A sporting goods store sells 100 pool tables per year. It costs $20 to store one pool table for a year, based on the average inventory on hand. It costs $40 for each delivery of pool tables. How many times per year and in what lot size should the store order pool tables to minimize its inventory costs?
derivative way, help please?
To minimize inventory costs, we can use the Economic Order Quantity (EOQ) model. This model helps determine the optimal order quantity by considering the ordering cost and carrying cost.
Let's define the variables first:
D = Annual demand = 100 pool tables
S = Ordering cost per order = $40
H = Carrying cost per unit per year = $20
The formula for EOQ is given by:
EOQ = sqrt((2DS)/H)
Substituting the values:
EOQ = sqrt((2 * 100 * 40)/20)
Simplifying:
EOQ = sqrt(8000/20)
EOQ = sqrt(400)
EOQ = 20
So, the optimal order quantity is 20 pool tables per order.
To calculate the number of orders per year, we can divide the annual demand by the order quantity:
Number of orders = D / EOQ
Number of orders = 100 / 20
Number of orders = 5
Therefore, the store should place 5 orders per year, with a lot size of 20 pool tables per order, to minimize its inventory costs.
To find the lot size and the number of times the store should order pool tables per year to minimize inventory costs, we can use derivatives. Here's how:
Let's assume the store orders pool tables 'n' times per year, and the lot size of each order is 'Q'.
The carrying cost is the cost of storing a pool table for a year, which is $20.
The ordering cost is the cost of each delivery, which is $40.
The annual carrying cost can be calculated by multiplying the average inventory by the carrying cost, and the annual ordering cost can be calculated by multiplying the number of orders (n) by the ordering cost:
Annual carrying cost = n * (Q/2) * $20
Annual ordering cost = n * $40
The total annual inventory cost is the sum of the annual carrying cost and the annual ordering cost:
Total annual cost = Annual carrying cost + Annual ordering cost
To minimize the total annual cost, we need to find the optimal values of 'n' and 'Q'.
Now, we'll take the derivative of the total annual cost with respect to 'n' and 'Q', set them equal to zero, and solve the resulting equations.
1. Let's find the derivative with respect to 'n':
d(Total annual cost)/dn = d(Annual carrying cost)/dn + d(Annual ordering cost)/dn
Since the annual ordering cost is constant, the derivative of the annual ordering cost with respect to 'n' is zero.
d(Annual ordering cost)/dn = 0
Now, let's find the derivative of the annual carrying cost with respect to 'n':
d(Annual carrying cost)/dn = (Q/2) * $20
Setting the derivative equal to zero, we have (Q/2) * $20 = 0. Solving for 'Q', we get Q = 0.
From this equation, we see that 'Q' does not affect the number of times pool tables should be ordered. Therefore, to minimize the total annual cost, we should order all 100 pool tables in a single lot size.
2. Now, let's find the derivative with respect to 'Q':
d(Total annual cost)/dQ = d(Annual carrying cost)/dQ + d(Annual ordering cost)/dQ
The derivative of the annual carrying cost with respect to 'Q' is:
d(Annual carrying cost)/dQ = n * $20
The derivative of the annual ordering cost with respect to 'Q' is zero since the ordering cost is constant.
d(Annual ordering cost)/dQ = 0
Setting the derivative equal to zero, we have n * $20 = 0. Solving for 'n', we get n = 0.
From this equation, we see that 'n' does not affect the lot size of pool tables. Therefore, to minimize the total annual cost, we should order all 100 pool tables in a single lot size.
In conclusion, to minimize inventory costs, the store should order 100 pool tables once per year in a single lot size.