**2. Single Stage Acceptance Sampling Plan based on the Control Threshold Policy [1]**

We suppose a batch of size *n* is received which its proportion of the defectives items is equal to*p*. For a batch of size*n*, random variable *Y* is defined as the number of inspected items and *z* is defined as the number of items classified as 'defective' after inspection. The number of inspected items has an upper threshold equal to*m*. For *Y* =1, 2, ..., *m* inspected items (*m*≤*n*) the batch will be rejected if *x* ≤ *z*where *x* is the upper control level for batch acceptance. In the other words, when the number of defective items in the inspected items gets more than the control threshold *x* then decision making process stops and the batch is rejected.

The probability distribution function of *Y* is determined by the following equations,

$$\Pr\left\{Y\right\}=\begin{cases} \left\{\sum\_{z=0}^{x} \Pr\left\{z\right\}=\\ \Pr\left\{z \le x-1\right\} + \Pr\left\{z=x\right\}\\ \vdots\\ \sum\_{z=0}^{x-1} \binom{m}{z} p^z \left(1-p\right)^{n-z} + \\ \left\lfloor \begin{pmatrix} m-1\\x-1 \end{pmatrix} p^x \left(1-p\right)^{n-x} \right\} \\ \left\lfloor \begin{pmatrix} Y-1\\x-1 \end{pmatrix} p^x \left(1-p\right)^{n-x} \right\} & \text{x } \le Y < m \end{cases} \tag{1}$$

Pr{*<sup>Y</sup>* }=*Poisson*(*λ*)= *<sup>e</sup>* <sup>−</sup>*λλ <sup>Y</sup>* <sup>−</sup>*<sup>x</sup>*

*<sup>z</sup>* ) *<sup>p</sup> <sup>z</sup>*(1<sup>−</sup> *<sup>p</sup>*)*m*−*<sup>z</sup>* <sup>+</sup> <sup>∑</sup>

of this approximation, *m*and *x*should be sufficiently large numbers. Using the above ap‐

Now, let *Px* denotes the probability of rejecting the batch. The batch is rejected if the number of defective items is more than or equal to *x* thus the value of *Px* is determined by the fol‐

In order to calculate the total cost, including the cost of rejecting the batch, the cost of in‐ spection and the cost of defective items, assume *R* is the cost of rejecting the batch, *c*is the inspection cost of one item and *c* ' is the cost of one defective item, so the total cost, *Cx*, is determined by conditioning *Cx*on two events of rejecting or accepting the batch, thus the ob‐

*<sup>p</sup>* is the parameter of Poisson distribution. In order to improve the accuracy

*<sup>Y</sup> <sup>e</sup>* <sup>−</sup>*λλ <sup>Y</sup>* <sup>−</sup>*<sup>x</sup>*

*Y* =*x m*

where *λ* = *x*

lowing equation,

Thus we have,

1− *p*

proximation method, following is concluded,

jective function is written as follows:

*x x*

+*c*∑ *Y* =*x m*

+∑ *z*=0 *x*−1 ( *m*

*C EC P*

*E Y <sup>x</sup>* ≃*m*∑

*z*=0 *x*−1 ( *m*

> *Px* <sup>=</sup>∑ *z*=*x m* ( *m*

( ) ( )

= +

Reject the batch Reject the batch

*E C P P R cE Y*

( [ ] )( ) ( ) [ ] ' '

*npc cE Y P P R npc P cE Y*

+ -= + -+

*Cx* <sup>=</sup>*PxR* <sup>+</sup> *npc* '(1−*Px*) <sup>+</sup> *mc*∑

*<sup>Γ</sup>*(*<sup>Y</sup>* <sup>−</sup> *<sup>x</sup>* <sup>+</sup> 1) <sup>=</sup>*R*∑

*<sup>z</sup>* ) *<sup>p</sup> <sup>z</sup>*(1<sup>−</sup> *<sup>p</sup>*)*m*−*z*(*npc* ' <sup>+</sup> *mc*) <sup>+</sup> *<sup>c</sup>*<sup>∑</sup>

*<sup>Y</sup> <sup>e</sup>* <sup>−</sup>*λλ <sup>Y</sup>* <sup>−</sup>*<sup>x</sup>*

1 1

Accept the batch Accept the batch

( ) ( ) ( [ ] )

*x x x*

*x x x x x*

*z*=0 *x*−1 ( *m*

*z*=*x m* ( *m* *<sup>z</sup>* ) *<sup>p</sup> <sup>z</sup>*(1<sup>−</sup> *<sup>p</sup>*)*m*−*<sup>z</sup>*

