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1、2. TIME VALUE OF MONEY 2.1 Theory of Equivalence 2.2 Equivalence Relationship 2.3 Nominal and Effective Interest Rates 2.4 Continuous Payments 2.5 Gradient Equations for Continuous Payment,2.1 Theory of Equivalence（等值） In any decision making process, we have to account for the benefits and the costs
2、 of a project. In a typical project, the costs occur at the beginning of the project and the benefits accur over a period of time. For example, installation of a waterflood project results in benefits in terms of additional oil recovery over several future years.,However, to install waterflood, we m
3、ay have to incur sufficient costs at the present time. This money has to come from internal capital of a corporation or from a lending institute. Either way, by using internal capital or by borrowing money, we will either lose the opportunity to invest the money somewhere else or we will have to pay
4、 interest to the lending institution.,That is, instead of investing the money in a waterflooding project, we could have earned interest from a bank by investing that money in the bank; or, if we have borrowed the money to invest in this project, we have to pay interest on the borrowed amount. This l
5、ost opportunity (opportunity cost) or the interest payment has to be accounted for in our cost benefit analysis. One way to do this is through the understanding of the time value of money.,Money is a valuable commodity. People will pay you to use your money. You will have to pay someone to use their
6、 money. The cost of money is established and measured by an interest rate. Interest rate is periodically applied and added to(额外的) the amount (the principal) over a specific period of time.,For example, depositing $100 in the bank for one year may generate an interest of $6 at the end of the year. T
7、hat is, the bank has paid you 6% interest to use your money. In the same vane, the bank will turn around and lend that money to another individual and charge 10% interest. The borrower will have to pay back $110 at the end of one year. We can also say that, for you, $100 today is equivalent to $106
8、one year from now. For the bank, $100 today is equivalent to $110 one year from now. This principal is called a theory of equivalence.,This simple example clearly illustrates that money has an earning power. Just like any other commodity, money can be put to work and earn more money for its owner. B
9、ecause of this, a dollar received today is worth more than a dollar received one year from now. Todays dollar can be invested to earn more than one dollar one year from now.,2. l Theory of Equivalence Once we know that money is a valuable commodity and it has an earning power, we need to establish a
10、 method to compare the monies collected at different times. For example, if you have the option of receiving $100 today or $110 one year from now, what would you prefer? If you can only invest $100 in a bank at 5% interest rate, you can only earn $5 one year from now. That will total up to $100 + $5
11、 = $105 one year from now. Obviously, $105 is less than $110 you could receive one year from now. Therefore, you would prefer $110 one year from now.,The economic equivalence is also affected by the time. Consider the extension of the previous example. If some one offers you $100 today, $110 one yea
12、r from now or $121 two years from now, which one would you prefer? As before, It depends on what you can do with the $100 you will have today. If you invest it at 5% interest rate, after one year you will have $100 + $100(.05) = $105 After two years, you will have $105+ $105(.05) = $110.25 Since $11
13、0.25 is smaller than $121, you would prefer $121 two years from today. On the other hand, if you can earn 25% interest rate, you will have, after one year $100 + $100(0.25) = $125 and after two years $125 + $125(0.25) = $156.25,Think about why this principle is so important. When conducting the cost
14、 benefit analysis of any project, if the benefits are received in the future, we cannot directly compare the present costs to the future benefits unless we can convert the future benefits to equivalent present benefits; or, in the alternative, we will have to convert the present cost to equivalent f
15、uture costs. In this chapter, we will learn how we can accomplish this task.,2.2 Equivalence Relationships To understand the theory of equivalence in a more rigorous way, we need to establish certain relationships starting from some basic parameters. We can define these parameters as: P = present su
16、m of money F = future sum of money A = end of period cash payment or receipt i = interest rate per period n = number of periods,In addition to defining the necessary parameters, we will also define the cash flow diagrams we will be using in this book. The cash flow diagram represents the cash flow p
17、rofile during the life of the project. For example, after investing $10,000 at the beginning of the project, if we receive $3,000 each year in benefits for five years, we can draw the cash diagram as in Figure 2.1.,In this cash flow diagram, the horizontal line represents the time scale; the arrows
18、signify cash flows and are placed at the end of period. Upward arrows represent cash receipts (benefits), whereas the downward arrows represent expenses (costs). Note that cash flow diagram is a function of whose point of view it represents.,0 1 2 3 4 5,2.2.1 Relationship Between P and F The cash fl
19、ow diagram for establishing the relationship between the present sum and the future sum can be drawn as in Figure 2.4.,0 1 2 3 n-1 n,P,F,Figure 2.4: Relationship Between P and F,(1) P, i, n-F=? after one period P+Pi=P(1+i) If we continue to invest the new principal, p(l + i), for another period , af
20、ter n periods, (2.1),(2) F, i, n- P =?,0 1 2 3 n-1 n,P,F,Figure Relationship Between P and F,Example 2.2 If you need $10,000 after 5 years, how much should you invest today at an interest rate of 10%? Solution Given: F = $10.000, n = 5 years, i = 10% Find: P,Eq. 2. I can also be written as,Using Eq.
