Question

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)

Answer 1

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.