*<sup>Y</sup> <sup>e</sup>* <sup>−</sup>*λλ <sup>Y</sup>* <sup>−</sup>*<sup>x</sup> Γ*(*Y* − *x* + 1)

*<sup>z</sup>* ) *<sup>p</sup> <sup>z</sup>*(1<sup>−</sup> *<sup>p</sup>*)*m*−*<sup>z</sup>*

*Y* =*x m*

*<sup>Γ</sup>*(*<sup>Y</sup>* <sup>−</sup> *<sup>x</sup>* <sup>+</sup> 1) (3)

New Models of Acceptance Sampling Plans http://dx.doi.org/10.5772/50835 57

*<sup>Γ</sup>*(*<sup>Y</sup>* <sup>−</sup> *<sup>x</sup>* <sup>+</sup> 1) (4)

*<sup>z</sup>* ) *<sup>p</sup> <sup>z</sup>*(1<sup>−</sup> *<sup>p</sup>*)*m*−*<sup>z</sup>* (5)

=+ +

(6)

(7)

In Eq. (1), *Y* =*m*indicates that all items are inspected therefore, the number of defective items has been less than *x*or *xth* defective item has been *mth* inspected item. For the case*x* ≤*Y m*, *xth* defective item has been *Yth* inspected item thus, the probability distribution function of *Y* follows a negative binomial distribution. The expected mean of the number of inspected items is determined as follows:

$$\begin{aligned} E\{Y\}\_\mathbf{x} &= m \sum\_{z=0}^{\mathbf{x}-1} \binom{m}{z} p^{z} (1-p)^{m-z} + m \binom{m-1}{\mathbf{x}-1} p^{-\mathbf{x}} (1-p)^{m-\mathbf{x}} \\ &+ \sum\_{\mathbf{y}=\mathbf{x}}^{\mathbf{m}-1} Y \binom{\mathbf{y}-1}{\mathbf{x}-1} p^{\mathbf{x}} (1-p)^{\mathbf{y}-\mathbf{x}} = m \sum\_{z=0}^{\mathbf{x}-1} \binom{m}{\mathbf{y}} p^{z} (1-p)^{m-z} + \\ &\sum\_{\mathbf{y}=\mathbf{x}}^{\mathbf{m}} Y \binom{\mathbf{y}-1}{\mathbf{x}-1} p^{\mathbf{x}} (1-p)^{\mathbf{y}-\mathbf{x}} \end{aligned} \tag{2}$$

Since Pr{*Y* }=( *Y* −1 *<sup>x</sup>* <sup>−</sup><sup>1</sup> ) *<sup>p</sup> <sup>x</sup>*(1<sup>−</sup> *<sup>p</sup>*)*<sup>Y</sup>* <sup>−</sup>*<sup>x</sup> <sup>x</sup>* <sup>≤</sup>*<sup>Y</sup> <sup>m</sup>* is a negative binomial distribution thus using the approximation method of estimating negative binomial probabilities with Poisson distribu‐ tion [2], following is concluded,

$$\Pr\{Y\} = Poisson(\lambda) = \frac{e^{-\lambda}\lambda^{\prime\prime - \chi}}{\Gamma(Y - \chi + 1)}\tag{3}$$

where *λ* = *x* 1− *p <sup>p</sup>* is the parameter of Poisson distribution. In order to improve the accuracy of this approximation, *m*and *x*should be sufficiently large numbers. Using the above ap‐ proximation method, following is concluded,

$$\mathbb{E}\mathbb{E}\mathbb{Y}\mathbb{J}\_{\mathbf{x}} \simeq m \sum\_{z=0}^{x-1} \binom{m}{z} p^{z} (1-p)^{m-z} + \sum\_{Y=x}^{m} Y \frac{e^{-\lambda} \lambda^{Y-\mathbf{x}}}{\Gamma(Y-\mathbf{x}+1)} \tag{4}$$

Now, let *Px* denotes the probability of rejecting the batch. The batch is rejected if the number of defective items is more than or equal to *x* thus the value of *Px* is determined by the fol‐ lowing equation,

$$P\_{\mathbf{x}} = \sum\_{z=\mathbf{x}}^{m} \binom{m}{z} p^{z} (\mathbf{1} - p)^{m-z} \tag{5}$$

