MECH 431

# Lesson 7 Practice Problems

Updated 2018-06-10

# Lesson 7 Practice Problems1

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)? Answer: 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 is85,000, 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. Answer: 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 was60,000, 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}

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? Answer: 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 of463,000 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:

$A=\frac{P}{i}=\frac{463,000}{0.125}=3,704,000$

## 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.