Question

In: Economics

The following annual costs are associated with two new extruder machines being considered for use in...

The following annual costs are associated with two new extruder machines being considered for use in a Styrofoam cup plant.

Data

X

X-Trud

Useful Life, Years

9

13

First Cost

$2,300,000

$2,780,000

Salvage Value

$82,000

$118,000

Annual Benefit ($/year)

$580,000

$670,000

Operations and Maintenance Costs, $/year

$65,000

$78,000

Operations & Maintenance Gradient, $/year (starting at year 2)

$11,000

$15,000

The company's interest rate (MARR) is 12%. Which extruder should the Styrofoam company choose? Use Annual Cash Flow Analysis.

Complete the problem using the following methods:

  1. By hand calculations on Engineering Computation Paper
    1. IPE Approach, Interpretation will be a cash flow diagram for each
    2. Communicate your formulas correctly, showing the variables; followed by the data; followed by the value you used. (Example: In separate lines of the solution your equations should read as follows:
      1. … + F(p/f,i%,n) …
      2. ... + $40,000(p/f,4%,5) …
      3. … + $40,000(0.8219)

Solutions

Expert Solution

We have the following information

X

X-Trud

First cost ($)

2,300,000

2,780,000

Annual benefit ($)

580,000

670,000

Operation and maintenance cost ($)

65,000 in year 1and then increases by 11,000 every year

78,000 in year 1 and then increases by 15,000 every year

Salvage value ($)

82,000

118,000

Life (n)

9 years

13 years

MARR (i)

12% or 0.12

12% or 0.12

Option X

First we will calculate annual equivalent amount of the operation and maintenance cost

A = A1 + G[((1 + i)n – in – 1))/i(1 + i)n – i]

A = A1 + G(A/G, i, n)

Where (A/G, i, n) is called uniform gradient series factor

A = Annual equivalent amount

A1 = Amount at the end of the first year = 65,000

G = Equal increment amount = 11,000

n = Number of interest periods = 9 years

i = Interest rate = 12% or 0.12

A = 65,000 + 11,000[((1 + 0.12)9 – (0.12×9) – 1))/0.12(1 + 0.12)9 – 0.12]

A = 65,000 + 11,000(A/G, 12%, 9)

A = 65,000 + 11,000[((1.12)9 – (0.12×9) – 1))/0.12(1.12)9 – 0.12]

A = 65,000 + (11,000 × 3.257)

A = $100,831.60

This is equivalent to paying an equivalent amount of $100,831.60 at the end of every year for the next 9 years.

Annual Worth (AW) = – First Cost(A/P, i, n) – Annual cost + Annual benefit + Salvage value(A/F, i, n)

AW = – 2,300,000(A/P, 12%, 9) – 100,831.60 + 580,000 + 82,000(A/F, 12%, 9)

AW = – 2,300,000[0.12(1 + 0.12)9/((1 + 0.12)9 – 1)] – 100,831.60 + 580,000 + 82,000[0.12/((1 + 0.12)9 – 1)]

AW = – 431,661.44 – 100,831.60 + 580,000 + 5,549.67

AW = – 532,493.04 + 585,549.67

Annual Worth of Option X = $53,056.62

Option X-Trud

First we will calculate annual equivalent amount of the operation and maintenance cost

A = A1 + G[((1 + i)n – in – 1))/i(1 + i)n – i]

A = A1 + G(A/G, i, n)

Where (A/G, i, n) is called uniform gradient series factor

A = Annual equivalent amount

A1 = Amount at the end of the first year = 78,000

G = Equal increment amount = 15,000

n = Number of interest periods = 13 years

i = Interest rate = 12% or 0.12

A = 78,000 + 15,000[((1 + 0.12)13 – (0.12×13) – 1))/0.12(1 + 0.12)13 – 0.12]

A = 78,000 + 15,000(A/G, 12%, 13)

A = 78,000 + 15,000[((1.12)13 – (0.12×13) – 1))/0.12(1.12)13 – 0.12]

A = 78,000 + (15,000 × 4.468)

A = $145,024.60

This is equivalent to paying an equivalent amount of $145,024.60 at the end of every year for the next 13 years.

Annual Worth (AW) = – First Cost(A/P, i, n) – Annual cost + Annual benefit + Salvage value(A/F, i, n)

AW = – 2,780,000(A/P, 12%, 13) – 145,024.60 + 670,000 + 118,000(A/F, 12%, 13)

AW = – 2,780,000[0.12(1 + 0.12)13/((1 + 0.12)13 – 1)] – 145,024.60 + 670,000 + 118,000[0.12/((1 + 0.12)13 – 1)]

AW = – 432,782.60 – 145,024.60 + 670,000 + 4,209.91

AW = – 577,807.20 + 674,209.91

Annual Worth of Option X-Trud = $96,402.71

Since, the annual worth of X-Trud is higher so it should be selected.


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