In order to calculate the total cost, including the cost of rejecting the batch, the cost of in‐ spection and the cost of defective items, assume *R* is the cost of rejecting the batch, *c*is the inspection cost of one item and *c* ' is the cost of one defective item, so the total cost, *Cx*, is determined by conditioning *Cx*on two events of rejecting or accepting the batch, thus the ob‐ jective function is written as follows:

$$\begin{aligned} C\_x &= E\left(C\_x \, \middle| \, \text{Reject the batch} \right) P\left(\text{Reject the batch} \right) + \\ E\left(C\_x \, \middle| \, \text{Accept the batch} \right) P\left(\text{Accept the batch} \right) &= P\_x \left(R + cE\left[Y\right]\_x \right) + \\ E\left(npc\, ^\circ + cE\left[Y\right]\_x\right) \left(1 - P\_x \right) &= P\_x R + npc \, ^\circ \left(1 - P\_x \right) + cE\left[Y\right]\_x \end{aligned} \tag{6}$$

Thus we have,

**2. Single Stage Acceptance Sampling Plan based on the Control**

control threshold *x* then decision making process stops and the batch is rejected.

{ }

*z*

Pr 1 Pr

ïï £- + = ïï

ïï= -+ ç ÷ <sup>=</sup> <sup>í</sup> è ø <sup>ï</sup>

0

*z*

ìì

ïï

=

*x*

å

1

Pr

ï = ï

0

*p p <sup>Y</sup> <sup>z</sup>*

*z*

=

*m*

*x Y*

*x*

è ø - î

ï

1

ïïæ ö - ïïç ÷ - <sup>ï</sup> è ø - <sup>î</sup>

1 1

1

{ }

*E Y <sup>x</sup>* =*m*∑

+∑ *Y* =*x m*−1 *Y* ( *Y* −1

∑ *Y* =*x m Y* ( *Y* −1

*Y* −1

tion [2], following is concluded,

Since Pr{*Y* }=(

*z*=0 *x*−1 ( *m*

Pr

items is determined as follows:

The probability distribution function of *Y* is determined by the following equations,

{ } { }

*zx zx*

1

ïí æ ö =

( )

*m x x*

1

*p p*

*<sup>x</sup> m z <sup>z</sup>*


( )

*<sup>z</sup>* ) *<sup>p</sup> <sup>z</sup>*(1<sup>−</sup> *<sup>p</sup>*)*m*−*<sup>z</sup>* <sup>+</sup> *<sup>m</sup>*(

*<sup>x</sup>* <sup>−</sup><sup>1</sup> ) *<sup>p</sup> <sup>x</sup>*(1<sup>−</sup> *<sup>p</sup>*)*<sup>Y</sup>* <sup>−</sup>*<sup>x</sup>* <sup>=</sup>*m*<sup>∑</sup>

*<sup>x</sup>* <sup>−</sup><sup>1</sup> ) *<sup>p</sup> <sup>x</sup>*(1<sup>−</sup> *<sup>p</sup>*)*<sup>Y</sup>* <sup>−</sup>*<sup>x</sup>*

*Y x x*

ïæ ö - ç ÷ - £< <sup>ï</sup>

In Eq. (1), *Y* =*m*indicates that all items are inspected therefore, the number of defective items has been less than *x*or *xth* defective item has been *mth* inspected item. For the case*x* ≤*Y m*, *xth* defective item has been *Yth* inspected item thus, the probability distribution function of *Y* follows a negative binomial distribution. The expected mean of the number of inspected

> *z*=0 *x*−1 ( *m*

approximation method of estimating negative binomial probabilities with Poisson distribu‐

1


( )

*m Y m*

<sup>å</sup> (1)


*p p xY m*

*m*−1

*<sup>x</sup>* <sup>−</sup><sup>1</sup> ) *<sup>p</sup> <sup>x</sup>*(1<sup>−</sup> *<sup>p</sup>*)*<sup>Y</sup>* <sup>−</sup>*<sup>x</sup> <sup>x</sup>* <sup>≤</sup>*<sup>Y</sup> <sup>m</sup>* is a negative binomial distribution thus using the

*<sup>x</sup>* <sup>−</sup><sup>1</sup> ) *<sup>p</sup> <sup>x</sup>*(1<sup>−</sup> *<sup>p</sup>*)*m*−*<sup>x</sup>*

(2)

*<sup>z</sup>* ) *<sup>p</sup> <sup>z</sup>*(1<sup>−</sup> *<sup>p</sup>*)*m*−*<sup>z</sup>* <sup>+</sup>

