Perfect repair for observed nonconforming items

2.3 Complete Inspection and Adopting Two Actions for Observed Nonconforming Items

In complete inspection plan, assume that the producer takes perfect repair or sells at low price for nonconforming items. In order to compare the expected profits per unit time of using complete inspection plan and process control, we need to derive their profit models.

2.3.1 Perfect repair for observed nonconforming items

If the item does not satisfy producer specifications then it will be perfect repaired.. We assume that the repaired product would be equal to the target value. The repaired cost per item is assumed to be Cpr. Because producer only knows the observed quality, the observed quality per item will conform the target value after repair (i.e. Yr=T). Let Xr be the true quality distribution of the repaired

nonconforming item. Since

Y

_p

= X

_p

+ ε

_p , and after repair Yr=T, so

X

_r

= − T ε ~ N T , σ

_p

(

²_pe

)

.

Figure 2.1 Producer specification based on Y

P

, nonconforming item

may perfect repaired with repair cost C

pr.

‧

Figure 2.2 Quadratic loss function based on X

P

Assumptions, and also assume the pdf is the same under process is in-control and out-of-control.

For in-control Xr , its

the pdf is,

2.3.1-1 Derivation of the profit model per unit time

For in-control process, the expected profit per item contains, (1) The sale price for each conforming item is, PPM,

(2) The expected loss of a conforming item is,

‧

(3) The expected loss of a nonconforming item is,

1

^p

( ) ( )

2

( )

(4) Inspection cost of an item for producer is, IC.

Hence, the expected in-control profit per item for producer is E(PI), that is, producer selling price minus the cost of producer, The expected out-of-control profit per item contains

(1) The sale price for each conforming item is PPM

(2) The expected loss of a conforming item is,

( )

(3) The expected loss of a nonconforming item is,

1

^p

( ) ( )

2

( )

(4) Inspection cost of an item for producer is, IC.

Hence, the expected out-of-control profit per item for producer E(PO), is producer selling price minus the cost of producer. That is,

‧

To compare the profits for producer using complete inspection plan and that of using process control, we first derive the profit model for complete inspection plan. Assume that the time between the occurrences of a shift to the out-of-control state is an exponential distribution with mean θ, and take m unit time as comparison criteria for comparing the profit of PC with the profit of complete inspection, where m should be larger than the expected per cycle times of PC.

Figure 2.3 Total profit in m unit time for producer take complete inspection

The expected profit of per time is the sum of the in-control profit in m unit time and out-of-control profit in m unit time divided by m unit time, that is,

1

Since we also want to compare the yield of perfect repair action and that of sell low price action,

we need to calculate the yield. The yield calculation of perfect repair is using the mixture distribution

‧

of Xr and XP within specification limits. Therefore, we illustrate the derivation of the mixture

distribution as follows, and provide a simulation example to verify the theoretical mixture distribution.

2.3.1-2 The mixture distribution (X

m

) of the true quality

(1). The distribution of Xm

For

X ~ N µ ,σ

_p

(

_{x x}²

)

and

ε ~ N 0, σ

_p

(

²_pe

)

, then

Y

_p

= X

_p

+ ε ~ N µ , σ

_p

(

_x ²_x

+ σ

²_pe

)

.

The true quality distribution for repaired nonconforming item is,

(

²

)

r pe

X ~ N T, σ

Then the mixture distribution (Xm) is,

( )

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Figure 2.5 illustrates the histograms of XP within (LSLP, USLP) and Xr , and Figure 2.6 illustrates the histogram and theorical curve of mixture distribution.

Table 2.1 Numeric example for finding true quality distribution (X

M

) of products received by middleman

No.(i) Xpi εpi Ypi Within (LSLP, USLP) Mixture distribution (Xm)

1

349.706 0.197 349.903 Y Xp1 = 349.706

2

350.534 -0.035 350.499 Y Xp2 = 350.534

M M M M M M

62

351.977 0.288 352.265 N XR62 =T-εp62 = 349.712

63

349.669 0.05 349.719 Y Xp63 = 349.669

64

351.712 -0.277 351.435 Y Xp64 = 351.712

65

352.879 -0.124 352.755 N XR65 =T-εp65 = 350.124

M M M M M M

1999

351.229 -0.397 350.832 Y Xp1999 = 351.229

2000

351.239 -0.249 350.99 Y Xp2000 = 351.239

Figure 2.5 histograms of X

P

within (LSL

P

, USL

P

) and X

r

Figure 2.6 histogram of mixture distribution and theorical curve

‧

Now we apply the goodness of fit to prove that the data in Table 3.1 has the same distribution as in (2.6).

First we states the null hypothesis and the alternate hypothesis,

H0︰The data follows the mixture distribution

H1︰The data do not follows the mixture distribution And we select the significant levels, α = 0.05 in this problem.

Then we cut the data into eight intervals, and count the number in each interval, also use the mixture distribution to calculate the expected frequency in each interval.

