Transcript Inventory Management Self Study Exercisesx

Solved Problem 1
Nelson’s Hardware Store stocks a 19.2 volt cordless drill that is a popular
seller. Annual demand is 5,000 units, the ordering cost is $15, and the
inventory holding cost is $4/unit/year.
a. What is the economic order quantity?
b. What is the total annual cost for this inventory item?
SOLUTION
a. The order quantity is
EOQ =
2DS
H
=
2(5,000)($15)
$4
=
37,500 = 193.65 or 194 drills
b. The total annual cost is
Q
194
5,000
D
C = 2 (H) +
(S) =
($4) +
($15) = $774.60
Q
2
194
Solved Problem 2
For a product managed according to the Economic Order Quantity method,
the table shows monthly demand rates for the last two years
Year 1
Year 2
Jan Feb Mar Apr May June July Aug Sept Oct Nov Dec
80 70 110 100 110 90 80 120 100 130 110 100
120 130 50 70 120 100 90 70 100 80 140 140
The product is bought at 20 Euros per unit. Set up cost is 105 Euros and
holding cost is 35% of the purchasing price.
Service level decided by management is 95%.
Delivery lead time is 1 month.
•Calculate EOQ, Number of orders and time between orders
•Calculate the reorder point ROP
•Trace the inventory profile
Solved Problem 2
Solution:
• Total demand of the 1st year: 1200 (sum of the 1st row)
• Total demand of the 2nd year: 1210 (sum of the 2st row)
• Mean year demand: D = (1200 + 1210)/2 = 1205 per year
• Mean monthly demand: d = (1200+1210)/24 = 100,4167 ≈ 100
prod/month
Economic Order Quantity:
2  1205  105
• EOQ = 2HDS = 20  0,35 = 190,1315 ≈ 190
• Number of orders = D/EOQ = 1205/190 ≈ 6,3 orders per year
• Time between orders = EOQ / D = 0,15 year or similarly EOQ/D*(12
months/year) = 1,89 months ≈ 2 months
Solved Problem 2
Calculation of Reorder Point
• ROP = Average demand during lead time +
Safety Stock
• The lead time L is 1 month, and we also know
the mean monthly demand d. Therefore, the
average demand during lead time is d*L = 100.
Solved Problem 2
• We also need to determine the safety stock S. The formula for
the safety stock is S = z * standard deviation of the demand
during lead time (sd).
• The term z can be obtained from normal distribution tables. For
a given service level of 95% z = 1,645. On the other hand, the
monthly standard deviation of the demand can be calculated
from the table of the monthly demand rates using the formula
below.
1 n
1
2
sd 
(d i  d ) 
13095,83  23,86177

n  1 i 1
23
Thereafter, S = 1,645*23,86177 = 39,2526 ≈ 40 products
As such, ROP = 100 + 40 = 140
To trace the inventory, we start at the replenishment level (which is
EOQ + safety stock) and we draw a graph similar to that of the slides.
Solved Problem 3
Our company manages an inventory of light bulbs. A particular lamp is
bought at 10 cents. Ordering cost is 150 cents and holding cost is 20% of
unit price per unit and year.
The factory works 12 months per year, steady production and
management requires an 80% service level for this particular item.
Delivery lead time from the supplier is 1,5 months.
The monthly demand over the last 24 months is as follows:
500, 500, 425, 425, 450, 425, 425, 475, 475, 425, 425, 425, 475, 525, 425,
425, 450, 475, 425, 425, 475, 450, 425, 425.
Compute the following:
• Economic order quantity
• Reorder point
Solved Problem 3
SOLUTION
• Total demand: 10800 (sum of 24months)
• Mean year demand: D = (10800)/2 = 5400 lamps per year
• Mean monthly demand: d = 5400/12 = 450 lamps/month
Economic Order Quantity:
• EOQ =
2 DS
H
=
2  5400  150
10  0,2
= 900 lamps
• Number of orders = D/EOQ = 5400/900 ≈ 6 orders per year
• Time between orders = EOQ / D = 0,167 year or approximately ≈ 2
months
Solved Problem 3
Calculation of Reorder Point
• ROP = Average demand during lead time + Safety Stock
• The lead time L is 1,5 month. The mean monthly demand d is 450.
Therefore, the average demand during lead time is d*L = 675.
Solved Problem 3
We also need to determine the safety stock S. The formula for the safety
stock is S = z * standard deviation of the demand during lead time (sd). The
term z can be obtained from normal distribution tables. For a given service
level of 80% z = 0,84. On the other hand, the monthly standard deviation of
the demand can be calculated from the table of the monthly demand rates
using the formula below.
1 n
2
sd 
(
d

