MECH 431

Lesson 7 Practice Problems

Updated 2018-06-10

Lesson 7 Practice Problems1

Problem 1

You work for a company that provides maintenance for marine diesel engines in large container ships. The main product you sell is a five-year service contract for a specific type of engine. You charge $85,000 each year, payable at the end of the year.

One of your customers wants to pre-pay the entire five-year contract. If your company’s interest rate is 12.5% compounded annually, how much should you charge them (to the nearest dollar)?


We are interested in the present value of a uniform series for 5 years. Our known variables: \(A=85000\), \(i=12.5\%\), \(n=5\).

Apply present worth factor formula:

\[\begin{aligned} P&=A\left(\frac{(1+i)^n-1}{i(1+i)^n}\right)\\ &=85000\left(\frac{(1+0.125)^{5}-1}{0.125(1+0.125)^5}\right)\\ &=302,648 \end{aligned}\]

Problem 2

With another customer, your company has included a riser to account for increasing costs over time.  The first-year payment is $85,000, with the price increasing $2,500 each year thereafter.  If the customer wanted to just pay a constant amount for the five years, how much should that be? Use the same 12.5% interest rate, round to the nearest dollar.


This time, since the cost is increasing linearly each period, we have a arithmetic series. In this case, the arithmetic series follows \(P_i=A+nG\), where \(A=85000\) and \(G=2500\).

We could get the arithmetic gradient uniform series factor, which gives us the equivalent \(A_{eq}\) and thus we can solve this like a uniform series.

\[\begin{aligned} A_{eq}&=G\left(\frac{1}{i}-\frac{n}{(1+i)^n-1}\right)\\ &=2500\left(\frac{1}{0.125}-\frac{5}{(1+0.125)^5-1}\right)\\ &=4415 \end{aligned}\]

\(4415 is the amount need to paid in addition to the \\)85000 base cost. Thus the equivalent total annuity is

\[A=85,000 +4,415=89,415\]

Problem 3

A third, and somewhat less helpful customer, has a seven-year contract.  The first-year payment was $60,000, with a clause that payments would increase by 4% per year.  The contract has now just expired (the seven years are up), and they have not yet payed a single bill.  What amount should you claim is owing when you sue them?


Since payments increase by 4% each year, this is an indication of geometric series. Assuming 12.5% interest rate still. Here are the givens: \(g=0.04\), \(i=0.125\), \(A_1=60,000\), \(n=7\).

We are looking future values now, so we shall use the compound amount factor to compute \(F\):

\[\begin{aligned} F&=A_1\left(\frac{(1+i)^n-(1+g)^n}{i-g}\right)\\ &=60000\left(\frac{(1+0.125)^7-(1+0.04)^7}{0.125-0.04}\right)\\ &=681,011 \end{aligned}\]

Problem 4

For your fourth customer, you estimate the present value of the work to be done is $1.2 million over six years. To avoid another situation like customer three, you are making this customer pay in advance.  Still with 12.5% interest, what should their (uniform) annual payments be?


We are back to working with uniform payments, and knowing present value \(P=1.2\text M\), we shall work with the capital recover factor to solve for the annuity \(A\) over the next \(n=6\) years.

Because we ask that they pay up front (annuity due), we append the \(1+i\) term. Since this is capital recovery factor, it is inversed.

\[\begin{aligned} A&=P\left(\frac{i(1+i)^n}{(1+i)^n-1}\right)(1+i)^{-1}\\ &=1,200,000\left(\frac{0.125(1+0.125)^6}{(1+0.125)^6-1}\right)(1+0.125)^{-1}\\ &=263,125 \end{aligned}\]

Problem 5

Your fifth customer is a government operator.  They expect to operate a fleet of five ships for a long time (assume forever), and want to capitalize the maintenance costs.  If the expected costs for the fleet are a total of $463,000 per year, how much should the government pay you now to contract you to maintain their ships for their lifetime?


This is a perpetuity situation. Assuming interests are still 12.5% and the cost each period are uniform, then the annuity can be given simply:


Problem 6

Your sixth and final customer is a small shop. Their contract is only for four years, and is $20,000 per year.  You charge them 12.5% compounded annually, however, they want to make quarterly payments. How much should their payments be?


This problem can be split into two parts:

  1. Find the present worth of all the costs.
  2. Use present worth find payment per adjusted quarterly period.

The present worth of all costs:

\[\begin{aligned} P&=A\left(\frac{(1+i)^n-1}{i(1+i)^n}\right)\\ &=20000\left(\frac{(1+0.125)^{4}-1}{0.125(1+0.125)^4}\right)\\ &=60,113 \end{aligned}\]

Now we find the equivalent interest rate because payment period is different than compounding period: (where \(c=1\) is the number of compounding periods per year and \(p=4\) is the number of payment periods per year)

\[\begin{aligned} i_{eq}&=(1+i)^\frac{c}{p}-1\\ &=(1+0.125)^\frac{1}{4}-1\\ &=0.03 \end{aligned}\]

Applying capital recovery factor, we get the amount of payment per period, where \(n=4\times 4\):

\[\begin{aligned} A&=P\left(\frac{i_{eq}(1+i_{eq})^n}{(1+i_{eq})^n-1}\right)\\ &=60,113\left(\frac{0.03(1+0.03)^{16}}{(1+0.03)^{16}-1}\right)\\ &=4781 \end{aligned}\]
  1. M. Hollett, MECH 431 001 Lesson 7 Practice Problems. Web. Accessed 2018-06-10.