Results for Pu Let R be the true Pu to Am ratio, i.e., Pu = Rx Am. R s(Am). Using Equation Bl for s(Am) we obtain Therefore, s(Pu) = s(Pu) = R aw? . (B7) Hence, v5 = [s(Pu)]7/n = Rea*/nw-P . (B8) Suppose the following linear cost equation is applicable: Cc = n(% + Bw) , (B9) where 2 + Bw is the Pu analysis cost for a single aliquot of size w (in grams). © and 8 are parameters to be determined on the basis of cost information from the analytical laboratory. Then n = C/(a + Bw) and v, Rea- (a + Bw)/cwe? . (B10) If b = 0.5 as suggested by our Am data and by theoretical considerations (Grant and Pelton, 1973, p. 20), then 2.2 2.2 v5 ~Raa,Rab ote . (B11) Solving Equation (B10) for w gives we Raa (CV, - R-a“p) 3 (B12) where CVy must be greater than Rae, Hence if estimates of R and a are available from the analysis of prior samples, if cost parameters g, 8, and C are specified, and the desired precision of the mean Pu concentration per sample (Vp) is agreed upon, then Equation (B12) may be solved to obtain the aliquot size w. An estimate of n may be obtained from Equation (B9). If Pu analyses are available that relate s(Pu) to w, then this functional form should be used for s(Pu) in Equation (B8). eliminate the need for using Equation B7 in B8. 444 This would