
212 Supply Chain Management and Logistics
Because the above
n
is difcult to analyze, we shall start to make the
simplifying assumption:
rx x()= . (8.28)
Now, from Equation 8.24, we will have
jx
n
n
()=
2
−
. (8.29)
Hence, from Equations 8.23 and 8.25, we have
hzxnnnz
zL
nL L
(, )= 12 ((2 1) (8 8))
[(1 2)
0
−+ −+
×+ ⋅
//α
α
/
//
()
],
(, )= 12 ((2 1) (8 8
0
nx Hnx
hzxnnn
nH H
+
−+ −+
αα
α
z
zHnx Lnx
0
0
))
[(1 2) ()
],×+ ⋅+
(8.30)
from which we get
xz nLHLzn z
nL
*
()=(21)( 2)(16(
)
0
2
0
23
0
2
−× ++
+
/ ,,
*
()=(21)( 2)(16(
0
2
0
23
0
xz nHLHzn z
nH
−× ++
+
/
22
).
(8.31)
Plugging Equations 8.28 and 8.31 into 8.27, we obtain a closed-form expres-
sion for