Please answer within 30 minutes and round correctly. 1.Bob\'s Bumpers has a repe
ID: 368870 • Letter: P
Question
Please answer within 30 minutes and round correctly.
1.Bob's Bumpers has a repetitive manufacturing facility in Kentucky that makes automobile bumpers and other auto body parts. The facility operates 300 days per year and has annual demand of 55,000 bumpers. They can produce up to 320 bumpers each day. It costs $53 to set up the production line to produce bumpers. The cost of each bumper is $95 and annual holding costs are $25 per unit. Setup labor cost is $23 per hour.
a.Based on the above information, what is the optimal size of the production run for bumpers? (Display your answer to two decimal places.)
b.Based on your answer to the previous question, and assuming the manufacturer holds no safety stock, what would be the average inventory for these bumpers? (Display your answer to two decimal places.)
c.Based on your answer two questions back, how many production runs would be required each year to satisfy demand? (Display your answer to two decimal places.)
d.Suppose the customer (an auto manufacturer) wants to purchase these bumpers in lots of 200 and that Bob's Bumpers is able to reduce setup costs to the point where 200 is now the optimal production run quantity. How much will they save in annual holding costs with this new lower production quantity? (Display your answer to twodecimal places.)
e.How much will they save in annual setup costs with this new lower production quantity? (Display your answer to two decimal places.)
Explanation / Answer
This problem will be solved using Economic Production Quantity ( EPQ) model. As per EPQ model, optimal production size ( EPQ) will be defined as per following :
EPQ = Square root ( 2 x Co x D / Ch x ( 1 – d/p))
Where,
D = Annual demand = 55,000 bumpers
Co = Set up cost = $53
Ch = Annual unit holding cost = $25
‘d = Daily demand = 55,000/300
‘p = Daily production capacity = 320
Therefore ,
EPQ = Square root ( 2 x 53 x 55000 / 25 x ( 1 – 55000/( 300 x 320))
= Square root ( 2 x 53 x 55000/ 25 x 0.4271)
= 738.92
OPTIMAL SIZE OF PRODUCTION = 738.92
Minimum inventory for this production run = 0
Maximum inventory for this production run
= EPQ x ( 1 – d/p)
= 738.92 x ( 1 – 55000/(300 x 320))
= 738.92 x 0.4271
= 315.59
Thus Average inventory
= ( Minimum inventory + Maximum Inventory ) / 2
= 315.59/2
= 157.79
AVERAGE INVENTORY = 157.79
Number of production run required to satisfy annual demand
= Annual demand / EPQ
= 55000/738.92
= 74.43
NUMBER OF PRODUCTION RUN REQUIRED = 74.43
Let the corresponding set up cost for EPQ to be 200 is Cs
Therefore ,
200 = Square root ( 2 x Cs x 55000 / 25 x ( 1 – 55000/300 x 320))
Or, 40000 = 110,000x Cs / 25 x 0.4271
Or, Cs = ( 40,000 x 25 x 0.4271/ 110,000)
Or , Cs = 3.88
Annual holding cost when EPQ is 738.92
= Ch x Average inventory
= $ 25 x 157.79
= $3944.75
Annual holding cost when EPQ = 200 will be
= Ch x Average Inventory
= Ch x EPQ x ( 1 – d/p)/ 2
= $25 x 200 x 0.4271/2
= $1067.75
Thus, amount which will be saved = $3944.75 - $1067.75 = $2877
SAVINGS IN ANNUAL HOLDING COST = $2877
Annual set up cost
= Set up cost x Number of set ups
= Set up cost x Annual demand / EPQ
Thus,
Initial set up cost with EPQ 738.92
= $53 X 55000/ 738.92
= $3944.94
Revised set up cost with EPQ = 200
= 3.88 X 55000/ 200
= $1067
Savings in annual set up cost = $3944.94 - $1067 = $2877.94
SAVINGS IN ANNUAL SET UP COST = $2877.94
OPTIMAL SIZE OF PRODUCTION = 738.92