3: Balance Equations

Transcription

1 3.1 Balance Equations Accounts with Constant Interest Rates 15 3: Balance Equations Investments typically consist of giving up something today in the hope of greater benefits in the future, resulting in a growth of capital. Personal investments frequently take the form of placing money in saving accounts, stocks, bonds, or real estate. Industry uses these methods, and it also gives up capital in the form of money to acquire labor, machines, technical knowledge, and managerial skills that will produce income for several years. Similarly, governments make investments, such as in education, highways, or other projects designed to provide benefits over time in exchange for an initial investment. This chapter is the first of several that progressively develop models for examining investments in which the time-value of money is important. It studies the situation in which funds are placed in a single savings account at stated rates of interest. An account can pay a constant rate, such as 10% per year; alternatively, the rate can change in a well defined manner, such as 8% the first year, 9% the second, and so forth. The objective of the chapter is to obtain equations that can be used to calculate future balances (amounts) in saving accounts as a function of interest rates, deposits, and withdrawals. In the following sections, accounts with constant interest rates are described and then a procedure for computing account balances is illustrated by a numerical example. This procedure is subsequently simplified via the development of the constant rate balance equation. Then this model is extended to apply to accounts with interest rates that change over time. The following chapters examine different applications of balance equations in a banking environment, and then extend the results to industry and government. 3.1 Accounts with Constant Interest Rates Many accounts have a constant rate of interest per period for the duration of an investment, such as 10% per year or 1% per month. The length of the period can be yearly, monthly, daily, or any other convenient time interval. A transaction or cash flow occurs when funds come into or out of an account. Deposits and withdrawals are transactions in a saving account, and loans and repayments are the transactions in an account for a loan. The amount of savings or debt at any time is known as the balance of an account. All transactions earn interest as if they were made at the beginning or end of a period. For example, a mid-month deposit to a saving account with a monthly period typically does not start earning interest until the beginning of the next month. Conversely, if a mid-month withdrawal is made, then the withdrawal is usually treated as having occurred at the beginning of the month, so it would earn no interest during the month it is withdrawn. The advantage goes to the banker, but choosing an account that computes interest on a daily basis reduces the extent of this problem to no more than one day of lost interest. The following subsections explain how to compute balances of a constant rate account, provide an example, contrast project economics versus accounting profit, introduce future value, and then discuss concepts involving the duration of alternatives.

2 3.1 Balance Equations Accounts with Constant Interest Rates 16 Simple Interest If simple interest is used, then balances are based on the product of each transaction, the interest rate, and the number of periods that have elapsed since the transaction. For example, if $100 is borrowed at 10% per year simple interest, then the simple interest owed after 3 years is $100(0.1)(3) or $30. The component of a simple interest balance for each transaction is: Component = Transaction [1 + (Simple Interest Rate Elapsed Periods)]. (3-1) If the only transaction is a loan of $100 that occurred 3 years ago, then the balance would be 100( ) or $130, in this case. If a payment of $50 is made to the lender at year 1, then the simple interest balance at year 3 would be $70, 70 = 100( ) 50( ), (3-2) since 2 periods (years, in this case) elapsed since the $50 payment. Simple interest rarely occurs today, so no further treatment of it is provided. The remainder of the text assumes the type of interest described in the next section, compound interest. Compound Interest and the Recursive Balance Formula Compound interest uses a recursive balance formula shown below to compute the balance each period: New Balance = Old Balance + (Old Balance Interest Rate) + New Cash Flows (3-3) The new balance each period equals old balance from last period, plus interest, and plus or minus the total of any new transactions, the net cash flow. For example, if $100 is borrowed at 10% per year and nothing ($0) is repaid, then $110 will be owed after 1 year, 110 = ( ) + 0, (3-4) $121 after 2 years, 121 = ( ) + 0, (3-5) and $ after 3 years instead of the $130 owed using simple interest, = ( ) + 0. (3-6) Unlike simple interest, compound interest is paid on the original amount plus any interest that has already accumulated. For example, after 1 year, interest is paid on the original $100 and on the $10 interest. This is known as compounding, and periods are sometimes called compounding periods. Transactions plus their accumulated interest are referred to as compound amounts, such as the $110, $121, or $ Transactions or cash flows can have either a positive or a negative sign, where positive cash flows increase the balance and negative ones decrease it. A value of $0 means that no cash flow occurs. The reinvestment assumption states that interest continues to accrue on the earlier transactions even when a $0 cash flow occurs. The following example uses the end-of-period formula to compare two investments, illustrating how to compute balances and the use of the reinvestment assumption. Example 3.1 Investments Using Funds from a Constant Rate Saving Account An investor currently has $10,000 in a saving account paying 10% per year. Other