d
)
 30,82134

i
n  1 i 1
• Thereafter, S = 0,84*sd* = 31,70 ≈ 32 lamps
• As such, ROP = 900 + 32 = 932
Solved Problem 4
Grey Wolf Lodge is a popular 500-room hotel in the North Woods.
Managers need to keep close tabs on all room service items, including a
special pine-scented bar soap. The daily demand for the soap is 275 bars,
with a standard deviation of 30 bars. Ordering cost is $10 and the
inventory holding cost is $0.30/bar/year. The lead time from the supplier is
5 days, with a standard deviation of 1 day. The lodge is open 365 days a
year.
a. What is the economic order quantity for the bar of soap?
b. What should the reorder point be for the bar of soap if
management wants to have a 99 percent cycle-service level?
c. What is the total annual cost for the bar of soap, assuming a
Q system will be used?
Solved Problem 4
SOLUTION
a. We have D = (275)(365) = 100,375 bars of soap; S = $10; and H = $0.30.
The EOQ for the bar of soap is
EOQ =
2DS
=
H
=
2(100,375)($10)
$0.30
6,691,666.7 = 2,586.83 or 2,587 bars
Solved Problem 4
b. We have d = 275 bars/day, σd = 30 bars, L = 5 days,
and σLT = 1 day.
σdLT =
Lσd2 + d2σLT2 =
(5)(30)2 + (275)2(1)2 = 283.06 bars
Consult the body of the Normal Distribution appendix for
0.9900. The closest value is 0.9901, which corresponds to
a z value of 2.33. We calculate the safety stock and reorder
point as follows:
Safety stock = zσdLT = (2.33)(283.06) = 659.53 or 660 bars
Reorder point = dL + Safety stock = (275)(5) + 660 = 2,035 bars
Solved Problem 4
c. The total annual cost for the Q system is
Q
D
C = 2 (H) + Q (S) + (H)(Safety stock)
2,587
100,375
C=
($0.30) +
($10) + ($0.30)(660) = $974.05
2
2,587
Solved Problem 5
ABC Classification: Example
PART
UNIT COST
ANNUAL USAGE
1
2
3
4
5
6
7
8
9
10
$ 60
350
30
80
30
20
10
320
510
20
90
40
130
60
100
180
170
50
60
120
Solved Problem 5
ABC Classification: Example (cont.)
PART
9
8
2
1
4
3
6
5
10
7
TOTAL
PART
VALUE
$30,600
1
16,000
2
14,000
3
5,400
4
4,800
5
3,900
3,600
6
CLASS
3,000
7
2,400
A8
1,700
B
9
C
$85,400
10
% OF TOTAL % OF TOTAL
UNIT COST QUANTITY
ANNUAL
USAGE
VALUE
% CUMMULATIVE
35.9
6.0
$ 60
18.7
5.0
350
16.4
4.0
30
6.3
9.0
5.680
6.0
4.630
10.0
4.220 % OF TOTAL
18.0
ITEMS
VALUE
3.510
13.0
12.0
9, 8,2.8
2320
71.0
17.0
1, 4,2.0
3
16.5
5107
6, 5, 10,
12.5
20
6.0
90
11.0
A40
15.0
130
24.0
B60
30.0
100
40.0
% OF TOTAL
58.0
180
QUANTITY
71.0
170
C
83.0
50 15.0
100.0
25.0
60 60.0
120
Example 10.1