APPENDIX 



FISHERY BULLETIN: VOL. 71, NO. 3 



Derivations of Text Equations 12-15 



As pointed out in tiie text, 



-6Mi 



The probability of ^1 : ^i = me "*''''!. (1) 



The probability of C: ^2 " 



If + My) 



-4.5(F + M2) 



(2) 



The probability of £^2- ^3 " 



and maximum likelihood estimators of the 0,- are: 

 L -ll. or^i = me'^^i from (1). (4) 



02 = C/No or 02 = (I-'") 



^-6Mi^-7.5M2 / F \ 

 \F + iWo/ 



-4.5(F+M2) f ,„, 



e from (2). 



(5) 



03 = E^INq or 03 = 



(1 -m)e'^'^lc- ^-5^26- 4.5 (F + M2) 



from (3). (6) 



Then for m = m (fixed) (0j < m < 1), text 

 Equation 12, by rearranging Equation 4 and 

 taking natural logarithms we obtain 



= e 



= k< 



-(4.5F + I2M2) 



The natural logarithm of k2 



ln/e2 = -(4.5F+ 12i\/2) 

 which can be solved for F as follows: 



-4.5F = In /e2 + I2A/2 

 In /e2 + 12Af2 



(7) 



4.5 



(text Equation 

 J 15). 



(8) 



Equation 5 can be written 



09 RM, .7 !SMr. / F 



2 6M1 



e 



-l.bMof 



-e ^( 



(1-M) 



-4.5 (F 



F + M, 



and since 



-4.5 (F + A/2) 



(9) 



= e 



-4.5F- I2M2 + 7.5M2 

 -(4.5F + I2A/2) + 7.5M2 



In ;;2 + 7.5M2 ,- t? *.- n\ 



= e (from Equation 7) 



and 



^In^^M^ 

 6  m 



Ml (text Equation 13). 



(in ^2 + 12M2\ 

 . 4.5 / 



\F+M2/ /lnte2+12M2 \ 

 \ 4.5 / 



(from Equation 8) 



A/2 



Then, Equation 6 can be rewritten as 



-(In/; 2 + I2M2) 







(1-m) 



3 6M1 -7.5M9 - 4.5 (F + Mo) 



-(In ^2 + 12M2) + 4.5M2 



In ^2 + 12^2 



-7.5M2 - 4.5F - 4.5F - 4.5Mo 



In /e2 + I2M2- 4.5M2 



692 