3 3.1 Balance Equations Accounts with Constant Interest Rates 17 investments can be considered, as long as they do not last longer than the planning horizon of 5 years when the investor needs the funds to purchase a car. The investment objective is to have as much money as possible after 5 years. Three mutually exclusive investments are shown in Table 3.1. Funds for the alternatives come from and return to the savings account. Negative cash flows decrease its balance, and positive ones increase it. Investment A requires a withdrawal of $4,000 at year 0 (today). Its returns to the account begin with $1,200 at year 1 and decline by $100 per year for the next 4 years. Investment B requires a withdrawal of $3,000 from the savings account today, and returns $1,300 per year for 3 years. The investor also can choose to reject both investments. Rejecting all investments is referred to as the null or do nothing alternative. It has no cash flows, so each of its entries in the table is $0. Figure 3.1 shows the cash flow diagrams for the first two alternatives. The directions of the arrows indicate whether the funds are going from or returning to the account. The end-of-period formula provides the account balance at the end of the first year for investment A. The amount in the account prior to any investment is Table 3.1 Investments A, B, and the Null Alternative Year A B Null 0 4,000 3, ,200 1, ,100 1, ,000 1, , ,100 1, Investment A , Investment B 4,000 3,000 Figure 3.1 Cash Flow Diagrams for Investments A and B

4 3.1 Balance Equations Accounts with Constant Interest Rates 18 $10,000, so the $4,000 cost at year 0 produces a balance of $6,000. This earns interest at 10%, and then there is revenue of $1,200, so the new balance is given by the new balance formula as: 7,800 = 6, ,000(0.1) + 1,200. (3-7) Factoring the old balance of $6,000 facilitates computations: 7,800 = 6,000(1.1) + 1,200. (3-8) In general, factoring allows the new balance formula to be written as: New Balance = Old Balance (1 + Interest Rate) + New Cash Flows (3-9) For investment B, the first period has a $7,000 beginning-of-period balance (10,000 3,000) and an end-of-period revenue of $1,300, so its end-ofperiod balance is: 9,000 = 7,000(1.1) + 1,300. (3-10) The null alternative involves no transactions, so its cash flows are always $0. The reinvestment assumption indicates that it still earns interest, so the end-ofperiod formula can be used to compute its balance at the end of year 1: 11,000 = 10,000(1.1) + 0. (3-11) The balances for the remaining years are calculated as shown in Table 3.2. Notice that the reinvestment assumption is used for the last two years of alternative B, since the account continues to pay interest after its last transaction at time 3. The amounts in the saving account at the end of the planning horizon are $15,884 for investment A, $16,480 for investment B, and $16,105 for the null alternative. If all alternatives should be equally safe, then the preferred alternative would be investment B, followed by the null alternative, then investment A. 3.2 Final Balance and Project Selection This section makes two observations regarding selecting projects based on the previous example. Standard bookkeeping procedures that ignore the time-value of money should not be used to select projects, and there are situations where reinvestment makes it unnecessary for all alternatives to have the same duration. Project Economics versus Bookkeeping Profit Choosing alternative B instead of A increases the investor s capital by $596 (16,480 15,884) after five years. However, bookkeeping procedures typically ignore reinvestment and consider only the amounts of the costs and revenues. This leads to a bookkeeping profit of $1,000 for A, $1,000 = 1, , , ,000, (3-12) compared to only $900 for B, $900 = 1, , ,300 3,000. (3-13) Project economics recognizes that reinvestment does occur, so both the timing and amounts of cash flows are important. Alternative B is the preferred alternative.

