OCTOBER 19-20, 2012 - YMCA University of Science & Technology

β−1
−α
β
t
αβt
e
β −1
= = αβt
, 0
− α
β
t
Proceedings of the National Conference on
Trends and Advances in Mechanical Engineering,
YMCA University of Science & Technology, Faridabad, Haryana, Oct 19-20, 2012
Z()
t
t > . (5)
e
Eq. (5) will be used in the model development in this chapter. When β > 1, deteriorating rate increase with
time;
when β < 1deteriorating rate decreases with time; and when β = 1, deteriorating rate is constant.
1−n
1
n
n
7. The demand up to time t is assumed to be D()
t = dt nT . Where d is the demand size during the fixed
cycle time T and n ∈(1, ∞)
is the pattern index. Such pattern in the demand rate is called power demand
pattern.
8. During the shortage period, the backlogging rate is variable and is dependent on the length of the waiting time
for the next replenishment. We have defined the backlogging rate to be11 + δ ( T −t)
where inventory is
negative. The backlogging parameter δ is a positive constant, t 1
≤t ≤ T .
4 Model developments
The inventory system during a given cycle is depicted in Fig.1. At t = 0 , an initial replenishment of
Inventory
Level
I
max
Q
0 t T time
1
Lost sale
Figure 1. Inventory system for Weibull distribution deteriorating items with partial backorder
Q units are made, of which S units are delivered towards backorders, leaving a balance of I max
units in the initial
inventory. From t = 0 to t 1
= 0 time units, the inventory level depletes due to both demand and deterioration. At t 1
,
the inventory level is zero. During the time ( T − t ) part of the shortage is backlogged and part of it is lost sales.
1
Only the backlogging items are replaced by the next replenishment.
The differential equation describing I()
t over the length t 1
is given as follow:
dI () t
β 1
+ αβt −
It () = − Dt () ; 0 ≤t ≤ t .
1
dt
(6)
The boundary conditions are:
I(0)
= I and I( t ) = 0.
max
1
The approximate solution of eqs. (6) by neglecting higher order term ofα is
1 1
⎡
+ β + β
n n ⎤
d 1 n 1 n α ( t − t )
1
t
β
− α
I() t = ⎢( t − t ) +
⎥e
; 0 ≤t ≤ t . (7)
1 n 1
1
T ⎢
(1 + nβ
) ⎥
⎢⎣
⎥⎦
Now, again taking the first two terms of the exponential series and neglecting the terms containingα 2
the
equation (7) becomes
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