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# Math 1300: Section 3-2 Compound Interest

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Lecture for Section 3-2 of Barnett's "Finite Mathemaitcs."

Lecture for Section 3-2 of Barnett's "Finite Mathemaitcs."

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## Math 1300: Section 3-2 Compound InterestPresentation Transcript

• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Math 1300 Finite Mathematics Section 3.2 Compound Interest Jason Aubrey Department of Mathematics University of Missouri university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield If at the end of a payment period the interest due is reinvested at the same rate, then the interest as well as the original principal will earn interest at the end of the next payment period. Interest payed on interest reinvested is called compound interest. university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Suppose you invest \$1,000 in a bank that pays 8% compounded quarterly. How much will the bank owe you at the end of the year? university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Suppose you invest \$1,000 in a bank that pays 8% compounded quarterly. How much will the bank owe you at the end of the year? A = P(1 + rt) university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Suppose you invest \$1,000 in a bank that pays 8% compounded quarterly. How much will the bank owe you at the end of the year? A = P(1 + rt) 1 = 1, 000 1 + 0.08 4 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Suppose you invest \$1,000 in a bank that pays 8% compounded quarterly. How much will the bank owe you at the end of the year? A = P(1 + rt) 1 = 1, 000 1 + 0.08 4 = 1, 000(1.02) = \$1, 020 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Suppose you invest \$1,000 in a bank that pays 8% compounded quarterly. How much will the bank owe you at the end of the year? A = P(1 + rt) 1 = 1, 000 1 + 0.08 4 = 1, 000(1.02) = \$1, 020 This is the amount at the end of the ﬁrst quarter. Then: university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Suppose you invest \$1,000 in a bank that pays 8% compounded quarterly. How much will the bank owe you at the end of the year? Second quarter: = \$1,020(1.02) = A = P(1 + rt) \$1,040.40 1 = 1, 000 1 + 0.08 4 = 1, 000(1.02) = \$1, 020 This is the amount at the end of the ﬁrst quarter. Then: university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Suppose you invest \$1,000 in a bank that pays 8% compounded quarterly. How much will the bank owe you at the end of the year? Second quarter: = \$1,020(1.02) = A = P(1 + rt) \$1,040.40 1 Third quarter: = 1, 000 1 + 0.08 4 = \$1,040.40(1.02) = = 1, 000(1.02) = \$1, 020 \$1,061.21 This is the amount at the end of the ﬁrst quarter. Then: university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Suppose you invest \$1,000 in a bank that pays 8% compounded quarterly. How much will the bank owe you at the end of the year? Second quarter: = \$1,020(1.02) = A = P(1 + rt) \$1,040.40 1 Third quarter: = 1, 000 1 + 0.08 4 = \$1,040.40(1.02) = = 1, 000(1.02) = \$1, 020 \$1,061.21 Fourth quarter: This is the amount at the end = \$1,061.21(1.02) = of the ﬁrst quarter. Then: \$1,082.43 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Look at the pattern: university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Look at the pattern: First quarter: = \$1, 000(1.02) university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Look at the pattern: First quarter: = \$1, 000(1.02) Second quarter: = \$1, 000(1.02)2 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Look at the pattern: First quarter: = \$1, 000(1.02) Second quarter: = \$1, 000(1.02)2 Third quarter: = \$1, 000(1.02)3 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Look at the pattern: First quarter: = \$1, 000(1.02) Second quarter: = \$1, 000(1.02)2 Third quarter: = \$1, 000(1.02)3 Fourth quarter: = \$1, 000(1.02)4 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Deﬁnition (Amount: Compound Interest) A = P(1 + i)n university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Deﬁnition (Amount: Compound Interest) A = P(1 + i)n where i = r /m and A = amount (future value) at the end of n periods university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Deﬁnition (Amount: Compound Interest) A = P(1 + i)n where i = r /m and A = amount (future value) at the end of n periods P = principal (present value) university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Deﬁnition (Amount: Compound Interest) A = P(1 + i)n where i = r /m and A = amount (future value) at the end of n periods P = principal (present value) r = annual nominal rate university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Deﬁnition (Amount: Compound Interest) A = P(1 + i)n where i = r /m and A = amount (future value) at the end of n periods P = principal (present value) r = annual nominal rate m = number