Dimensión 4: Estrategias de apoyo al procesamiento
1.3.2. Teorías relacionadas al rendimiento académico
2.4.9 Summary of Interest Factors
The interest factors developed in this section are known as discrete interest factors because the interest compounding occurs over discrete time periods.
Table 2.2 summarizes the interest factors developed in this section.
2.5 Time Value of Money Examples
In this section, several examples will be presented that demonstrate how the discrete interest factors developed in Section 2.4 can be applied in various analyses.
Table 2.2
Interest Factors for Discrete Compounding
Name Converts Symbol Computed by
Single payment compound amount P to F (F/P, i, n) (1+ i)n
Present worth F to P (P/F, i, n) (1+i)−n
Uniform series sinking fund F to A (A/F, i, n)
1 1
With installment loans, it is interesting to see how the monthly payment is distributed between payment of interest and repayment of the principal.
Many banks offer a payment schedule known as an amortization schedule.
The following example shows how to develop an amortization schedule.
EXAMPLE 2.2
A bank is advertising new car loans at 3.5%. Consider a situation where $20,000 is borrowed from this bank to buy a new car. The loan is to be paid back in equal monthly installments over a 6-year period. Determine the monthly payment and the effective interest rate of the loan.
SOLUTION: In most economic analysis problems, it is a good idea to start with a cash flow diagram to help visualize the problem. The cash flow diagram for this problem is shown in Figure E2.2
FIGURe e2.2
The $20,000 that the bank lends for purchase of the car is viewed as an income received. The monthly payments are expenses. Notice that there are 72 periods in this cash flow diagram because the life of the loan is 6 years, which is equivalent to 72 months. On the basis of this scenario, the interest rate paid per month is i = 0.035/12. The monthly payment can be found by applying the capital recovery factor
, , $20, 000 ,0.035
12 , 72 $308.37
0.0154184
( )
=
=
A P A =
P i n A
P
The capital recovery factor was calculated using the expression developed in Section 2.4.7 and shown in Table 2.2. There are many online loan calculators. There are also many loan calculator applications for smart phones and tablets. These cal-culators use the expressions shown in this example to determine loan payments.
For this loan, the bank advertises an interest rate of 3.5%. This is the nominal annual interest rate. The effective annual rate of this loan is
=
− =
− = =
i F
P i
m m F
, , 1 P,0.035
12 , 12 1 0.03557 3.56%
eff
1.03557
As expected, the interest rate advertised by the bank is slightly lower than the actual interest rate being paid because of the monthly compounding. If the com-pounding period is yearly, then the effective rate is equal to the nominal rate.
0 1 2 3 4 5 6 7
A
8 9 10 70 71 72= (6)(12)
P = $20,000
EXAMPLE 2.3
Develop an amortization schedule for the car loan described in Example 2.2.
SOLUTION: From Example 2.2, the monthly payment is $308.37. In the first month, the interest paid is the initial loan amount multiplied by the monthly interest rate,
12 $20, 000 0.035
12 $58.33
1= =( ) =
I P i
The balance of the payment is used to pay the principal,
$ 308.37 58.33 $250.04
1= − =1 ( − )=
P A I
Therefore, at the end of the first month the principal balance on the loan is Bal1=$20, 000− =P1 $ 20, 000 250.04( − )=$19, 749.96
The second payment results in
12 $19, 749.96 0.035
12 $57.60
$ 308.37 57.60 $250.77 Bal $ 19749.96 250.77 $19, 499.19
2 1
This process repeats for subsequent months. For any time period m, during the loan period,
This type of calculation can be easily adapted to a spreadsheet. The various columns can be totaled to determine the total payments made and the interest paid over the life of the loan. Figure E2.3 is an abridged table that shows the resulting amortization schedule for this example.
FIGURe e2.3
The following example shows how interest factors are used to manipulate a complex cash flow scenario.
The interest paid each month decreases while the amount paid to the prin-cipal increases. The life of the loan, and thus the total amount of interest paid, can be reduced significantly by making extra payment to the principal early in the life of the loan.
The total amount of interest paid for this loan is $2,202.49. The longer the term of the loan, the more the interest paid. However, the longer loan term allows for a smaller payment. The consumer must balance both these issues when deciding what loan term to select.
EXAMPLE 2.4
A company is considering investing in some new equipment to enhance one of their production lines. It is estimated that the addition of this new equip-ment will increase annual profits by $50,000. The annual cost to operate the equipment is $1,000. After the first year, maintenance costs are anticipated to be $1,000, increasing by $1,000 each subsequent year. The equipment is expected to last 10 years, at which time it can be sold for $5,000. The com-pany’s board of directors has specified that a minimum rate of return of 20% annually must be realized from the equipment investment. Determine the maximum amount of money that the company can spend now to meet this investment scenario. Assume annual compounding.
SOLUTION: On the basis of the information given in the problem, the cash flow diagram is shown in Figure E2.4.
FIGURe e2.4 0
P =?
1 2 3 4 5 6 7 8 9 10
Ap= $50,000 S=$5,000
G=$1,000 Ao=$1,000
In certain instances, it is desirable to determine the interest rate or the rate of return for a given scenario. This is demonstrated in the following problem.
EXAMPLE 2.5
Suppose the equipment cost for the processing facility in Example 2.4 has an initial cost of $225,000. Keeping all other parameters as defined in Example 2.4, determine the corresponding rate of return on this equipment investment.
SOLUTION: The cash flow diagram for this scenario is the same as that in Example 2.4 with the exception that the initial cost of the equipment is known, There are several ways to determine P, which depend on the common point in the cash flow diagram where all incomes and expenses are referred to. In this case, the present value is desired. Therefore, bringing all incomes and expenses to the present (year 0) results in
, , , , , , , ,
The interest rate and number of periods are known; therefore, the interest fac-tors can be found. Solving for P,
( )
This calculation indicates that an investment in the equipment of $193,352 will meet the company’s minimum required rate of return of 20%. If the equip-ment is less expensive than this, the rate of return would be higher than 20%.
Conversely, if the equipment is more expensive than $193,352 then the rate of return drops below 20% and the improvement is viewed as financially unacceptable.