Homework#20

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San Diego State University *

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E330

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Industrial Engineering

Date

Jan 9, 2024

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4

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CON E330 Fall 2023 Engineering Economy. Homework #20. Given Class Example: Production of Small Items for Holidays MARR = 10% Method 1: Auto-Fed Machine (Challenger) Method 2: Manual-Fed Machine (Defender) 1. F.C = 23k [$] 2. S = 4k [$] 3. n = 10 [yrs.] 4. 1 person needed 5. Operator Salary = 24 [$/hr.] 6. A (O+M) Cost = 3,500 [$/yr.] 7. Output = 8 [lb./hr.] 8. Working Day= 8 [hrs.] 9. Work Days: M-F 1. F.C = 8k [$] 2. S = 0k [$] 3. n = 5 [yrs.] 4. 3 people needed 5. Workers Salary = 12 [$/hr.] 6. A (O+M) Cost = 1,500 [$/yr.] 7. Output = 6 [lb./hr.] (per team) 8. Working Day = 10 [hrs.] 9. Work Days: M-Saturday Method 1: 23k(A/P, 10%,10)[$/yr.] - 4k(A/F,10%,10) + 3.5k[1/yr.] + 24[$/hr.] ⅛[hr./lb.] * Q[lb./yr.] 23k(0.1627) - 4k(0.0627) Method 2: 8k(A/P, 10%,5)[$/yr.] - 0 + 1.5k[1/yr.] + 12[$/hr.] ⅙ [hr./lb.] *3[workers]* Q[lb./yr.] 8k(0.2638) Q = 1127 [lb./yr.] If Q ≥ 1127 Go for Challenger If Q < 1127 Stay with Defender
CON E330 Fall 2023 Engineering Economy. Homework #20. Self-Created Problem 1: There are 2 methods for the Production of Small Items for Holidays. Method 1 involves an Auto-Fed machine while Method 2 involves a Manual-Fed machine. The values corresponding to each method are listed below. For both ways, the MARR is 10% and the production level has been set to 1140 [lb./yr.]. However, both methods cannot be compared as there is a missing value for one of the methods. As seen, Method 1 is missing the value for its F.C. Therefore, with the information provided find the missing value and compare the Methods. Method 1: Auto-Fed Machine (Challenger) Method 2: Manual-Fed Machine (Defender) 1. F.C =?[$] 2. S = 4k [$] 3. n = 10 [yrs.] 4. 1 person needed 5. Operator Salary = 24 [$/hr.] 6. A (O+M) Cost = 3,500 [$/yr.] 7. Output = 8 [lb./hr.] 8. Working Day= 8 [hrs.] 9. Workdays: M-F 10. Production Level (Q) = 1140 [lb./yr.] 1. F.C = 8k [$] 2. S = 0k [$] 3. n = 5 [yrs.] 4. 3 people needed 5. Workers Salary = 12 [$/hr.] 6. A (O+M) Cost = 1,500 [$/yr.] 7. Output = 6 [lb./hr.] (per team) 8. Working Day = 10 [hrs.] 9. Workdays: M-Saturday 10. Production Level (Q) = 1140 [lb./yr.] Solution to Problem 1: To solve we must first get each method's full equations. For method 1 the equation is: X (A/P, 10%,10)[$/yr.] - 4k(A/F,10%,10) + 3.5k[1/yr.] + 24[$/hr.] ⅛[hr./lb.] * 1140[lb./yr.] = X(0.1627) - 4k(0.0627) + 3.5k[1/yr.] + 24[$/hr.] ⅛[hr./lb.] * 1140[lb./yr.] The Equation for the second method is: 8k(A/P, 10%,5)[$/yr.] - 0 + 1.5k[1/yr.] + 12[$/hr.] ⅙ [hr./lb.] *3[workers]* 1140[lb./yr.] = 8k(0.2638)[$/yr.] - 0 + 1.5k[1/yr.] + 12[$/hr.] ⅙ [hr./lb.] *3[workers]* 1140[lb./yr.] Then we make them equal to one another to solve for X. X(0.1627) - 4k(0.0627) + 3.5k + 24*⅛ * 1140 = 8k(0.2638) - 0 + 1.5k + 12*⅙ *3* 1140
? = (8𝑘(0.2638) − 0 + 1.5𝑘 + 12 ∗ ⅙ ∗ 3 ∗ 1140) + (4𝑘(0.0627) + 3.5𝑘 + 24 ∗ ⅛ ∗ 1140) (0.1627) = $23,240 = $23.24𝑘 Method 1 F.C = 23.24 k [$] Self-Created Problem 2: There are 2 methods for the Production of Small Items for Holidays. Method 1 involves an Auto-Fed machine while Method 2 involves a Manual-Fed machine. The values corresponding to each method are listed below. For both ways, the MARR is 10% and the production level has been set to 1000 [lb./yr.]. However, both methods cannot be compared as there is a missing value for one of the methods. As seen, Method 1 is missing the value for its Working Hours. Therefore, with the information provided find the missing value and fully compare the methods. Method 1: Auto-Fed Machine (Challenger) Method 2: Manual-Fed Machine (Defender) 1. F.C = 23k[$] 2. S = 4k [$] 3. n = 10 [yrs.] 4. 1 person needed 5. Operator Salary = 24 [$/hr.] 6. A (O+M) Cost = 3,500 [$/yr.] 7. Output = 8 [lb./hr.] 8. Working Hours = ? [hrs.] 9. Workdays: M-F 10. Production Level (Q) = 1000 [lb./yr.] 1. F.C = 8k [$] 2. S = 0k [$] 3. n = 5 [yrs.] 4. 3 people needed 5. Workers Salary = 12 [$/hr.] 6. A (O+M) Cost = 1,500 [$/yr.] 7. Output = 6 [lb./hr.] (per team) 8. Working Hours = 10 [hrs.] 9. Workdays: M-Saturday 10. Production Level (Q) = 1000 [lb./yr.] Solution to Problem 2: To solve it we must first get each method's full equations. For method 1 the equation is: 23k (A/P, 10%,10) [$/yr.] - 4k(A/F,10%,10) + 3.5k[1/yr.] + 24[$/hr.] 1/W[hr./lb.] * 1000[lb./yr.] = 23k(0.1627) - 4k(0.0627) + 3.5k[1/yr.] + 24[$/hr.] 1/W[hr./lb.] * 1000[lb./yr.] The Equation for the second method is: 8k (A/P, 10%,5)[$/yr.] - 0 + 1.5k[1/yr.] + 12[$/hr.] ⅙ [hr./lb.] *3[workers]* 1140[lb./yr.] = 8k(0.2638)[$/yr.] - 0 + 1.5k[1/yr.] + 12[$/hr.] ⅙ [hr./lb.] *3[workers]* 1140[lb./yr.] Then we make them equal to one another to solve for W. 23k (0.1627) - 4k(0.0627) + 3.5k + 24*1/W * 1000 = 8k(0.2638) - 0 + 1.5k + 12*⅙ *3* 1000
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Rewrite into: ? = 24000 (8𝑘(0.2638) − 0 + 1.5𝑘 + 12 ∗ ⅙ ∗ 3 ∗ 1000) − (23𝑘(0.1627) − 4𝑘(0.0627) + 3.5𝑘 ) = 24000 (9610.4) − (6991.3) = 9.16345309457 Method 1 Working Hours ≈ 9.16 [hrs.]