SAMPLING INSPECTION TABLES 



27 



it can be readily solved only for small values of ci and c^,. It is usually easier, 

 particularly for the larger values of ci and cz, to determine the maximum value 

 of pA (i.e., Pl) by trial and error, using work charts for estimating the region 

 in which px will be found. 



The procedure used in preparing the tables and in finding the solution for a 

 specific set of conditions is probably best illustrated by working out an actual 

 example. In this procedure, use is made of known relationships between pt 

 and pL values as given by the DL tables, where an initial risk of 0.06 and a Con- 

 sumer's Risk of 0.10 are associated with pt as outlined on page 11. For a given 

 lot size, a work chart is prepared on which points corresponding to associated 

 pL and pt values are plotted for each pair of ci, c^ values given in Fig. 7, A line 

 drawn through all points for a single pair, such as ci = 0, C2 = 1, indicates what 

 pt value should be associated with any pL value specified. Fig. 10 indicates the 

 nature of the work charts and the following example illustrates its use. 



4 5 6 8 10 



LOT TOLERANCE (p.) IN PER CENT 



20 



Fig. 10 — Work chart giving ^t values corresponding to pi, values for given pairs of ci, 

 c-i values — lot size, N = 500 



Example: Given: N = 500, ^^ = .01, ^ = .004. 



To find: «i, «2, ci and cg that will minimize average amount of 

 inspection per lot. (Condition: For the associated lot 

 tolerance value, pt, the initial risk is 0.06 and the Con- 

 sumer's Risk Pc = 0.10). 

 Solution: Step 1 — Consult work chart. Fig. 10 for N = 500. Try 

 ci = 0, C2 = 1, and corresponding to pL = .01, read pt 

 = .054. 



Step 2 — To determine if first choice of ci, cz was the best. 



M = PtN = 0.054 (500) = 27; yfe = 



0.004 



= 0.074. 



pt 0.054 

 Consult Fig. 7, giving best ci, C2 values for given M and 



