3: Balance Equations 3.1 Accounts with Constant Interest Rates. Terms. Example. Simple Interest

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1 3: Balance Equations 3.1 Accounts with Constant Interest Rates Example Two different accounts 1% per year: earn 1% each year on dollars at beginning of year 1% per month: earn 1% each month on dollars at beginning of month Transaction or cash flow Money entering or leaving an account Timing must be consistent with rate 1% per year: at year, 1, 2, 1% per month: at month, 1, 2, Simple Interest Balance: amount of savings or debt in an account Updating depends on rate 1% per month: updated monthly 1% per year: updated yearly Each balance component depends linearly on time Component = Transaction [1 + (Simple Interest Rate Elapsed Periods)]

2 Example Borrowed $1 3 years ago at 1% per year simple interest and paid $5 2 years ago $1 Component Debt = $13 = 1( ) $5 Component Reduction = $6 = 5( ) Balance at year 3 = 7 = 13 6 Compound Interest and the Recursive Balance Formula Compound interest uses a recursive balance formula: New Balance = Old Balance + (Old Balance Interest Rate) + New Cash Flows Simple Examplet. Borrow $1 at 1% per year and repay nothing ($) One year: $11 = 1 + (1.1) + Two years: $121 = 11 + (11.1) + Simple vs. Compound Interest Debt is $13 with simple interest and $133.1 with compound interest Simple interest is rare. Only consider compound interest from now on Three years: $133.1 = (121.1) + Earning compound interest is called compounding If i = 1% per year, then the compounding period is 1 year If i = 1% per month, then the compounding period is 1 month Compound amounts equal the original amount plus interest. The compound amount of $1 at 1% per year after 1 year is $11 after 2 years is $121 after 3 years is $133.1

3 Transactions or cash flows have signs Positive: increases a balance Negative: decreases a balance Reinvestment assumption Compounding continues even when there is no transaction If there is no transaction, then a cash flow of $ occurs Example 3.1 Investment Example $1, in saving 1% / yr Planning horizon: 5 years to buy car Objective: Max $ at end of horizon Three mutually exclusive investments Cash Flows Table 3.1 Investments A, B, and the Null Alternative Year A B Null 4, 3, 1 1,2 1,3 2 1,1 1,3 3 1, 1, Cash Flow Diagrams Year Balances 1,2 1,1 1,9 Investment A 8 1,3 Investment B Balances for year A: $6, = 1, 4, B: $7, = 1, 3, Null: $1, = 1, 4, 3,

4 Year 1 Balances Balances for year 1 A: $7,8 = 6, + 6,(.1) + 1,2 = 6,(1.1) + 1,2 B: $9, = 7, + 7,(.1) + 1,3 = 7,(1.1) + 1,3 Balance Formula New Balance = Old Balance (1 + Interest Rate) + New Cash Flows Null: $11, = 1, + 1,(.1) + = 1,(1.1) + Year 2 Balances Balances for year 2 A: $9,68 = 7,8(1.1) + 1,1 B: $11,2 = 9,(1.1) + 1,3 Null: $12,1 = 11,(1.1) + Balances for A Investment A Year End of Year Balance 1 $7,8 = 6,(1.1) + 1,2 2 $9,68 = 7,8(1.1) + 1,1 3 $11,648 = 9,68(1.1) + 1, 4 $13,713 = 11,648(1.1) $15,884 = 13,713(1.1) + 8 Balances for B Investment B Year End of Year Balance 1 $9, = 7,(1.1) + 1,3 2 $11,2 = 9,(1.1) + 1,3 3 $13,62 = 11,2(1.1) + 1,3 4 $14,982 = 13,62(1.1) + 5 $16,48 = 14,982(1.1) + Balances for C Null Alternative Year End of Year Balance 1 $11, = 1,(1.1) + 2 $12,1 = 11,(1.1) + 3 $13,31 = 12,1(1.1) + 4 $14,641 = 13,31(1.1) + 5 $16,15 = 14,641(1.1) +

