## Problem 6-8 (Algo) Solution Let's break down and answer each part in Chegg India style, step by step. --- ### **Given Data** - **Unit 1:** - Hours: 5,000 - Material, equipment, supplies: $250,000 - **Unit 2:** - Hours: 3,250 - Material, equipment, supplies: $187,500 - **Labor Cost:** $30/hour - **Learning curve:** Use Exhibit 6.5 (Assume 80% learning curve, as commonly referenced) - **Company expects “learning” to occur relative to labor and pricing of parts from suppliers** --- ## **a. Cost for 10 more units (after 2nd unit, immediate start)** ### **Step 1: Find Learning Curve Rate** From the data: - Time to produce 2nd unit = 3,250 hours - Time to produce 1st unit = 5,000 hours Learning curve formula for cumulative average time per unit when doubling quantity: \[ \text{Learning rate (LR)} = \frac{\text{Time for 2nd unit}}{\text{Time for 1st unit}} \] \[ \text{LR} = \frac{3,250}{5,000} = 0.65 \] This suggests a **65% learning curve**, not 80%. We'll use 65% for calculations. --- ### **Step 2: Calculate Average Hours for 12 Units** The learning curve cumulative average time formula: \[ Y = aX^b \] Where: - \( Y \) = Average time per unit for X units - \( a \) = Time for first unit = 5,000 hours - \( X \) = Cumulative number of units = 12 - \( b \) = log(learning curve rate) / log(2) Calculate \( b \): \[ b = \frac{\log(0.65)}{\log(2)} \approx \frac{-0.1871}{0.3010} \approx -0.6218 \] Calculate average time per unit for 12 units: \[ Y_{12} = 5,000 \times (12)^{-0.6218} \approx 5,000 \times 0.233 \approx 1,165 \text{ hours (average per unit)} \] Total hours for 12 units: \[ \text{Total hours} = Y_{12} \times 12 = 1,165 \times 12 = 13,980 \text{ hours} \] Already produced 2 units, so hours for next 10 units: \[ \text{Total hours for 2 units:} \] - 1st unit: 5,000 - 2nd unit: 3,250 - Total = 8,250 hours \[ \text{Hours for next 10 units} = 13,980 - 8,250 = 5,730 \text{ hours} \] --- ### **Step 3: Calculate Total Cost for 10 Additional Units** #### **Labor cost:** \[ \text{Labor cost} = 5,730 \text{ hours} \times \$30/\text{hour} = \$171,900 \] #### **Material, Equipment, Supplies:** - 1st unit: $250,000 - 2nd unit: $187,500 - Assume same learning curve for materials. Learning curve for cost: \[ b = \frac{\log(0.75)}{\log(2)} \approx \frac{-0.1249}{0.3010} \approx -0.415 \] But using data, 2nd/1st = 187,500/250,000 = 0.75 (so 75% learning curve for materials). For 12 units: \[ Y_{12} = 250,000 \times (12)^{-0.415} \approx 250,000 \times 0.425 \approx \$106,250 \text{ (average per unit)} \] \[ \text{Total for 12 units} = 106,250 \times 12 = \$1,275,000 \] \[ \text{Already spent for 2 units} = 250,000 + 187,500 = \$437,500 \] \[ \text{Cost for next 10 units} = \$1,275,000 - \$437,500 = \$837,500 \] #### **Total Bid for 10 More Units:** \[ \text{Total Cost} = \text{Labor} + \text{Material, Equipment, Supplies} = \$171,900 + \$837,500 = \$1,009,400 \] --- ### **a. Final Answer:** \[ \boxed{\$1,009,400} \] **(This is the total bid for 10 additional units, if production starts immediately after the 2nd unit.)** --- ## **b. New Bid if There is a Significant Delay (No Learning Effect)** If there is a significant delay, learning is lost. Each unit will take the same as the 2nd unit. - **Labor per unit:** 3,250 hours - **Material per unit:** $187,500 For 10 units: - **Labor:** \(3,250 \times 10 = 32,500\) hours - **Material, equipment, supplies:** \(187,500 \times 10 = \$1,875,000\) **Labor Cost:** \[ 32,500 \text{ hours} \times \$30/\text{hour} = \$975,000 \] **Total Bid:** \[ \text{Total Cost} = \$975,000 + \$1,875,000 = \$2,850,000 \] --- ### **b. Final Answer:** \[ \boxed{\$2,850,000} \] --- ## **Summary Table** | Case | Total Cost for 10 Units | |------------------------------|------------------------| | **Immediate Start (Learning)**| \$1,009,400 | | **Significant Delay (No Learning)**| \$2,850,000 | --- ### **Final Summary** - **If production starts immediately, the cost for 10 more units is \$1,009,400 due to the learning effect.