On the Determination
of Single Sampling
Attribute
Plans Based upon a Linear Cost
Model and a Prior Distribution
WILLIAM
C.
GUENTHERDepartment of Statistics, University of Wyoming and
University
of
OsloThe linear cost model previously formalized by Hald [4], [5], [9] is reviewed. Tech- niques are described which permit easy determination of sampling plans based on that model. The degenerate, the beta, and the two point distributions are considered as prior distributions of p, the process fraction defective. For calculations only standard tables and a desk calculator are required.
1. I?~TR~DUCTI~X
In recent years a number of papers have appeared concerning sampling in- spection models which are constructed under the assumption that both costs and a prior distribution of p, the process fraction defective, should be incor- porated into the model. Hald [4], [5], [9] has done extensive work in this area and is responsible for the notation of Section 2. This paper presents elementary procedures by which these sampling plans can be determined and through which properties of the plans can be investigated.
In our discussion and evaluations we will need some of the well known discrete probability functions. For a binomial random variable X we will use the notation
wx; 12,
PI = 0
; PV - PI”-“, 5=0, 1,2;..,n (1.1) for the probability function and cumulative sums will be denoted byand
(1.3) The most convenient table to use is the one published by the Ordnance Corps [16] which gives (1.3) to seven decimal places for p = .01(.01).50 and n = 1(1)150. The Harvard [12] table gives the same sum to five decimal places for
Received Mar. 1969; revised Aug. 1969. 483
n’s up to 1000 but in jumps ranging from 2 units between n = 50 and n = 100 to 50 units between n = 500 and n = 1000. The probability function of a hy- pergeometric random variable will be denoted by
tK:~,u<~<, p(N, n, k, d = N
0
- - (1.4)
n
where a = max [0, n - N + k], b = min [k, n] and cumulative sums by
PP, n, k 4 = 2 PW, n, k 4. (1.5)
z-a
Lieberman and Owen [13] have given both (1.4) and (1.5) to six decimal places for N = (1)50(10)100. For other N we use
or I P(N, n, k, r) C B r; n, k ( > , n 5 k, f+ 2 .lO (1.6) P(N, n, k, r) s B r; k, n_ ( N ’ > k < n, $ 5 .lO (1.7) approximations which appear to be quite adequate. For the Poisson probability function we use
and
for cumulative sums. Both Molina [14] and General Electric [3] have tabulated (1.8) and (1.9) to at least six decimal places for a considerable range of ~1. The former table is probably better if p > 2 and the latter if p _< 2. Finally, we also need the negative binomial probability function
f(x* k p) = ‘(’ + k, ~“(1 - p)” 7 , -__ x! l?(k) 7 2 = 0, 1, 2, **a (1 JO) with cumulative sums
Williamson and Bretherton [17] have given both (1.10) and (1.11) to six decimal places for p mostly in steps of .02 and IC mostly in steps of .l, a table that is very adequate for our purposes.
Terminology for single sampling attribute plans is fairly standard. Items pro- duced by a process are assembled at random into lots of size N. From each lot a random sample of size n is selected and X, the number of defective items in the
sample, is observed. If 2 5 c, where c is called the acceptance number, the lot is accepted and if 2 > c the lot is rejected. One course of action (sometimes called rectifying inspection) which is frequently followed calls for replacement of defective items found in sampling accepted lots and for total inspection of rejected lots with all defectives in these lots being replaced.
Let Y be the number of defective items in the lot. We assume that the behavior of Y is governed by a binomial distribution with parameters N and p. A given lot will have a fixed number of defectives, say Y = y = k, and the probability function of X given k is the hypergeometric (1.4). Further, if we let U = Y - X be the number of defective items in the remaining N - n items in a lot, it is well known (i.e., see Hald [lo], pp. 401-402) that the unconditional distributions of X and U are independent and binomially distributed with parameters n, p and
N - n, p respectively.
The probability of accepting a lot is called the operating characteristic (OC) and is
OC = P(N, n, k, c) (1.12)
a sum of hypergeometric probabilities. Since (1.12) is a conditional probability depending on k, we may be more interested in the average operating character- istic (AOC), the expected value of (1.12) taken over Y, given by
AOC = B(c; n, p). (1.13)
From a practical standpoint usually it makes little difference which view is adopted since, according to (1.6)) (1.13) approximates (1.12) with p = k/N.
