Chapter 13 – Inventory Management
13–21
Education.
10. Given:
A chemical firm produces 100-pound bags. Demand for the product = 20 tons per day. The
capacity = 50 tons per day. Setup cost = $100, and storage and handling costs = $5 per ton a year.
The firm operates 200 days a year. Note: 1 ton = 2,000 pounds.
p = 50 tons per day * 2,000 pounds per ton = 100,000 pounds per day = 100,000 pounds per day /
100 pounds per bag = 1,000 bags per day
u = 20 tons per day * 2,000 pounds per ton= 40,000 pounds per day = 40,000 pounds per day /
100 pounds per bag = 400 bags per day
a.
bags 10,32897.327,10
400000,1
000,1
25.0
100)000,80(22
up
p
H
DS
Qp
328,10
Q
000,1
p
d. Runs per year:
7.75
328,10
000,80
Q
D
runs per year
e. S = $25:
bags 164,598.163,5
400000,1
000,1
25.0
25)000,80(22
up
p
H
DS
Qp
3,098.4 )400000,1(
000,1
164,5
)(
max up
p
Q
Ip
bags
Savings when S = $25 = $1,549.19 – $774.60 = $774.59 per year.
Chapter 13 – Inventory Management
13–22
Education.
11. Given:
Assembly takes place 5 days a week, 50 weeks a year. It will take a full day to get the machine
ready for a production run of the component for the new product.
S = $300
H = $10.00
a. Optimal run quantity to minimize total annual costs:
414,121.414,1
80200
200
10
300)000,20(22
up
p
H
DS
Qp
units
b. Days to produce the optimal run quantity:
days 07.7
200
414,1
p
Qp
c. Average amount of inventory:
units
units
d. The manager would like to run another job between runs of the component for the new
product and needs a minimum of 10 days per cycle (including setup) for the other job:
How much time is available to run the other job? The job must be finished during the pure
consumption time for the component for the new product. The end of the pure consumption
time is when inventory of the component for the new product falls to 0 units. If the other job
takes longer than the pure consumption time, we will run out of inventory of the component
for the new product.
Plugging in values and solving for Pure Consumption Time:
13–23
e. Three options that the manager could consider that will allow this other job to be performed:
1) Try to shorten the setup time of the component for the new product.
f. Determine the additional units to produce of the component for the new product and the
increase in total annual cost from this new Q:
We know the following:
The Pure Consumption Time for the component for the new product must equal 10 days to
allow the other job to be run.
Plugging in values and solving for Qp:
Using the least common denominator:
The additional units per run = 1,467 – 1,414 = 53 units per run.
Chapter 13 – Inventory Management
Increase in total cost:
Q = 1,467:
units
Increase = $8,490.98 – $8,485.28 = $5.70 per year
12. Given:
p = 800 units per day
13–25
Education.
d. The other component requires 4 days (including setup). Setup time for the heating element =
0.5 days. Is there enough time to run the other component between batches of heating
elements?
How much time is available to run the other component? The other component must be
finished during the pure consumption time for the heating element. The end of the pure
Chapter 13 – Inventory Management
13. Given:
D = 18,000 boxes/year
S = $96
H = $.60/box/year
13–27
14. Given:
D = 25 stones/day * 200 days/year = 5,000 stones/year
S = $48
Price Schedule:
Number of Stones
Price per Stone (P)
1-399
$10
400-599
$9
600+
$8
a. H = $2. Determine the optimal order quantity:
Step 1:
Compute the common minimum point.
49090.489
2
48)000,5(22 H
DS
Q
stones
This quantity is feasible in the range 400-599.
Step 2:
Determine total cost for the common minimum point and for the price breaks of all lower unit
44721.447
)8(30.
48)000,5(22
H
DS
Not feasible
Chapter 13 – Inventory Management
Minimum point P = $9:
42264.421
)9(30.
48)000,5(22
H
DS
Feasible
Step 2:
Compare the total cost at Q = 422 to Q = 600.
15. Given:
D = 4,900
S = $50
H = 40% of purchase cost
Chapter 13 – Inventory Management
Step 2:
Compare the total cost at Q = 495 to Q = 1,000, 4,000, & 6,000.