Discrete Cash Flow

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Continuous Compounding

• Discrete Cash Flow
• Continuous Compounding (DC)
• Continuous Cash Flow
• Continuous Compounding (CC)

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Discrete Cash Flow Continuous Compounding (DC)

• i er - 1 then follow the DD case.
• Suppose that one has a present loan of 1,000 and
  desires to determine what equivalent uniform
  end-of-year payments, A, could be obtained from
  it for 10 years if the nominal interest rate is
  20 compounded continuously
• i er - 1 e0.2 - 1 0.2214
• A P(A/P,r,n) 1,000(A/P,0.2214,10)
• 1,0000.22147.389/6.389 256

3
Another Example

• An individual needs 12,000 immediately as a down
  payment on a new home. Suppose that he can borrow
  this money from his insurance company. He must
  repay the loan in equal payments every six months
  over the next eight years. The nominal interest
  rate being charged is 7 compounded continuously.
  What is the amount of each payment?
• i er - 1 e0.035 - 1 0.0356
• A P(A/P, 0.419, 16) 12,0000.03561.75/0.75
• 997

4
Continuous Cash Flow Continuous Compounding

• If there is a continuous cash flow A each year
  for n years with nominal interest rate per year
  r.
• then
• (P/A, r, n)
• (F/A, r, n)

5
Example

• What will be the future equivalent amount at the
  end of five years of a uniform, continuous cash
  flow, at the rate of 500 per year for five
  years, with interest compounded continuously at
  the nominal annual rate of 8?
• Sol
• F A(F/A,8,5) 500 x 6.1478 3,074

6
Example

• What is the future equivalent of 10,000 per year
  that flows continuously for 8.5 years if the
  nominal interest is 10 compounded continuously?
• Sol
• F A(F/A,5, 17) 133,964.5
• or
• F A 10,000

7
Discussion Examples

• Suppose the 8 nominal interest rate. If
  compounding occurs monthly, what is the effective
  annual interest rate?

8
Example

• With the minimum number of interest factors, find
  the value of X below so that the two cash flow
  diagrams are equivalent when the interest rate is
  10 per year.

9
Example

• Set up an expression for the value of Z on the
  left-hand cash flow diagram that establishes
  equivalence with the right-hand cash flow
  diagram. The nominal interest rate is 12
  compounded quarterly.

10
Example

• A student decides to make semi-annual payments of
  500 each into a bank account that pays an APR
  (nominal interest) of 8 compounded weekly. How
  much money will this student have accumulated in
  this bank account at the end of 20 years? Assume
  only one (the final) withdrawal is made.

11
Example

• Consider an end-of-year (EOY) geometric gradient,
  which lasts for eight years, whose initial value
  at EOY one is 5,000, and f 6.04 per year
  thereafter. Find the equivalent uniform gradient
  amount over the same time period if the initial
  value of the uniform gradient at the end of year
  one is 4,000. The nominal rate is 8 compounded
  semi-annually. What is iCR? P0? and What is G?

12
Example

• An individual makes five annual deposits of
  2,000 in a savings account that pays interest at
  a rate of 4 per year. One year after making the
  last deposit, the interest rate changes to 6 per
  year. Five years after the last deposit the
  accumulated money is withdrawn from the account.
  How much is withdrawn?

13
Example

• Some future amount, F, is equivalent to 2,000
  being received every 6 months over the next 12
  years. The nominal interest rate is 20
  compounded continuously. What is the value of F?

14
Example

• What is the value of P that is equivalent to A
  800/yr (800 flowing continuously each year) for
  11.2 years? The nominal rate of interest is 10,
  continuously compounded.

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Discrete Cash FlowsContinuous Compounding

• i ( 1 r/M)M - 1
• If M -gt ? then i er - 1
• F/P (1i)n gt F/P ern
• P/F e-rn
• F/A (ern - 1) / (er - 1)

16
Example

• Suppose that one has a present loan of 1,000 and
  desires to determine what equivalent uniform
  end-of-year payments, A, could be obtained from
  it for 10 years if the nominal interest rate is
  20 compounded continuously.
• A P (A/P, r, n) Pern (er - 1)/ (ern - 1)
• ern e0.2 (10) er e0.2

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Homework 3 - Chapter 3

• Problem Number
• 93, 95, 98, 99, 101, 102, 105, 106
• Due Date Oct. 8 (Thursday)

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Engineering Economy