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α
(
μ, r
+1)=
β
1
;
μ
2=
μ
2
−
1;
end
One can also use an alternative algorithm to compute
d
(
j
) as given by
equation (7.29). Now, before we describe the algorithmic steps, we first con-
sider the underlying background of it. By equation (7.25),
μ
d
(
j
)=
α
i,k
(
j
)
P
i
i
=
μ−k
+1
μ
=
[(
t
j
+
k−
1
−
τ
i
)
β
i,k−
1
(
j
)+(
τ
i
+
k
−
t
j
+
k−
1
)
β
i
+1
,k−
1
(
j
)]
P
i
.
i
=
μ
−
k
+1
Since
β
μ−k
+1
,k−
1(
j
)=
β
μ
+1
,k−
1
(
j
)
= 0 by (1) in the corollary, we get,
μ
α
i,k−
1
(
j
)
P
[2]
d
(
j
)=
i,j
,
i
=
μ
−
k
+2
where
P
[2]
i,j
=[(
t
j
+
k−
1
−
τ
i
)
P
i
+(
τ
i
+
k−
1
−
t
j
+
k−
1
)
P
i−
1
]
/
(
τ
i
+
k−
1
−
τ
i
)
.
In general, for
r
=1
,
2
,
···
,k
,
μ
α
i,k−r
+1
(
j
)
P
[
r
]
d
(
j
)=
i,j
,
i
=
μ
−
k
+
r
with
P
[1]
i,j
=
P
i
and
P
[
r
+1]
i,j
τ
i
)
P
[
r
]
t
j
+
k−r
)
P
[
r
]
=[(
t
j
+
k−r
−
i,j
+(
τ
i
+
k−r
−
1
,j
]
/
(
τ
i
+
k−r
−
τ
i
)
.
i
−
Therefore, when
r
=
k
,wehave
d
(
j
)=
α
μ,
1
(
j
)
P
[
k
]
μ,j
=
P
[
k
]
μ,j
.
With this, we write down Algorithm 2 to compute
d
(
j
).
Algorithm 2:
Step 1:
μ
2=
μ
k
+1;
Step 2: for
i
=
μ
2
,μ
2
+1
,
−
···
,μ
do
begin
P
[1
i
=
P
i
;
Step 3: for
r
=1
,
2
,
···
,k
−
1do
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