2. THE LINEAR COST MODEL
Suppose that a single sampIe rectifying inspection plan as described in the introduction is being used on lots of size N. We will assume that associated with the sampling plan is a cost function which can be expressed in the form h(x, k, P; N, n, 4 = n& + x& + (N - 44 + (k - x)A, , x s c (2.1)
= n& + xS, + (N - n)R, + (k - x)R, , 2 > c. Hald [4] has suggested the following interpretation for the constants: S1 Cost per item of sampling and testing
S, Repair cost for a defective item found in sampling
A, Cost per item associated with handling the N - n items not inspected in an accepted lot (frequently is 0).
A2 Cost associated with a defective item which is accepted (may be quite large) RI Cost per item of inspecting the remaining N - n items in a rejected lot R, Repair cost associated with a defective item in the remaining N - n items
of a rejected lot.
Logically we would expect that S, 1 R, , S, 2 R, (with equality frequently holding) since it should be no more expensive to sample or repair on a large scale than on a small scale.
As an example suppose that a manufacturer of boys jackets assembles lots of size 100 for shipment. An item is regarded as defective if it is judged to require repair before it would be acceptable to a customer. Let us assume that R, = S, = .lO, R2 = S, = 2.00, A, = 0, A, = 4.00, figures determined on the basis of accounting records. The cost A, arises from the fact that the buyers also use sampling inspection procedures and may return individual items or an entire lot.
The plans which we intend to discuss are based upon the average cost per lot. To obtain that average we take the expected value of the random variable associated with (2.1). Recall that U and X are independent so that E(U 1x) = (V - n)p. Hence, if we take the conditional expectation with respect to U given X = x followed by taking the expectation with respect to X, we get (for fixed p) that the average cost is
K(N, n, c, p> = n&(p) + W - n)K(p)P(p) +
K(p)&(p)1
(2.2) where K,(p) =S,
+&I?,
K,(p) = Al + A,P, K,(p) = RI + R,p, P(P) = B(c; n, p), Q(p) = 1 - P(p). Alternatively we may write (2.2) using only one binomial sum asKC% n, c, PI = 6(p) + W - n)IK(p) + [K(P) - UP)IQ(P)I. (2.3) Denote the solution of K,(p) - K,(p) = RI - A, - (A, - RJp = 0 by p, . Usually we will have 0 < p, < 1. We see that
P < Pr , KO - KaO > 0 (2.4)
P > Pr 7 K,Cd - K(P) < 0. 3. PRIOR DISTRIBUTIOS OF p DEGENERATE
We first consider the case in which the distribution of p concentrates all of its probability at one point. In other words p does not really have a distribution but assumes some unknown value with probability 1 (the situation of classical statistics). If in (2.2) we let RI = X1 = 1 and all other cost constants are set equal to zero, then
KW, n, c, p> = n + W - n)E(c 4 1; n, p> (3.1) the cost function for rectifyin g inspection proposed by Dodge and Romig [l], [2] as early as 1929. Hald [7] has already mentioned the generalization of (3.1) considered in this section.
If we knew p, then we could always minimize (2.2). If p < p, , then we would always accept without sampling so that P(p) = 1, n = 0 and
K(N, n, c, PI = ~liK,(p). (3.2)
On the other hand if p > p, we would always reject without sampling so that P(p) = 0, n = 0 and
K(N, n, c, p) = NK,(p). (3.3)
Since p is generally unknown, a reasonable type of problem is to minimize K(N) n, c, @), where @ is our guess for the true p, subject to one or two conditions
on the OC (or AOC) curve. For example, we could require WV, n, k, , c> I PI
(as Dodge and Romig have done) or
P(N, n, Lo ) c) 2 1 - a0 or both (3.4) and (3.5).
