Mgmt 120
Technique Review for Materials Management
This section contains study problems for the following topics.
•Inventory Measures
•Bill of Materials Concepts
•Independent Demand Inventory Models
•Materials Requirements Planning
Answers for all the problems are found in the back of this section.
Materials Management Problems
Materials Management Problems
1. Joan Pontius, the materials manager at Money Enterprises, is beginning to look for
ways to reduce inventory. A recent accounting statement shows inventories at the
following levels.
Raw materials:
$2,345,000
Work-in-progress: $5,670,000
Finished Goods:
$2,161,000
This year's cost of goods sold should be about $29.4 million. Assuming 52 business
weeks per year, express total inventory as:
a. weeks of supply
b. inventory turns
2. One product line is experiencing 7 turns per year, and its annual sales volume (at
cost) is $750,200. How much inventory is being held, on the average?
3. MINC, Inc. keeps 4 items in stock, three machines and a repair kit. The current
inventory level and last years demand are shown below along with the value of the items.
(Assume 52 weeks per year.)
Last year's Currently
Value
demand
on hand
of item
Model 1
17,000
2500
$200
Model 2
15,000
1500
$300
Model 3
8,500
2000
$150
Repair kits
450
100
$24.00
a. How many weeks of supply do they currently have on hand?
b. If the current inventory level is characteristic, how many turns per year does MINC
Inc. have?
Inventory Measures - 1
Materials Management Problems
Inventory Measures - 2
Materials Management Problems
4. A subassembly is produced in lots of 450 units. It is assembled from two
components worth $50. The value added (for labor and variable overhead) in
manufacturing one unit from its two components is $40, bringing the total cost per
completed unit to $90. The typical lead time for the item is 5 weeks and its annual
demand is 1872 units. There are 52 business weeks per year.
a. How many units of cycle inventory are held, on average, for the subassembly?
What is the dollar value of this cycle inventory?
b. How many units of pipeline inventory are held, on average, for the subassembly?
What is the dollar value of this inventory? (Assume that the typical job in pipeline
inventory is 50 percent completed. Thus one half the labor and overhead costs have
been added, bringing the total unit cost to $70 (or $50 + $40/2).
5. The bill of materials from item A shows that it is made from two units of B. Item
B, in turn is made from two units of purchased item C. The per unit purchase price of
item C is $10. The table below gives additional information. No anticipation inventory
is held.
a. what is the dollar value of inventory held, on average, for each item, broken down
by cycle, safety stock, and pipeline inventory?
b. How many weeks of supply are held to support demand for item A?
Value
added
($/unit)
Lot
size
(units)
Safety
stock
(units)
Lead
time
(wk)
Item
Demand
(units/wk
)
A
40
50
80
20
1
B
80
20
160
0
2
C
160
-
400
100
1
Inventory Measures - 3
Materials Management Problems
6. Consider the bill of materials diagram below.
a. How many immediate parents does item I have?
item B
item E?
b. How many unique components does item A have at all levels?
c. How many unique purchased items does item A have at all levels?
d. How many of item I are used in each item A?
e. How many unique intermediate items does product A have at all levels?
f. The items have the following lead times.
A: 2 weeks
D: 2 weeks
G: 5 weeks
B: 1 week
E: 4 weeks
H: 3 weeks
C: 3 weeks
F: 3 weeks
I: 2 weeks
What is the customer response time if
i. all items are specials (made to order)
ii. only items E and I are standard (made to stock)
iii. only items E, F, G, H, and I are standard
A
B(2)
E(1)
I(1)
Bill of Materials - 4
C(1)
F(1)
E(1)
I(1)
D(1)
G(1)
E(1)
I(1)
H(1)
Materials Management Problems
7. Item A is made from components B, C, and D. Item B, in turn, Is made from C.
Item D is also an intermediate item, made from B. Usage quantities are all one except
that two units of item C are needed to make one unit of B. Draw the bill of materials for
item A.
8. Consider the BOM below, which includes lead times.
a. What would be the customer response time if every item is a special?
b. What would be the customer response time if items D and F are made standards?
c. List each item and whether it is an end item, a purchased part, or an intermediate
item.
d. Are there any subassemblies in this BOM? If so, which item(s)?
Bill of Materials - 5
Materials Management Problems
A
B(2) 2 wk
D(1) 2 wk
E(1) 3 wk
G(1) 1 wk
Bill of Materials - 6
1 wk
C(1) 5 wk
F(1) 4 wk
Materials Management Problems
9. A discount appliance store sells combination radio and tape cassette players for
only $60 per unit. These hand held units have exceptional sound quality and are in great
demand. For these units:
• Demand = 80 units/wk
• Order cost = $70/order
• Annual holding cost rate = 25% of selling price
• Desired cycle-service level = 75%
• Lead time = 2 wk
• standard deviation of weekly demand = 20 units
• current on-hand inventory is 183 units, with no open orders or backorders.
