b)
The holding cost are 9% off the value, with a value off $100 per pound for the
breakdown
and
$125 per pound for the
finish
.
This leads to:
9% × ($100×526,540+$125×182,837)/(526,540+182,837) = $9.58
The fixed cost are 25$x8x2= $400
The total demand is 526540 + 182387 = 750377 units per year
Q
*
=
√
2DK
h
=
√
2×400×750377
9,58
= 7515,931 = 7516
Average throughout the process: 570 + 450 = 1020.
7916/1020 =
7,761
Quantity for
breakdown:
570*7,761 = 4423,608
Quantity for
finish
: 450 * 7,761 = 3492,45
526540 /4423,608 *55 minutes = 6546,62 minutes = 109 uur (optimal use
Breakdown
)
182837 /3492,45* 120 minutes = 6282,249 minutes = 104,7 hours = 105 hours
(109+104)/24 = 8,90 days + 2*8 hours (
changeover shifts
) = 9,566 hours
This is exactly two weeks
a)
The large mill has an uptime of 85%, therefore the total uptime per year is:
(524)= 48 weeks with a changeover time of 8 hours. The machine operates 5 days a week:
47*5 = 240 days
hence, the uptime is: 0,85*240= 204 days
The amount of cycles per year are: 709,377/7697= 92,16 cycles = 93 cycles
93 cycles *1,46 days = 134,57 days, which is less than 204 days resulting in the large mill having enough
capacity.
d) The manufacturing leadtime would thus significantly reduce. When an order arrives, the
breakdown phase is a lot faster compared to the twoweek cycle. The order can go through the process
of breakdown faster and go to the finish phase faster, as the products don’t have to wait anymore.
7
6.
Consider material 1003 in Figure 331. This product has high sales volumes with a reasonably low
variability since its monthly sales has an average of 629.0 units and a standard deviation of 235.2
units. As a result, H.C. Starck decided to keep inventory for this final product. For simplicity, they
assume the following:

the sales numbers represent the actual demand,

the demand follows a normal distribution,

the replenishment lead time is given by Figure 326 (that is, the average is 2.3 weeks and the
standard deviation is 1.8 weeks),

there is a twoweek production cycle and

there are 4.3 weeks in one month.
a)
Which type of inventory replenishment policy would you recommend for this final product?
b)
What should the value(s) of the policy parameter(s) be to achieve a cycle service level equal to
90%? Include the formula that you used.
c)
For the value(s) of the policy parameter(s) found in Question 7b, what is safety stock level?
d)
For the value(s) of the policy parameter(s) found in Question 7b, what is the average fill rate?
Include the formula that you used.
a)
That I would recommend for H.C. Starck is continuous reviews. There is a lot off variation in
the demand off Starck as the average demand is 629 units and the standard deviation is
235,2. Therefore it is necessary to pay attention on the stock. This won’t be necessary if the
company would have a high safety stock, however the goal is too minimize stock making it no
option.
b) AVG = 629
STD = 235,2
8
AVGL = 2,3/4,3 = 0,5349
STDL = 1,8/4,3 = 0,4186
Z value = 1.29
R
=
AVG× AVGL
+
z ×
√
AVGL×STD
2
+
AVG
2
×STDL
2
.
R
=
629
×
0,5349
+
1,29
×
√
0,5349
×
235,2
2
+
629
2
×
0,4186
2
.
R = 742,17
c) Safety Stock =
z ×
√
AVGL × STD
2
+ AVG
2
× STDL
2
Safety stock =
1,29
×
√
0,5349
×
235,2
2
+
629
2
×
0,4186
2
.
Safety Stock = 405,718
d) Z= 1.29, L(z)= 0.0465 (in Table)
STD = 253,2
L = 0,5349
Q = 629
Fill rate= 1 STD*
√L* L(z)/Q
Fill rate = 1 – 235,2 *
√0,5349 * 0,0465/629
Fill rate = 0,987 = 99%
9
10
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 Fall '16
 B. Beije
 H.C. Starck