# 91 9091year a rate of inﬂation of 10year is

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given by our growth rate formula: (0.90909 – 1)/1 = –0.09091 = –9.091%/year. A rate of inﬂation of 10%/year is equivalent to a rate of reduction in purchasing power of 9.091%/year. You may want to calculate the rate of change in purchas- ing power for the other years in the table. (87,000 – 100,000)/1 100,000 MATH MODULE 9: GROWTH RATES, INTEREST RATES, AND INFLATION: THE ECONOMICS OF TIME M9-3

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E XAMPLE M.9.4 The interest rate i can also be interpreted as a growth rate. If you borrow \$1000 at an interest rate i of 8 %/year, for example, then at the end of the year you will owe the lender \$1080: the principal P of the loan (\$1000) plus \$80 ( i x P = 0.08 x 1000 = \$80) as an interest payment for the use of the money over the year. At the end of the year, just before you repay the loan plus interest, your indebtedness is now P + iP = P (1 + i ). Your indebtedness has grown at a rate of P (1 + P i ) – P = i P P = i , or 8%/year. If you were the lender in this situation, so that the loan (the money owed to you) was an asset rather than a liability, then the rate of interest would measure the rate of growth over the year in the value of your asset. 9.2 COMPOUND INTEREST AND MULTI-PERIOD GROWTH If a variable X grows at rate g 01 from period 0 to period 1, and at rate g 12 from period 1 to period 2, then in period 2 its value will be X 2 = X 0 (1 + g 01 )(1 + g 12 ). (M.9.3) For example, if you lend out \$1000 today (at the start of period 0), and the interest rate this year, i 01 , is 10%/year, while the interest rate next year, i 12 , is 8%/year, then one year from today (at the beginning of period 1), your (perfectly secure) loan will be worth \$1000(1 + .1) = \$1100. If you loan out the entire amount of \$1100 (principal plus accrued interest) at 8%/year for another year, then exactly 2 years from today (at the start of period 2) it will be worth (1000)(1 + .1)(1 + .08) = 1100(1.08) = \$1188. (Note that you would end up with the same result if i 01 were 8%/year and i 12 were 10%/year.) If you loan it out on the same basis over a period of t years, then at the beginning of period t its value Vt will be V t = V 0 (1 + i 01 )(1 + i 12 ) (1 + i 23 ) ..... (1 + i ( t –1) t ) (M.9.4) If the interest rate i is constant over the entire period, then this formula simplifies to V t = V 0 (1 + i ) t . (M.9.5) E XAMPLE M.9.5 Suppose that you borrow \$1000 today for a period of 5 years. At the end of exactly 5 years, how much will you owe if the interest rate i is: (a) 5%/year? (b) 10%/year? (c) 20%/year? What would the results be in each case if the loan period were 10 years? If the loan period is 5 years, then at the end of 5 years you will owe: (a) V 5 = 1000(1.05) 5 = 1000(1.27628) = \$1276.28; (b) V 5 = 1000(1.1) 5 = \$1610.51: (c) V 5 = 1000(1.2) 5 = \$2488.32. M9-4 MATH MODULE 9: GROWTH RATES, INTEREST RATES, AND INFLATION: THE ECONOMICS OF TIME
If the loan period is 10 years, then at the end of 10 years you will owe: (a) V 10 = 1000(1.05) 10 = 1000(1.62889) = \$1628.89; (b) V 10 = 1000(1.1) 10 = \$2593.74: (c) V 10 = 1000(1.2) 10 = \$6191.74. [In solving Example M.9.5, you will need to use the y x (or x y ) button on your calculator. To calculate (1.05) 5 , on most calculators we enter “1.05” y x (or x y ) 5 “=.” Experiment with your own calculator to verify the correct sequence.] 9.3 CALCULATING THE PRESENT DISCOUNTED VALUE OF A FUTURE INCOME STREAM Suppose that someone guarantees to pay you an amount of \$1610.51, exactly 5 years from today, and the interest rate over the period is a constant 10%/year. What is the pre-

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