute
of
Technology
1.203J/6.281J/13.665J/15.073J/16.76J/ESD.216J
Logistical
and
Transportation
P
lanning
Methods
Fall
2002
Consider
link
1.
Define
X
2
to
be
the
distance
from
the
right
most
point
of
link
1.
Then
the
conditional
travel
distance,
given
that
X
1
is
defined
to
be
the
distance
on
link
7
from
its
left
most
point
and
that
X
2
is
on
link
1,
is
(
D

1
,
7)
=
X
1
+
X
2
+
1.
Let’s
define
V
=
X
1
+
X
2
.
Note
that
X
1
, X
2
∼
U(0
,
1)
i.i.d.
Either
by
convolution,
or
by
the
“never
fail”
sample
space
method
using
cumulative
distribution
functions,
or
by
recalling
problem
2(e)(i)
of
HW1,
we
find
d,
d
∈
[0
,
1]
f
V
(
d
)
=
2
−
d,
d
∈
[1
,
2]
0
,
otherwise
Now
the
conditional
pdf
we
want
for
link
1
is
f
V
(
d
)
“shifted
to
the
right”
by
one
unit
of
distance.
Call
this
conditional
pdf
f
(
D

1
,
7)(
d
).
Then
we
have
for
link
1,
f
(
D

1
,
7)(
d
) =
f
V
(
d
−
1).
By
inspection
we
also
have
f
(
D

3
,
7)(
d
) =
f
(
D

8
,
7)(
d
) =
f
(
D

11
,
7)(
d
) =
f
(
D

1
,
7)(
d
) =
f
V
(
d
−
1).
For
the
remaining
links
in
Set
1,
links
4,
5,
6,
9
and
10
“touch”
link
7,
so
there
is
no
shifting
of
the
pdf
by
one.
That
is,
there
is
no
intermediate
link
between
them
that
would
add
1.0
km
to
the
travel
distance.
Hence,
f
(
D

4
,
7)(
d
) =
f
(
D

5
,
7)(
d
) =
f
(
D

6
,
7)(
d
) =
f
(
D

9
,
7)(
d
) =
f
(
D

10
,
7)(
d
) =
f
V
(
d
).