1. Future Value: The value at a future date of dollar today Period 0Period 1Period 2 $1 $1*(1+r){$1*(1+r)}*(1+r) $1*(1+r) 2 Rate of time preference: proportional increase in income required next period to compensate for the loss of $1 today :personal interest rate Compounding: sequential computation of future value

2. Future Value: The value at a future date of dollar today Period 0Period 1Period 2Period 3Period T $1 $1*(1+r)$1*(1+r) 2 $1*(1+r) 3 $1(1+r) T r = 0.10; T=40 $1 $1*(1.1)$1*(1.1) 2 $1*(1.1) 3 $1(1.1) 40 $1 $1.10$1.21 $1.33$45.26

3. Future Value: The value at a future date of dollar today Period 0Period 1Period 2Period 3Period T $1 $1*(1+r)$1*(1+r) 2 $1*(1+r) 3 $1(1+r) T r = 0.10; T=40 $1 $1*(1.1)$1*(1.1) 2 $1*(1.1) 3 $1(1.1) 40 $1 $1.10$1.21 $1.33$45.26

4. Present Value: The current value of a stream of future income Period 0Period 1 $1 $1*(1+r) $1/(1+r)$1

5. Present Value: The current value of a stream of future income Period 0Period 1Period 2 $1 $1*(1+r) $1*(1+r) 2 $1/(1+r) 2 $1/(1+r)$1 r = 0.1 $1/(1.1) 2 $1/(1.1)$1 $0.83$0.91$1

6. Present Value: The current value of a stream of future income Period 0Period 1Period 2Period 3Period T $1 $1*(1+r)$1*(1+r) 2 $1*(1+r) 3 $1(1+r) T $1/(1+r) T $1 r = 0.10; T=40 $1/(1.1) 40 $1 $0.022$1

7. Present Value: The current value of a stream of future income Stream of $1 every year for T periods, starting next period T Σ t=1 1/(1+r) t = $1/(1+r) + $1/(1+r) 2 + $1/(1+r) 3 + … + $1/(1+r) T < T

8. Present Value: The current value of a stream of future income Stream of $1 every year for T periods, starting next period T Σ t=1 1/(1+r) t = $1/(1+r) + $1/(1+r) 2 + $1/(1+r) 3 + … + $1/(1+r) T < T If r = 0.1; T = 40; Present Value = $9.78 If r = 0.05; T=40; Present Value = $17.16

9. Present Value: The current value of a stream of future income $2000 furnace saves $200/year Good deal?

10. Present Value: The current value of a stream of future income $2000 furnace saves $200/year Good deal? T Σ t=1 1/(1+r) t If r = 0.1; T = 40; Present Value = $1956 Breakeven: NEVER If r = 0.05; T=40; Present Value = $3432 Breakeven: 15 years