(3.4)
(3.5) For the minimization procedure it is convenient to rewrite (2.2) as
K(N, n, c, p) = n[K,(p) - K,(p)] + (N - n)PL(p) - fL(P)lJ%) + NK(P) (3.6) If R, = S, , R, = S, so that K,(p) - K,(p) = 0, then let
w, n, c, P> = VW,
n, c, P) - ~K(p~lIKdP~ - adI
(3 *7)= (N - n)P(p). (3.8)
To minimize (3.6) maximize or minimize (3.8) according to whether the denomi- nator of (3.7) is negative or positive. If either IS’, > R, or Sz > R, (or both) let
RW, n, c, P) = VW,
n, c, P) - ~K(P~IIPLcP~ - mP)1
(3 9
= n + (N - n)rPbl
(3 .lO) where y = [K,(p) - K,(p)]/[K,(p) - K,(p)] and minimize (3.10). In all situ- ations either the minimum is obvious or minimization is easily accomplished by first finding the minimum with c = 0, then the minimum with c = 1, then with c = 2, etc., terminating when it is obvious that the absolute minimum has been found. We illustrate by examples.Example 3.1
Using the cost constants of the jacket example of Section 2, minimize the average cost when p = .02 subject to the condition (a) OC I .lO if k = Icl = 10, (b) OC 2 .95 if k = k, = 1, (c) both conditions (a) and (b).
Solution
(a) Since RI = S, = .lO, Rz = S, = 2.00, K.J.02) - KJ.02) = -.lO + 2(.02) = -.06, we maximize R(lOO, n, c, .02) = (100 - n)[l - E(c $ 1; n, .02)] subject to P(100, n, 10, c) 5 .lO. Using the Lieberman and Owen [13] table we observe that the side condition is satisfied if n > 20 with c = 0, if n 2 33 with c = 1, if n 2 44 with c = 2, if n > 55 with c = 3. With each value of c we need consider only the smallest value of n since both 100 - n and 1 - E(c + 1; n, .02) decrease with increasing n.
Using the Ordnance Corps [16] table for binomial sums we obtain R(100, 20, 0, .02) = 80(.66761) = 53.41, R(100, 33, 1, .02) = 67(.85917) = 57.56, R(lOO, 44, 2, .02) = 56(.94223) = 52.76
and the desired plan is n = 33, c = 1 with K(100, 33, 1, .02) = (-.06)(57.56) + 100(.14) = 10.55.
(b) The side condition P(100, n, 1, c) 2 .95 is satisfied for n < 5 with c = 0 and for all n 5 100 with c 2 1. Now R(lOO, n, c, .02) is maximized if we always accept with no sampling and K(100, 0, c, .02) = 8.00.
(c) With both conditions we see from parts (a) and (b) that solutions are possible for c = 1, 33 5 n 5 100, for c = 2, 44 I n I 100, for c = 3, 55 5 n _< 100 and the desired plan is n = 33, c = 1.
Example 3.8
Repeat Example 3.1 with N = 1000, k, = 10, lc, = 100. Solution
Now the side conditions are expressed in terms of the binomial approximation and are E(c + 1; n, .lO) 2 .90, E(c + 1; n, .Ol) < .05. Otherwise, the solution is obtained the same way and the reader may verify that in (a), (c) we get n = 78, c = 4, K(1000, 78, 4, .02) = 95.80 while in (b) we accept with no sampling and K(lOOO, 0, c, .02) = 80.00.
4. THE BETA PRIOR DISTRIBUTION
In many situations it is probably not realistic to assume that p remains con- stant from lot to lot. For example, suppose that a machine is used to produce a particular item. After every lot the machine is checked and reset at the correct value. We could assume that during the production of a particular lot p remains constant (and the other binomial conditions are satisfied) but for each lot p is determined according to some distribution. That distribution is called the prior distribution of p.
Perhaps one of the most reasonable choices for a prior distribution of p is the beta distribution with density
fcP; a1 I 4 = a(ul1, u,) P”‘-‘(l - PY, O<p<l (4.1) where a, > 0, az > 0 are constants suitably chosen to fit a given problem.