5 Choice? End of Planning Horizon Balances Alt Balance A $15,884 B $16,48 Null $16, Final Balance and Project Selection Project Economics vs. Bookkeeping Profit Bookkeeping profit just considers the transactions and ignores interest. A: $1, = 1,2 + 1,1 + 1, , B: $9 = 1,3 + 1,3 + 1,3 3, Null: $ Makes the project with smallest final balance look best in this case. Duration of Alternatives Alternatives A (5 years) and B (3 years) do not have the same lives $ cash flows for B s last 2 years Reinvestment assumption reasonable Once B ends, no more flows Still earn interest Careful! Sometimes more flows implied 3 yr pump vs 5 yr pump Need for 5 years replace 3 yr More later Notation 3.3 Constant Rate Balance Equation Easier than year-by-year calculations j = time (, 1, 2, ) B j = Balance at time j A = Initial assets ($1,) γ = Growth factor = 1 + i (1 +.1) c j = cash flow at time j ( 4, 12, 1,1, 1,, 9, 8)

6 Constant Rate Equation B = A γ t + c γ t + c t c 2 γ t-2 t γ + + c γ t-1 + ct Balance j cj Balance -4, 6, = 1, - 4, 1 1,2 7,8 = 6,(1.1) + 1,2 2 1,1 9,68 = 7,8(1.1) + 1,1 3 1, 11,648 = 9,68(1.1) + 1, ,713 = 11,648(1.1) ,884 = 13,713(1.1) + 8 j Factored Factored Balance 6, = 1, 4, 1 7,8 = 1,(1.1) 4,(1.1) + 1,2 2 9,68 = 1,(1.1)2 4,(1.1)2 +1,2(1.1) + 1,1 3 11,648 = 1,(1.1) 3 4,(1.1)3 +1,2(1.1)2 + 1,1(1.1) + 1, 4 13,713 = 1,(1.1) 4 4,(1.1)4 +1,2(1.1)3 + 1,1(1.1)2 + 1,(1.1) ,884 = 1,(1.1)5 4,(1.1)5 +1,2(1.1) ,1(1.1)3 + 1,(1.1)2 + 9(1.1) + 8 Symbols Symbols B = A + c B 1 = A γ + c γ + c 1 B 2 = A γ 2 + c γ 2 + c 1 γ + c 2 B 3 = A γ 3 + c γ 3 + c 1 γ 2 + c 2 γ + c 3 B 4 = A γ 4 + c γ 4 + c 1 γ 3 + c 2 γ 2 + c 3 γ + c 4 B 5 = Aγ 5 + c γ 5 + c 1 γ 4 + c 2 γ 3 + c 3 γ 2 + c 4 γ + c 5 Example 3.2 Previous Final Balances Alternative A 15,884 = 1,(1.1)5 4,(1.1)5- + 1,2(1.1) ,1(1.1) ,(1.1) (1.1) + 8 Alternative B 16,48 = 1,(1.1)5 3,(1.1)5- + 1,3(1.1)5-1+ 1,3(1.1) ,3(1.1) 5-3 Null alternative 16,15 = 1,(1.1)5 Example 3.3 Change in Balance c 3 for A increases by $4 16,368 = 1,(1.1) 5-4,(1.1) 5 + 1,2(1.1) ,1(1.1) (1, + 4)(1.1) (1.1) + 8 Balance increases to $16,368 from $15,884, for a change of $ = 4(1.1)5-3 Easier than redoing year-by-year computations

