Answer:
A) 208
B) 104
C) 5.77 days
Explanation:
EOQ:
[tex]Q_{opt} = \sqrt{\frac{2DS}{H}} \sqrt{\frac{p}{p-u}}[/tex]
Where:
D = monthly demand = 360
S= setup cost = ordering cost = 60
H= Holding Cost = $2 per month
p = production rate: 36
u = daily usage
[tex]Q_{opt} = \sqrt{\frac{2(360)(60)}{2}}\sqrt{\frac{36}{36-18}}[/tex]
EOQ: 208
Maximum Inventory
[tex]\frac{EOQ}{p} (p-u)[/tex]
[tex]\frac{208}{36} (36-18)[/tex]
Max Inventory: 104
days per run:
[tex]\frac{EOQ}{p}[/tex]
[tex]\frac{208}{36}[/tex]
days per run: 5.77 (almost 6 days)
The best order amount for a corporation to purchase to reduce inventory costs such as holding charges, shortage costs, and order fees is the economic order quantity (EOQ).
a. approximately 208 boxes
b. 104 boxes
c. 5.8 days
The Given information
A = Annual demand = [tex]360 \text{ x } 12[/tex]= 4,320 boxes
p = 36 boxes per day
Operating days per month = 20 days
d = daily usage rate = [tex]\frac{360}{20}[/tex] = 18 boxes
S = setup cost per lot = $60 per lot
H = Holding cost ($) = $2 per box per month = $24 per box per year
[tex]Q = \sqrt{\frac{2AS}{H}(\frac{p}{p-d})}\\\\\\Q = \sqrt{\frac{2 \text{ x } 4,320 \text{ x } 60}{24}(\frac{36}{36-18})}\\\\\\Q=208 \text{ boxes}[/tex]
[tex]\text{Maximum inventory} = \text{I}_{max} = (1 - \frac{d}{p})}Q\\\\= (1 -\frac{18}{36} ) 208 \\\\= 104 \text{boxes}[/tex]
[tex]\text{The number of days in a run} = \text{T}_{p} = (\frac{Q}{p} ) \\\\= \frac{208}{36} \\\\= 5.8 \text{days}[/tex]
Learn more about EOQ computations, refer below
https://brainly.com/question/14120284