The store operates 52 weeks per year. It has a continuous inventory review system.
a. What is the EOQ?
b. What would be the average time between orders (in weeks) using this EOQ?
c. What should R be?
d. An inventory withdrawal 0f 10 units has just occurred. Is it time to reorder?
Independent Demand - 7
10. Suppose that you were in charge of the inventory of an item worth $45 per unit.
Every 4 weeks you place an order. To determine the order quantity, you take the current
on-hand quantity and subtract it from 400. The result is the amount you order. The
weekly demand for this product is 60, with a standard deviation of 8. The cost to place an
order is $35, and the holding cost rate is $10 per unit per year. You desire to avoid
having a stock out 92% of the time. Assume there are 52 weeks in a year. The lead time
is a constant 1 week.
a. What is the current annual inventory cost (holding and ordering) for this item?
b. If you cannot change the frequency of orders, could you save money by changing
your target amount from 400?
c. If you could change your frequency of orders, what would be the "optimal" ordering
policy (Use periodic review system and round P to nearest week)? How much
money would this save over the current strategy?
11. Given the following data, what is the EOQ(optimal Q) and R?
• Annual demand is 20,800 units
• Ordering cost is $40 per order
• Holding cost rate is $2 per unit per year.
• Lead time = 2 weeks
• Safety stock level, set by company policy, is 1000 units.
• Assume 52 weeks in a year.
Independent Demand - 8
Materials Management Problems
12. Given the following data, compute the parameters for a periodic review system (P
and T) for a P that would provide an average order quantity approximately equal to the
EOQ.
• Annual demand is 20,800 units
• Ordering cost is $40 per order
• Holding cost rate is $2 per unit per year.
• Lead time = 2 weeks
• Safety stock level, set by company policy, is 1000 units.
• Assume 52 weeks in a year.
Independent Demand - 9
13. a. Complete the record using an FOQ of 45 units
Item:
Lot size
Desc.
Lead time
Parents:
12 13 14 15 16 17 18
19
Gross requirements
25
25
Scheduled receipts
45
30
40
25
units
weeks
4
20
21
30
Projected on hand | 25
Planned receipts
Planned order releases
b. Complete the record using the L4L lot sizing rule.
Item:
Lot size
Desc.
Lead time
Parents:
12 13 14 15 16 17 18
19
Gross requirements
25
25
Scheduled receipts
45
30
40
25
units
weeks
4
20
21
30
Projected on hand | 25
Planned receipts
Planned order releases
c. Complete the record using the POQ, with P=3 lot sizing rule
Item:
Lot size
units
Desc.
Lead time
4 weeks
Parents:
12 13 14 15 16 17 18 19 20 21
Gross requirements
25
Scheduled receipts
45
Projected on hand | 25
Planned receipts
Planned order releases
MRP - 10
30
40
25
25
30
Materials Management Problems
14. The MPS for product A calls for 70 units to be completed in week 4 and 80 units
in week 7 (the lead time is one week.) The MPS for product B calls for 150 units in week
7 (the lead time is 2 weeks). Develop the material requirements plan for the next six
weeks for items C, D, E, and F. Identify any action notices that will be generated.
Data Category
Lot-size rule
Lead time
Scheduled receipts
On-hand inventory
C
FOQ = 100
1
100(week
1)
35
Item
D
L4L
2
None
0
E
F
FOQ = 150 POQ (P=3)
3
2
150(week
None
3)
125
400
A
B
C(1)
D(1)
D(1)
E(1)
F(1)
F(1)
E(1)
F(1)
F(1)
E(1)
F(1)
MRP - 11
Product
Lead
Time
1
2
MPS
4
3
Item:
Desc.
Parents:
Quantities
5
6
Lot size
Lead time
1
2
3
units
weeks
4
5
6
Gross requirements
Scheduled receipts
Projected on hand |
Planned receipts
Planned order releases
Item:
Desc.
Parents:
Lot size
Lead time
1
2
3
units
weeks
4
5
6
Gross requirements
Scheduled receipts
Projected on hand |
Planned receipts
Planned order releases
Item:
Desc.
Parents:
Lot size
Lead time
1
2
3
units
weeks
4
5
6
Gross requirements
Scheduled receipts
Projected on hand |
Planned receipts
Planned order releases
Item:
Desc.
Parents:
Lot size
Lead time
1
Gross requirements
Scheduled receipts
Projected on hand |
Planned receipts
Planned order releases
MRP - 12
2
3
units
weeks
4
5
6
Materials Management Problems
Answers:
1. average aggregate inventory value = $10,176,000
29.4 million
weekly sales = 52 weeks per year = .565 mill/week
10.176 million
weeks of supply = .565million/week = 17.99 ≈ 18 weeks of supply
29.4 million
inventory turns = 10.176 million = 2.89 ≈ 2.9 turns per year
annual sales
2. turns = ave. aggr. inventory value
750200
7 = ave. aggr. inventory value
Thus ave. aggr. inventory value =
750200
= 107,171 ≈ 107,000
7
3.
Model 1
Model 2
Model 3
Repair
kits
Last year's
demand
Currently
on hand
Value
of item
Value on
hand
17,000
15,000
8,500
2500
1500
2000
$200
$300
$150
$500,000
$450,000
$300,000
450
100
$24.00
$2400
Yearly
value in
sales
$3.4 mill
$4.5 mill
$1.275
mill
$.0108
mill
Ave. aggr. inventory value = $1.2524 mill.
Annual sales = $9.185 mill.
9.185
weekly sales = 52 = $.176650 mill
1.2524
weeks of supply = .176650 = 7.09 ≈ 7 weeks of supply
9.1858
turns = 1.2524 = 7.33 turns per year
Answers - 13
450
4. cycle inventory = 2 = 225 units
$ value = (225)($90) = $20,250
1872
pipeline inventory = dL =  52 (5) = 180 units


