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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 = e = A = P(A/P,r,n) = 1,000(A/P,0.2214,10) = 1,000*0.2214*7.389/6.389 = 256 Sep

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**Another Example i = er - 1 = e0.035 - 1 = 0.0356**

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 = e = A = P(A/P, 0.419, 16) = 12,000*0.0356*1.75/0.75 = 997 Sep

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**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) = er n - 1 rern ern - 1 r Sep

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**Example Sol: F = A(F/A,8%,5) = $500 x 6.1478 = $3,074**

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 = $3,074 Sep

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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 ern - 1 r e0.1(8.5) - 1 0.1 Sep

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Discussion Examples Suppose the 8% nominal interest rate. If compounding occurs monthly, what is the effective annual interest rate? Sep

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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. Sep

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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. Sep

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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. Sep

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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? Sep

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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? Sep

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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? Sep

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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. Sep

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**Discrete Cash Flows Continuous Compounding**

i = ( 1 + r/M)M - 1 If M -> then i = er - 1 F/P = (1+i)n => F/P = ern P/F = e-rn F/A = (ern - 1) / (er - 1) Sep

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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) = P[ern (er - 1)]/ (ern - 1) ern = e0.2 (10) er = e0.2 Sep

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**Homework # 3 - Chapter 3 Problem Number:**

93, 95, 98, 99, 101, 102, 105, 106 Due Date: Oct. 8 (Thursday) Sep

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Engineering Economy

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(c) 2002 Contemporary Engineering Economics

(c) 2002 Contemporary Engineering Economics

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