We suppose a batch of size *n* is received which its proportion of the defectives items is equal to*p*. For a batch of size*n*, random variable *Y* is defined as the number of inspected items and *z* is defined as the number of items classified as 'defective' after inspection. The number of inspected items has an upper threshold equal to*m*. For *Y* =1, 2, ..., *m* inspected items (*m*≤*n*) the batch will be rejected if *x* ≤ *z*where *x* is the upper control level for batch acceptance. In the other words, when the number of defective items in the inspected items gets more than the

**Threshold Policy [1]**

56 Practical Concepts of Quality Control

$$\begin{aligned} \mathbf{C}\_{x} &= P\_{x}R + mpc^{\cdot} \{1 - P\_{x}\} + mc \sum\_{z=0}^{x-1} \binom{m}{z} p^{z} \{1 - p\}^{m-z} \\ &+ c \sum\_{Y=x}^{m} Y \frac{e^{-\lambda} \lambda^{Y-x}}{\Gamma(Y-x+1)} = R \sum\_{z=x}^{m} \binom{m}{z} p^{z} \{1 - p\}^{m-z} \\ &+ \sum\_{z=0}^{x-1} \binom{m}{z} p^{z} \{1 - p\}^{m-z} \{mpc^{\cdot} + mc\} + c \sum\_{Y=x}^{m} Y \frac{e^{-\lambda} \lambda^{Y-x}}{\Gamma(Y-x+1)} \end{aligned} \tag{7}$$

In Eq. (7), *cE Y <sup>x</sup>*is the total cost of inspection and *npc* ' is the total cost of defective items. The optimal value of *x*is determined by minimizing the value of objective function*Cx*. Using the optimization methods, it is concluded that,

*ΔCx*+1 =(*mc* + *npc* '−*R*)(

*ΔCx* =(*mc* + *npc* '−*R*)(

Now If*mc* + *npc* '<*R*, then,

*m*

*m x* −1

( )

*m Y x*

( )

l


*m Y x*

l

*e*

1

å

*c*

using the following formula,

*x Min*

( ( )) ( )

*Y x m x x*

= + -

1 '

1

*Y x m p p R mc npc <sup>x</sup>*

G -+ æ ö

*xx m p p p*

l

*e*

1

*p p*

l


1

å

*c*

( ( )) ( )

*Y x m x x*

= + -

1

*Y x m p p R mc npc <sup>x</sup>*

G -+ æ ö > - ç ÷ - + è ø

1 1

*x m x Y x*


è ø - - +

*m Y x*

æ ö G -+ ç ÷ -> >

*x R mc npc*

*<sup>x</sup>* ) *<sup>p</sup> <sup>x</sup>*(1<sup>−</sup> *<sup>p</sup>*)*m*−*<sup>x</sup>* <sup>+</sup> *<sup>c</sup>* <sup>∑</sup>

) *p <sup>x</sup>*−1(1− *p*)*m*−(*x*−1)

( ) ( ) ( )

Since with increasing the value of *x* the value of binomial distribution with parameters *m* and *p* decreases thus according to the properties of binomial distribution, it is concluded that *x* >(*m* + 1)*p* therefore, the optimal value of *x*is determined using the following formula,

( ) ( ) ( ) ( )

; 1; 1 <sup>1</sup> '

è ø - - + <sup>=</sup> í ý

1 1

ì ü ï ï æ ö G -+ > + ç ÷ - > >

ï ï

> - ç ÷ î þ - + è ø

Also The objective function, *Cx*, should be minimized regarding two constraints on Type-I and Type-II errors associated with the acceptance sampling plans. Type-I error is the proba‐ bility of rejecting the batch when the nonconformity proportion of the batch is acceptable. Type-II error is the probability of accepting the batch when the nonconforming proportion of the batch is not acceptable. Then, in one hand, if*p* =*δ*1, the probability of rejecting the batch should be less than*α*. On the other hand, in case where*p* =*δ*2, the probability of accept‐ ing the batch should be less than*β* where *δ*1 is the AQL (Accepted Quality Level ) and *δ*2is the LQL (Limiting Quality Level) and*α* is the probability of Type-I error and *β* is the proba‐ bility of Type-II error in making a decision, therefore, the optimal value of *x*is determined

*x m x Y x*


*m Y x*

*x R mc npc*

*c*

å

1 '

1 '

