Question

A local distributor for a national tire company expects to sell approximately 10,160 tires of a certain size and tread design next year. Annual carrying cost is $14 per tire and ordering cost is $76. The distributor operates 287 days a year.
a. What is the EOQ
b. How many times per year does the store reorder?
c. What is the length of an order cycle?
d. What is the total annual cost if the EOQ quantity is ordered?

Answer 1

Answer:

Following are the solution to the given points:

Explanation:

Given:

$(D) = 10,160\ tires / year\\\\(H) = \ 14 / tire\\\\(S) = \ 76\\\\work\ days\ number = 287 \ \frac{days}{year}$

For point a:

$EOQ = \sqrt{(\frac{2DS}{H})}$

         $=\sqrt{(\frac{2\times 10,160\times 14 }{14})}\\\\=\sqrt{({2\times 10,160})}\\\\=\sqrt{20320}\\\\=142.548$

For point b:

Calculating the order of number of per year $= \frac{D}{EOQ}$

                                                                          $=\frac{10,160}{142.548}\\\\=71.27\approx 71$

therefore, the reorded store 71 times per year

For point c:

Calculating the order cycle length $= (\frac{EOQ}{D}) \times \text{work days number in a year}$

                                                         $= (\frac{142.548}{10,160}) \times287\\\\= 0.0140\times287\\\\=4.018$

For point d:

$\text{Total annual cost = carrying cost + ordering cost}$

Carrying cost:

$= (\frac{EOQ}{2}) \times H \\\\= (\frac{142.548}{2}) \times 14 \\\\= 71.274 \times 14 \\\\= \ 997.836 \approx 998$

Ordering cost:  

$= (\frac{D}{EOQ}) \times S \\\\ = (\frac{10160}{142.548}) \times 76 \\\\ = 71.274\times 76\\\\ = \ 5416.824$

$\therefore\\\\\text{Total annual cost = Carrying cost + Holding cost}$

                             $=998+5416.824\\\\=6414.824$