Baker Mfg. Inc inventory turnover


Baker Mfg. Inc (see table 11.9) wishes to compare its inventory turnover to those of industry leaders wou have turnover of about 13 times per year and 8% of their assets invested in inventory
Arrow Distributing Corp
NET Rev. $16,500
Cost of sales $13,500
inventory $1,000
Total Assets $8,600
Baker MFG. INC
Net rev $27,500
Cost of sales $21,500
inventory $1,250
total assets $16,600
A) What is Baker’s Inventory Turnover?
Inventory turnover ratio = Cost of good sold/ inventory investment
Inventory turnover = $21,500/$1,250 = 17.2
B) What is Baker’s percent of assets committed to inventory?
Percent invested in inventory = (total inventory investment/total assets) * 100
Percent invested in inventory = ($1,250/$16,600) * 100 = 7.53 %
C) How does Baker’s performance compare to the industry leaders?
8%-7.53% 0.47
Compared to the industry leaders, Baker’s performance is very low because they are .47% lower than the industry leaders performance
Good job


Johnson Chemicals is considering two options ofr its supplier portfolio. Option 1 uses two local suppliers. Each has a “unique event” risk of 5%, and the probability of a “super event”
that would disable both at the same time is estimated to be 1.5%. Option 2 uses two suppliers located in different countries.
Each has a “unique event” risk of 13% and the probability of a “Super event” that would disable both at the same time is estimated to be .2%.
P(n) = S+(1-S)U^n
A) What is the probability that both suppliers would be disrupted using option 1?
S = 1.50%
U = 5%
P(n) = 0.0175
B) What is the probability that both suppliers would be disrupted using option 2
S= 0.20%
U= 13%
P(n) = 0.0189
C) Which option would provide the lowest risk of total shutdown?
Option 1 has the lowest risk of total shutdown
because it has a lower probability than option 2


Consider a three firm supply chain consisting of a retailer, manufacturer, and a supplier. Ther retailer’s demand over an 8 week period was 100 units each of the first 2 weeks,
200 units of each of the second week, 300 units of the third 2 weeks and 400 units of the fourth 2 weeks. The following table represents the orders placed by each irm in the supply chain.
Notice, as is often the case in supply chains due to economies of scale, that total units are the same in each case, but firms further up the supply chain (away form the reatailer) place larger, less requent, orderes
Week Retailer Manufacturer supplier
1 100 200 600
2 100 0 0
3 200 400 0
4 200 0 0
5 300 600 1400
6 300 0 0
7 400 800 0
8 400 0 0
Recall that the same sample variance of a data set can be found by using VAR.S function in excel or by plugging each x value of the data set into the formula:
Variance = (x-x)^2/(n-1), where x is the mean of the data set and n is the nubmer of values in the set.
Variance = 14286
Variance = 100000
Variance = 260000
A) What is the bullwhip measure for the retailer?
Bullwhip measure = Variance of orders/ Variance of demand
Bullwhip measure for retailer = 14286/14286 1 correct
B) What is the bullwhip measure for manufacturer?
Bullwhip measure for manufacturer = 100,000/14,285 7 correct
C) What is the bullwhip measure for supplier?
Bullwhip measure for supplier = 260,000/100,000 2.6 correct
D) What conclusions can you draw regarding the impact that economies of scale may have on the bullwhip effect?
We can see that the economies of scale seem to contribute to the bullwhip effect. As less frequent, large orders occur, the variance is larger as it moves up the supply chain toward the supplier


Monczka-Trent Shipping is the logistics vendor for Handfield Manufacutring Co. in Ohio. Handfield has daily shipments of a power steering pup from its Ohio plant to an auto assembly line in Alabama.
The value of the standard shipment is $250,000. Monczka-Traent has two options: 1: its standard 2 day shipment or 2: a subcontractor who will team drive overnight with an effective delivery of one day.
The extra driver cost $175. Hndfield’s holding cost is 35% annually for this kind of inventory.
Standard shipping $250,000
Holding cost 35% $240
extra driver $175
A) Which option is more economical?
Option 2 with the subcontractor is more economical than the standard shipping.
The holding cost is $240 with and extra $175 for the driver rather than $250,000.
B) What production issues are not included in the data presented?
The issue that is not included in the data is that the Alabama’s assembly line is receiving its inputs from Ohio.
Changing the delivery pattern might affect the work of the assembly line for example the 1 day shipping can overstock the assembly line.
This can also effect the production process or issues with the parts and cause delays in the production.