5 3.2 Balance Equations Final Balance and Project Selection 19 Table 3.2 End of Year Account Balances Investment A Year End of Year Balance 1 $7,800 = 6,000(1.1) + 1,200 2 $9,680 = 7,800(1.1) + 1,100 3 $11,648 = 9,680(1.1) + 1,000 4 $13,713 = 11,648(1.1) $15,884 = 13,713(1.1) Investment B Year End of Year Balance 1 $9,000 = 7,000(1.1) + 1,300 2 $11,200 = 9,000(1.1) + 1,300 3 $13,620 = 11,200(1.1) + 1,300 4 $14,982 = 13,620(1.1) $16,480 = 14,982(1.1) + 0 Null Alternative Year End of Year Balance 1 $11,000 = 10,000(1.1) $12,100 = 11,000(1.1) $13,310 = 12,100(1.1) $14,641 = 13,310(1.1) $16,105 = 14,641(1.1) + 0 Duration of Alternatives The reinvestment assumption is used for investment B. After its last transaction occurs at time 3, no other investment is necessary and the funds in the saving account continue earning interest. This allows investment B which lasts only 3 years to be compared to investment A which lasts 5 years, the length of the investor s planning horizon. A planning horizon is the period of time for which estimates of future events (the cash flows) are made. The reinvestment assumption allows alternatives with unequal lives to be compared if it is reasonable to assume that proceeds from shorter alternatives will be reinvested until the end of the planning horizon. Reinvestment after the last cash flow is reasonable in the preceding example, but this is not always the case. For example, consider comparing a pump that lasts for 3 years to a pump that lasts for 5 years. If the pump with the shorter life has to be replaced, then future balances cannot be computed unless a replacement for the shorter lived pump is considered. This topic is explored in more detail later in the text.

6 3.3 Balance Equations Constant Rate Balance Equation Constant Rate Balance Equation Balances can be computed on a year-by-year basis, but it is easier to use a formula that directly computes any year's balance without knowing the previous ones. This section first develops the constant rate balance equation and applies it to the previous example, and then provides four more examples illustrating its use. The following section treats the situation in which interest rates vary from one period to the next. Development Table 3.3 shows the balances for alternative A in the preceding example. The first column indicates time, and the second column shows the cash flow at that time. The third and fourth columns contain the same balances computed before using the end-of-period formula. However, the fourth column factors terms to reveal the individual contributions of the initial assets and each cash flow. The last column repeats the fourth column except that it uses symbols instead of numbers: B j is the balance at time j, A is the initial assets, γ is the growth factor equal to 1 plus the interest rate, and cj is the cash flow at time j. The exponents in the last two columns always equal the number of compounding periods between the balance and the assets or cash flow, so the following constant rate balance equation emerges for time t: B t = A γ t + c 0 γ t + c 1 γ t-1 + c 2 γ t c t-1 γ + c t (3-14) This formula clearly shows that c j plus its interest contributes c j γ t-j to the balance at time t. This is the compound amount of c j. Example 3.2 Final Balances for a Constant Interest Rate An easy way to determine the best alternative in the preceding example is to use the constant rate balance equation to compute the balance in the saving account at the end of period 5 for each alternative. For alternative A, it is: 15,884 = 10,000(1.1) 5-4,000(1.1) 5 + 1,200(1.1) ,100(1.1) 5-2 (3-15) + 1,000(1.1) (1.1) The balance for alternative B equals: 16,480 = 10,000(1.1) 5 3,000(1.1) 5 + 1,300(1.1) 5-1 (3-16) + 1,300(1.1) ,300(1.1) 5-3. The values of c 4 and c 5 are $0 for alternative B, so they are not shown. All of the cash flows equal 0 for the null alternative, so its balance is: 16,105 = 10,000(1.1) 5. (3-17) All of these values agree with those shown in Table 3.2, but it is no longer necessary to compute year-by-year balances. Applications and Use of the Balance Equation The first two examples of this section demonstrate applications of the balance equation, and the second two examples show how to avoid two common difficulties in its use.