(Estimates for a, , a2 are given later in the section.) We will be interested in distributions which concentrate the probability about small values of p. Since the expected value of a beta random variable is ~,/(a, + a,), this means that az will be somewhat larger than a, . No matter how estimates of the u’s are obtained we will round al and a2 to the nearest integers if the estimate of a, is greater than or equal to 1 and a, to the nearest .l if the estimate is between 0 and 1 (a, need not be rounded), keeping the expectation of p as close as possible to the estimate of the average value of p. This convention allows us to use stand- ard tables for the determination of sampling plans and, due to the apparent insensitivity of plans to small changes in the parameters, without undue restric- tions on the choice of priors available. Even with these limitations we still have available three types of beta distributions:
1. Those with a density which is 0 at p = 0 and p = 1 and rises to a
single
2. Those with a density which is 1 at p = 0 and decreases monotonically (a, = 1, a, > 1).
3. Those with a density which is monotonic decreasing with no maximum (0 < a, < 1, a, > 1).
As we already mentioned in Section 2, the plans which we discuss are based upon the average cost per lot. With a non-degenerate prior we have three random variables to consider, those associated with x, k, and p. We already averaged over the first two and obtained (2.2). Now we average K(N, n, c, p) over all values of p. Denote that result by K(N, n, c).
We first assume that both a, and CL* are integers. For the purpose of finding the expected value with respect to p, rewrite (2.2) as
KW, n, c, p> = n(S1 + SJp
+ (N - n> {[(Al - Rd + (A2 - R,)plP(p) + (RI +
JLp)].
(4.2)
The four expectations arising from (4.2) are E[n(S, + &p)] = n (‘I + 1:: t u’s’ ,
2
1
E[(iV - n)(R1 + R2p)] = (N - n)
(4 -3)
We next show that (4.5) and (4.6) can be expressed in terms of hypergeometric sums. To do this consider the inverse hypergeometric distribution (the termi- nology of Patil and Joshi [15]). F rom N items, k of which are defective, drawings are made one at a time at random and without replacement. The probability that u draws are required to obtain d defectives is
g(u) = N - (u - 1) ’ k-d+1
u=d, d+l,...,N--k+d
and the probability that r or less draws are required to obtain the dth defective is (4.7)
Now suppose we set as a goal the obtaining of d defectives but agree to limit the number of draws to r. Then the goal will be achieved if the dth defective is obtained on the dth draw, the d + lth draw, . . . , or the rth draw. Hence the probability of achieving the goal is (4.7). Alternatively, the goal is achieved if we make r draws and obtain d, d + 1, . . . , or r defectives. The probability of the later event is
2 (:K I :)/(9
= 1 - P(N, r, k, d - 1). Hence we havewhich enables us to evaluate inverse hypergeometric probabilities from a hyper- geometric table.
In (4.5) let u = y + a, . Then, using (4.8) and identifying d = a, , k = a, + a, - 1, r = c + a, , N = n + a, + a2 - 1 we get
E[P(p)] = 1 - P(n -I- a, + a, - 1, c + al , a, -l- a2 - 1, a, - 1). (4.9) Similarly, letting u = y + a, + 1 in (4.6) yields
E[PWI = A P - f’(n + al + a2 , c + a, + 1, a, + a2 , a,)]. (4.10) 1
Now define
where
R(N, n, c) = [K(N, n, c) - NK,]/(K. - K,) (4.11)
K, = [(a, + G)& + a,&ll(a, + a,>,
(4.12) Km = I(a, + ~01 + dLl/‘(~, + a,>. (4.13) Then, using (4.3), (4.4), (4.9), (4.10) to evaluate the expectation associated with(4.2) yields
R(N, n, 4 = n + (N - n>Ml - P(n + al + a2 , c + al + 1, al + a2 , aI>1 + Y2Pbl + a1 + a2 - 1, c + a1 , a, + a2 - 1, a, - 1)) (4.14) where
~1 = (A, - R,)[u,,‘(a, + u,)]j(K, - Km), (4.15)
-/z = (R, - AJ,‘(K, - Km). (4.16)
(We note that if S, = R2 , then K, - K, = S, - Al .)
