Question

Question1

Frank signed up with a new movie service. It costs $832 per year, with a one-time set-up fee of $205 due today. The movie service requires a 4-year contract. If movies typically cost $9.32 per movie, what is the minimum number of movies Frank would need to see in a year to make the service worthwhile? Use an interest rate of 5% to make the calculation.

Enter your answer as follows: 12345

Round your answer. Do not use any commas (","), or a decimal point (".")

Note: Frank makes the payments at the end of the year. Frank makes the first payment at the end of year 1 and the last payment at the end of year 4.

Question 2

Flo's Frozen Yogurt shop is evaluating new dispensing machines.

The Flownator has a first cost of $31172 and annual expenses of $9551 that will increase by $408 per year. It will require an overhaul at the end of year 4 at a cost of $6854. The Flownator will save $19142 per year in labor costs. The Flownator has a salvage value of $9779 and a lifespan of 10 years.

The YogGoo300 has a first cost of $21336 and annual expenses of $3994 that will increase by 2% per year. The YogGoo300 will save Flo's $11358 per year in labor costs. The YogGoo300 has a lifespan of 5 years.

What must the salvage value of the YogGoo30 be to make it equally desirable to the Flownator? Use a MARR of 2% to make your calculation.

Enter your answer as follows: 12345

Round your answer. Do not use a dollar sign ("$"), any commas (","), or a decimal point (".").

Answer 1

Q1

We need to find annual worth of the movie service

i = 5%

First cost = 205, annual payment = 832 for 4 years

Price per movie = 9.32

Annual worth of movie service = 205 *(A/P, 5%,4) + 832

=  205* 0.282011 + 832 = 889.812

No of movies per year = Annual worth of service / price per movie = 889.812 / 9.32 = 95.47

= 96 movies per year to make the service worthwhile. we are rounding up as seeing only 95 movies will not make the service worthwhile

Q2

MARR = 2%

Flownator option

first cost = 31172

annual expenses = 9551 increase by 408 per year.

overhaul at the end of year 4 at a cost of 6854.

Annual savings = 19142 per year in labor costs.

salvage value = 9779

life = 10 years.

Annual worth = -31172 *(A/P, 2%,10) - 9551 - 408* (A/G, 2%, 10) - 6854 *(P/F, 2%, 4) *(A/P,2%,10) + 19142+ 9779* (A/F, 2%, 10)

= -31172 *0.1113265 - 9551 - 408*4.336736 - 6854 *0.9238454*0.1113265 + 19142+ 9779* 0.0913265

= 4539.50

YogGoo300 option

first cost = 21336

annual expenses = 3994 that will increase by 2% per year.

here g= 2% = 0.02 and i =2% too

Annual savings = 11358 per year

life = 5 years.

Let salvage value = S

Present value of geometric series, when g=i is given by C*n / (1+i)

where C is first payment and n is no of years and i is interest rate

Present value of annual cost = 3994 * 5 / (1+0.02) = 19578.43

Annual worth = -21336*(A/P,2%,5) - 19578.43*(A/P,2%,5) + 11358 + S * (A/F, 2%, 5)

= -21336*0.2121583 - 19578.43*0.212158+ 11358 + S * 0.192158

= 2677.68 + S*0.192158

As per the condition given in the ques

2677.68 + S*0.192158 = 4539.50

S = (4539.50 - 2677.68) /0.192158

S = 9689.02= 9689 (rounding off)