## Problem 6-8 (Algo) Solution Lambda Computer Products produced two prototype units: - **1st unit:** 5,000 hours, $250,000 materials, $30/hour labor - **2nd unit:** 3,250 hours, $187,500 materials, $30/hour labor We are to use the **learning curve** to estimate costs for: a. **10 additional units (immediate production)** b. **Significant delay (no learning effect, costs reset to the first unit)** Let's solve step by step: --- ### Step 1: Find the Learning Curve Rate **Learning curve formula:** \[ Y = aX^b \] Where: - \( Y \) = average labor hours per unit for X units - \( a \) = labor hours for the first unit - \( X \) = cumulative number of units - \( b \) = log(learning rate)/log(2) Given: - 1st unit: 5,000 hours - 2nd unit: 3,250 hours Set up the learning curve equation: \[ \frac{3,250}{5,000} = (2)^b \] \[ 0.65 = 2^b \] Take log on both sides: \[ \log(0.65) = b\log(2) \] \[ b = \frac{\log(0.65)}{\log(2)} = \frac{-0.1871}{0.3010} \approx -0.622 \] --- ### a. **Cost for 10 more units (Units 3 to 12)** #### **Step 2: Calculate Total Labor Hours for 12 Units** \[ \text{Total labor hours for } n \text{ units} = a \cdot n^{b+1} \] But more commonly, use the cumulative average: \[ \text{Cumulative average hours per unit for 12 units:} \] \[ Y = a \cdot n^b \] \[ Y_{12} = 5,000 \times 12^{-0.622} \] Calculate \( 12^{-0.622} \): \[ \ln(12) = 2.4849 \] \[ -0.622 \times 2.4849 = -1.545 \] \[ e^{-1.545} = 0.213 \] So, \[ Y_{12} = 5,000 \times 0.213 = 1,065 \text{ hours (average per unit for 12 units)} \] **Total labor hours for 12 units:** \[ 1,065 \times 12 = 12,780 \text{ hours} \] **Labor hours for first 2 units:** \[ 5,000 + 3,250 = 8,250 \text{ hours} \] **Labor hours for next 10 units:** \[ 12,780 - 8,250 = 4,530 \text{ hours} \] #### **Step 3: Compute Labor Cost** \[ \text{Labor cost} = 4,530 \times \$30 = \$135,900 \] #### **Step 4: Materials and Equipment Cost** - Average materials/equipment cost per unit: - 1st unit: $250,000 - 2nd unit: $187,500 - **Learning also applies to materials per problem statement.** So, average for 12 units: \[ \text{Avg material cost per unit for 12 units} = \$250,000 \times 0.213 = \$53,250 \] \[ \text{Total material/equipment cost for 12 units} = \$53,250 \times 12 = \$639,000 \] \[ \text{For 10 additional units: } 639,000 - (250,000 + 187,500) = 639,000 - 437,500 = \$201,500 \] #### **Step 5: Total Cost for Next 10 Units** \[ \text{Total cost} = \text{Labor} + \text{Materials/Equipment} \] \[ = \$135,900 + \$201,500 = \boxed{\$337,400} \] --- ### b. **Cost If There Was a Significant Delay (No Learning, All at First Unit Cost)** - Each new unit will cost the same as the **first unit**. \[ \text{Labor for 10 units} = 5,000 \times 10 = 50,000 \text{ hours} \] \[ \text{Labor cost} = 50,000 \times \$30 = \$1,500,000 \] \[ \text{Materials/Equipment for 10 units} = \$250,000 \times 10 = \$2,500,000 \] \[ \text{Total cost} = \$1,500,000 + \$2,500,000 = \boxed{\$4,000,000} \] --- ## **Final Summary** | Scenario | Total Cost for 10 Units | |---------------------------------------------------------|------------------------| | a. Immediate production (learning curve applies) | **\$337,400** | | b. Significant delay (no learning curve, reset to first unit) | **\$4,000,000** | - **With learning**, cost for 10 units is much lower due to increased efficiency. - **With delay**, learning is lost and costs reset, leading to much higher expenses.

## Summary of Cost Analysis ### a. Immediate Production (Learning Curve Applies) When Lambda Computer Products produces 10 additional units immediately, leveraging the learning curve, the estimated total cost is **$337,400**. This reduced cost reflects the efficiency gained from prior production experience. ### b. Significant Delay (No Learning Effect) If there is a significant delay between contracts, resulting in no learning effect, the estimated total cost for the same 10 units rises sharply to **$4,000,000**. This increase is due to the loss of cost efficiencies, as all subsequent units are priced at the initial unit's cost.