21、 2.2, You need to invest $6,209 today.,Example 2.3 If you invest $4,000 in a bank at an interest rate 6.25%,the money would you have at the end of three years? Solution Given: P = $4,000, n = 3 years, i = 6.25%, Find: F Using Eq. 2.1 You will have $4,798 after three years.,Example 2.4 If you want to
22、 invest $3,000 at an interest rate of 7%, how long will it take to double the initial investment? Solution Given: P = $3,000, F = $6,000, i = 7% Find: n Using Eq. 2.1, Taking log on both sides The amount will double in 10.2 years.,Thumb of rule (rule of 72),2.2.2 Relationship Between A and F Let us
23、extend the previous relationships to a case where a payment is made at the end of each period. We would like to calculate the future value of these payments at the end of the total period. Cash flow diagram for this arrangement is shown in Fig. 2.5.,Figure 2.5: Periodic Payments,F n,0 1 2 3 n-1,A A
24、A A A,(3) A, i, n-F=? Considering that for the first payment, we earned interest for (n - l) periods (see Fig. 2.6), and for the last payment we earned no interest, using Eq. 2.l, we can write, (2 3) multiplying Eq. 2.3 by (l + i), (2.4) subtracting Eq. 2.3 from Eq. 2.4 and rearranging, we obtain,Th
25、erefore, (2.5) (4) F, i, n-A=?,Figure Periodic Payments,F n,0 1 2 3 n-1,A A A A A,(4) F, i, n-A=? We can rewrite q. 2.5 as, (2.6) Eq. 2.5 and Eq. 2.6 establish the relationships between A and F.,Example 2.5 If you deposit $10,000 at the end of each year, how much would you accumulate at the end of f
26、ive years at an interest rate of 6%? Solution Given: A = $10,000, n = 5 years, i = 6% Find: F Using Eq. 2.5, You would have $56,371 at the end of 5 years.,Example 2.6 If you need $100,000 at the end of 10 years for a college education, how much should you invest at the end of each year at an interes
27、t rate of 8%? Solution Given: F = $100,000, n = 10 years, i = 8% Find: A Using Eq. 2.6, You should invest $6,903 at the end of each year to receive $100,000 at the end of 10 years.,Example 2.7 You intend to invest $1,000 per year in mutual funds. If the average annual yield from this fund is expecte
28、d to be 12%, how long will it take before you would have accumulated $15,000 in your account? Solution Given: A=$1,000, F = $15,000, i = 12% Find: n Using Eq. 2.5, Substituting, Therefore, n = 9.1 years It will take approximately 9. 1years before $15,000 would be accumulated.,Example 2.8 After gradu
29、ating from college, Betty desires to buy a house worth $150,000 with a 20% down payment after 5 years. Bettys father gives her $10,000 as a graduation gift. If Betty invests that money at 6% interest rate, how much additional annual savings will she have to invest at the same interest rate to accumu
30、late the desired 20% down payment at the end of 5 years? Solution Given: F = 20% of $150,000, P = $10,000, i = 6%, n = 5 years Find: A,In this example, Betty is investing $10,000 at the beginning of year l plus additional annual investments to get $30,000 at the end of 5 years. The cash flow diagram
31、 can be drawn as shown in Fig. 2.7.,A A A A A,0 1 2 3 4 5,As a first step, we can calculate the future value (F1 ) of $10,000 after 5 years. Using Eq. 2. l, The remaining future value has to be the result of annual investments. We can calculate the remaining future value Using Eq. 2.6. That is, Bett
32、y will have to invest $2,948 at the end of each year.,2.2.3 Relationship Between A and P Let us extend the relationship one step further by relating the present value to the periodic payments. As shown in Fig. 2.8, we want to calculate the present value of future periodic payments.,0 1 2 3 n-1 n,A A
33、 A A A,P=?,Figure 2.8: Relationship Between P and A,(5) A, i, n-P=? From Eq. 2.l, we know that From Eq. 2.5, we know that Substituting Eq. 2. l in Eq. 2.5, we can write Simplifying, (2.7),(6) P, i, n-A=? Eq.2.7 can also be written as (2.8) Eq. 2.7 and Eq. 2.8 establish the relationship between the p
34、eriodic payment (A) and the present worth ( P) .,Example 2.9 If you take a home improvement loan of $10,000 to be paid over a five year period, what would be the yearly payment if the interest rate is 12% per year? Solution Given: P = $10,000, n = 5 years, i = 12% Find: A Using Eq. 2.8, The yearly p