There are, of course, different types of sampling plans which can be determined incorporating the average cost function K(N, n, c). If n and c are determined so that K(N, n, c) is a minimum, then (n, c) is called a Bayesian sampling plan. When n and c are determined so that K(N, n, c) is minimized subject to side conditions on either the OC or the AOC curve, then (n, c) is called a restricted
Bayesian plan. One reason for using a restricted Bayesian plan is that a
Bayesian
solution requiring no sampling may be unsatisfactory. We will discuss both the Bayesian and the restricted Bayesian cases.Let us first consider Bayesian plans. If the coefficient of Q(p) in (2.3) is always negative, as it is if p, < 0, the K(N, n, c) is minimized by taking Q(p) = 1, n = 0. The minimum value under these circumstances is
K(N, n, c) = NK, . (4.17)
On the other hand, if the coefficient of Q(p) is always positive, as it is if p, > 1, the minimum is obtained by taking Q(p) = 0, n = 0 and now the minimum value is
K(N, n, c) = NK, . (4.18)
In the minimization process it is more convenient to work with R(N, n, c),
given by (4.14). Consequently, it is useful to know that corresponding to (4.17) we have
R(N, n, c) = N-/s (4.19)
and to (4.18)
R(N, 7~, c) = Ny, . (4.20)
If 0 < p, < 1, R(N, n, c) may achieve its smallest value at min (N-r1 , N-y*). Hence this minimum should be computed early in a problem. The remainder of the procedure for minimizing (4.14) consists of determining by trial the smallest R(N, n, 0), then the smallest R(N, n, l), then the smallest R(N, n, 2), etc., terminating when the absolute minimum has been found.
Before we demonstrate the minimization procedure several comments seem to be in order. For fixed N and c we take R(N, 0, c) = min (Nrl , NY,). Numerical results indicate that for fixed N and c, R(N, n, c) behaves in one of the following ways:
(a) It drops to a relative minimum below min (~1~7~ , NT,) and then increases monotonically.
(b) It is a monotonic increasing function of n.
(c) It rises to a relative maximum, then drops to a relative minimum below min (NT, , N-rJ, then increases monotonically.
(d) It behaves as in (c) except that the relative minimum is above min (h , NrA.
Further, these minima over n when plotted against c have a unique minimum which may or may not occur at c = 0. It can be added that any such “curves” (actually a set of isolated points) can be drawn as accurately as desired by using more values of n or c, whichever is appropriate.
Intuitively, we might expect that the minimum of R(N, n, c) is near Nrl if there is little of the p distribution above p, (small probability of exceeding pr). Similarly, if there is very little of the p-distribution below p, we
might expect
Example
4.1
Let the cost constants be those of the jacket example, RI = Xl = .lO, R, = S, = 2.00, A, = 0, A, = 4.00. Suppose that it is decided that p has a beta distribution with a, = 1, az = 19 (so that the mean of the distribution is .05). Using N = 100 find (a) the Bayesian plan, (b) the restricted Bayesian plan satisfying OC 5 .lO if k = k, = 10, (c) the restricted Bayesian plan satisfying OC 2 .95 if k = k, = 1, (d) the restricted Bayesian plan satisfying both the OC conditions given in (b) and (c).
Solution
We get K, = .20, K, = .lO, K, = .20, y1 = yz = 1, NrI = Nrz = 100, p, = .05 and
R(lOO, n, c) = n + (100 - n)[l - P(n 4 20, c + 2, 20, 1)
+ P(n + 19, c + 1, 19, 0)l. (a) We first try c = 0. Then with the table of Lieberman and Owen [13] we find
R(lOO, 5, 0) = 5 + 95[1 - P(25, 20, 2, 1) + P(24, 19, 1, O)] = 5 + 95[.633333 + .203333] = S5.96.
Similarly R(lOO, 10, 0) = 80.34, R(lOO, 15, 0) = 79.64, R(lOO, 20, 0) = 80.51, R(100, 14, 0) = 79.61, R(lOO, 13, 0) = 79.65, R(lOO, 12, 0) = 79.76 and the minimum occurs at n = 14.
With c = 1 we get R(lOO, 20, 1) = 80.51, R(lO0, 25, 1) = 79.91, R(lOO, 30, 1) = 80.32, R(lOO, 26, 1) = 79.93, R(lOO, 24, 1) = 79.93. The minimum occurs at n = 25. Since this minimum is larger than the one obtained with c = 0, we ter- minate calculations.
Thus the desired plan is n = 14, c = 0 with R(lOO, 14, 0) = 79.61. From (4.11) we find K(lOO, 14, 0) = 17.96.
