![Interest
• % interest
• Time over which it is compounded
– Day, Week, Month, quarter or year
• Two types of interest
– Simple Interest – rarely used
– Compound Interest
• Be careful with interest
– Credit card statement 1.9% per month = 22.8% per
year simple interest, IS=ni
– Credit card statement 1.9% per month = 25.34% per
year compound interest, IC=[(1+i)n
-1]](https://image.slidesharecdn.com/7-l3-timevalueofmoney-241115121351-1e27bf91/85/7-L3-Time-Value-of-Money-and-interest-ppt-3-320.jpg)
![Annuity
• Series of Single payments, A, made at fixed time periods
• Examples – Installment Loans
– Student Loan Repayment
– Mortgage Loan
– Car Loan
– Retirement – old system
• Assumes periodic Compound Interest and payment at end
of first period
– discrete uniform-series compound-amount factor
– F=A[(1+i)n
-1]/i
• Present Worth of Annuity
– P=F/(1+i)n](https://image.slidesharecdn.com/7-l3-timevalueofmoney-241115121351-1e27bf91/85/7-L3-Time-Value-of-Money-and-interest-ppt-9-320.jpg)
![Retirement Annuity
• Monthly payments into 401k Account $200/mo at
5%/y interest. After working 25 years, what is
value?
• A= 12*$200
• N=25
• i=0.05
• F=A[(1+i)n
-1]/i= $1,145,000
• Present value of all that investment on your first day of work
• P=F/(1+i)n
=$33,830](https://image.slidesharecdn.com/7-l3-timevalueofmoney-241115121351-1e27bf91/85/7-L3-Time-Value-of-Money-and-interest-ppt-15-320.jpg)
![Future Value
• A=Rs. 5691.60
• i=15%/yr
• N=9yr
• F=A[(1+i)n
-1]/I = Rs 1,155,62.25
• Use this where inflation is figured into the
annual maintenance cost of pumps](https://image.slidesharecdn.com/7-l3-timevalueofmoney-241115121351-1e27bf91/85/7-L3-Time-Value-of-Money-and-interest-ppt-22-320.jpg)
7-L3-Time Value of Money and interest.ppt
- 1. Time Value of Money Chapter 23.5 -23.9 ChEn 4253 Terry A. Ring
- 2. Examples of Time Value of Money • Saving Account – Interest increases the amount with time • Loan – Payment amount • Retirement Annuity – Pays out constant amount per month – Pays out an amount that increases with inflation per month
- 3. Interest • % interest • Time over which it is compounded – Day, Week, Month, quarter or year • Two types of interest – Simple Interest – rarely used – Compound Interest • Be careful with interest – Credit card statement 1.9% per month = 22.8% per year simple interest, IS=ni – Credit card statement 1.9% per month = 25.34% per year compound interest, IC=[(1+i)n -1]
- 4. Some Nomenclature • F= Future value • P=Present value • i= interest rate for interest period • r=nominal interest rate (%/yr) • ny= no. of years • n= no. of interest periods
- 5. Interest • Simple interest – F=(1+n*i)P • Compound Interest – F=(1+i)n P • Allows present or future value to be determined • Can be inverted to give present value associated with a discount factor • Nominal Interest (simple interest when period is not 1 yr) – r =i*m • m= periods per year • Effective Interest Rate (compound interest when period is not 1 yr) – ieff= (1+r/m)m -1 • Continuous Compounding – ieff==exp(r) - 1
- 7. Present Value/Future Value • Determine the Present Value of an investment (or payment) in the Future. – You are due a $10,000 signing bonus to be paid to you after you have completed 2 yrs of service with your new company. What is the present value of that bonus given 7% interest? • Determine the Future Value of an investment made today – What is $10,000 worth if kept in a bank for 10 years at 3%/yr (compound) interest – Present value of retirement fund is $300,000. What will it be worth when I am 64 years old.
- 8. Student Loan • Get $10,000 in August 2009. Collects interest at 5% until graduation August 2013. What amount do you owe upon graduation? • F=(1+i)n P =(1+0.05)4 $10,000=$12,160
- 9. Annuity • Series of Single payments, A, made at fixed time periods • Examples – Installment Loans – Student Loan Repayment – Mortgage Loan – Car Loan – Retirement – old system • Assumes periodic Compound Interest and payment at end of first period – discrete uniform-series compound-amount factor – F=A[(1+i)n -1]/i • Present Worth of Annuity – P=F/(1+i)n
- 10. Annuity Types • Mix and match interest and payment schedules • Compound Interest – Discrete – monthly, quarterly, semi-annually annually – Continuous • Payments – Discrete – monthly, quarterly, semi-annually, annually – Continuously
- 11. Annuity Table i=r/m=periodic interest rate, A = payment per interest period, n=mny number of interest periods, Ā=pÂ=total annual payments per year, p=payments per year, r nominal annual interest rate.
- 12. See Article • Engineering Economics-FE Exam.pdf
- 13. Payment for Student Loan • Loan amount =$12,160 • What is payment if annual interest rate is 5% and loan is to be paid off over 10 years using monthly payments? • Do this for practice example for practice. Answer is $128.98 (see next slide) • Principle is being charged interest each month • Each payment pays interest and lowers principle so interest is less • Fix payment – Shifts from mostly paying interest to – Mostly paying principle as time goes on
- 15. Retirement Annuity • Monthly payments into 401k Account $200/mo at 5%/y interest. After working 25 years, what is value? • A= 12*$200 • N=25 • i=0.05 • F=A[(1+i)n -1]/i= $1,145,000 • Present value of all that investment on your first day of work • P=F/(1+i)n =$33,830
- 16. Compare two alternative pumps Pump A Pump B Installed Cost $ 20,000.00 $ 25,000.00 Yearly maintenance $ 4,000.00 $ 3,000.00 Service Life (yr) 2 3 Salvage Value $ 500.00 $ 1,500.00 Interest Rate 6.8% 6.8% Life of Plant (yr) 6 6
- 17. Determine Present Value • Each Purchases • Each Sale of Salvage Equipment • All Annual Payments to for Maintenance • Add them up – Purchases are negative – Sales are positive
- 18. Pump A Pump B Installed Cost 20,000.00 $ 25,000.00 $ Yearly maintenance 4,000.00 $ 3,000.00 $ Service Life (yr) 2 3 Salvage Value 500.00 $ 1,500.00 $ Interest Rate 6.8% 6.8% Life of Plant (yr) 6 6 Calculation of Present value of future purchases (-) and sales (+) of salvage equipment 1st Pump (20,000.00) $ (25,000.00) $ Purchase Price is present value Annual Maintenance (19,184.45) $ (14,388.34) $ Present Value of Annuity for Annual Maintenance 2nd Pump - Salvage (17,095.91) $ (19,290.97) $ Present value of future purchase 3rd Pump - Salvage (14,988.20) $ Present value of future purchase Slavage value 336.93 $ 1,010.80 $ Present value of future sale Total Present Value (70,931.63) $ (57,668.51) $
- 19. Uniform Gradient Rs = Rupee A= equivalent annual payment for an annuity
- 22. Future Value • A=Rs. 5691.60 • i=15%/yr • N=9yr • F=A[(1+i)n -1]/I = Rs 1,155,62.25 • Use this where inflation is figured into the annual maintenance cost of pumps