Continuous Compounding

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1/20/2013 1 Continuous Compounding Continuous Compounding Based on the assumption that cash payments occur once per year but compounding is continuous throughout the year. Review: Future Amount, F Future Amount or Future Worth F = P(1 + i) n Where: P = principal i = interest per period (in decimal) n = number of interest periods (1 + i) n = single payment compound amount factor. PW FW 0 1 2 3 4 … n

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Continuous Compounding lecture and sample problems

Transcript of Continuous Compounding

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Continuous Compounding

Continuous Compounding

• Based on the assumption that cash payments occur once per year but compounding is continuous throughout the year.

Review: Future Amount, F

Future Amount or Future Worth • F = P(1 + i)n

Where: • P = principal • i = interest per period (in

decimal) • n = number of interest

periods • (1 + i)n = single payment

compound amount factor.

PW

FW

0 1 2 3 4 … n

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For m periods of year…

Future Amount or Future Worth

• F = P(1 + i)n

• F = P(1 + 𝑁𝑅

𝑚)mn

PW

FW

0 1 2 3 4 … m

Future Worth of Continuous Compounding

Let x = 𝑚

𝑁𝑅

• F = P(1 + i)n

• F = P(1 + 𝑁𝑅

𝑚)mn

• F = P(1 + 1

𝑥)x(NR)n

• F = P[(1 + 1

𝑥)x](NR)n

• But lim𝑥→∞

(1 + 1

𝑥)x = e

Thus, • F = Pe(NR)n

Where: • P = principal • e = 2.71828… • NR = nominal rate • n = number of years • e(NR)n = continuous

compounding amount factor

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Present Worth of Continuous Compounding

Present Amount or Present Worth

• P = F

e(NR)n

PW

FW

0 1 2 3 4 … m

Problem #15

• Money is deposited in a certain account for which the interest is compounded continuously.

• If the balance doubles in 6 years, what is the annual percentage rate?

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Solution

• F = Pe(NR)n

• 2P = Pe(NR)6

• 2 = e6(NR)

Take ln on both sides

• ln 2 = ln e6(NR)

• ln 2 = 6(NR) ln e

• ln 2 = 6(NR)

• NR = ln 2

6

• NR = 0.1155

• NR = 11.5524%

ME Board Problem #16

• A nominal interest of 3% compounded continuously is given on the account.

• What is the accumulated amount of Php 10,000 after 10 years?

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Solution

• F = Pe(NR)n

• F = (10,000) e(0.03)10

• F = (10,000) e(0.3)

• F = Php 13,498.5880

Nominal and Effective Rate of Interest

Rate of Interest

• The cost of borrowing money.

• Also refers to the amount earned by a unit principal per unit time.

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Two Types of Rate of Interest

Two Types of Rate of Interest

1. Nominal Rate of Interest

• The basic annual rate of interest.

2. Effective Rate of Interest

• The actual or the exact rate of interest earned on the principal during a one-year period .

Problem #17

• A principal is invested at 5% compounded quarterly.

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Solution

• Nominal Rate of Interest = 5%

• Effective Rate of Interest => 5%, because of the compounding which occur four times a year.

Effective Rate of Interest

Effective Rate of Interest

• ER = (1 + i)m – 1 or ER = (1 + 𝑁𝑅

𝑚)m – 1

Where: • m = number of interest period per year • i = interest per period • NR = nominal rate of interest

Note: • NR is the mode of compounding is annually

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Solution

• Nominal Rate of Interest = 5%

• Effective Rate of Interest => 5.0945%, because of the compounding which occur four times a year.

• ER = (1 + 𝑁𝑅

𝑚)m – 1

• ER = (1 + 0.05

4)4 – 1

• ER = 0.0509

• ER = 5.0945%

ECE Board Problem #18

• What is the effective rate corresponding to 18% compounded daily?

• Take 1 year is equal to 360 days.

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Solution

• ER = (1 + i)m – 1

• ER = (1 + 0.18

360)360 – 1

• ER = 0.1971

• ER = 19.7163%

ECE Board Problem #19

• What nominal rate, compounded semi-annually, yields the same amount as 16% compounded quarterly?

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Solution

• ER1

2

= ER1

4

• (1 + 𝑁𝑅

2)2 – 1 = (1 +

𝑁𝑅

4)4 – 1

• (1 + 𝑁𝑅

2)2 = (1 +

0.16

4)4

• (1 + 𝑁𝑅

2)2 = 1.1698

• 1 + 𝑁𝑅

2 = 1.0816

•𝑁𝑅

2 = 0.0816

• NR = 0.1632 • NR = 16.32%

ECE Board Problem #20

• What rate of interest compounded annually is the same as the rate of interest of 8% compounded quarterly.

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Solution

• ER1 = ER1

4

• (1 + i)1 – 1 = (1 + 𝑁𝑅

4)4 – 1

• (1 + i) = (1 + 0.08

4)4

• (1 + i) = 1.0824

• i = 0.0824

• i = 8.2432%

ECE Board Problem #21

• Find the nominal rate, which if converted quarterly could be used instead of 12% compounded semi-annually?

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Solution

• ER1

4

= ER1

2

• (1 + i)4 – 1 = (1 + 𝑁𝑅

2)2 –

1

• (1 + i)4 = (1 + 0.12

2)2

• (1 + i)4 = 1.1236

• i = 0.0295

• i = 2.9563%

But :

• i = 𝑁𝑅

𝑚

• 0.0295 = 𝑁𝑅

4

Thus,

• NR = 0.1182

• NR = 11.8252%

ECE Board Problem #22

• What is the corresponding effective rate of 18% compounded semi-annually?

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Solution

• ER = (1 + i)m – 1

• ER = (1 + 0.18

2)2 – 1

• ER = 0.1881

• ER = 18.81%

ME Board Problem #23

• What is the corresponding effective interest rate of 18% compounded quarterly?

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Solution

• ER = (1 + i)m – 1

• ER = (1 + 0.18

4)4 – 1

• ER = 0.1925

• ER = 19.2518%

ME Board Problem #24

• Compute the equivalent rate of 6% compounded semi-annually to a rate compounded quarterly.

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Solution

• ER quarterly = ER annually

• (1 + 𝑖

4)4 – 1 = (1 +

0.06

2)2 – 1

• (1 + 𝑖

4)4 = (1 +

0.06

2)2

• i = 0.0595

• i = 5.9556%

EE Board Problem #25

• A man borrowed Php 100,000 at the interest rate of 12% per annum, compounded quarterly.

• What is the effective rate?

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Solution

• ER = (1 + i)m – 1

• ER = (1 + 0.12

4)4 – 1

• ER = 0.1255

• ER = 12.5508%

ME Board Problem #26

• An interest rate is quoted as being 7.5% compounded quarterly.

• What is the effective annual interest rate?

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Solution

• ER = (1 + i)m – 1

• ER = (1 + 0.075

4)4 – 1

• ER = 0.0771

• ER = 7.7135%

ME Board Problem #27

• A bank pays one percent interest on savings account four times a year.

• The effective annual interest rate is

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Solution

• ER = (1 + i)m – 1

• ER = (1 + 0.01)4 – 1

• ER = 0.0406

• ER = 4.0604%