Question

You are interested in manufacturing a new product and wish to purchase entry-level equipment. You have identified two alternative sets of equipment and gear. Package A has a first cost of $160,000, an operating cost of $8000 per quarter, and a salvage value of $40,500 after its 2-year life. Package B has a first cost of $210,000 with a lower operating cost of $5000 per quarter, and an estimated $25,000 salvage value after its 4-year life. Which package offers the lower present worth analysis at an interest rate of 8% per year, compounded quarterly?

Answer 1

The given question is:	Package A	Package B
First Cost	1,60,000	2,10,000
Annual operating cost	32,000	20,000
Salvage value	40,500	25,000
Life(n)	2	4
Interest rate(i)	0.08	0.08

As the interest rate compounded quaterly we will calculate effective annual rate of return from compound interest formula and it will come out for both package is 8.24% . We will use this interest rate for the calculation.

Annual operating cost is arrived by converting quaterly operating cost into annual. i.e.

For package A = 8000*4= 32000

For package B= 5000*4= 20000

Now, In the given question, the life of equipment has different life alternatives. So, we have to calculate the value of each machine using the LCM (Least common multiple) method.

The life of package a is 2 years and life of package B is 4 years.

The LCM of both package will come out 4 years. So we will calculate the present value at 4 years study period.

The Present value of package A for 2 years

PW_A = -160000-32000(P/A, i,n) + 40,500(P/F, i,n)

= -160000-32000((1+i)ⁿ-1/(i(1+i)ⁿ)) +40500(1/(1+i)ⁿ

=-160000-32000((1+0.0824)²)-1/(0.0824(1+0.0824)²))+ 40500(1/(1+0.0824)²

= -160000-32000(1.171-1)/0.0824(1.171)) + 40500(1/1.171)

=-160000-32000(1.21)+34568

=-164151

For 4 years

=-164151-164151(P/F,i,n)

=-164151-164151((1/(1+0.0824)²)

=-304260

The Present value of package B for 4 years

PW_B = -210000-20000(P/A, i,n) + 25000(P/F, i,n)

= -210000-20000((1+i)ⁿ-1/(i(1+i)ⁿ)) +25000(1/(1+i)ⁿ

=-210000-20000((1+0.0824)⁴)-1/(0.0824(1+0.0824)⁴))+ 25000(1/(1+0.0824)⁴

^{= -210000-20000(1.37-1)/0.0824(1.37)) + 25000(1.37)}

^{=-210000-20000(3.27)+29250}

^=-246150

Clearly we can see PW_A<PW_B

_{So, the package B offers the lower present worth analysis.}