of compounding periods per year university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Deﬁnition (Amount: Compound Interest) A = P(1 + i)n where i = r /m and A = amount (future value) at the end of n periods P = principal (present value) r = annual nominal rate m = number of compounding periods per year i = rate per compounding period university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Deﬁnition (Amount: Compound Interest) A = P(1 + i)n where i = r /m and A = amount (future value) at the end of n periods P = principal (present value) r = annual nominal rate m = number of compounding periods per year i = rate per compounding period n = total number of compounding periods university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: P = 800; i = 0.06; n = 25; A =? university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: P = 800; i = 0.06; n = 25; A =? A = P(1 + i)n university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: P = 800; i = 0.06; n = 25; A =? A = P(1 + i)n A = \$800(1.06)25 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: P = 800; i = 0.06; n = 25; A =? A = P(1 + i)n A = \$800(1.06)25 A = \$3, 433.50 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A = \$18, 000; r = 8.12% compounded monthly; n = 90; P =? university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A = \$18, 000; r = 8.12% compounded monthly; n = 90; P =? r 0.0812 First we compute i = m = 12 = 0.0068. Then, university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A = \$18, 000; r = 8.12% compounded monthly; n = 90; P =? r 0.0812 First we compute i = m = 12 = 0.0068. Then, A = P(1 + i)n university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A = \$18, 000; r = 8.12% compounded monthly; n = 90; P =? r 0.0812 First we compute i = m = 12 = 0.0068. Then, A = P(1 + i)n \$18, 000 = P(1.0068)90 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A = \$18, 000; r = 8.12% compounded monthly; n = 90; P =? r 0.0812 First we compute i = m = 12 = 0.0068. Then, A = P(1 + i)n \$18, 000 = P(1.0068)90 \$18, 000 ≈ P(1.84) university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A = \$18, 000; r = 8.12% compounded monthly; n = 90; P =? r 0.0812 First we compute i = m = 12 = 0.0068. Then, A = P(1 + i)n \$18, 000 = P(1.0068)90 \$18, 000 ≈ P(1.84) P ≈ \$9, 782.61 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Deﬁnition (Amount - Continuous Compound Interest) If a principal P is invested at an annual rate r (expressed as a decimal) compounded continuously, then the amount A in the account at the end of t years is given by A = Pert university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: P = \$2, 450; r = 8.12%; t = 3 years; A =? university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: P = \$2, 450; r = 8.12%; t = 3 years; A =? A = Pert university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: P = \$2, 450; r = 8.12%; t = 3 years; A =? A = Pert A = \$2, 450e(0.0812)(3) = \$3, 125.79 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A = \$15, 875; P = \$12, 100; t = 48 months; r =? university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A = \$15, 875; P = \$12, 100; t = 48 months; r =? A = Pert university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A = \$15, 875; P = \$12, 100; t = 48 months; r =? A = Pert \$15, 875 = \$12, 100e4r university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A = \$15, 875; P = \$12, 100; t = 48 months; r =? A = Pert \$15, 875 = \$12, 100e4r 1.311 = e4r university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A = \$15, 875; P = \$12, 100; t = 48 months; r =? A = Pert \$15, 875 = \$12, 100e4r 1.311 = e4r ln(1.311) = 4r university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A = \$15, 875; P = \$12, 100; t = 48 months; r =? A = Pert \$15, 875 = \$12, 100e4r 1.311 = e4r ln(1.311) = 4r ln(1.311) =r 4 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A = \$15, 875; P = \$12, 100; t = 48 months; r =? A = Pert \$15, 875 = \$12, 100e4r 1.311 = e4r ln(1.311) = 4r ln(1.311) =r 4 r = 0.068 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: If an investment company pays 6% compounded semiannually, how much should you deposit now to have \$10,000? 5 years from now? 10 years from now? university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: If an investment company pays 6% compounded semiannually, how much should you deposit now to have \$10,000? 5 years from now? 10 years from now? i = r /m = 0.06/2 = 0.03 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: If an investment company pays 6% compounded semiannually, how much should you deposit now to have \$10,000? 5 years from now? 10 years from now? i = r /m = 0.06/2 = 0.03 n = 5x2 = 10 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: If an investment company pays 6% compounded semiannually, how much should you deposit now to have \$10,000? 