$ value = (180)($70) = $12,600
5. The inventory items are valued, per unit, as follows:
Item
Unit value upon
Unit value as WIP
completion
C
$10
$10
B
2($10)+$20 = $40
2($10)+$20/2 = $30
A
2($40)+ $50 = $130
2($40)+$50/2 = $105
a. Average inventory in dollars
Item
cycle
safety
pipeline
A
$130(80/2) $130(20) = $105(1)(40
= $5,200
$2,600
)= $4,200
B
$40(160/2)
0
$30(2)(80)
= $3,200
= $4,800
C
$10(400/2) $10(100)= $10(1)(160
= $2,000
$1000
)= $1,600
total
$12,000
$8,000
$4,600
Ave. aggr. inventory value = $24,600
$24600
b. weeks of supply = (40 units/week)($130/unit) = 4.7 weeks
6. a. I: 1 ; B: 1 ; E: 3
b. 8
c. 4
d. 4
e. 4: E,B,C,D
f. i. 11
ii. 10
iii. 5
Answers - 14
Materials Management Problems
7.
A(1)
B(1)
C(1)
D(1)
B(1)
C(2)
C(2)
8. a. 10
b. 7
c. A - end, B - intermediate, C - intermediate, D - purchased, E - intermediate,
F - purchased, G - purchased
d. B is a subassembly
9. a. d = 80 ; D = (80)(52) = 4160
H = i*price = (.25)(60) = 15
(2)(4160)(70)
= 197.04 => 197
15
Q 197
b. Time between orders is = d = 80 = 2.46 ≈ 2.5 weeks
EOQ =
c. R = d(LT) + zL
z, from normal table with cycle service level = 75%, = .67
demand is variable, leadtime is constant: L = L ( d ) = 2 (20)
R = (80)(2) + (.67) 2 (20) = 179 units
d. 183 - 10 = 173. This is lower that the reorder point figured in part c, so an order
should be placed.
10. a. Current approach: P = 4, T = 400
Safety stock = T - d(P+L) = 400 - (60)(4+1) = 100
(4)(60)

