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

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m β (c 1 , c 2 , ρ) = m ′ ...(1.3.7)<br />

Note that the smallest value of n (as a real number) is obtained as n = m β (c)/p ′ as the plan<br />

giving the smallest sample size.<br />

1.3.3 Construction of C kind <strong>plans</strong> of given strength<br />

The Problem<br />

We consider the case r = 2 and we want to find a C kind plan satisfying the equations:<br />

G(c 1 , np 1 ).G(c 2 , np 2 ) = 1 − α<br />

G(c 1 , np ′ 1).G(c 2 , np ′ 2) = β<br />

...(1.3.5)<br />

Restricting to the situation where p 1 /(p 1 + p 2 ) = p ′ 1/(p ′ 1 + p ′ 2) = ρ we write p = p 1 +<br />

p 2 , p ′ = p ′ 1 + p ′ 2, m = np, m ′ = np ′ . For a given ρ, we define m P (c 1 , c 2 , ρ) as the value of m<br />

satisfying the equation<br />

G(c 1 , mρ)G(c 2 , m(1 − ρ)) = P<br />

...(1.3.6)<br />

Note: The function m P (c 1 , c 2 , ρ) used for the C kind MASSP is similar to the function<br />

m P (c) defined in the context to the single <strong>sampling</strong> plan for single attribute.<br />

We must have n, c 1 , c 2 such that<br />

m 1−α (c 1 , c 2 , ρ) = m<br />

Lemma<br />

Let<br />

G(c 1 , M 1 ρ)G(c 2 , M 1 (1−ρ)) = G(c 1 , M 2 ρ)G(c 2 +1, M 2 (1−ρ)) = G(c 1 +1, M 3 ρ).G(c 2 , M 3 (1−ρ)) = P.<br />

Then M 2 ≥ M 1 and M 3 ≥ M 1<br />

...(1.3.8)<br />

Proof<br />

65

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