Homework#20
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E330
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Industrial Engineering
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Jan 9, 2024
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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.]