L. Houts Plastics is a large manaufacturer of injection molded plastics in North Carolina. An investigation of the company’s Manufacturing facility in Charlotte yields the information presented in the table below.
How would the plant classify these items according to an ABC classification system?
Item Code Average inventory units Value of unit Total value % of total value % of total quality % of valuation/ quantity Classification
2347 300 $4 $1,200 35.59 28.90 1.23 A
1289 400 $3.75 $1,500 44.49 38.54 1.15 A
9111 6 $3.00 $18 0.53 0.58 0.92 B
2349 120 $2.50 $300 8.90 11.56 0.77 B
7844 12 $2.05 $25 0.73 1.16 0.63 B
2395 30 $2.00 $60 1.78 2.89 0.62 B
8310 7 $2.00 $14 0.42 0.67 0.62 B
8210 8 $1.80 $14 0.43 0.77 0.55 C
2394 60 $1.75 $105 3.11 5.78 0.54 C
2363 75 $1.50 $113 3.34 7.23 0.46 C
6782 20 $1.15 $23 0.68 1.93 0.35 C
Total 1038 Total $3,372
A Category comprised of 20-30% of total quantity & contributes to 70-80% of $ valuation
B Category comprised of 20-30% of total quantity & contributes to 15-20% of $ valuation
C Category comprised of 45-50% quantity & contributes to 5-8% of $ valuation


Lindsay Electronics, a small manufacturer of electronic research equipment, has approximately 7,000 items in its inventory and has hired Joan Blasco-Paul
to manage its inventory. Joan has determined that 10% of the items in inventory are A items, 35% are B items, and 55% are C items.
She would like to set up a system in which all A itmes are counted monthy (every 20 working days), all B items are counted quarterly (every 60 working days), and all C items are counted semiannually
(every 120 working days). How many items need to be counted each day?
Inventory 7,000 working days
A 10% 20
B 35% 60
C 55% 120
A items per day = 35
B Items per day = 41
C items per day = 32
Total = 108 Items


Joe Henry’s machine shop uses 2,500 brackets during the course of a year. These bractets are purchased from a supplier
90 miles away. The following information is known about the brackets:
Annual demand 2,500
Holding cost per bracket per yr $1.50
oder cost per order $18.75
lead time 2 days
working days per yr 250
A) What would be the economic order quantity? (EOQ)
EOQ =√2CD/h
EOQ = 250
B) Given the EOQ, what would be the average inventory? What would be the annual inventory holding cost?
AVG Inv = EOQ/2
AVG INV = 125
Annual holding cost = AVG INV * HOLDING COST
Annual holding cost = $187.50
C) Given the EOQ, how many orders would be made each year? What would be the annual order cost?
# of orders = D/working days in a year
# of orders = 10
Annual order cost = (D/EOQ)*C
Annual order cost = $187.50
D) Given the EOQ, what is the total annual cost of managing the inventory?
Annual managing cost = √2CD*h
annual managing cost = $ 375.00
E) What is the time between orders?
Time between orders = 250/number of orders
Time between orders = 25 days
F) What is the reorder point? (ROP)
ROP = LEAD TIME * Daily Demand
ROP = 20


Arthur Meiners is the production manager of Wheel-Rite, a small producer of metal parts. Wheel-Right supplies Caltex, a larger assembly company, with 10,000 wheel bearings each Wheel-right is $40, and holding cost is $.60 per wheel bearing per year.
Wheel-Rite can produce 500 wheel bearings per day. Caltex is a just-in-time manufacturer and requires that 50 bearings be shipped to it each business day.
Annual Demand 10,000
Setup cost $40
holding cost $0.60
daily production rate 500 per day
daily demand rate 50 per day
Length of production 1 day
days in a year 365
A) What is the optimum production quantity?
Qp = √2DS/H(1-(d/p)
Optimum production quantity = 1217.16 correct
B) What is the maximum number of wheel bearings that will be in inventory at Wheel-Rite?
Q = 500
Maximum inventory level = (total production during the production run) – (Total used during the production run) Q(1-(d/p))
Maximum inventory level = 450 1095
C) How many production runs of wheel bearings will Wheel-Rite have in a year?
Production runs = 365 in a year 8.22
D) What is the total setup + holding cost for Wheel-Rite?
Setup = (D/Q) * S
Holding cost = 1/2*H*Q(1-(d/p))
Setup cost = $800
Holding cost = $ 135
Setup cost + holding

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