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# Compound Interest Calculations

Trong tài liệu Construction Financial Management (Trang 35-43)

### 3.2 Compound Interest Calculations

In the discussion of compound interest mathematics, we want to make an assumption – an inflation- free assumption, for the time being. We assume that there is no inflation in this world, and hence i is equal to i'. This assumption is of course not true in the real world. The purpose of making such an assumption temporarily is that we do not want to create confusion between nominal rate and real rate at the time being. We will remove this unrealistic assumption later after the next chapter and go back to live in the real world.

3.2.1 Uniform Series Compound Amount

In section 3.1, we have only considered a principal sum of single payment. Now we begin to see a series of uniform payments. For a series of uniform payments, let A = the periodic uniform payment made at the end of each period continuing for n periods to accumulate a sum S. The situation can be presented diagrammatically as shown in Fig. 3.1.

Fig. 3.1 – Sum of money accumulated after n periods due to uniform periodic payments.

Compound amount of the first A at period n = A(1 + i)n-1 Compound amount of the second A at period n = A(1 + i)n-2

: :

Compound amount of the (n – 1)th A at period n = A(1 + i)1 Compound amount of the nth A at period n = A

Total = S

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Compound Interest, Net Present Value (NPV), Equivalent Annual Cost and Loan Redemption

So it is a geometric series. We can apply geometric series formula to obtain that

S = A ×



 

 + −

i i n 1

1 (Eq 3.3)

The values of

### ()



 

 + −

i i n 1

1 are shown in the Appendix for different values of i and n.

Example 3.2

A construction company will replace an excavator after 5 years. A new one costs \$250,000. How much is the end-of-year annual uniform payment the company has to put into a bank in order to save enough money in five years’ time for purchasing the equipment if the bank is offering an interest rate of 4% per annum?

.

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Compound Interest, Net Present Value (NPV), Equivalent Annual Cost and Loan Redemption The problem can be presented diagrammatically as follows:

### \$250,000 5

Fig. 3.2 Sum of \$250,000 accumulated by 5 uniform periodic (annual) payments.

We have to bear in mind that the excavator always costs \$250,000, whether now or after five years, as the inflation-free assumption has been made.

Applying Equation 3.3, 250,000 = A ×

### ()



 

 + −

i i n 1

1 = A × 5.4163

(5.4163 is found by substituting i = 0.04 and n = 5 into the formula, or from Appendix) Hence, A = 5.4163

000 ,

250 = \$46,157

3.2.2 Uniform Series Sinking Fund

It is a little troublesome to obtain \$46,157 in Example 3.2. First, we have to use Equation 3.3. Second, we have to rearrange terms to calculate A (A is usually referred to as Sinking Fund). In fact, we can calculate the sinking fund A in a more direct (quicker) way.

Equation 3.3 can be rearranged as follows:

A = S ×

### ()

1+i n−1

i (Eq 3.4)

Equation 3.4 allows us to calculate the sinking fund \$46,157 (i.e. A) in Example 3.2 directly and more quickly. It is left to the readers as an exercise to try it.

The values of

### ()

 

+ 1

1 i n

i are shown in the Appendix for different values of i and n.

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Compound Interest, Net Present Value (NPV), Equivalent Annual Cost and Loan Redemption

3.2.3 Uniform Series Capital Recovery

In the previous two subsections, the relationships between the final accumulated sum and a principal sum investment or a series of uniform installment investment are discussed. This subsection discusses the relationship between an initial loan P and the subsequent uniform series of payments to offset against the loan P. The situation can be presented in Fig. 3.3 diagrammatically.

### 0 1 2 n

Fig. 3.3 Uniform periodic payments for recovering a loan P

Since Equation 3.1 gives S = P × ( 1 + i ) n and Equation 3.3 gives S = A ×

### ()



 

 + −

i i n 1 1

Solving the above two equations by eliminating S, we obtain

A = P ×

 

− +

+ 1 1

1

n n

i i

i (Eq 3.5)

The values of

### ()

 

− +

+ 1 1

1

n n

i i

i are shown in the Appendix for different values of i and n.

We are going to see an example how we can apply Equation 3.5.

Example 3.3

If the excavator of Example 3.2 suddenly breaks down now so the company needs to purchase a new one immediately and not 5 years later, and therefore has to borrow from a bank \$250,000 at an interest rate of 4% per annum. What will be the annual installment for repaying the loan in 5 years?

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Compound Interest, Net Present Value (NPV), Equivalent Annual Cost and Loan Redemption The problem can be presented diagrammatically as follows:

### 5

Fig. 3.4 Uniform annual installments for recovering a loan of \$250,000.

A = 250,000 ×

### ()

 

− +

+ 1 1

1

n n

i i

i = 250,000 × (0.22463)

= \$56,157

(0.22463 is found by substituting i = 0.04 and n = 5 into the formula, or from Appendix)

It should be interesting to note that the answer of Example 3.3, which is \$56,157, is exactly \$10,000 more than the answer of Example 3.2, which is \$46,157. This is not a coincidence. The later part of this chapter will give hints to the explanation. Readers may know why it is so after they have finished reading Example 3.6, particularly the last part of it.

