Multiattribute acceptance sampling plans - Library(ISI Kolkata ...

Proof:
Denoting the regret function value of the 1 st plan and 2 nd plan by R 1 & R 2 respectively we get
a −
R1 − R2
= γ
2
g ( a r −1
, m′
( r −2
) ) ( 0,
−1
(
′
−1 ) )
r
ar
g m
r ∑ g
( x , ′ )
x = 0
r
mr
a −a
- γ
1g
( a r −1
, m(
r −2
) ) ( r r
g 0, m( r −1 ) )
x = 0
To show that R 1 –R 2 > 0 we show that,
( γ / )
2 γ − ( m′−m
e
) ⎡
/
⎤
a
m ′ m
r−1
. F 1 > 1
1 ⎢⎣ ( r−2)
( r−2)
⎥⎦
where,
ar
− ar−1
ar
− a
x
r−1
1 = ∑ [ r
x
F
m' !] / ∑ [ r
r /x r
m r /x r !]
x = 0
x = 0
r
r
r
∑ − 1
r
g ( x , )
r m r
In order to prove theorem 2.3.2, we first prove the lemma stated below.
… (2.3.5)
Lemma:
c
∑ ( m ' / x!
) / ∑ ( m ′ / x!)
> ∑ ( m / x!)
/ ∑ ( m / x!)
; for c ≥ 1 and m′ > m .
x = 0
x
c −1
x = 0
x
c
x=
0
x
c−1
x=
0
x
….(2.3.6)
c−1
x−c
Proof of the lemma: The function ∑ m / x!
c ≥ 1, decreases as m increases . Thus,
x=
0
c
x
c−1
x
1
∑ ( m / x!) / ∑ ( m / x!)
= 1 +
is an increasing function of m and
x=
0
x=
0
c−1
x−c
c!
∑ m / x!
x=
0
therefore (2.3.6) holds.
Let us come back to theorem 2.3.2. We continue with the proof in the following lines:
( r 1)
Since optimal a
r −1
must satisfy the inequality ∆ R1
> 0 for the first optimal plan,
( r −1)
a r
− a r −1
where ∆ R1
is the standard notation for the increase in R 1
achieved when
−1
replaced by a 1
r −1 + keeping all other parameters unaltered.
The condition:
a is
r −1
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