Table 2.2 The number of simulate data in each interval and expected frequency

<349 349~349.5 349.5~350 350~350.5 350.5~351 351~351.5 351.5~352 >352

Oi

44 120 313 374 405 378 260 106

Pi

0.0218 0.0497 0.145 0.203 0.197 0.191 0.14 0.05

Ei

43.689 99.345 290.8 406.869 394.155 382.539 280.345 102.256

The test statistic is,

( ) (

²

) (

²

) (

²

)

²

406 869 394 155 382 539

. . .

And the rejection region is,

{

^{2 0 05 (7)2. ,}

^{14 067} }

R.R.= χ > χ = .

Since

χ ² ≈

10 612

. ∉ R.R.

, so we do not reject H0, that is, we do not have sufficient evidence to say this simulate data do not follows the mixture distribution.

‧

After having the mixture distribution, then we may calculate its yield (see (2.7)). .

(

^{*p m *p}

)

T d^{T d*}_p^{*p Xm( )}

2.3.1-3 Determine the optimal inspection specification limits and data analyses

Table 2.3 Three Levels of Each Parameters

Level

Since 8 parameters each with 3 levels, the parameters could be assigned to each combination of orthogonal array table

^L

²⁷

( ) ³

¹³ (see table 2.4).

‧

Let the expected profit of per time under producer taking perfect repair for nonconforming item is

IS1

EPR

P . To determine optimal producer specification limits, we maximize

IS1

EPR

P with the constraint

2 2

p

3

x pe

0≤d ≤ σ +σ by using “optim” routine in R program, where σ²_x+σ²_pe is the standard deviation of YP, and here use three times σ²_x +σ²_pe is because it can more easy to find profit difference in the two actions than use six times σ²_x+σ²_pe.

Table 2.4 The 27 combinations of these parameters by using an orthogonal array table L

₂₇

(3 )

¹³

_.

No. δ δ1 δ2 δ3 σx (θ, m) (R, Cpr, IC, PPM) kp

‧

Table 2.5 The optimal solutions of the 27 combinations of parameters

No.

d^*_P

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The response figure and table are applied to find the significant parameters of

d

^*_P and

EPR*

PIS₁.

(I) Response figure and table of

d

^*_P

Figure 2.7 Response figure of d

^*_P

for each parameter

Table 2.6 Table Response table of d

^*_P

δ δ1 δ2 δ3 σx (θ, m) (R, Cpr, IC, PPM) kp

level1 2.82 2.844 2.677 4.343 1.289 2.297 2.705 3.003

level2 3.003 2.54 2.83 2.503 3.194 3.232 2.649 2.848

level3 2.389 2.829 2.705 1.367 3.73 2.684 2.859 2.361

diff 0.614 0.304 0.153

2.976 2.441

0.935 0.21 0.642

Based on the Table 2.6, if the difference between maximal and minimal values of the three levels is larger than 1.5, the parameters, δ3 and σx are determined to be significant in relation to

d

^*_P .

The optimal producer specification is determined by the constraint of the variance of observed quality characteristics,

^σ

²_x

+ ^σ

²_pe. Therefore, δ3 and σx are significant parameters of

d

^*_P , and the δ3 and σx trends are,

(1) when δ3 increases then

d

^*_P decreases.

(2) when σx increases then

d

^*_P increases.

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(II) Response figure and table of

EPR*

PIS₁

Figure 2.8 Response figure of

IS1

EPR*

P for each parameter

Table 2.7 Response table of

IS1

EPR*

P

δ δ1 δ2 δ3 σx (θ, m) (R, Cpr, IC, PPM) kp

level1 ^{10552.01 10294.78 9490.604 9508.316 10818.77 10372.61 22105.46 10382.6} level2 ^{9896.423 9948.406 10586.46 10431.54 9877.818 9767.996 5544.863 9888.289} level3 ^{9518.19 9723.445 9889.566 10026.77 9270.042 9826.019 2316.309 9695.736} diff ^{1033.823 571.333 1095.853 923.227 1548.726 604.616 19789.15 686.865}

Based on Table 2.7, if the difference between the maximal and minimal values of the three levels is larger than 3000, the parameter (R, Cpr, IC, PPM) is determined to be significant for

EPR* .

PIS₁

When (R, Cpr, IC, PPM), increases

EPR* decreases. Although P

PIS₁ PM increases between the three levels, but the difference in PPM value is not large. Moreover, R decreases between the levels, but the difference in R value is large. Therefore

EPR* is significantly affected by the amount of production

PIS₁

(R). This means when per item profit is small, the producer should adopt mass production to increase profit.

‧

In 2.3.1-3 data analyses, the parameter combination of (θ, m) may relate to the time of in-control and out-of-control process, and the ratio of 1/θ to m is the proportion of expected in-control time to the m unit time, here denoted by η. The expected in-control time can’t be shorter than out-of-control time, that is, η should be larger than 0.5. But in the data analysis there are two parameter combination of (θ, m), (0.05, 150) and (0.1, 120) are not appropriate, since the η values are 0.14 and 0.08 respectively, so here we change the η value to be 0.85 and 0.95 to find whether the optimal specification limits and the

IS1

EPR*

P may be changed or not.

Here we also revise the profit model to be:

( )

Table 2.8 The optimal solutions of the 18 combinations of parameters under different η value

No. η d^*P

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