7 Example 3.4 Internal Rate of Return The internal rate of return of an investment equals the interest rate that a savings account would pay if it were initially empty and then had the same cash flows as the investment, resulting in a final balance of $. Alternative B Alternative B is like a saving account that starts with initial assets of $, receives $3, at time and returns $1,3 for 3 years, when the last withdrawal empties the account. c equals $3,. c 1, c 2, and c 3 all equal $1,3. Solve by trial and error = + 3,(1+i)3-1,3(1+i)3-1 1,3(1+i) 1,3 i = 14.36% Using Signs from Cash Flow Diagram Obtain same solution if multiply by -1 and use signs of diagram 1,3 = 3,(1+i)3 + 1,3(1+i) ,3(1+i) + 1,3 i = 14.36% Investment B 3, Alternative A 1,2 4, 1,1 1,9 Investment A 8 = 4,(1+ i)5 + 1,2(1+i) ,1(1+i) ,(1+i) (1+i) + 8 i = 8.55% Using Excel s IRR Function A B C 1 Yr Alt A Alt B IRR =IRR(B2:B7,.1) =IRR(C2:C7,.1) Example 3.5 Arrows and Signs A person currently owes $3 and plans to borrow $1 at time 2 and make payments of $4 and $55 at times 3 and 4. The problem is to determine the balance at time

8 Debt Defined as Negative or Positive If a negative balance is defined as a debt: B4 = $ = 3(1.1)4-1(1.1) (1.1) If the focus is on the amount of debt, can define debt as positive and switch signs of flows: B4 = $ = 3(1.1)4- + 1(1.1)4-2 4(1.1) Observations B4 = $ = 3(1.1)4- + 1(1.1)4-2 4(1.1) The $1 increases debt, so it is positive. The $4 and $55 decrease the debt, so the are negative. Get same result as long as use signs consistently within the balance equation. Easier to call a debt negative when first getting started. Example 3.6 Labeling the Time Axis Exponents in the balance equation always equal the number of periods between a cash flow and the final balance If we say that the starting balance equal to the debt of $3 occurred at time : B4 = $ = 3(1.1)4-1(1.1) (1.1) Multiple Rate Balance Equation Different Labeling, Same Results If we say that the increase in debt of $1 occurs today (usually called time ), then the original debt was 2 years ago, at time B2 = $ = 3(1.1) 2-(-2) + 1(1.1) 2-4(1.1) Values of exponents same as before, equal to the periods between the flows and balance Illustration i 1 i 2 i 3 i 4 i 5 i 1 i 2 i 3 i 4 i 5 9.% 9.5% 1.% Let γ j = 1 + i j and consider CA of c 1 CA at t = 2 is c 1 (1.9) or c 1 γ 2 CA at t = 3 is c 1 (1.9)(1.95) or c 1 γ 2 γ 3 CA at t = 4 is c 1 (1.9)(1.95)(1.1) or c γ γ γ , and so forth

9 General Balance Equation In general, a cash flow is multiplied by the growth factors for the periods following its occurrence. This leads to: B t = Aγ γ γ γ t + c γ γ γ γ t + c 1 γ 2 γ 3 γ t + c 2 γ 3 γ t + + c t-1 γ t + c t A very bad looking formula Example 3.7 Multiple Interest Rates 1,2 1,1 1, % 9.5% 1.% 1.5% 11.% Alternative A 4, 15,945 = 1,(1.9)(1.95)(1.1)(1.15)(1.11) 4,(1.9)(1.95)(1.1)(1.15)(1.11) + 1,2(1.95)(1.1)(1.15)(1.11) + 1,1(1.1)(1.15)(1.11) + 1,(1.15)(1.11) + 9(1.11) + 8 Other Alternatives Similarly, for alternative B: 16,541 = 1,(1.9)(1.95)(1.1)(1.15)(1.11) 3,(1.9)(1.95)(1.1)(1.15)(1.11) + 1,3(1.95)(1.1)(1.15)(1.11) + 1,3(1.1)(1.15)(1.11) + 1,3(1.15)(1.11) Null alternative: 16,13 = 1,(1.9)(1.95)(1.1)(1.15)(1.11) Alternative B maximizes the final balance