April - June 2007 - Kasetsart University

c
≤
c2 jq2 j
n
q p k p
*
2,
j
, ,
p s2 w ∏
j2= j+
1
2,
j2
Kasetsart J. (Nat. Sci.) 41(2) 385
∀ j = {, 12K , , n}
(21)
* * *
The reference costs, c1, i cw and c2, j were then analyzed against all the unit costs in the model
to identify when the corresponding delivery performances and safety stocks should be set to their lower
or upper bounds. If the opportunity cost was high, the manufacturer should hold safety stocks to prevent
the products shortages. In contrary, it would not be economical to stock the materials, when the inventory
costs (and hence the reference costs) were costly. The optimal solution of the presented optimization
model could be derived as follows:
Stage-1 raw materials:
⎧ *
c then k 1 1 and x
* ⎪ o ≤ c1, i ,i = p ,i 1,i
= 0
(i) If c 1,i
≤ min( cw, cp)
and ⎨ *
⎩⎪ c o > c1, i then k 1,i = 1and x1,i
= q1,i 1−
p1,i
*
(ii) If c 1,i > min( cw, cp) then k 1,i = p1,i
and x1,i
= 0
Work-in-process:
⎧
m
*
⎪c
o ≤ cw then kw = ps
∏ k 1,i
and xw
= 0
1
* ⎪
i= 1
(iii) If c w ≤ cp
and ⎨
m
⎪ *
⎛
⎞
⎪
c o > cw then k w = 1and xw = qw⎜1− ps
k
1∏
1,i⎟
⎩
⎝ i= 1 ⎠
m
*
(iv) If c w > cp then kw = ps
∏ k 1,i
and xw
= 0
1
i= 1
Stage-2 raw materials:
*
⎧c
then k and x
* ⎪ o ≤ c2, j 2, j = p2,
j 2,
j = 0
(v) If c 2,
j ≤ cp
and ⎨ *
c o > c , then k = 1and x = q 1−
p
⎩
⎪
( )
2 j 2, j 2, j 2, j 2,
j
(vi)
*
If c 2, j > cp then k 2, j = p2,
j and x2,
j = 0
Finished product:
(vii)
n
c o > cp then kp = ps
kw∏k j and x p =
2 2, 0
j= 1
⎛
n ⎞
(viii) c o > cp then k p = 1and xp = qp⎜1− ps kw∏k
2 2,
j⎟
⎝
j= 1 ⎠
( )