*Y* =*x*+1

+ *c*∑ *Y* =*x*

( ( ))

*m Y x*

l

l


*e*

*<sup>m</sup> e* <sup>−</sup>*λλ <sup>Y</sup>* <sup>−</sup>(*x*+1)

*<sup>m</sup> e* <sup>−</sup>*λλ <sup>Y</sup>* <sup>−</sup>*<sup>x</sup> <sup>Γ</sup>*(*<sup>Y</sup>* <sup>−</sup> *<sup>x</sup>* <sup>+</sup> 1) <0

1

( ( ))

*m Y x*

l

l


*e*

*c*

å

1

*<sup>Γ</sup>*(*<sup>Y</sup>* <sup>−</sup>(*<sup>x</sup>* <sup>+</sup> 1) <sup>+</sup> 1) >0

New Models of Acceptance Sampling Plans http://dx.doi.org/10.5772/50835

(12)

59

(13)

(14)

$$\Delta C\_{x} = C\_{x} - C\_{x-1} = R \sum\_{z=x}^{m} \binom{m}{z} p^{z} \left(1 - p\right)^{m-z} + \left(mc + npc'\right) \sum\_{z=0}^{x-1} \binom{m}{z} p^{z} \left(1 - p\right)^{m-z} + \left(mc + npc'\right) \left(mc + npc'\right) \left(mc + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right) \left(mc' + npc'\right)$$

To evaluate above equation, following equality is considered,

$$\begin{aligned} \sum\_{\mathbf{Y}=\mathbf{x}}^{m} Y \frac{e^{-\lambda} \lambda^{\mathbf{Y}-\mathbf{x}}}{\Gamma(\mathbf{Y}-\mathbf{x}+1)} - \sum\_{\mathbf{Y}=\mathbf{x}-1}^{m} Y \frac{e^{-\lambda} \lambda^{\mathbf{Y}-\mathbf{(}\mathbf{x}-1)}}{\Gamma(\mathbf{Y}-(\mathbf{x}-1)+1)} &= \\ \sum\_{\mathbf{Y}=\mathbf{x}}^{m} \frac{e^{-\lambda} \lambda^{\mathbf{Y}-\mathbf{x}}}{\Gamma(\mathbf{Y}-\mathbf{x}+1)} - m \frac{e^{-\lambda} \lambda^{\mathbf{m}-(\mathbf{x}-1)}}{\Gamma(m-(\mathbf{x}-1)+1)} \end{aligned} \tag{9}$$

Since *m* is a sufficiently large number thus the value of *m e* <sup>−</sup>*λλ <sup>m</sup>*−(*x*−1) *<sup>Γ</sup>*(*m*−(*<sup>x</sup>* <sup>−</sup>1) <sup>+</sup> 1) is approximately equal to zero therefore it is concluded that,

$$\begin{split} \Delta \, \prescript{}{\mathbf{x}}{}{\mathbf{C}}\_{\mathbf{x}} &= -R \binom{m}{\mathbf{x}-1}{\mathbf{1}-\mathbf{1}} p^{\mathbf{x}-1} (\mathbf{1}-p)^{m-\mathbf{f}(\mathbf{x}-\mathbf{1})} \\ &+ (mc+npc^{\prime}) \binom{m}{\mathbf{x}-1}{\mathbf{1}-\mathbf{1}} p^{\mathbf{x}-1} (\mathbf{1}-p)^{m-\mathbf{f}(\mathbf{x}-\mathbf{1})} + c \sum\_{\mathbf{Y}=\mathbf{x}}^{m} \frac{e^{-\lambda} \lambda^{\mathbf{Y}-\mathbf{x}}}{\Gamma(\mathbf{Y}-\mathbf{x}+\mathbf{1})} = \\ & \left(mc+npc^{\prime}-R\right) \binom{m}{\mathbf{x}-1}{\mathbf{1}-\mathbf{1}} p^{\mathbf{x}-1} (\mathbf{1}-p)^{m-\mathbf{f}(\mathbf{x}-\mathbf{1})} + c \sum\_{\mathbf{Y}=\mathbf{x}}^{m} \frac{e^{-\lambda} \lambda^{\mathbf{Y}-\mathbf{x}}}{\Gamma(\mathbf{Y}-\mathbf{x}+\mathbf{1})} \end{split} \tag{10}$$

