Respuesta :
Answer:
The minimum cost will be "$214085".
Explanation:
[tex]D = 1700 units \\\\S = \$ 50 \\\\H= 20%\\[/tex]
i) When quantity = 1-1500, price = $ 12.50 , and holding price is $12.50 * 20 %= $2.50.
ii) When quantity = 1501 -10,000, price = $ 12.45 , and holding price is $12.45 * 20 %= $2.49.
iii) When quantity = 10,0001- and more, price = $ 12.40 , and holding price is $12.40 * 20 %= $2.48.
[tex]EOQ= \sqrt{\frac{2DS}{H}} \\\\EOQ1= \sqrt{\frac{2\times 17000\times 50}{2.50}} \\\\EOQ1=824.62 \ \ \ or \ \ \ 825\\[/tex]
[tex]EOQ2= \sqrt{\frac{2\times 17000\times 50}{2.49}} \\\\EOQ1=826.2T \ \ \ or \ \ \ 826\\[/tex]
[tex]EOQ3= \sqrt{\frac{2\times 17000\times 50}{2.48}} \\\\EOQ3=827.93 \ \ \ or \ \ \ 828\\[/tex]
know we should calculate the total cost of EOQ1 and break ever points (1501 to 10,000)units
[tex]total \ cost = odering \ cost + holding \ cost + \ Annual \ product \ cost\\\\total_c = \frac{D}{Q} \times S + \frac{Q}{2} \times H + (p \times D) \\\\T_c = \frac{17000}{825} \times 50+ \frac{825}{2} \times 2.50 + (12.50 \times 17000)\\\\T_c = 1030 .30 +1031.25+212500\\\\T_c =$ 214561.55\\\\[/tex]
[tex]T_c = \frac{17000}{1501} \times 50+ \frac{1501}{2} \times 2.49 + (12.45 \times 17000)\\\\T_c = 566.28 +1868.74+211650\\\\T_c =$ 214085.02 \ \ \ or \ \ \ $ 214085\\\\[/tex]
[tex]T_c = \frac{17000}{10001} \times 50+ \frac{10001}{2} \times 2.48 + (12.40 \times 17000)\\\\T_c = 84.99+ 12401.24+210800\\\\T_c =$ 223286.23 \\[/tex]
The total cost is less then 15001. So, optimal order quantity is 1501, that's why cost is = $214085.