(b) The condit’ion on OC is P(100, n, 10, c) _< .lO. As in Example 3.1 we need n 2 20 with c = 0, n 2 33 with c = 1, n 2 44 with c = 2. We found above that
with c = 0, R(lOO, n, 0) is minimized when n = 14 and is increasing when n 2 14. Hence, with the restriction on OC we minimize R(100, n, 0) with n = 20 and R(lOO, 20,O) = 80.51 as previously found. Similarly, with c = 1 the unrestricted minimum occurs at n = 25 and R(lOO, n, 1) is increasing for n > 25. Thus, with the restriction we minimize R(lOO, n, 1) with n = 33 and R(lOO, 33, 1) = 33 + 67[1 - P(53,3,20,1) + P(52,2, 19, O)]. Now we use the binomial approximation and P(53, 3, 20, 1) g 1 - E(2; 3, .377), P(52, 2, 19, 0) s 1 - E(1; 2, .365) where 20/53 = .377, 19/52 = .365. Then, usin g linear interpolation in the binomial table yields
R(lOO, 33, 1) = 33 + 6711 - .682 + .403] = 81.3.
We could try c = 2, 3, etc. but the cost function appears to be increasing. The
desired plan is n = 20, c = 0 with R(100, 20, 0) = 80.51, K(lOO, 20, 0) = 18.05. (c) To satisfy P(100, n, 1, c) 2 .95 we need n 5 5 with c = 0 and n _< 100
n = 14 and R(lOO, n, 0) gets larger as n is decreased. Hence with the restriction and c = 0 we minimize R(lOO, n, 0) with n = 5 and R(lOO, 5, 0) = 84.96.
For
c >_ 1 all n are allowable and the minima are the same as in the Bayesian case. Thus the desired plan is n = 25, c = 1. The minimum average cost per lot is K(lOO, 25, 1) = 17.99.
(d) Finally, to satisfy both conditions solutions are possible for c = 1 with 33 < n 5 100, for c = 2 with 44 2 n < 100, etc. With c = 1 and both restric- tions we minimize R(lOO, n, 1) with n = 33 and R(lOO, 33, 1) = 81.3. Similarly, we can verify that with c = 2 we take n = 44 and R(lOO, 44,2) = 84.2 and that larger values of c require a larger value of R(lOO, n, c). The desired plan is n = 33, c = 1 with K(lOO, 33, 1) = 15.13.
Example
4.2
Rework part (a) of Example 4.1 if a, = 3, a, = 57 (so that again the average of the p-distribution is .05). If we use the Bayesian plan of Example 4.1 (n = 14, c = 0) when actually a, = 3, a, = 57, what does the erroneous assumption cost on the average per lot?
Solution
The cost constants are the same so again yL = yz = 1. Then R(lOO, n, c)
= n + (100 - n)[l - P(n + 60, c + 4, 60, 3) + P(n + 59, c + 3, 59, 2)] = n + (100 - n)[P(n + 60, n, c + 4, c) + 1 - P(n + 59, n, c + 3, c)]. To evaluate the hypergeometric sums we use the binomial approximation and linear interpolation on p in the binomial table. hlore precise evaluations seem to be unwarranted.
It can be verified that the desired plan is n = 28, c = 1 with R(lOO, 28, 1) = 91.2, K(lOO, 28, 1) = 19.12.
If we use n = 14, c = 0, then R(lOO, 14, 0) = 91.8, K(100, 14, 0) = 19.18 and on the average the cost is .06 more per lot.