To ensure that *x* minimizes the objective function (7), it is necessary to find the value of *x* that satisfies following inequalities:

$$
\Delta \mathbf{C\_{x+1}} = \mathbf{C\_{x+1}} - \mathbf{C\_x} > \mathbf{0\_r} \text{ } \Delta \mathbf{C\_x} = \mathbf{C\_x} - \mathbf{C\_{x-1}} < \mathbf{0\_r} \tag{11}
$$

Hence,

#### New Models of Acceptance Sampling Plans http://dx.doi.org/10.5772/50835 59

$$\begin{aligned} \Delta \mathcal{C}\_{\mathbf{x}+1} &= (mc+npc^\circ - R) \binom{m}{\mathbf{x}} p^\times (1-p)^{m-\mathbf{x}} + c \sum\_{\mathbf{Y}=\mathbf{x}+1}^m \frac{e^{-\lambda} \lambda^{\mathbf{Y}-\{\mathbf{x}+1\}}}{\Gamma(\mathbf{Y}-(\mathbf{x}+1)+1)} > 0\\ \Delta \mathcal{C}\_{\mathbf{x}} &= (mc+npc^\circ - R) \binom{m}{\mathbf{x}-1} p^{\mathbf{x}-1} (1-p)^{m-\{\mathbf{x}+1\}} + c \sum\_{\mathbf{Y}=\mathbf{x}}^m \frac{e^{-\lambda} \lambda^{\mathbf{Y}-\mathbf{x}}}{\Gamma(\mathbf{Y}-\mathbf{x}+1)} < 0 \end{aligned} \tag{12}$$

Now If*mc* + *npc* '<*R*, then,

In Eq. (7), *cE Y <sup>x</sup>*is the total cost of inspection and *npc* ' is the total cost of defective items. The optimal value of *x*is determined by minimizing the value of objective function*Cx*. Using

*m m*

*z z*

è ø è ø

( )

æ ö æ ö ç ÷ ç ÷ - +

æ ö

( )

G - -+ è ø

l


( 1) 1

*e* <sup>−</sup>*λλ <sup>m</sup>*−(*x*−1) *Γ*(*m*−(*x* −1) + 1)

( ) ( )

0

l

*Y x*

*Y* =*x*−1 *m*

) *p <sup>x</sup>*−1(1− *p*)*m*−(*x*−1)

) *p <sup>x</sup>*−1(1− *p*)*m*−(*x*−1)

) *p <sup>x</sup>*−1(1− *p*)*m*−(*x*−1)

To ensure that *x* minimizes the objective function (7), it is necessary to find the value of *x*

*m Y x*

2 '

1

*z x z <sup>m</sup> m z <sup>z</sup>*

= =

æ ö æ ö D= - = - + + ç ÷ ç ÷ - +

å å

*C C C R p p mc npc p p*

1

*m R pp z*


=- + - + ç ÷ G -+ è ø

è ø

*z x m x Y x m z <sup>z</sup>*

= -

å

*e m c Y mc npc p p Y x <sup>z</sup>*

1

*<sup>e</sup> c Y*

*<sup>Γ</sup>*(*<sup>Y</sup>* <sup>−</sup> *<sup>x</sup>* <sup>+</sup> 1) <sup>−</sup> ∑

*Y x*

To evaluate above equation, following equality is considered,

*<sup>Y</sup> <sup>e</sup>* <sup>−</sup>*λλ <sup>Y</sup>* <sup>−</sup>*<sup>x</sup>*

*<sup>Γ</sup>*(*<sup>Y</sup>* <sup>−</sup> *<sup>x</sup>* <sup>+</sup> 1) <sup>−</sup>*<sup>m</sup>*

*<sup>m</sup> e* <sup>−</sup>*λλ <sup>Y</sup>* <sup>−</sup>*<sup>x</sup>*

Since *m* is a sufficiently large number thus the value of *m*

*m x* −1

> *m x* −1

*m x* −1 = -

å

( ) ( ) ( )

1 1

*m x m z m z z z*


1

*<sup>Y</sup> <sup>e</sup>* <sup>−</sup>*λλ <sup>Y</sup>* <sup>−</sup>(*x*−1) *<sup>Γ</sup>*(*<sup>Y</sup>* <sup>−</sup>(*<sup>x</sup>* <sup>−</sup>1) <sup>+</sup> 1) <sup>=</sup>

> + *c*∑ *Y* =*x*

> > + *c*∑ *Y* =*x*

*<sup>m</sup> e* <sup>−</sup>*λλ <sup>Y</sup>* <sup>−</sup>*<sup>x</sup> <sup>Γ</sup>*(*<sup>Y</sup>* <sup>−</sup> *<sup>x</sup>* <sup>+</sup> 1) <sup>=</sup>

