A car repair shop requires various supplies for routine maintenance of cars. One
ID: 416085 • Letter: A
Question
A car repair shop requires various supplies for routine maintenance of cars. One such supply is windshield washer fluid. The average rate of washer fluid usage throughout the year is constant, with 120 gallons of usage forecasted per month. Fluid costs $2 per gallon, and the holding rate is 40% annually. The fixed ordering cost is $65.
a. Calculate the optimal economic order quantity (EOQ). About how many orders will be placed in a year and how often (in days) will orders be placed?
b. If the holding rate was actually 20% (not 40%), what would be the optimal economic order quantity, the number of orders per year, and the length of time between orders (in days)?
c. Assume that quantity discounts were available as follows: $1.90 per gallon if more than 500 gallons are ordered, $1.80 per gallon if more than 1000 gallons are ordered, and $1.80 per gallon if more than 2000 gallons are ordered. Calculate the optimal order quantity (assume the annual holding rate is 40%).
I need the answers to all the parts
Explanation / Answer
Answer to question a :
Economic order quantity ( EOQ ) = square root ( 2 x Co x D / Ch )
D = Annual demand = 120 Gallons / month x 12 months = 1440 gallons
Co = Fixed ordering cost = $65
Ch = Annual holding cost per unit = 40% of fluid cost of $2/ gallon = $0.8
Thus optimal order quantity ( i.e. EOQ )
= Square root ( 2 x 65 x 1440/ 0.8)
= 483.73 ( 484 rounded to nearest whole number )
Number of orders to be placed in a year = 1440/484 = 2.97 average
Frequency of order placement = Optimum order quantity / annual demand x 365 days = 484/1440 x 365 = 122.68 days on average
Answer to question b :
Revised annual holding cost per unit = 20% of fluid cost of $2 / Gallon = 20% of $2 = $0.40
Revised optimal order quantity ( EOQ )
= Square root ( 2 x 65 x 1440 / 0.4 )
= 684.10 ( 684 rounded to nearest whole number )
Number of orders to be placed in a year = 1440 / 684 = 2.10 orders on average
Frequency of order placement = Optimum order quantity / annual demand x 365 days = 684/1440 x 365 = 173.37
days
Answer to question c :
Since annual holding cost per unit (Ch) is proportional to unit cost of washer fluid, value of Ch will be different for different quantity slabs with their corresponding price levels
Following table presents different values of Ch and corresponding derived EOQs for different quantity slabs :
Derived EOQ for a certain quantity slab = Square root ( 2 x 65 x 1440/ Ch for that quantity slab)
Quantity slab
Price / Unit ( $)
Ch ( 40% of Price / unit) , $
Derived EOQ ( rounded to nearest whole number )
501 – 1000
1.90
0.76
496
1001 – 2000
1.80
0.72
510
2001 and above
1.80
0.72
510
It can however be noticed that none of the derived EOQ values match with their corresponding quantity slabs and hence none of the qualify as optimum order quantity.
Thus optimum order quantity remains unchanged at 484 ( as answered in question a )
Quantity slab
Price / Unit ( $)
Ch ( 40% of Price / unit) , $
Derived EOQ ( rounded to nearest whole number )
501 – 1000
1.90
0.76
496
1001 – 2000
1.80
0.72
510
2001 and above
1.80
0.72
510
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.