52
C =  2 + 100 (10) +  4  (35) = $2,655


 
b. With a service level desired of 92%, the safety stock can be reduced. z = 1.41
Answers - 15
The new target inventory would be:
T = d(P+L) + zd P  L = (60)(4+1) + 1.41(8) 4+1 = 300 + 25.2 = 325.2
Safety stock would be:
= T - d(P+L) = 325.2 - (60)(4+1) ≈ 25 units
(4)(60)

52
C =  2 + 25 (10) +  4  (35) = $1,905


 
Savings = $2,655 - $1,905 = $750
c. D = (d)(ppy) = (60)(52) = 3120 per year
1
optimal P = d
(ppy = period per year)
(2)(3120)(35)
= 2.46 weeks ≈ 2 weeks
(10)
new T = (60)(2+1) + 1.41(8) 2+1 = 180 + 19.5 = 199.5 ≈ 200
safety stock = 200 - (60)(2+1) = 20 units.
(2)(60)

52
new C =  2 + 20 (10) +  2  (35) = $1,710 for a savings of $945.


 
(2)(20800)(40)
= 912
2
Usually, R = d(L) + zL , but in this case safety stock is set at 1000.
20800
That means R = d(L) + safety stock or R =  52 (2) + 1000 = 1800


11. EOQ =
D
20800
12. d = ppy = 52 = 400
1
P = 400
(2)(20800)(40)
= 2.28 ≈ 2.3 weeks
2
As in problem 11, safety stock is given, thus the revised formula for T is
T = d(OI + LT) + safety stock = (400)(2.3 + 2) + 1000 = 2720
Answers - 16
Materials Management Problems
13. a.
Item:
Desc.
Parents:
Lot size
Lead time
12
Gross requirements
25
Scheduled receipts
45
Projected on hand | 25
45
13
14
15
30
45
15
15
16
17
40
25
2
4
0
Planned receipts
4
5
Planned order releases
45
45
0
18
45
4
19
units
weeks
20
25
4
0
21
30
1
1
5
5
4
3
0
4
5
5
4
5
b.
Item:
Desc.
Parents:
Lot size
Lead time
12
Gross requirements
25
Scheduled receipts
45
Projected on hand | 25
45
13
14
15
30
45
15
15
Planned receipts
16
17
40
25
0
0
2
2
5
Planned order releases
25
25
25
18
19
L4L
units
4
weeks
20
25
0
0
30
0
2
5
21
0
3
5
0
3
0
c. Complete the record using the P=3 lot sizing rule
Item:
Lot size
Desc.
Lead time
Parents:
12 13 14 15 16 17 18
19
Gross requirements
25
25
Scheduled receipts
45
Projected on hand | 25
45
30
45
15
15
40
25
2
0
5
Planned receipts
5
0
Planned order releases
50
0
P = 3 units
4 weeks
3
0
20
21
30
3
0
0
5
5
55
Answers - 17
Answers - 18
Materials Management Problems
14.
Item:
C
Desc.
Parents:
Lot size
Lead time
1
1 per A,
3
4
1 per B,
100
135
135
65
6
80
65
65
85
100
100
1 per B
1
2
3
4
70
0
0
70
0
70
150
0
1 per D
1
2
70
55
55
150
150
1 per E
1
2
L4L
2
units
weeks
5
6
150
80
0
150
150
3
units
weeks
6
3
4
5
150
150
55
80
150
3
0
80
80
125
150
Lot size
Lead time
1 per D,
5
70
Lot size
Lead time
Gross requirements
Scheduled receipts
Projected on hand | 125
Planned receipts
Planned order releases
Item:
F
Desc.
Parents:
2
Lot size
Lead time
Gross requirements
Scheduled receipts
Projected on hand | 0
Planned receipts
Planned order releases
Item:
E
Desc.
Parents:
units
weeks
1 per A
Gross requirements
Scheduled receipts
Projected on hand | 35
Planned receipts
Planned order releases
Item:
D
Desc.
Parents:
100
1
4
125
150
125
P=3
2
units
weeks
5
6
Gross requirements
220 150
150
80
Scheduled receipts
Projected on hand | 400 180
30
80
0
0
0
Planned receipts
200
Planned order releases
200
Action notices: Order 70 Ds now, order 150 Es now, order 200 Fs now
Delay shipment of 100 Cs until period 3, if possible.
Answers - 19