inventory guide

M96 Materials Booklet

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.

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 demandCurrently on handValue

of item

Model 117,0002500$200

Model 215,0001500$300

Model 38,5002000$150

Repair kits450100$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?

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?

ItemDemand (units/wk)Value added

($/unit)Lot size

(units)Safety stock

(units)Lead time

(wk)

A405080201

B802016002

C160-4001001

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 weeksD: 2 weeksG: 5 weeks

B: 1 weekE: 4 weeksH: 3 weeks

C: 3 weeksF: 3 weeksI: 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

EMBED Word.Picture.8

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)?



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?

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.

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.

13. a. Complete the record using an FOQ of 45 units

Item:Lot sizeunits

Desc.Lead time4weeks

Parents:

12131415161718192021

Gross requirements253040252530

Scheduled receipts45

Projected on hand | 25

Planned receipts

Planned order releases

b. Complete the record using the L4L lot sizing rule.

Item:Lot sizeunits


Parents:

12131415161718192021




Planned receipts


c. Complete the record using the POQ, with P=3 lot sizing rule

Item:Lot sizeunits


Parents:

12131415161718192021




Planned receipts


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.

Item

Data CategoryCDEF

Lot-size ruleFOQ = 100L4LFOQ = 150POQ (P=3)

Lead time1232

Scheduled receipts100(week 1)None150(week 3)None

On-hand inventory350125400


LeadMPSQuantities

ProductTime123456

Item:Lot sizeunits

Desc.Lead timeweeks

Parents:

123456

Gross requirements

Scheduled receipts

Projected on hand |

Planned receipts


Item:Lot sizeunits

Desc.Lead timeweeks

Parents:

123456

Gross requirements

Scheduled receipts

Projected on hand |

Planned receipts


Item:Lot sizeunits

Desc.Lead timeweeks

Parents:

123456

Gross requirements

Scheduled receipts

Projected on hand |

Planned receipts


Item:Lot sizeunits

Desc.Lead timeweeks

Parents:

123456

Gross requirements

Scheduled receipts

Projected on hand |

Planned receipts


Answers:

1. average aggregate inventory value = $10,176,000

weekly sales = EQ \F(29.4 million,52 weeks per year) = .565 mill/week

weeks of supply = EQ \F(10.176 million,.565million/week) = 17.99 18 weeks of supply

inventory turns = EQ \F(29.4 million,10.176 million) = 2.89 2.9 turns per year

2. turns = EQ \F(annual sales,ave. aggr. inventory value) 7 = EQ \F(750200,ave. aggr. inventory value) Thus ave. aggr. inventory value = EQ \F(750200,7) = 107,171 107,000

3.

Last year's demandCurrently on handValue

of itemValue on handYearly value in sales

Model 117,0002500$200$500,000$3.4 mill

Model 215,0001500$300$450,000$4.5 mill

Model 38,5002000$150$300,000$1.275 mill

Repair kits450100$24.00$2400$.0108 mill

Ave. aggr. inventory value = $1.2524 mill.

Annual sales = $9.185 mill.

weekly sales = EQ \F(9.185,52) = $.176650 mill

weeks of supply = EQ \F(1.2524,.176650) = 7.09 7 weeks of supply

turns = EQ \F(9.1858,1.2524) = 7.33 turns per year

4. cycle inventory = EQ \F(450,2) = 225 units

$ value = (225)($90) = $20,250

pipeline inventory = dL = EQ \B(\F(1872,52))(5) = 180 units

$ value = (180)($70) = $12,600

5. The inventory items are valued, per unit, as follows:

ItemUnit value upon completionUnit value as WIP

C$10$10

B2($10)+$20 = $402($10)+$20/2 = $30

A2($40)+ $50 = $1302($40)+$50/2 = $105

a. Average inventory in dollars

Itemcyclesafetypipelinetotal

A$130(80/2) = $5,200$130(20) = $2,600$105(1)(40)= $4,200$12,000

B$40(160/2)= $3,2000$30(2)(80)= $4,800$8,000

C$10(400/2)= $2,000$10(100)= $1000$10(1)(160)= $1,600$4,600

Ave. aggr. inventory value = $24,600

b. weeks of supply = EQ \F($24600,(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

7.


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

EOQ = EQ \R(\F((2)(4160)(70),15)) = 197.04 => 197

b. Time between orders is = EQ \F(Q,d) = EQ \F(197,80) = 2.46 2.5 weeks

c. R = d(LT) + zL z, from normal table with cycle service level = 75%, = .67

demand is variable, leadtime is constant: L = = EQ \R(2) (20) R = (80)(2) + (.67) EQ \R(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

C = EQ \B(\F((4)(60),2) + 100) (10) + EQ \B(\F(52,4)) (35) = $2,655

b. With a service level desired of 92%, the safety stock can be reduced. z = 1.41

The new target inventory would be:

T = d(P+L) + zd= (60)(4+1) + 1.41(8) EQ \R(4+1) = 300 + 25.2 = 325.2

Safety stock would be:

= T - d(P+L) = 325.2 - (60)(4+1) 25 units

C = EQ \B(\F((4)(60),2) + 25) (10) + EQ \B(\F(52,4)) (35) = $1,905

Savings = $2,655 - $1,905 = $750

c. D = (d)(ppy) = (60)(52) = 3120 per year (ppy = period per year)

optimal P = EQ \F(1,d)\R(\F((2)(3120)(35),(10))) = 2.46 weeks 2 weeks

new T = (60)(2+1) + 1.41(8) EQ \R(2+1) = 180 + 19.5 = 199.5 200

safety stock = 200 - (60)(2+1) = 20 units.

new C = EQ \B(\F((2)(60),2) + 20) (10) + EQ \B(\F(52,2)) (35) = $1,710 for a savings of $945.

11. EOQ = EQ \R(\F((2)(20800)(40),2)) = 912

Usually, R = d(L) + zL , but in this case safety stock is set at 1000.

That means R = d(L) + safety stock or R = EQ \B(\F(20800,52))(2) + 1000 = 1800

12. d = EQ \F(D,ppy) = EQ \F(20800,52) = 400

P = EQ \F(1,400)\R(\F((2)(20800)(40),2)) = 2.28 2.3 weeks

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

13. a.

Item:Lot size45units


Parents:

12131415161718192021




Planned receipts454545

Planned order releases4545 45

b.

Item:Lot sizeL4Lunits


Parents:

12131415161718192021





Planned order releases25252530

c. Complete the record using the P=3 lot sizing rule

Item:Lot sizeP = 3units


Parents:

12131415161718192021






14.

Item:CLot size100units


Parents: 1 per A

123456




Planned receipts100


Item:DLot sizeL4Lunits


Parents:1 per A,1 per B

123456


Scheduled receipts




Item:ELot size150units


Parents:1 per B, 1 per D

123456






Item:FLot sizeP=3units


Parents:1 per D, 1 per E

123456


Scheduled receipts


Planned receipts200


Action notices: Order 70 Ds now, order 150 Es now, order 200 Fs now

Delay shipment of 100 Cs until period 3, if possible.

Materials Management Problems

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inventory guide

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