-- 120 -


     5           K1
C e =    iZ d. + C
        0i  1   1 Ce


(6.37)


Solving for C1 from (6.36) and (6.37) and equating the results,


     5                      5
Ce     Zoi d.      1 - C -   d
  e                           1le
          i=1 ~ji~l
             = C     1 =
   K /Ce             1 + K*I/Ce


(6.38)


Using equations (6.7) through (6.10) to obtain expressions for d1, d3, d4, and d., equation (6.38) reduces to


ad2 + bd2 + C = 0



   pCe + (p-1)K04
      C K* e 01


      (p-1)K02
b =p+         +
         C e


            c = p(Ce - 1) + Ce

                 K*
            p=
                C + K*
                e    1

The solution to (6.39) is


        d2 = 2l - 4ac/b2)


             (6.39) (6.40)



   e         (6.41)

K03


(6.42) (6.43)


(6.44)


where