Engineering Economics Blog

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GreenTree Financing lent an engineering company $100,000 to retrofi t an environmentally unfriendly building. The loan is for 3 years at 10% per year simple interest. How much money will the fi rm repay at the end of 3 years?

Solution

The interest for each of the 3 years is

Total interest for 3 years from Equation [1.7] is


The amount due after 3 years is

                                             Total  due = $100,000 + 30,000 = $130,000  

The interest accrued in the fi rst year and in the second year does not earn interest. The interest due each year is $10,000 calculated only on the $100,000 loan principal.


In most fi nancial and economic analyses, we use   compound interest   calculations.

For compound interest,  the interest accrued for each interest period is calculated on the principal plus the total amount of interest accumulated in all previous periods.   Thus, compound interest means interest on top of interest.
  
Compound interest refl ects the effect of the time value of money on the interest also. Now the
interest for one period is calculated as


In mathematical terms, the interest I t for time period t may be calculated using the relation.

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The  terms    interest,     interest period,  and   interest rate  (introduced in Section 1.4) are useful in calculating equivalent sums of money for one interest period in the past and one period in the future.

However, for more than one interest period, the terms   simple interest  and   compound interest   become important.
  
Simple interest  is calculated using the principal only, ignoring any interest accrued in preceding interest periods. The total simple interest over several periods is computed as where I is the amount of interest earned or paid and the interest rate I is expressed in decimal form.

EXAMPLE 1.14 - Simple and Compound Interest

EXAMPLE 1.15 - Simple and Compound Interest

EXAMPLE 1.16 - Simple and Compound Interest

 

 

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Howard owns a small electronics repair shop. He wants to borrow $10,000 now and repay it over the next 1 or 2 years. He believes that new diagnostic test equipment will allow him to work on a wider variety of electronic items and increase his annual revenue. Howard received 2-year repayment options from banks A and B.


After reviewing these plans, Howard decided that he wants to repay the $10,000 after only 1 year based on the expected increased revenue. During a family conversation, Howard’s brother-in-law offered to lend him the $10,000 now and take $10,600 after exactly 1 year.

Now Howard has three options and wonders which one to take. Which one is economically
the best?
 
Solution

The repayment plans for both banks are economically equivalent at the interest rate of 5% per year. (This is determined by using computations that you will learn in Chapter 2.) Therefore, Howard can choose either plan even though the bank B plan requires a slightly larger sum of money over the 2 years.

The brother-in-law repayment plan requires a total of $600 in interest 1 year later plus the principal of $10,000, which makes the interest rate 6% per year. Given the two 5% per year options from the banks, this 6% plan should not be chosen as it is not economically better than the other two. Even though the sum of money repaid is smaller, the timing of the cash fl  ows and the interest rate make it less desirable.   The point here is that cash fl  ows themselves, or their sums, cannot be relied upon as the primary basis for an economic decision. The interest rate, timing, and economic equivalence must be considered. 

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Manufacturers make backup batteries for computer systems available to Batteries + dealers through privately owned distributorships. In general, batteries are stored throughout the year, and a 5% cost increase is added each year to cover the inventory carrying charge for the distributorship owner. Assume you own the City Center Batteries + outlet. Make the calculations necessary to show which of the following statements are true and which are false about battery costs.
 
  (a)    The amount of $98 now is equivalent to a cost of $105.60 one year from now. 
  (b)    A truck battery cost of $200 one year ago is equivalent to $205 now. 
   (c)    A $38 cost now is equivalent to $39.90 one year from now. 
  (d)    A $3000 cost now is equivalent to $2887.14 one year earlier. 
  (e)    The carrying charge accumulated in 1 year on an investment of $20,000 worth of  batteries  is  $1000.  

Solution 

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Economic equivalence is a fundamental concept upon which engineering economy computations are based. Before we delve into the economic aspects, think of the many types of equivalency we may utilize daily by transferring from one scale to another. Some example transfers between scales are as follows:

Often equivalency involves two or more scales. Consider the equivalency of a   speed  of 110 kilometers per hour (kph) into miles per minute using conversions between distance and time scales with three-decimal accuracy.



Four scales—time in minutes, time in hours, length in miles, and length in kilometers are combined to develop these equivalent statements on speed. Note that throughout these statements, the fundamental relations of 1 mile 1.609 kilometers and 1 hour 60 minutes are applied. If a fundamental relation changes, the entire equivalency is in error.

Now  we  consider  economic  equivalency.  
  

Economic equivalence  is a combination of   interest rate  and   time value of money  to determine the different amounts of money at different points in time that are equal in economic value.  

As an illustration, if the interest rate is 6% per year, $100 today (present time) is equivalent to $106 one year from today.     


If someone offered you a gift of $100 today or $106 one year from today, it would make no difference which offer you accepted from an economic perspective. In either case you have $106 one year from today. However, the two sums of money are equivalent to each other   only  when the interest rate is 6% per year. At a higher or lower interest rate, $100 today is not equivalent to $106
one year from today.

In addition to future equivalence, we can apply the same logic to determine equivalence for previous years. A total of $100 now is equivalent to $100 /1.06 = $94.34 one year ago at an interest rate of 6% per year. From these illustrations, we can state the following: $94.34 last year, $100 now, and $106 one year from now are equivalent at an interest rate of 6% per year.

The fact that these sums are equivalent can be verifi ed by computing the two interest rates for 1-year interest periods.

and


The cash fl ow diagram in  Figure 1–10  indicates the amount of interest needed each year to make these three different amounts equivalent at 6% per year.

Figure 1–10 Equivalence of money at 6% per year interest.

EXAMPLE 1.12 - Economic Equivalence

EXAMPLE 1.13 - Economic Equivalence

 

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A rental company spent $2500 on a new air compressor 7 years ago. The annual rental income from the compressor has been $750. The $100 spent on maintenance the fi rst year has increased each year by $25. The company plans to sell the compressor at the end of next year for $150. Construct the cash fl ow diagram from the company’s perspective and indicate where the present worth now is located.


Solution

Let now be time t = 0. The incomes and costs for years 7 through 1 (next year) are tabulated below with net cash fl ow computed using Equation [1.5]. The net cash fl ows (one negative, eight positive) are diagrammed in  Figure 1–9 . Present worth   P  is located at year 0.

Figure 1–9 Cash fl ow diagram, Example 1.11.

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An electrical engineer wants to deposit an amount   P  now such that she can withdraw an equal annual amount of  A1 = $2000 per year for the fi rst 5 years, starting 1 year after the deposit, and a different annual withdrawal of A2 = $3000 per year for the following 3 years. How would the cash fl ow diagram appear if i = 8.5% per year?
 
Solution

The cash fl  ows are shown in Figure 1–8. The negative cash outfl  ow   P  occurs now. The withdrawals (positive cash infl ow) for the A1 series occur at the end of years 1 through 5, and A2 occurs in years 6 through 8.

Figure 1–8  Cash  fl  ow diagram with two different   A  series, Example 1.10.