** - **If there is a significant delay and learning is lost, the cost rises to \$2,850,000.** --- **If you have Exhibit 6.5, you can use the exact learning curve rate from the table for more accurate results.**

# Problem 6-8 (Algo) Solution Let's break down the solution step by step. ## Given Data - **Unit 1:** - Hours: 5,000 - Material, equipment, supplies: $250,000 - **Unit 2:** - Hours: 3,250 - Material, equipment, supplies: $187,500 - **Labor Cost:** $30/hour - **Learning Curve:** Assume 80% learning curve as a baseline. --- ## Part a: Cost for 10 Additional Units (Immediate Start) ### Step 1: Determine the Learning Curve Rate Using the data: - Time for 2nd unit = 3,250 hours - Time for 1st unit = 5,000 hours Calculate the learning curve: \[ \text{Learning rate} = \frac{3,250 \text{ hours}}{5,000 \text{ hours}} = 0.65 \] This indicates a **65% learning curve**. ### Step 2: Calculate Average Hours for 12 Units Using the learning curve formula: \[ Y = aX^b \] Where: - \( Y \) = Average time per unit for \( X \) units - \( a \) = Time for the first unit = 5,000 hours - \( X \) = Cumulative number of units = 12 - \( b = \frac{\log(0.65)}{\log(2)} \approx -0.6218 \) Calculating \( Y_{12} \): \[ Y_{12} = 5,000 \times (12)^{-0.6218} \approx 1,165 \text{ hours} \] Total hours for 12 units: \[ \text{Total hours} = 1,165 \times 12 = 13,980 \text{ hours} \] Hours required for the next 10 units: \[ \text{Hours for next 10 units} = 13,980 - (5,000 + 3,250) = 5,730 \text{ hours} \] ### Step 3: Calculate Total Cost for 10 Additional Units #### Labor Cost: \[ \text{Labor cost} = 5,730 \text{ hours} \times \$30/\text{hour} = \$171,900 \] #### Material, Equipment, Supplies Cost: Using the learning curve for materials: \[ b = \frac{\log(0.75)}{\log(2)} \approx -0.415 \] Calculating the material cost: \[ Y_{12} = 250,000 \times (12)^{-0.415} \approx 106,250 \] \[ \text{Total for 12 units} = 106,250 \times 12 = 1,275,000 \] \[ \text{Cost for next 10 units} = 1,275,000 - 437,500 = 837,500 \] #### Total Bid for 10 More Units: \[ \text{Total Cost} = 171,900 + 837,500 = 1,009,400 \] --- ## Final Answer for Part a: \[ \boxed{\$1,009,400} \] --- ## Part b: New Bid if There is a Significant Delay If there is a significant delay, the learning effect is lost. Each unit will take the same time as the 2nd unit. ### Labor Cost for 10 Units: \[ \text{Labor} = 3,250 \text{ hours/unit} \times 10 \text{ units} = 32,500 \text{ hours} \] \[ \text{Labor Cost} = 32,500 \text{ hours} \times \$30/\text{hour} = \$975,000 \] ### Material Cost for 10 Units: \[ \text{Material Cost} = 187,500 \text{ per unit} \times 10 = 1,875,000 \] ### Total Cost: \[ \text{Total Cost} = 975,000 + 1,875,000 = 2,850,000 \] --- ## Final Answer for Part b: \[ \boxed{\$2,850,000} \] --- ## Summary Table | Case | Total Cost for 10 Units | |----------------------------------|--------------------------| | **Immediate Start (Learning)** | \$1,009,400 | | **Significant Delay (No Learning)**| \$2,850,000 | --- ## Final Summary - **Cost for 10 additional units if production starts immediately is \$1,009,400.** - **If there is a significant delay and learning is lost, the cost increases to \$2,850,000.**