7 3.3 Balance Equations Constant Rate Balance Equation 21 Table 3.3 Balance Equation Development j c j Balance Factored Balance Symbols 0-4,000 6,000 = 10,000-4,000 6,000 = 10,000 4,000 B 0 = A + c 0 1 1,200 7,800 = 6,000(1.1) + 1,200 7,800 = [10,000 4,000](1.1) + 1,200 = 10,000(1.1) 4,000(1.1) + 1, ,100 9,680 = 7,800(1.1) + 1,100 9,680 = [10,000(1.1) 4,000(1.1) + 1,200] (1.1) + 1,100 =10,000(1.1) 2 4,000(1.1) 2 +1,200(1.1) + 1, ,000 11,648 = 9,680(1.1) + 1,000 11,648 = [10,000(1.1) 2 4,000(1.1) 2 +1,200(1.1) + 1,100](1.1) + 1,000 =10,000(1.1) 3 4,000(1.1) 3 +1,200(1.1) 2 + 1,100(1.1) + 1, ,713 = 11,648(1.1) ,713 = [10,000(1.1) 3 4,000(1.1) 3 +1,200(1.1) 2 + 1,100(1.1) + 1,000] (1.1) =10,000(1.1) 4 4,000(1.1) 4 +1,200(1.1) 3 + 1,100(1.1) 2 + 1,000(1.1) ,884 = 13,713(1.1) ,884 = [10,000(1.1) 4 4,000(1.1) 4 +1,200(1.1) 3 + 1,100(1.1) 2 + 1,000(1.1) + 900] (1.1) =10,000(1.1) 5 4,000(1.1) 5 +1,200(1.1) 4 + 1,100(1.1) 3 + 1,000(1.1) (1.1) B 1 = [A + c 0 ] γ + c 1 = A γ + c 0 γ + c 1 B 2 = [A γ + c 0 γ + c 1 ] γ + c 2 = A γ 2 + c 0 γ 2 + c 1 γ + c 2 B 3 = [A γ 2 + c 0 γ 2 + c 1 γ + c 2 ]γ + c 3 = A γ 3 + c 0 γ 3 + c 1 γ 2 + c 2 γ + c 3 B 4 = [A γ 3 + c 0 γ 3 + c 1 γ 2 + c 2 γ + c 3 ] γ + c 4 = A γ 4 + c 0 γ 4 + c 1 γ 3 + c 2 γ 2 + c 3 γ + c 4 B 5 = [A γ 4 + c 0 γ 4 + c 1 γ 3 + c 2 γ 2 + c 3 γ + c 4 ] γ + c 5 = A γ 5 + c 0 γ 5 + c 1 γ 4 + c 2 γ 3 + c 3 γ 2 + c 4 γ + c 5

