ABC Inventory Analysis and Economic Order Quantity
Problem #12-1:
The Welsh Corporation uses 10 key components in one of its manufacturing plants. Perform an ABC analysis from the data shown below. Explain your decisions and logic.
SKU | Item Cost $ | Annual Demand |
WC219 | 0.10 | 12,000 |
WC008 | 1.20 | 22,500 |
WC916 | 3.20 | 700 |
WC887 | 0.41 | 6,200 |
WC397 | 5.00 | 17,300 |
WC654 | 2.10 | 350 |
WC007 | 0.90 | 225 |
WC419 | 0.45 | 8,500 |
WC971 | 7.50 | 2,950 |
WC713 | 10.50 | 1,000 |
Problem #12-2:
The following table contains figures on the monthly volume and unit costs for a random sample of 16 items from a list of 2,000 inventory items at a health care facility. Develop an A-B-C classification for these items.
Item | Unit Cost | Usage |
K34 | 10.00 | 200 |
K35 | 25.00 | 600 |
K36 | 36.00 | 150 |
M10 | 16.00 | 25 |
M20 | 20.00 | 80 |
Z45 | 80.00 | 200 |
F14 | 20.00 | 300 |
F95 | 30.00 | 800 |
F99 | 20.00 | 60 |
D45 | 10.00 | 550 |
D48 | 12.00 | 90 |
D52 | 15.00 | 110 |
D57 | 40.00 | 120 |
N08 | 30.00 | 40 |
P05 | 16.00 | 500 |
P09 | 10.00 | 30 |
Problem #12-3:
A large bakery buys flour in 25-pound bags for $30 per bag. The bakery uses an average of 4,860 bags a y
ear. Preparing an order and receiving a shipment of flour involves a cost of $10 per order. Annual holding costs are $75 per bag.
a. Determine the Economic Order Quantity?
b. What is the average number of bags on hand?
c. How many orders per year will there be?
d. Compute the total costs of ordering and holding flour.
Problem #12-4:
Garden Variety Flower Shop uses 750 clay pots a month. The pots are purchased at $2 each. Annual carrying costs per pot are estimated to be 30 percent of costs, and ordering costs are $20 per order. The manager has been using an order size of 1,500 flower pots.
a.What additional annual cost is the shop incurring by staying with this order size?
b.Other than cost savings, what benefit would using the optimal order quantity yield?
Problem #12-5:
A mail-order house uses 18,000 boxes a year. Carrying costs are 60 cents per box a year, and ordering costs are $96. The following price schedule applies. Determine
a. the optimal order quantity
b. the number of orders per year.
Number of Boxes | Price per Box |
1000 to 1999 | 1.25 |
2000 to 4999 | 1.20 |
5000 to 9999 | 1.15 |
10000 or more | 1.10 |
Problem #12-6:
The friendly Sausage Factory (FSF) can produce hot dogs at a rate of 5,000 per day. FSF supplies hot dogs to local restaurants at a steady rate of 250 per day. The cost to prepare the equipment for producing hot dogs is $66. Annual holding costs are 45 cents per hot dog. The factory operates 300 days a year. Find:
a.the optimal run size
b.the number of runs per year
c.the length (in days) of a run
Problem #12-7:
A company is about to begin production of a new product. The manager of the department that will produce one of the components for the product wants to know how often the machine used to produce the item will be available for other work. The machine will produce the item at a rate of 200 units per day. Eighty units will be used daily in assembling the final product. Assembly will take place five days a week, 50 weeks a year. The manager estimates that it will take almost a full day to get the machine ready for a production run, at a cost of $300. Inventory holding costs will be $10 a year.
a. what run quantity should be used to minimize total annual costs?
b. what is the length of a production run in days?
c. during production, at what rate will inventory build up?