35、ayment would be $2,774.,Example 2.10 If you want to invest sufficient money in the bank such that at an interest rate of 8%, you will receive $20,000 per year for the next 10 years, how much should you invest in the bank? Solution Given: A= $20,000, n = 10 years, i = 8% Find: P Using Eq. 2.7, You wi
36、ll have to invest $ 134,202 today to earn $20,000 per year for the next ten years.,Example 2.11 Able borrows $1,000 from a loan shark. In return, the loan shark demands that Able pay $100 per month for a one year period. What is the monthly interest rate the loan shark is charging? Solution Given: P
37、 = $l.000, A = $100 per month, n = 12 months Find: i Notice that the periodic payment and the number of periods are given in terms of monthly units. Using Eq. 2.7. Substituting,There is no explicit solution for i. By trial and error, For Therefore, the interest rate charged is 2.9% per month.,Exampl
38、e 2.12 Betty buys a new computer at a price of $10,000. Betty expects that the use of the computer should result in an annual income of $2,500. If Betty wants to earn at least a 15% return on her investment at what price would the computer have to be sold after 4 years? Solution Given: P = $10,000,
39、A = $2,500, i = 15%, n = 4years Find: Salvage value (resale price) of the computer We can draw a cash flow diagram for this example as shown in Fig. 2.9.,A A A,0 1 2 3 4,A=$2500,P=$10,000,F=? A,Figure 2.9: Cash Flow Diagram for Example 2.12,Using Eq. 2.7, we can calculate the present value of period
40、ic payments as Substituting The remaining present investment will have to be recovered by the future resale price. The remaining present investment is 10,000-7,137=$2,863 Using Eq. 2.l, The computer would have to be sold at a price of $5,007 at the end of 4 years.,In solving these and the other exam
41、ples and problems, please remember that the units of the periodic payment, the interest rate per period and the number of periods have to be consistent. For example, if the payment is paid per month, then the number of periods have to be in months and the interest rate has to be defined per month. S
42、imilarly, if the payment is per year, the interest rate has to be defined per year and the number of periods have to be in years.,2.2.4 Gradient Equations（梯度方程） Arithmetic Gradient The arithmetic gradient series is graphically represented in Fig. 2. 10. In this series, the initial payment is A at th
43、e end of period 1. The payment changes by a constant sum during each period. An example would be maintenance cost of a computer. The maintenance cost is, say, $500 in the first year. It increases by $50 each year. This can be represented by arithmetic gradient series where A is $500 and G is $50.,A+
44、(n-2)G,0 1 2 3 n-1,F n,A,A+G,A+2G,A+(n-1)G,Figure 2.10: Arithmetic Gradient Series,0 1 2,0 1 2,A,A+G,A+(n-1)G,F n,=,Figure 2.111 Equivalent Constant Value Payment for Arithmetic Gradient Series,n,Example 2.13 If you invest $1,000 in the bank in the first year followed by a steady increase of $50 per
45、 year over the next ten years, how much money will you get back at the end of the ten year period if the interest rate is 8%? Solution Given: A = $l,000, G = $50, i = 8%, n= 10years Find: F Using Eq. 2.13, You will get $17,291 at the end of ten years.,Example 2.14 A company buys a new computer. The
46、maintenance agreement requires that the company pay $2,000 in the first year for the maintenance cost to be increased at a rate of $200 per year. If the company intends to keep the computer for six years, how much money should it set aside to cover the maintenance costs if the money is invested at a
47、 6% interest rate? What would be an equivalent constant annual maintenance cost for the above cost schedule? Solution Given: A = $2,000, G = $200, n = 6 years, i= 6% Find: P, Aeq Using Eq. 2.14.,A total sum of $ 12,127 has to be set aside to cover the maintenance costs. To calculate an equivalent an
48、nual cost, we use Eq. 2. 15, The equivalent, constant annual maintenance cost is $2,466.,Example 2.15 Able intends to invest $200 in the first month followed by an increase of $3 per month in a mutual fund account for his daughters education. His daughter is 8 years old and he intends to have $100,000 at the time she is 18 years old. What interest rate does he need to earn on his investment? Solution Given: A = $200, G = $3, n = 120 months, F = $100,000 Find: i Using Eq. 2.13,Substituting, There is no