HW4 sol

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Economics

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Apr 3, 2024

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SE261 HW4 02-17-2022 Siddh Saraogi Section G Effective Annual Interest Rate, I =(1+r/m)^m−1 where: r=Nominal interest rate m=Number of periods Effective rate Nomina l rate Compoundin g annually Compoundin g semi- annually Compoundin g quaterly Compoundin g monthly Compoundin g daiy Continuous Compoundin g 6% 6% 6.09% 6.13% 6.16% 6.18% 6.18% 8.4% 8.4% 8.58% 8.66% 8.73% 8.76% 4.21% 4.21% 4.25% 4.27% 4.3% 6.97% 7.2% 7.22% 7.22% Problem #2 a) You have $35,000 and want to buy a house. After a search you find your dream house but the owner is asking for $80,000. The owner will not bargain but he is willing to wait for you to get the money together. Your only choice is to put the money in the bank with a 10% interest rate. How many years do you have to wait to buy your dream house if the interest is compounded – 1) Quarterly 2) Monthly 3) Continuously b) If the seller says that he will only wait a maximum of 4 years, how much money do you need to put in your bank account today for the same 10% interest rate when compounded – 1) Quarterly 2) Monthly 3) Continuously to purchase the house in 4 years? Using the formula A=P(1+r/4)^4m   where    A=future value   P=present value       r=rate of interest t=time period. 80,000=35000*(1+0.1/4)^4t (80,000/35000)=(1+0.025)^4t
2.28571429=(1.025)^4t Taking log on both sides; log 2.28571429=4t*log (1.025) 0.359021943=4t*0.0107238654 t=1/4*(0.359021943/0.0107238654) =8.37 years b. Using the formula A=P(1+r/12)^12n   where    A=future value   P=present value       r=rate of interest t=time period. 80,000=35000*(1+0.1/12)^12t (80,000/35000)=(1+0.00833333333)^12t 2.28571429=(1.00833333333)^12t Taking log on both sides; log 2.28571429=12t*log (1.00833333333) 0.359021943=12t*0.00360412427 t=1/12*(0.359021943/0.00360412427) =8.30 years c.We use the formula:   A=P(e)^rt   where    A=future value   P=present value       r=rate of interest t=time period.    e=2.71828 80,000=35000*(2.71828)^(0.1t) (80,000/35000)=(2.71828)^(0.1t) 2.28571429=(2.71828)^(0.1t) Taking log on both sides; log 2.28571429=(0.1t)*log (2.71828)
0.359021943=(0.1t)*0.43429419 t=1/0.1*(0.359021943/0.43429419) =8.27 years 2.a.Using the formula A=P(1+r/4)^4t  where    A=future value   P=present value       r=rate of interest t=time period.    80,000=P*(1+0.1/4)^(4*4) 80,000=P*(1+0.025)^(16) 80,000=P*(1.025)^(16) P=80,000/(1.025)^(16) =80,000*0.673624934 =$53889.99 b.Using the formula:   A=P(1+r/12)^12t   where    A=future value   P=present value       r=rate of interest t=time period.    80,000=P*(1+0.1/12)^(12*4) 80,000=P*(1+0.00833333333)^(48) 80,000=P*(1.00833333333)^(48) P=80,000/(1.00833333333)^(48) =80,000*0.671431999 =$53714.56 c.We use the formula A=P(e)^rt where    A=future value   P=present value       r=rate of interest t=time period.    e=2.71828 80,000=P*(2.71828)^(0.1*4)
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