IEOR 150, Fall 2010
Suggested Solution to Sample Final
1. Let
c
= $15 be the unit purchasing cost,
K
= $50 be the ordering cost per order,
I
= 0
.
3 be the
annual interest rate,
h
=
cI
= $4
.
5 be the unit holding cost per year,
p
= $40 be the shortage
cost per unit, and
τ
=
2
12
= 0
.
167 year be the order lead time. Since the monthly demand
for the ﬁlter follows a normal distribution with mean 1800 and standard deviation 400, let
μ
= 1800
×
2 = 3600 units be the mean of the demand and
σ
=
√
2
×
400
2
= 565
.
69 units be
the standard deviation of the demand during a lead time. Finally, let
λ
= 1800
×
12 = 21600
units be the annual demand size.
Because we use the EOQ quantity as the order quantity, we have
Q
*
=
r
2
Kλ
h
= 692
.
82
as the order quantity. Then we compute
R
*
=
F

1
‡
1

Q
*
h
pλ
·
= 5119
.
81
,
z
=
R
*

μ
σ
= 2
.
69
,
and
n
(
R
*
) =
σL
(
z
) = 0
.
62
,
where the value of
L
(
z
) is looked up from Table A4 in the textbook.
R
*
is the optimal reorder
level. It then follows that the safety stock level is
R
*

λτ
= 1519
.
81 and the expected annual
stockout penalty cost is
λn
(
R
*
)
Q
*
= $776.
2. We have the production rate
P
= 250
,
000 units per year, demand rate
λ
= 120
,
000 units per
year, setup cost
K
= $3
,
000 per production run, production cost
c
= $16 per unit, annual
interest rate
I
= 0
.
25, holding cost
h
=
cI
= $16
×
0
.
25 = $4 per unit per year, and the
eﬀective holding cost
h
0
=
h
(1

λ
P
) = $2
.
08 per unit per year.
(a) The optimal production lot size is
Q
*
=
r
2
Kλ
h
0
= 18605
.
21 units
.
(b) Note that
P
= 250
,
000 and
λ
= 120
,
000, so the proportion of uptime is
λ
P
= 48% and
the proportion of downtime is
P

λ
P
= 52%.
(c) The total holding and setup cost is
G
*
=
√
2
Kλh
0
= $38698
.
84
.
Therefore, the annual holding cost and the annual setup cost are both
G
*
2
= $19349
.
42.
3. Let range 0 be [0
,
2000), range 1 be [2000
,
4000), and range 2 be [4000
,
∞
). Let
c
(0)
= $25,
c
(1)
= $23, and
c
(1)
= $22 be the unit purchasing cost for ranges 0, 1, and 2, respectively.
It follows that the unit holding costs are
h
(0)
=
Ic
(0)
= $5,
h
(1)
=
Ic
(1)
= $4
.
6, and
h
(0)
=
Ic
(2)
= $4
.
4. For an allunits quantity discount, we may directly apply
¯
Q
(
i
)
=
r
2
Kλ
h
(
i
)
,
i
∈ {
0
,
1
,
2
}
,
1