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Compound Interest, Net Present Value (NPV), Equivalent Annual Cost and Loan Redemption The value A in Example 3.3 is usually referred to as Equivalent Annual Cost as the period is defined as one year in the question. For example, the equivalent annual cost of owning the excavator in Example 3.3 is \$56,157 a year with i equal to 4% per annum. Comparatively, the annual cost of owning it is \$50,000 a year (i.e. 250,000 ÷ 5) if i is taken as 0% p.a. (i.e. zero interest rate; p.a. stands for per annum).

3.2.4 Present Value (Single payment and Uniform Series Payments)

Let us look at Fig. 3.5, which explains what Present Value is for a single payment.

### F = \$10,000 × (1+i)

10

= \$16,289 year

Fig. 3.5 – Present value of a single payment F.

F would be equal to \$16,289 if i is taken as 5% p.a. In other words, \$16,289 of 10 years later is equivalent to \$10,000 today, or we can say that the present value (i.e. today’s value) of \$16,289 of ten years later is

\$10,000, because:

\$10,000 × ( 1 + 0.05 ) 10 = \$16,289 Mathematically, the present value of F can be expressed in the form of:

Present value of

 

× + + =

= n n

F i i F F

1 1

1 (Eq 3.6)

The value of

### ()



 +i n 1

1 are shown in the Appendix for different values of i and n.

Now, let us consider a uniform series of payments as shown in Fig. 3.6:

### 0 1 2 n

Fig. 3.6 Present value of uniform periodic payments.

The equivalent sum of money today (the present value), which is equivalent to all these uniform series of payment for n periods, can be calculated from the equation:

i n F

i F i

F i F

+ + + +

+ +

+ + ... 1

1

1 1 2 3 = F ×

### ()



 +

− +

n n

i i

i 1

1

1 (Eq 3.7)

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Compound Interest, Net Present Value (NPV), Equivalent Annual Cost and Loan Redemption

The above equation is just a simple geometric series and its derivation is left to the readers.

The value of

### ()



 +

− +

n n

i i

i 1

1

1 are shown in the Appendix for different values of i and n.

Now, let us see an example that the application of present value is illustrated.

Example 3.4

The construction cost of a service reservoir for supplying water to a housing estate is estimated to be

\$60,000,000. The annual operation and maintenance cost is estimated to be \$2,500,000 per year. The annual income from the collection of water supply fees (for this service reservoir’s part) from the users will be \$7,000,000. Assuming a time horizon of 30 years and taking i as 5% p.a., find out if the project is financially feasible.

The problem can be presented diagrammatically in Fig. 3.7 as follows:

,QFRPH P P P

\HDU

2SHUDWLRQ PDLQWHQDQFH

FRVW P P P

&RQVWUXFWLRQ P FRVW

Fig. 3.7 – Cash flows presented in diagram for service reservoir project

a) Present Value Method

(i) Present value of benefits in 30 years

= 7,000,000 ×

### ()



 +

− +

n n

i i

i 1

1

1 for i = 0.05 and n = 30

= 7,000,000 × (15.3724) = \$107,606,800

(ii) Present value of construction cost = \$60,000,000 Present value of operation and maintenance cost

= 2,500,000 ×

### ()



 +

− +

n n

i i

i 1

1

1 for i = 0.05 and n = 30

= 2,500,000 × (15.3724) = \$38,431,000 \$98,431,000

(iii) Net present value (NPV) = \$9,175,800

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Compound Interest, Net Present Value (NPV), Equivalent Annual Cost and Loan Redemption The positive net present value (i.e. present value of total benefits – present value of total costs) indicates that the project is feasible. NPV is the short form of net present value.

b) Equivalent Annual Cost Method

(i) Annual benefit = \$7,000,000

(ii) Annual operation and maintenance cost = \$2,500,000 Equivalent annual cost of construction

= 60,000,000 ×

### ()

 

− +

+ 1 1

1

n n

i i

i for i = 0.05 and n = 30

= 60,000,000 × (0.06505) = \$3,903,000 \$6,403,000

(iii) Net annual benefit (NAB) = \$597,000

The positive net annual benefit (i.e. total annual benefits – total annual costs) indicates that the project is feasible. NAB is the short form of net annual benefit.

### AxA globAl grAduAte progrAm 2015

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Compound Interest, Net Present Value (NPV), Equivalent Annual Cost and Loan Redemption The two methods (a) and (b) must have consistent results. If the NPV is calculated to be positive, the NAB must also be calculated to be positive, and vice versa. Therefore, either one method only is sufficient to determine whether the project is feasible or not.

We should note that there exists a relationship between the results of methods (a) and (b):

LH u »

¼

« º

¬ ª

Q Q

L L

L IRUL DQGQ

139 1\$%

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