5 years from now? 10 years from now? i = r /m = 0.06/2 = 0.03 n = 5x2 = 10 A = P(1 + i)n university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: If an investment company pays 6% compounded semiannually, how much should you deposit now to have \$10,000? 5 years from now? 10 years from now? i = r /m = 0.06/2 = 0.03 n = 5x2 = 10 A = P(1 + i)n \$10, 000 = P(1.03)10 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: If an investment company pays 6% compounded semiannually, how much should you deposit now to have \$10,000? 5 years from now? 10 years from now? i = r /m = 0.06/2 = 0.03 n = 5x2 = 10 A = P(1 + i)n \$10, 000 = P(1.03)10 \$10, 000 ≈ P(1.3439) university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: If an investment company pays 6% compounded semiannually, how much should you deposit now to have \$10,000? 5 years from now? 10 years from now? i = r /m = 0.06/2 = 0.03 n = 5x2 = 10 A = P(1 + i)n \$10, 000 = P(1.03)10 \$10, 000 ≈ P(1.3439) P ≈ \$7, 441.03 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: If an investment company pays 6% compounded semiannually, how much should you deposit now to have \$10,000? 5 years from now? 10 years from now? i = r /m = 0.06/2 = 0.03 i = r /m = 0.06/2 = 0.03 n = 5x2 = 10 A = P(1 + i)n \$10, 000 = P(1.03)10 \$10, 000 ≈ P(1.3439) P ≈ \$7, 441.03 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: If an investment company pays 6% compounded semiannually, how much should you deposit now to have \$10,000? 5 years from now? 10 years from now? i = r /m = 0.06/2 = 0.03 i = r /m = 0.06/2 = 0.03 n = 5x2 = 10 n = 10x2 = 20 n A = P(1 + i) \$10, 000 = P(1.03)10 \$10, 000 ≈ P(1.3439) P ≈ \$7, 441.03 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: If an investment company pays 6% compounded semiannually, how much should you deposit now to have \$10,000? 5 years from now? 10 years from now? i = r /m = 0.06/2 = 0.03 i = r /m = 0.06/2 = 0.03 n = 5x2 = 10 n = 10x2 = 20 n A = P(1 + i) A = P(1 + i)n \$10, 000 = P(1.03)10 \$10, 000 ≈ P(1.3439) P ≈ \$7, 441.03 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: If an investment company pays 6% compounded semiannually, how much should you deposit now to have \$10,000? 5 years from now? 10 years from now? i = r /m = 0.06/2 = 0.03 i = r /m = 0.06/2 = 0.03 n = 5x2 = 10 n = 10x2 = 20 n A = P(1 + i) A = P(1 + i)n \$10, 000 = P(1.03)10 \$10, 000 = P(1.03)20 \$10, 000 ≈ P(1.3439) P ≈ \$7, 441.03 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: If an investment company pays 6% compounded semiannually, how much should you deposit now to have \$10,000? 5 years from now? 10 years from now? i = r /m = 0.06/2 = 0.03 i = r /m = 0.06/2 = 0.03 n = 5x2 = 10 n = 10x2 = 20 n A = P(1 + i) A = P(1 + i)n \$10, 000 = P(1.03)10 \$10, 000 = P(1.03)20 \$10, 000 ≈ P(1.3439) \$10, 000 ≈ P(1.806) P ≈ \$7, 441.03 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: If an investment company pays 6% compounded semiannually, how much should you deposit now to have \$10,000? 5 years from now? 10 years from now? i = r /m = 0.06/2 = 0.03 i = r /m = 0.06/2 = 0.03 n = 5x2 = 10 n = 10x2 = 20 n A = P(1 + i) A = P(1 + i)n \$10, 000 = P(1.03)10 \$10, 000 = P(1.03)20 \$10, 000 ≈ P(1.3439) \$10, 000 ≈ P(1.806) P ≈ \$7, 441.03 P ≈ \$5, 536.76 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A zero coupon bond is a bond that is sold now at a discount and will pay it’s face value at some time in the future when it matures - no interest payments are made. Suppose that a zero coupon bond with a face value of \$40,000 matures in 20 years. What should the bond be sold for now if its rate of return is to be 5.124% compounded annually. university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A zero coupon bond is a bond that is sold now at a discount and will pay it’s face value at some time in the future when it matures - no interest payments are made. Suppose that a zero coupon bond with a face value of \$40,000 matures in 20 years. What should the bond be sold for now if its rate of return is to be 5.124% compounded annually. Compounded annually means i = m = 0.05124 = 0.05124. Here r 1 A = \$40, 000 and we want to ﬁnd P. We also know that n = 20. university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A zero coupon bond is a bond that is sold now at a discount and will pay it’s face value at some time in the future when it matures - no interest payments are made. Suppose that a zero coupon bond with a face value of \$40,000 matures in 20 years. What should the bond be sold for now if its rate of return is to be 5.124% compounded annually. Compounded annually means i = m = 0.05124 = 0.05124. Here r 1 A = \$40, 000 and we want to ﬁnd P. We also know that n = 20. A = P(1 + i)n university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A zero coupon bond is a bond that is sold now at a discount and will pay it’s face value at some time in the future when it matures - no interest payments are made. Suppose that a zero coupon bond with a face value of \$40,000 