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Solution #12-1:
ABC Inventory Analysis – Welsh Corporation | |||||
Projected | Projected | Cumulative | Cumulative | ||
Item | Annual | Annual | Dollar | Percent | |
Number | Usage | Unit Cost | Dollar Usage | Usage | of Total |
397 | 17,300 | $5.00 | $86,500 | $86,500 | 55.14% |
008 | 22,500 | $1.20 | $27,000 | $113,500 | 72.35% |
971 | 2,950 | $7.50 | $22,125 | $135,625 | 86.46% |
713 | 1,000 | $10.50 | $10,500 | $146,125 | 93.15% |
419 | 8,500 | $0.45 | $3,825 | $149,950 | 95.59% |
887 | 6,200 | $0.41 | $2,542 | $152,492 | 97.21% |
916 | 700 | $3.20 | $2,240 | $154,732 | 98.64% |
219 | 12,000 | $0.10 | $1,200 | $155,932 | 99.40% |
654 | 350 | $2.10 | $735 | $156,667 | 99.87% |
007 | 225 | $0.90 | $203 | $156,870 | 100.00% |
One possible ABC classification scheme is A items (397, 008, 971) with 30% of items accounting for 86.5% of total inventory value; B items (713, 419) with 20% of items accounting for 9.1% of total inventory value; C items (887, 916, 219, 654, 007) with 50% of items accounting for 4.4% of total inventory value. Since there are no absolute guidelines on ABC analysis students might, for example, define A items as 397 and 008 (20% of the items and 72.4% of the value) and this is fine.
Solution #12-2:
Item | Unit Cost | Usage | Dollar Usage | Category |
F95 | 30 | 800 | 24,000 | A |
Z45 | 80 | 250 | 16,000 | A |
K35 | 25 | 600 | 15,000 | A |
P05 | 16 | 500 | 8,000 | B |
F14 | 20 | 300 | 6,000 | B |
D45 | 10 | 550 | 5,500 | B |
K36 | 36 | 150 | 5,400 | B |
D57 | 40 | 120 | 4,800 | B |
K34 | 10 | 200 | 2,000 | C |
D52 | 15 | 110 | 1,650 | C |
M20 | 20 | 80 | 1,600 | C |
F99 | 20 | 60 | 1,200 | C |
N08 | 30 | 40 | 1,200 | C |
D48 | 12 | 90 | 1,080 | C |
M10 | 16 | 25 | 400 | C |
P09 | 10 | 30 | 300 | C |
Solution #12-3:
D = 4,860 bags/yr.
S = $10
H = $75
a.
b. Q*/2 = 36/2 = 18 bags
c.
d.
Solution #12-4:
D = 750 pots/mo. x 12 mo./yr. = 9,000 pots/yr.
Price = $2/pot, S = $20 H = ($2)(.30) = $0.60/unit/year
a.
TC = 232.35 + 232.36
= 464.71
If Q = 1500
TC = 120 + 450 = $570
Therefore the additional cost of staying with the order size of 1,500 is:
$570 – $464.71 = $105.29
b. Only about one half of the storage space would be needed.
Solution #12-5:
D = 18,000 boxes/yr.
S = $96
H = $0.60/box per yr.
a.Q* =
Since this quantity is feasible in the range 2000 to 4,999, its total cost and the total cost of all lower price breaks (i.e., 5,000 and 10,000) must be compared to see which is lowest.
TC2,400 =
TC5,000 =
TC10,000 =
b.
Solution #12-6:
p = 5,000 hotdogs/day
|
u = 250 hotdogs/day
300 days per year
S = $66
H = $.45/hotdog per yr.
a.
b. D/Q* = 75,000/4,812 = 15.59, or about 16 runs/yr.
c. run length: Q*/p = 4,812/5,000 = .96 days, or approximately 1 day
Solution #12-7:
S = $300
D = 20,000 (250 x 80 = 20,000)
H = $10.00
p = 200/day
u = 80/day
Q* = (1,095.451) (1.2910) = 1,414 units
b. Run length =
c. 200 – 80 = 120 units per day