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HW04
‐
F08_Key.doc
1
MGMT 46000: HW 04
40 Points
(Due: November 3, 2008)
Question 1:
Part A:
Solve problem 25 / 253. In part (b), round off ROP value to 2 decimal places.
(a)
Lot Size [gallons]
1399
400799
> 800
C [$/pulley]
2
1.70
1.62
H [$/pulley/year]
3
3
3
Q
0
=
√
(2*D*S/H)
300 feasible
300 (infeasible)
300 (infeasible)
Lot Size, Q*
300
400
800
Purchase Cost [$/yr]
2*4500=9000
1.70*4500=7650
1.62*4500 =7290
Ordering cost [$/yr]
30*4500/300=450
30*4500/400 =337.5
30*4500/800=168.75
Holding Cost [$/yr]
300/2*3=450
400/2*3=600
800/2*3=1200
Total Annual Costs
9900
8587.50
8658.75
So order quantity is 400 units
(b) Service level=98.5%
Z
SL
=2.17.
So,
μ
LT
=4500/360*4*3/4=37.5 g.
σ
LT
=2
√
3
ROP =
μ
LT
+ Z
0.985
*
σ
LT
= 37.5 +
2.17* 2
√
3
≈
45.02 gallons
Part B:
Solve problem 21 parts “a” and “b” only from page 253.
a.
Weekly demand ~ N[21, 3.5].
SL
= 90%,
LT
= 2 days =2/7 weeks.
Z
0.90
= 1.29
ROP
= 21 * 2/7 + 1.29 * (
√
2/7) * 3.5 = 8.41 gallons.
Average daily demand is 3 gal. This supply (
= ROP
) will last for 8.41 / 3
≈
2.8 days.
b.
Calculate order size Q for OI = 10 days and SL = 90%.
OI = 10 days.
IP
= 8 gallons.
OUL
=
(OI+LT)
*
μ
+ Z
SL
* {
√
(OI+LT)
} *
σ
= [(10 + 2)/7] * 21 + 1.29 * {
√
(10 + 2) /7)} * 3.5 = 36 + 5.91
≈
42 gallons.
Q =
OUL – IP
= 42 – 8 = 34 gallons.
What is the probability of stockout before this order arrives.
This was not covered in the class and will not be graded. However, here is how we
could solve this.
LT = 2 days. Average demand during LT = 21 * 2/7 = 6. Std. dev = 3.5 *
√
(2/7). What
we want to find is Prob (X
≤
8), with mean and std. deviation [6, 3.5 *
√
(2/7)]. The answer
will work out to be 0.1423.
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 Fall '08
 panwalker
 Management

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