The previous discussion and examples were based upon the assumption that al , a, are integers. Now we would like to consider the case 0 < a, < 1. If the mean of the p-distribution is very small this may be exactly the type of prior which seems to describe the behavior of the p’s. Hald [II] has suggested that the binomial be approximated by the Poisson and the beta prior be replaced by a gamma prior which closely approximates the desired beta distribution. Hence we use
and the prior is
j(p; b, , b2) = bbz’pb’-‘e-b’“/r(b,), p > 0. (4.22) A random variable with density (4.22) has mean b,/bz . This leads to the sug- gestion that we set b, = a, , bz = a, + az . We observe that for 0 < b1 < 1
the gamma density has the same shape as the beta with corresponding 0 < aI < 1. Using (4.21) and (4.22) we find that (4.14) should be replaced by
R(N, n, c) = n + (N - n)
A)
+ ~,[l - F(c; b, , g&y-)]}
(4.23) where the F symbol is the negative binomial sum defined by (1.11) and y1 , yZ , K. , K, are given by (4.15), (4.16), (4.12), (4.13) wit,h a, replaced by b, , al + az by ba . Here the derivation is straight forward and no special device, as was required to obtain (4.8), is necessary. Except for the fact that a different table is required, the minimization procedure is unchanged.Example
4.3
Using the cost constants of Example 4.1 and aI = .4, az = 7.6 (so again the p-distribution has mean #OS), find the Bayesian sampling plan.
Solution
Now we use the approximation based on (4.21), (4.22) with b, = .4, bz = 8. Again y1 = yz = 1 and (4.23) is
R(lOO, n, c) = n + (100 - n) ) + 1 - F(c; .4, &)]a For evaluations we use the table of Williamson and Bretherton [17] and linear interpolation on p (when necessary).
With c = 0 we find
R(lOO, 8, 0) = S + 93[F(O; 1.4, .50) + 1 - F(0; .4, .50)] = 8 + 92[.3789 + .2421] = 65.1.
Similarly we get R(lOO, 12, 0) = 63.4, R(lOO, 16,0) = 63.9 so that the minimum value appears to occur for an n between 8 and 16. Further calculations yield R(lOO, 11, 0) = 63.5, R(lOO, 13, 0) = 63.4, R(lOO, 14, 0) = 63.6. Thus the minimum occurs at n = 12 or 13.
With c = 1 we get R(lOO, 19, 1) = 65.5, R(lOO, 20, 1) = R(lOO, 21, 1) = R(lOO, 22, 1) = 65.4, R(lOO, 23, 1) = 65.5. The minimum occurs at n = 20, 21, 22 and is larger than the minimum obtained with c = 0. Hence we terminate calculations.
The desired plan is n = 12 (or 13), c = 0 with R(lOO, 12, 0) = 63.4, K(lOO, 12, 0) = 16.34.
Example
4.4
Using the cost constants of Example 4.1 find the Bayesian solution using (4.23) with b, = 1, b, = 20. (This will allow us to compare the approximate solution with the exact solution.)
Solution
Again y1 = yz = 1. Now we minimize
~(100, n, c) = n + (100 - 4 c; 272% )+I-F(cA&)]-
It can be verified that with c = 0 the minimum occurs at n = 13, 14, 15, 16 (which includes the correct n = 14) yielding in each case R(lOO, n, 0) = 79.2. With c = 1 the minimum occurs at n = 25 (the same as the exact solution) and R(100, 25, 1) = 79.4. Thus, according to the approximation the Bayesian plan is n = 13, 14, 15, or 16, c = 0. The minimum value of R(100, n, C) is 79.2 com-
pared to 79.61 for the exact solution. We now get K(100, 14, 0) = 17.92 as compared with 17.96 before.
The approximation appears to work quite well in this example and should be even better when the prior has smaller mean value.
The constants a, , a, can be estimated in various ways. One possibility is to equate the mean and variance of the beta distribution
P = ~I/(~, + az), (4.24)
u2 = l.4 - cL>l(a, + a2 + 1) (4.25) to estimates based upon sampling and solve for a, and a, . Thus, if records are available for m samples of size n each of which yields an estimate pi , then we could calculate
P = 2 fL/m, (4.26)
i=l
s2 = g (2% - F)‘l(m - 1). (4.27)
No matter how fl and s2 are obtained, if we set (4.24) equal to fl, (4.25) equal to s2 and solve for aI , a, we get (leittng Q = 1 - IS)
li, = p(pj - s2)/s2, (4.28)
6, = (1 - p)@fj - S”)/S”. (4.29)
Then we would round these numbers according to the convention given in the second paragraph of this section.
Examination of Examples 4.1, 4.2, 4.3 illustrate that for a fixed mean II, the greater the variance of the p-distribution, the smaller the average cost. This implies that as the variance increases the more urgent it becomes to incorporate the prior into the model.