8 Balance Equations Multiple Rate Balance Equation Example 3.3 Change in Balance Suppose that the cash flow of $1,000 at time 3 for alternative A in the preceding examples increases by $400 to $1,400, and the change in the final balance is desired. One way to solve this problem is to compute the balance at time 5 again, $16,368 = 10,000(1.1) 5-4,000(1.1) 5 + 1,200(1.1) ,100(1.1) 5-2 (3-18) + 1,400(1.1) (1.1) and note that it increases by $484 from its former value of $15,884. However, the balance equation clearly shows the compound amounts contributed to the final balance by the initial assets and each cash flow. The contribution from c 3 had been $1,000(1.1) 5-3, and now it is ($ $400)(1.1) 5-3 or $1,000(1.1) 5-3 plus $400(1.1) 5-3 or $484 more. Nothing else changed, so $484 is the total increase. An advantage of the balance equation is that it isolates individual cash flows, rather than having to recursively perform all computations again. Example 3.4 Internal Rate of Return The balance equation provides a mechanism for computing the internal rate of return of an investment. This equals the interest rate that a constant rate savings account would pay if it were initially empty and the same cash flows as the investment produce a final balance to $0. For example, alternatives A and B are not actually saving accounts, but they behave like ones paying 9% and 14% per year. Computing the internal rate of return for alternative B begins by treating it as a saving account with initial assets A of $0. Then $3,000 is deposited at time 0, so c 0 equals $3,000. Next, $1,300 is withdrawn three times, so c 1, c 2, and c 3 all equal $1,300. The last withdrawal produces a balance of $0 at time 3 and thereafter, so the constant rate balance equation for these transactions is: 0 = 0 + 3,000(1+i) 3 1,300(1+i) 3-1 (1,300)(1+i) 1,300 (3-19) Trial and error must be used to find the solution that i equals 14% per year. Notice that both sides of equation (3-19) can be multiplied by 1 without changing its solution: 0 = 3,000(1+ i) 3 + 1,300(1+ i) (1,300)(1+ i) + 1,300 (3-20) This keeps the signs in the rate of return equation the same as the signs indicated by the cash flow diagram, and results in fewer mistakes. Using this sign convention for alternative A yields the following balance equation at time 5: 0 = 4,000(1+ i) 5 + 1,200(1+i) ,100(1+i) 5-2 (3-21) + 1,000(1+i) (1+i) It has a trial and error solution of 9%. Rates of return computed in this manner are called internal rates of return since the procedure uses only information internal to the problem, i.e., its cash flows. There are other types of rates of returns that will be seen later in the text.

9 Balance Equations Multiple Rate Balance Equation Example 3.5 Arrows and Signs Frequently down arrows in a cash flow diagram denote negative cash flows in the balance equation, and up arrows indicate positive flows. This is not a hard and fast rule. For example, suppose that the three cash flows in Figure 3.2 represent a loan at 10% interest. A person currently owes $300 and plans to borrow $100 at time 2 and make payments of $40 and $55 at times 3 and 4. The problem is to determine the balance at time 4. The balance equation can be written as: B 4 = $ = 300(1.1) (1.1) (1.1) (3-22) In this case, negative signs are used for the down arrows, and they indicate debt. The balance of $ corresponds to a debt of $ at time 4. However, the focus of this problem is on debt, so it can be convenient to reverse the assignment of signs to arrows and consider a debt to be positive. In this case, the balance equation becomes B 4 = = 300(1.1) (1.1) (1.1) , (3-23) so there is a debt of $ at time 4, as before. Following a simple rule, such as making down arrows negative and up arrows positive is a safe way to get started. After that, it is a matter of convenience. It is essential, however, to be consistent for any one problem. Example 3.6 Labeling of the Time Axis The exponent of the growth factor γ for a cash flow equals the number of periods between the cash flow and the balance. This does not depend on the labeling of the time axis. For example, the loan in the preceding example could have been represented as shown in Figure 3.3. In this case, the initial assets are at time -2, or two years before the new time 0. The time interval from the initial assets to the final balance is 2 (-2) or 4 years, just as before. The balance equation for this labeling of the time axis can be either or B 2 = $ = 300(1.1) 2-(-2) 100(1.1) (1.1) (3-24) B 2 = $ = 300(1.1) 2-(-2) + 100(1.1) (1.1) (3-25) Notice that the exponents remain the same as before, so the balance remains the same. The value of each exponent for each cash flow equals the difference between the time of the balance and the cash flow, and this is not affected by how the time axis is labeled. The two most common methods are to let time 0 be the time of the initial assets or to call today time Figure 3.2 Loan Figure 3.3 Time Labels