matures in 20 years. What should the bond be sold for now if its rate of return is to be 5.124% compounded annually. Compounded annually means i = m = 0.05124 = 0.05124. Here r 1 A = \$40, 000 and we want to ﬁnd P. We also know that n = 20. A = P(1 + i)n \$40, 000 = P(1.05124)20 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: A zero coupon bond is a bond that is sold now at a discount and will pay it’s face value at some time in the future when it matures - no interest payments are made. Suppose that a zero coupon bond with a face value of \$40,000 matures in 20 years. What should the bond be sold for now if its rate of return is to be 5.124% compounded annually. Compounded annually means i = m = 0.05124 = 0.05124. Here r 1 A = \$40, 000 and we want to ﬁnd P. We also know that n = 20. A = P(1 + i)n \$40, 000 = P(1.05124)20 \$40, 000 P= = \$14, 723.89 (1.05124)20 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Deﬁnition If a principal is invested at the annual (nominal) rate r compounded m times a year, then the annual percentage yield is r m APY = 1 + −1 m If a principal is invested at the annual (nominal) rate r compounded continuously, then the annual percentage yield is APY = er − 1 The annual percentage yield is also referred to as the effective rate or the true interest rate. university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: What is the annual nominal rate compounded daily for a bond that has an APY of 6.8%. r m APY = 1 + −1 m university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: What is the annual nominal rate compounded daily for a bond that has an APY of 6.8%. r m APY = 1 + −1 m r 365 0.068 = 1 + −1 365 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: What is the annual nominal rate compounded daily for a bond that has an APY of 6.8%. r m APY = 1 + −1 m r 365 0.068 = 1 + −1 365 r 365 1.068 = 1 + 365 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: What is the annual nominal rate compounded daily for a bond that has an APY of 6.8%. r m APY = 1 + −1 m r 365 0.068 = 1 + −1 365 r 365 1.068 = 1 + 365 √ 365 r 1.068 = 1 + 365 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: What is the annual nominal rate compounded daily for a bond that has an APY of 6.8%. r m APY = 1 + −1 m r 365 0.068 = 1 + −1 365 r 365 1.068 = 1 + 365 √ 365 r 1.068 = 1 + 365 r 0.00018 = 365 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: What is the annual nominal rate compounded daily for a bond that has an APY of 6.8%. r m APY = 1 + −1 m r 365 0.068 = 1 + −1 365 r 365 1.068 = 1 + 365 √ 365 r 1.068 = 1 + 365 r 0.00018 = 365 r = 0.0658 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: What is the annual nominal rate compounded continuously has the same APY as 6% compounded monthly? r m APY = 1 + −1 m university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: What is the annual nominal rate compounded continuously has the same APY as 6% compounded monthly? r m APY = 1 + −1 m 0.06 12 APY = 1+ − 1 = 0.0617 12 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: What is the annual nominal rate compounded continuously has the same APY as 6% compounded monthly? r m APY = 1 + −1 m 0.06 12 APY = 1 + − 1 = 0.0617 12 APY = er − 1 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: What is the annual nominal rate compounded continuously has the same APY as 6% compounded monthly? r m APY = 1 + −1 m 0.06 12 APY = 1 + − 1 = 0.0617 12 APY = er − 1 0.0617 = er − 1 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: What is the annual nominal rate compounded continuously has the same APY as 6% compounded monthly? r m APY = 1 + −1 m 0.06 12 APY = 1 + − 1 = 0.0617 12 APY = er − 1 0.0617 = er − 1 ln(1.0617) = r university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: What is the annual nominal rate compounded continuously has the same APY as 6% compounded monthly? r m APY = 1 + −1 m 0.06 12 APY = 1 + − 1 = 0.0617 12 APY = er − 1 0.0617 = er − 1 ln(1.0617) = r r = 0.05987 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: Which is the better investment and why: 9% compounded quarterly or 9.3% compounded annually? university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: Which is the better investment and why: 9% compounded quarterly or 9.3% compounded annually? m 4 r 0.09 APY1 = 1 + −1= 1+ − 1 = 0.09308 m 4 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: Which is the better investment and why: 9% compounded quarterly or 9.3% compounded annually? m 4 r 0.09 APY1 = 1 + −1= 1+ − 1 = 0.09308 m 4 APY2 = 0.093 university-logo Jason Aubrey Math 1300 Finite Mathematics
• Compound Interest Continuous Compound Interest Growth and Time Annual Percentage Yield Example: Which is the better investment and why: 9% compounded quarterly or 9.3% compounded annually? m 4 r 0.09 APY1 = 1 + −1= 1+ − 1 = 0.09308 m 4 APY2 = 0.093 The ﬁrst offer is better because its APY is larger. university-logo Jason Aubrey Math 1300 Finite Mathematics