5. THE Two POINT PRIOR DISTRIBUTION
Another prior distribution which has been given considerable attention by Hald [6], [7], [8], [9] is the two point distribution with probability function
f(P) = w , P = Pl (5.1)
where ulZ = 1 - ZD~ and the zu’s and p’s are assumed to be known. Hald’s justi- fication for use of the model appears to be that those involved with quality con- trol can frequently identify good quality (pJ and poor quality (pz) together with ZU~ and zuZ even if they cannot furnish any additional information. In terms of the jacket example we could have the following situation. Suppose that three shifts, two day shifts and one night shift, have made the same number of lots. The lots have been mixed together at random in a warehouse. Later it was deter- mined that day shifts make jackets of which .Ol are defective on the average while the night shift has .lO on the average of their items defective. Then we ident)ify w1 = $, wZ = 4, p, = .Ol, pZ = .lO.
With (5.1) it is a relatively simple problem to average the random variable associated with (2.2) over all values of p. If we let
K, = wXJ&) + mK(pJ, (5.2)
Km = wX,Cpo,) + wXApJ (5 *3)
and again define R(N, n, c) as in (4.11), then we obtain
WV, n, c> = n + W - 4hQbd
+ YZ(PJI
(5.4) where~1 = wlK(pd - Ka(pdlI(K - Km),
(5.5)~2 = w Kdpz) - K,@JIlW. - Km). (5.6)
Again we may be interested in Bayesian or restricted Bayesian plans. The minimization procedure described in Section 4 can be applied without modi- fication.
Example 5.1
Find the Bayesian sampling plan for the jacket example using the tmo point prior with w1 = $, w:, = $, pl = .Ol, p, = .lO and the cost constants given in Section 2.
Solution
Again p, = .05. This time we find y1 = 1, yZ = .625 and min (Nyl , NyJ = 62.5. For (5.4) we get
R(100, n, c) = n -I- (100 - n)[E(c + 1; n, .Ol) + .625B(c; n, .lO)]. With c = 0 we try n = 10 and obtain
fi(100, 10, 0) = 10 + 90[.09562 + .625(.34868)] = 35.22.
Similarly R(100, 12, 0) = 37.53, B(100, 13, 0) = 37.45, n(l00, 14, 0) = 37.5s. Thus n = 13 yields the minimum with c = 0.
With c = 1 we find R(100, 24, 1) = 39.71, R(100, 25, 1) = 39.65, R(100, 26, 1) = 39.67 and n = 25 yields the minimum.
Since the minima are increasing we can terminate calculations. The desired plan is n = 13, c = 0, with K(lOO, 13, 0) = 14.67.
Example 6.6
If pl = .Ol, p2 = .05, w1 = .85, to2 = .15, X, = .40, X2 = 0, R, = .30, R, = A, = 0 (Hald’s [9] figures), find the Bayesian plan for N = 1000. (In this example we encounter curves of type (a), (b), (d) as described preceding Example 4.1.) Solution
We verify that p, = .03, y1 = .6296, yz = .llll so that
R(lOOO, n, c) = n + (1000 - n)[.6296 E(c + 1; n, .Ol) + .llll B(c; n, .05)] Since min (Nrl , N7.J = 111.1, obviously n 5 111.
With c = 0 a few calculations verify that R(1000, n, 0) is a monotonically
increasing function of n with minimum value 111.1 at n = 0.
With c = 1 we find that R(1000, n, 1) drops to a minimum value of 110.29
at n = 23.
With c = 2 we find that R(1000, n, 2) increases, then drops to relative mini- mum of 115.32 at n = 50 before finally increasing monotonically.
Hence the desired plan is n = 23, c = 1. 6. COMMESTS
The linear cost model which we have considered seems to be very general and quite reasonable. Potential users may decide that some modification of (2.1) or the interpretation of the constants is necessary. If the material presented in this paper is available, perhaps such changes will be of a routine nature. Our evaluations have been subject to the limitations of current standard tables. In the examples no serious handicap was encountered. If trouble should arise working with the range of paramet,ers which have been tabulated, this would be good justification for extending the tables in their next editions. In the meantime we could use the various approximations which are available.
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