10 Balance Equations Multiple Rate Balance Equation 3.4 Multiple Rate Balance Equation Interest rates tend to rise and fall over time. Their exact values usually cannot be predicted, and the balance equation with a constant interest rate is frequently used with an estimate of the average interest rate. However, there are situations in which more precision is warranted. For example, a sensitivity analysis might be performed to examine the effects of different scenarios for interest rates. This section develops the multiple rate balance equation and then uses it to examine a scenario of rising interest rates. Development The logic used to develop the multiple rate balance equation is almost identical to that of the constant interest model. Only slight changes in notation are necessary to distinguish the interest rates in different periods. Let i j be the interest rate during period j, as illustrated in Figure 3.4. For example, the first period is from time 0 to time 1, and i 1 is the rate during that time; similarly, i 3 is the rate during the third period, from time 2 to time 3. Let γ j equal 1 + i j, and examine the balance for each period. The account starts with A in it and then c 0 occurs, so the amount in the account at the beginning of period 1 is A + c 0. This beginning-of-period amount earns interest at rate i 1, and then c 1 occurs at the end of period 1. The balance at the end of period 1 is or B 1 = A + c 0 + ( A + c 0 ) i 1 + c 1 (3-26) B 1 = A γ 1 + c 0 γ 1 + c 1. (3-27) The end of period 1 balance equals the beginning of period 2 balance, which increases by a factor of γ 2 and then is augmented by c 2 : B 2 = B 1 γ 2 + c 2 (3-28) Substitute the expression for B 1 to obtain or B 2 = (A γ 1 + c 0 γ 1 + c 1 ) γ 2 + c 2 (3-29) B 2 = A γ 1 γ 2 + c 0 γ 1 γ 2 + c 1 γ 2 + c 2. (3-30) Repeating this procedure several times results in the multiple rate balance equation: B t = Aγ 1 γ 2 γ t + c 0 γ 1 γ 2 γ t + c 1 γ 2 γ 3 γ t + c 2 γ 3 γ 4 γ t + + c t-1 γ t + c t (3-31) Each cash flow is multiplied by the growth factors for the periods following its occurrence. Example 3.7 Final Balances for Multiple Interest Rates i 1 i 2 i 3 i 4 i Suppose that a scenario of rising interest rates is applied to the investments previously considered. Let the interest rates paid by the saving account rise from a level of 9% today by 0.5% per year. This would make the growth factor during 0 Figure 3.4 Multiple Interest Rates

11 3.4 Balance Equations Multiple Rate Balance Equation year 1 equal to 1.090, during year 2, during year 3, and so forth. The equation for balance of investment A at the end of period 5 is: 15,945 = 10,000(1.090)(1.095)(1.100)(1.105)(1.110) (3-32) 4,000(1.090)(1.095)(1.100)(1.105)(1.110) + 1,200(1.095)(1.100)(1.105)(1.110) + 1,100(1.100)(1.105)(1.110) + 1,000(1.105)(1.110) + 900(1.110) The final balance for investment B is: 16,541 = 10,000(1.090)(1.095)(1.100)(1.105)(1.110) (3-33) 3,000(1.090)(1.095)(1.100)(1.105)(1.110) + 1,300(1.095)(1.100)(1.105)(1.110) + 1,300(1.100)(1.105)(1.110) + 1,300(1.105)(1.110) And the final balance corresponding to the null alternative is: 16,103 = 10,000(1.090)(1.095)(1.100)(1.105)(1.110) (3-34) This scenario of rising interest rates increases the advantage of alternative B relative to the others alternatives. 3.5 Summary This chapter first uses the end-of-period formula to compute balances for an account with a constant interest rate. Then balance equations are developed to directly compute any end-of-period balance without first calculating each of the preceding balances. The constant rate balance equation assumes a constant interest rate, and the multiple rate balance equation permits interest rates to vary. Examples illustrate the use of the balance equations, and show how to use the constant rate balance equation to compute an alternative s internal rate of return. The examples also illustrate the reinvestment assumption and show how to compute internal rates of return. Questions Section 1: Accounts with Constant Interest Rates 1.1 If $100 is borrowed at 7% per year simple interest and $30 is repaid after 3 years, then how much is owed after 5 years? ($100.80) 1.2 Suppose that there is $10,000 in a savings account paying 10% compound interest per year. One of two investments can be chosen, or the funds can be left in the account. Investment A requires an initial outlay of $3,000 and then returns $1,800 at the end of year 3 and $2,800 at the end of year 5. Investment B has an initial cost of $4,000 and returns $5,000 after 2 years. It is reasonable to assume that neither alternative needs to be replaced at the end of its life, costs and revenues for the investment come from and return to the account, and any funds in the account earn interest at 10%. a) What is the null alternative in this instance? Is it a feasible alternative? b) Draw the cash flow diagrams for each alternative. c) What is the investment objective? 25

