Provide a step-by-step, in-depth, and comprehensive solution in a structured format, ensuring completeness and clarity . The solution should address all requirements specified in the question without

Question

Provide a step-by-step, in-depth, and comprehensive solution in a structured format, ensuring completeness and clarity . The solution should address all requirements specified in the question without omitting any details. 
Include an Explanation block after each step, containing relevant supporting statements that clarify the methodology and reasoning used.
 
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·	Ensure utmost accuracy and strict adherence to the necessary concepts and correct textbook-based formulas required to solve the question.
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·	Include an Explanation block at the end of each step to provide context and clarify the methodology used.
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VIPSolutions · Accepted Answer

**Step-by-Step Comprehensive Solution with Structured Explanations**

---

**Given Data:**

- Annual demand ($D$) = 20,000 units/year  
- Fixed order cost ($S$) = $600  
- Inventory holding cost rate ($h$) = 0.4 per year  
- Price schedule:
  - For the first 300 units, unit price ($p_1$) = $10/unit
  - For additional units, unit price ($p_2$) = $5/unit
- For $Q \leq 300$, all units cost $10/unit
- For $Q > 300$, first 300 units at $10/unit, extra units at $5/unit
- Example: If $Q=301$, total purchase cost = $3,005

**Required:**
- Determine the **optimal order quantity**
- Find the **corresponding optimal total cost per unit time**

---

### **Step 1: Define Total Cost Function for Each Price Break**

#### **Case 1: Order Quantity $Q \leq 300$ (All units at $10/unit)**
$$
\text{Total Cost}_1(Q) = \text{Purchase Cost} + \text{Ordering Cost} + \text{Holding Cost}
$$
$$
\text{Purchase Cost} = D \times p_1
$$
$$
\text{Ordering Cost} = \frac{D}{Q} \times S
$$
$$
\text{Holding Cost} = \frac{Q}{2} \times h \times p_1
$$
Thus,
$$
\text{Total Cost}_1(Q) = 20,000 \times 10 + \frac{20,000}{Q} \times 600 + \frac{Q}{2} \times 0.4 \times 10
$$

#### **Case 2: Order Quantity $Q > 300$ (First 300 at $10, rest at $5)**
$$
\text{Unit Purchase Cost for Q units} = 300 \times 10 + (Q - 300) \times 5
$$
$$
\text{Average Unit Price} = \frac{300 \times 10 + (Q - 300) \times 5}{Q}
$$
$$
\text{Purchase Cost per year} = D \times \text{Average Unit Price}
$$
$$
\text{Ordering Cost} = \frac{D}{Q} \times S
$$
$$
\text{Holding Cost} = \frac{Q}{2} \times h \times \text{Average Unit Price}
$$
$$
\text{Total Cost}_2(Q) = D \times \left( \frac{300 \times 10 + (Q - 300) \times 5}{Q} \right) + \frac{D}{Q} \times S + \frac{Q}{2} \times h \times \left( \frac{300 \times 10 + (Q - 300) \times 5}{Q} \right)
$$

---

**Explanation Block (Step 1):**  
The total cost includes all relevant expenses: annual purchase, ordering, and holding costs. For $Q \leq 300$, the cost is straightforward as all units are purchased at the higher price. For $Q > 300$, the cost is split between two price levels, requiring weighted average pricing in both purchase and holding cost calculations.

---

### **Step 2: Compute EOQ for Each Price Option**

#### **Case 1: All Units at $10 (EOQ$_1$)**

Economic Order Quantity (EOQ) formula:
$$
EOQ_1 = \sqrt{ \frac{2DS}{h \times p_1} }
$$
$$
EOQ_1 = \sqrt{ \frac{2 \times 20,000 \times 600}{0.4 \times 10} }
$$
$$
EOQ_1 = \sqrt{ \frac{24,000,000}{4} }
$$
$$
EOQ_1 = \sqrt{6,000,000}
$$
$$
EOQ_1 = 2,449.49
$$

Since $EOQ_1 > 300$, it is not a feasible solution for the $Q \leq 300$ range. The largest possible order in this range is 300.

#### **Case 2: Mixed Pricing for $Q > 300$ (EOQ$_2$)**

Let the average unit price for $Q > 300$ be denoted as $p_{avg}(Q)$:
$$
p_{avg}(Q) = \frac{3000 + 5(Q-300)}{Q} = \frac{5Q + 1500}{Q}
$$

Plug into EOQ formula:
$$
EOQ_2 = \sqrt{ \frac{2DS}{h \times p_{avg}(Q)} }
$$
However, since $p_{avg}(Q)$ depends on $Q$, the EOQ formula must be solved iteratively.

---

**Explanation Block (Step 2):**  
The EOQ formula provides the order quantity that minimizes total ordering and holding costs. For the price break, the largest allowed $Q$ is used in the $10 range, while for the mixed price, the EOQ must be solved iteratively because the average price changes with $Q$.

---

### **Step 3: Evaluate Total Annual Cost at Critical Points**

#### **Critical Points:**
- $Q = 300$ (end of first price break)
- EOQ$_2$: Iteratively estimated for the lower price range, but must be at least 301

---

#### **A. Calculate Total Cost at $Q = 300$**

$$
\text{Total Cost}_1(300) = 20,000 \times 10 + \frac{20,000}{300} \times 600 + \frac{300}{2} \times 0.4 \times 10
$$
$$
= 200,000 + 40,000 + 600
$$
$$
= 200,000 + 40,000 + 600
$$
$$
= 240,600
$$

---

#### **B. Calculate EOQ for Mixed Price ($Q > 300$)**

Let $Q = X$, $p_{avg}(X) = \frac{5X + 1500}{X}$:

The EOQ formula:
$$
EOQ_2 = \sqrt{ \frac{2 \times 20,000 \times 600}{0.4 \times p_{avg}(X)} }
$$
But $EOQ_2$ appears on both sides. Use initial guess for $p_{avg}$:

Try $Q=1000$:

$$
p_{avg}(1000) = \frac{5 \times 1000 + 1500}{1000} = \frac{6500}{1000} = 6.5
$$
$$
EOQ_2 = \sqrt{ \frac{24,000,000}{0.4 \times 6.5} } = \sqrt{ \frac{24,000,000}{2.6} } = \sqrt{9,230,769.23} = 3,037
$$

Update $p_{avg}(3037)$:

$$
p_{avg}(3037) = \frac{5 \times 3037 + 1500}{3037} = \frac{15,185 + 1500}{3037} = \frac{16,685}{3037} \approx 5.49
$$
$$
EOQ_2 = \sqrt{ \frac{24,000,000}{0.4 \times 5.49} } = \sqrt{ \frac{24,000,000}{2.196} } = \sqrt{10,931,372.55} \approx 3,307
$$

Update again $p_{avg}(3307)$:

$$
p_{avg}(3307) = \frac{5 \times 3307 + 1500}{3307} = \frac{16,535 + 1500}{3307} = \frac{18,035}{3307} \approx 5.456
$$
$$
EOQ_2 = \sqrt{ \frac{24,000,000}{0.4 \times 5.456} } = \sqrt{ \frac{24,000,000}{2.1824} } = \sqrt{10,995,905.82} \approx 3,317
$$

Now, convergence is achieved at EOQ ≈ **3,317 units**

---

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