*ΔCx*+1 =*Cx*+1 −*Cx* >0, *ΔCx* =*Cx* −*Cx*−<sup>1</sup> <0 (11)

*<sup>m</sup> e* <sup>−</sup>*λλ <sup>Y</sup>* <sup>−</sup>*<sup>x</sup> Γ*(*Y* − *x* + 1)

+

1 '


0

*e* <sup>−</sup>*λλ <sup>m</sup>*−(*x*−1)

*<sup>Γ</sup>*(*m*−(*<sup>x</sup>* <sup>−</sup>1) <sup>+</sup> 1) is approximately

(8)

(9)

(10)

the optimization methods, it is concluded that,

1


*x xx*

58 Practical Concepts of Quality Control

( )

l

l

1

*Y x z*

∑ *Y* =*x m*

∑ *Y* =*x*

equal to zero therefore it is concluded that,

*ΔCx* = −*R*(

+(*mc* + *npc* ')(

(*mc* + *npc* '−*R*)(

that satisfies following inequalities:

Hence,

= =

å å

$$\begin{split} & \binom{m}{\chi-1} p^{\chi-1} \binom{\chi}{\chi-p}^{w-\{\chi-1\}} > \frac{c \sum\_{\chi=x}^{m} \frac{e^{-\lambda} \lambda^{\chi-x}}{\Gamma\left(\frac{\chi}{\chi}-\chi+1\right)}}{\left(R-\left(mc+npc^{\circ}\right)\right)} >\\ & c \sum\_{\chi=x+1}^{m} \frac{e^{-\lambda} \lambda^{\chi-x}}{\Gamma\left(\frac{\chi}{\chi}-\chi+1\right)} > \binom{m}{x} p^{x} \left(1-p\right)^{w-x} \end{split} \tag{13}$$

Since with increasing the value of *x* the value of binomial distribution with parameters *m* and *p* decreases thus according to the properties of binomial distribution, it is concluded that *x* >(*m* + 1)*p* therefore, the optimal value of *x*is determined using the following formula,

$$\mathbf{x} = \operatorname{Min} \left\{ \begin{aligned} &\mathbf{x}; \mathbf{x} > \left(m+1\right)p; \binom{m}{\mathbf{x}-1} p^{\mathbf{x}-1} \left(1-p\right)^{m-\left(x-1\right)} > \frac{c \sum\_{\mathbf{y}=x}^{m} \frac{e^{-\lambda} \lambda^{\mathbf{y}-x}}{\Gamma\left(\mathbf{y}-\mathbf{x}+1\right)}}{\left(R-\left(mc+npc^{\circ}\right)\right)} > \\ &c \sum\_{\mathbf{y}=x+1}^{m} \frac{e^{-\lambda} \lambda^{\mathbf{y}-x}}{\Gamma\left(\mathbf{y}-\mathbf{x}+1\right)} > \binom{m}{\mathbf{x}} p^{\mathbf{x}} \left(1-p\right)^{m-\mathbf{x}} \end{aligned} \right\} \tag{14}$$

Also The objective function, *Cx*, should be minimized regarding two constraints on Type-I and Type-II errors associated with the acceptance sampling plans. Type-I error is the proba‐ bility of rejecting the batch when the nonconformity proportion of the batch is acceptable. Type-II error is the probability of accepting the batch when the nonconforming proportion of the batch is not acceptable. Then, in one hand, if*p* =*δ*1, the probability of rejecting the batch should be less than*α*. On the other hand, in case where*p* =*δ*2, the probability of accept‐ ing the batch should be less than*β* where *δ*1 is the AQL (Accepted Quality Level ) and *δ*2is the LQL (Limiting Quality Level) and*α* is the probability of Type-I error and *β* is the proba‐ bility of Type-II error in making a decision, therefore, the optimal value of *x*is determined using the following formula,

