Question 1168521: A retail outlet for calculators sells 700 calculators per year. It costs $2 to store one calculator for a year. To reorder, there is a fixed cost of $5, plus $2.35 for each calculator. How many times per year should the store order calculators, and in what lot size, in order to minimize inventory costs? The store should order ___ calculators ___ times per year to minimize inventory costs.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem using the Economic Order Quantity (EOQ) model.
**1. Define Variables**
* D = Annual demand = 700 calculators
* H = Holding cost per unit per year = $2
* S = Ordering cost per order = $5 (fixed cost) + 2.35 * Q (where Q is the lot size)
**2. EOQ Formula**
The EOQ formula is:
EOQ = √(2DS / H)
However, we have an ordering cost that depends on the quantity, which is not the typical EOQ situation.
**3. Total Inventory Cost**
Total cost (TC) = Holding cost + Ordering cost
* Holding cost = (Q/2) * H = (Q/2) * 2 = Q
* Ordering cost = (D/Q) * S = (700/Q) * (5 + 2.35Q)
TC = Q + (700/Q) * (5 + 2.35Q)
TC = Q + 3500/Q + 1645
To minimize TC, we need to find the derivative of TC with respect to Q and set it to 0.
**4. Differentiate and Solve for Q**
* d(TC)/dQ = 1 - 3500/Q² = 0
* 1 = 3500/Q²
* Q² = 3500
* Q = √3500 ≈ 59.16
Since we must order whole calculators, we consider Q = 59 and Q = 60.
**5. Calculate Total Cost for Q = 59 and Q = 60**
* TC(59) = 59 + 3500/59 + 1645 ≈ 59 + 59.32 + 1645 ≈ 1763.32
* TC(60) = 60 + 3500/60 + 1645 ≈ 60 + 58.33 + 1645 ≈ 1763.33
Q = 59 minimizes the total cost.
**6. Calculate Number of Orders Per Year**
* Number of orders = D / Q = 700 / 59 ≈ 11.86
Since we can't have a fraction of an order, we round to 12.
**7. Final Answer**
The store should order 59 calculators 12 times per year to minimize inventory costs.
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