12 3: Balance Equations - Questions 26 d) What are the year-by-year balances for each alternative? (Final balances: A = $16,251.57, B = $16,318.06, Null = $16,105.10) 1.3 Suppose that there is uncertainty regarding the interest rate that will be paid by the saving account in problem 1.2. It is thought that the rate will be constant, but its value is unknown. Final balances are shown in Table 3.4 for several interest rates. Plot these final balances versus the interest rates, and clearly mark on the graph which alternative should be chosen for different interest rates. (Choose A up to i = 9.03%, then B up to 11.80%, then Null) Table 3.4 Final Balances for Different Interest Rates Rate A B Null 8% $15, $15, $14, , , , , , , , , , , , , , , , Section 2: Final Balance and Project Selection 2.1 Consider problem 1.2. a) What are the bookkeeping profits of the alternatives? (A: $1,600. B: $1,000. Null: $0) b) Which alternative should be chosen on an economic basis? Why? (B) c) The two alternatives do not last for the same length of time, so how can they be compared? Section 3: Constant Rate Balance Equation 3.1 Use the balance equation to compute the final balance at 10% for each of the three alternatives in problem 1.2. (See 1.2.d) 3.2 Consider the cash flow in year 3 for alternative A in problem 1.2. Suppose that it increases at year 3 by $200 from $1,800 to $2,000. Determine the change in B 5 using yearby-year balances. Then verify this result by recomputing the balance equation. ( increase) 3.3 An investment costs $5,000 today and returns $3,000 at the end of year 2 and also at the end of year 5. What is the rate of return on this investment? Hint: Try 5% and then 6% to bracket the solution, and then interpolate between those two values. (5.44%) 3.4 A new loan of $200 at 10% interest is made at year 3 and payments of $70 are made at years 4 and 5. The balance at year 5 must be determined. a) Write and solve the balance equation twice, using different sign conventions for the arrows each time. Are the results the same? (debt of $95) b) Relabel the time axis so the loan occurs at time 0 and write the balance equations again. Does the relabeling affect the solution of the problem? (still debt of $95) 3.5 Rework the preceding problem assuming an initial debt of $100 at 10% for 3 years before the $200 loan. (debt of $256.05)

13 3: Balance Equations - Questions 27 Section 4: Multiple Rate Balance Equation 4.1 Consider the investment discussed in problem 1.2, but now suppose that the following interest rates will be paid by the saving account: 8% during year 1, 9% during year 2, 10% during year 3, 11% during year 4, and 12% during year 5. a) Compute the year by year balances for alternative A, and check the final balance against the one obtained by using the multiple rate balance equation. ($16,306.67) b) Use the multiple rate balance equation to compute the balance of the other alternatives. (B: $16,496.67; null: $16,098.45) c) Which alternative should be chosen on an economic basis? (B) 4.2 Continue the development of the multiple rate balance equation for B 3 and B 4 so that the general pattern is clearly revealed. Use symbols, rather than numerical values.