$$\begin{aligned} \text{If } & x > \left( m + 1 \right) p^\*; \\ \text{If } & x = \lim\_{\begin{subarray}{c} x = 1 \\ x = 1 \end{subarray}} \left( \frac{m}{x - 1} \right) p^{x - 1} \left( 1 - p \right)^{m - \left( x - 1 \right)} > \frac{c \sum\_{\begin{subarray}{c} x = 1 \\ x = 1 \end{subarray}}^{m} \frac{e^{-\lambda} \lambda^{\text{y} - x}}{\Gamma \left( \frac{\lambda}{x} - \left( mc + npc^{\circ} \right) \right)} > \\ & c \sum\_{\begin{subarray}{c} x = 1 \\ x = 1 \end{subarray}}^{m} \frac{e^{-\lambda} \lambda^{\text{y} - x}}{\Gamma \left( \frac{\lambda}{x} - \left( mc + npc^{\circ} \right) \right)} > \binom{m}{x} p^{x} \left( 1 - p \right)^{m - x} \\ & \sum\_{z = \omega}^{m} \binom{m}{z} \delta\_{1}^{z} \left( 1 - \delta\_{1} \right)^{m - z} \le \alpha, \sum\_{z = 0}^{x - 1} \binom{m}{z} \delta\_{2}^{z} \left( 1 - \delta\_{2} \right)^{m - z} \le \mathcal{B} \end{aligned} \tag{15}$$

*E*(*TC*): The expected total cost of the system

*U* : The upper control threshold

*L* : The lower control threshold

pressed using Eq. (17).

Let *Yi*

State 1: *Yi*

State 2: *Yi*

accepted.

State 3: *Yi*

rejected.

ess continues.

*E*(*AC*): The expected total cost of accepting the batch

*E*(*RP*): The expected total cost of rejecting the batch

*E*(*I*): The expected total cost of inspecting the items of the batch

Consider an incoming batch of *N* items with a proportion of nonconformities*p*, of which items are randomly selected for inspection and based on the number of conforming items between two successive nonconforming items, the batch is accepted, rejected, or the inspec‐ tion continues. The expected total cost associated with this inspection policy can be ex‐

forming items, *U* the upper and *L* the lower control thresholds. Then, if *Yi* ≥*U* the batch is accepted, if *Yi* ≤ *L* the batch is rejected. Otherwise, if *L* <*Yi* <*U* the process of inspecting

falls within two control thresholds L, i.e.,*L* <*Yi* <*U* , thus the inspection proc‐

is more than or equal the upper control threshold, i.e.,*Yi* ≥*U* , hence the batch is

is less than or equal the lower control threshold, i.e.,*Yi* ≤ *L* , hence the batch is

where the probabilities can be obtained based on the fact that the number of conforming

distribution with parameter*p*, i.e., Pr(*Yi* <sup>=</sup>*r*) =(1<sup>−</sup> *<sup>p</sup>*)*<sup>r</sup> <sup>p</sup>*;*<sup>r</sup>* =0, 1, 2, ...Then, the transition proba‐

*th* and *<sup>i</sup> th* nonconforming items, *Yi*

be the number of conforming items between the successive (*i* −1)

items continues. The states involved in this process can be defined as follows.

The transition probabilities among the states can be obtained as follows.

Probability of inspecting more items=*p*<sup>11</sup> =Pr{*L* <*Yi* <*U* }

Probability of accepting the batch=*p*<sup>12</sup> =Pr{*Yi* ≥*U* }

Probability of rejecting the batch=*p*<sup>13</sup> =Pr{*Yi* ≤ *L* }

items between the successive (*i* −1)

bility matrix is expressed as follows:

*E*(*TC*)= *E*(*AC*) + *E*(*RP*) + *E*(*I*) (17)

*th* and *i th* noncon‐

New Models of Acceptance Sampling Plans http://dx.doi.org/10.5772/50835 61

, follows a geometric

When*mc* + *npc* '>*R*, It is concluded that Eq. (16) is positive for all values of *x*so*x* =0. In this case, if one defective item is found in an inspected sample then the batch would be rejected. In this case, the rejection cost *R* is less than the total cost of inspecting *m* items and the cost of defective items, hence rejecting the batch would be the optimal decision. However, in practice the rejection cost *R* is usually big enough so that, we overlooked that case.

$$\Delta \, \mathcal{C}\_{\mathbf{x}} = (mc + np\mathbf{c}^\prime - R) \binom{m}{\mathbf{x} - \mathbf{1}} p^{\mathbf{x} - \mathbf{1}} (\mathbf{1} - p)^{m - (\mathbf{x} - \mathbf{1})} + c \sum\_{\mathbf{y} = \mathbf{x}}^m \frac{e^{-\lambda} \lambda^{\mathbf{y} - \mathbf{x}}}{\Gamma(\mathbf{y} - \mathbf{x} + \mathbf{1})} \tag{16}$$
