Chapter 21: Savings Models

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Chapter 21: Savings Models

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Last Time

A Model for Saving

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Problems

Question 1: Suppose that you want to save up $2000 for a semester abroad two years from now. How much do you have to put away at the end of each month in a savings account that earns 2 % interest compounded monthly?

Question 2: A colleague feels that he will need $1 million in savings to afford to retire at age 65 and still maintain his current standard of living. Younger colleague, age 30, decides to begin savings for retirement based on that advice. How much does the younger colleague need to save per month to have $ 1 million at retirement if the fund earns a steady 3% annual interest

compounded monthly?

Question 3: Suppose you start saving for retirement at age 45. How much do you have to save per month, with a steady return of 6% compounded monthly, to accumulate $250,000 by age 65?

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Answers

Question 1: Suppose that you want to save up $2000 for a semester abroad two years from now. How much do you have to put away at the end of each month in a savings account that earns 2 % interest compounded monthly?

Payment Formula d = A i (1 + i )n_{− 1} = A r /m (1 + _mr )mt_{− 1} Answer: d = 2000 .02/12 (1 + (.02/12))24_{− 1} = 81.75

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Answers

Question 2: A colleague feels that he will need $1 million in savings to afford to retire at age 65 and still maintain his current standard of living. Younger colleague, age 30, decides to begin savings for retirement based on that advice. How much does the younger colleague need to save per month to have $ 1 million at retirement if the fund earns a steady 3% annual interest

compounded monthly? Answer: d = 1000000 .03/12 (1 + (.03/12))35(12)_{− 1} = 1348.50 If he started at age 45: d = 1000000 .03/12 (1 + (.03/12))20(12)_{− 1} = 3045.98

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Answers

Question 3: Suppose you start saving for retirement at age 45. How much do you have to save per month, with a steady return of 6% compounded monthly, to accumulate $250,000 by age 65? Answer: d = 250000 .06/12 (1 + (.06/12))20(12)_{− 1} = 541.08

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Problems

Question 4: What is the present value of $10,000, 4 years from now, at an APY of 5%?

Question 6: Suppose that inflation proceeds at a constant rate of 2% per year from mid- 2012 through mid 2015.

a) Find the cost in mid-2015 of a basket of goods that cost $1 in mid-2012.

b) What will be the value of a dollar in mid-2015 in constant mid-2012 dollars?

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Answers

Question 4: What is the present value of $10,000, 4 years from now, at an APY of 5%? Present Value PV = A (1 + i )n = A (1 + r /m)mt Answer: APY = 10000 1.054 = 8, 227.02

Answer:

APY = 15000

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Answers

Question 6: Suppose that inflation proceeds at a constant rate of 2% per year from mid- 2012 through mid 2015.

a) Find the cost in mid-2015 of a basket of goods that cost $1 in mid-2012.

Answer:

Annual Rate of Inflation

The annual rate of inflation, a (= 100a%). Goods that cost $ 1 in the base year will then cost $ (1+a).

cost = (1 + 0.02)3 = $1.06

b) What will be the value of a dollar in mid-2015 in constant mid-2012 dollars?

Answer:

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This Time

The Consumer Price Index Real Growth

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Motivating Question

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The Consumer Price index

The official measure of inflation is the Consumer Price Index (CPI) which is the determined by the Bureau of Labor Statistics (BLS).

CPI for other year

100 =

cost of market basket in other year cost of market basket in base period The base period used to calculated the CPI-U is 1982-1984

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Question

Question: If someone bought a house in mid-1990 for 150,000. What would be the equivalent cost in mid-2012 dollars? Answer:

The CPI in 1990 was 130.7 The CPI in 2012 was 229.594

cost in 2012 cost in 1990 = CPI in 2012 CPI in 1990 so cost in 2012 150000 = 229.594 130.7 cost in 2012 = 150000229.594 130.7 = 263, 497.32

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Question

A 1963 Chevy Bel Air, a classic car today, cost $ 2,400 new in mid-1963. How much would that be in 2012 dollars?

Answer:

cost in 2012 2400 = 229.594 30.6 cost in 2012 = 2400229.594 30.6 = 18, 007.40

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Question

A 1963 Chevy Bel Air, a classic car today, cost $ 2,400 new in mid-1963. How much would that be in 2012 dollars?

Answer:

cost in 2012 2400 = 229.594 30.6 cost in 2012 = 2400229.594 30.6 = 18, 007.40

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Motivating Question

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Real Growth Under Inflation

Real rate of Growth

The real annual rate of growth of an investment at annual interest rate r with annual inflation rate a is

g = r − a

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Question

Question: In mid 2013 you put a $1000 into a savings account with APY 1 %. Assuming there is a constant inflation rate of 2 % for the next 3 years, how much money will you have in the account in mid 2016 in constant mid-2013 dollars?

Answer: g = .01 − .02 1.02 = − .1 1.02 = −0.0980392 A = 1000(1 − 0.0980392)3 = 970.876

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Question

Question: In mid 2013 you put a $1000 into a savings account with APY 1 %. Assuming there is a constant inflation rate of 2 % for the next 3 years, how much money will you have in the account in mid 2016 in constant mid-2013 dollars?

Answer: g = .01 − .02 1.02 = − .1 1.02 = −0.0980392 A = 1000(1 − 0.0980392)3 = 970.876

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Question

Question: In mid 2013 you put a $1000 into a savings account. Assuming there is a constant inflation rate of 2 % for the next 3 years, what would the APY of the savings account have to be in order to have $1100 dollars in constant mid-2012 dollars in 3 years?

Answer: 1100 = 1000 1 +r − .02 1.02 3 r = 0.0529257

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Question

Question: In mid 2013 you put a $1000 into a savings account. Assuming there is a constant inflation rate of 2 % for the next 3 years, what would the APY of the savings account have to be in order to have $1100 dollars in constant mid-2012 dollars in 3 years? Answer: 1100 = 1000 1 +r − .02 1.02 3 r = 0.0529257

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Next time

Chapter 22 Borrowing models