钱&183;帕克工程经济学原理Lecture No2

钱·帕克《工程经济学原理》课件

Time Value of Money

Lecture No.2

钱·帕克《工程经济学原理》课件

Chapter 2 Time Value of Money

Interest: The Cost of Money Economic Equivalence Interest Formulas – Single Cash Flows Equal-Payment Series Dealing with Gradient Series Composite Cash Flows.

Power-Ball Lottery

钱·帕克《工程经济学原理》课件

Decision Dilemma—Take a Lump Sum or Annual Installments

A suburban Chicago couple won the Power-ball. They had to choose between a single lump sum $104 million, or $198 million paid out over 25 years (or $7.92 million per year). The winning couple opted for the lump sum. Did they make the right choice? What basis do we make such an economic comparison?

钱·帕克《工程经济学原理》课件

Option A (Lump Sum) 0 1 2 3 25 $104 M

Option B (Installment Plan)

$7.92 M $7.92 M $7.92 M$7.92 M

钱·帕克《工程经济学原理》课件

What Do We Need to Know?

To make such comparisons (the lottery decision problem), we must be able to compare the value of money at different point in time. To do this, we need to develop a method for reducing a sequence of benefits and costs to a single point in time. Then, we will make our comparisons on that basis.

钱·帕克《工程经济学原理》课件

Time Value of Money

Money has a time value because it can earn more money over time (earning power). Money has a time value because its purchasing power changes over time (inflation). Time value of money is measured in terms of interest rate. Interest is the cost of money—a cost to the borrower and an earning to the lender

钱·帕克《工程经济学原理》课件

Delaying Consumption

钱·帕克《工程经济学原理》课件

Account Value

Cost of Refrigerator

Case 1: Inflation exceeds earning power Case 2: Earning power exceeds inflation

N = 0 $100 N = 1 $106 N = 0 $100 N = 1 $106

N = 0 $100 N = 1 $108 N = 0 $100 N = 1 $104

(earning rate =6%) (inflation rate = 8%)

(earning rate =6%) (inflation rate = 4%)

钱·帕克《工程经济学原理》课件

钱·帕克《工程经济学原理》课件

Which Repayment Plan?End of Year Receipts Payments Plan 1 $200.00 5,141.85 5,141.85 5,141.85 5,141.85 5,141.85 Plan 2 $200.00 0 0 0 0 30,772.48

Year 0 Year 1 Year 2 Year 3 Year 4 Year 5

$20,000.00

The amount of loan = $20,000, origination fee = $200, interest rate = 9% APR (annual percentage rate)

钱·帕克《工程经济学原理》课件

Cash Flow Diagram

钱·帕克《工程经济学原理》课件

End-of-Period Convention

0

1

Beginning of Interest period

End of interest period

0

1

钱·帕克《工程经济学原理》课件

Methods of Calculating Interest

Simple interest: the practice of charging an interest rate only to an initial sum (principal amount). Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.

钱·帕克《工程经济学原理》课件

Simple Interest

P = Principal amount i = Interest rate End of Year N = Number of 0 interest periods 1 Example:

Beginning Balance

Interest earned

Ending Balance $1,000

$1,000 $1,080 $1,160

$80 $80 $80

$1,080 $1,160 $1,240

P = $1,000 i = 8% N = 3 years

2 3

钱·帕克《工程经济学原理》课件

Simple Interest FormulaF P (iP ) N where P = Principal amount i = simple interest rate N = number of interest periods F = total amount accumulated at the end of period NF $1, 000 (0.08)($1, 000)(3) $1, 240

钱·帕克《工程经济学原理》课件

Compou

nd Interest

Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.

钱·帕克《工程经济学原理》课件

Compound Interest

P = Principal amount End i = Interest rate of N = Number of Year interest periods 0 Example:

Beginning Balance

Interest earned

Ending Balance $1,000

P = $1,000 i = 8% N = 3 years

1

$1,000

$80

$1,080

23

$1,080$1,166.40

$86.40$93.31

$1,166.40$1,259.71

钱·帕克《工程经济学原理》课件

Compounding Process$1,080 0 1 $1,000 2 3 $1,080 $1,166.40

$1,166.40

$1,259.71

钱·帕克《工程经济学原理》课件

$1,259.71

0

1

2

3

F $1, 000(1 0.08)3$1,000

$1, 259.71

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