Cash-flow models.
r = annual interest rate
x = current investment
After n years, discrete annual compounding yields
Net value of the investment = NFV = Net Future Value
yn  x(1  r ) n
Examples. Compute NFV for a cash-flow x after every t
units of time for a finite period.
x
t=0
x
t=1
x
t=2
x
t=3
Interest rate: r per annum investment: x every year
Compute the worth of the investment after n years.
NFV of the cash-flow = NFV (stream) 
x(1  r )n 1  x(1  r )n  2  ...  x(1  r )  x
x
 (1  r ) n  1
r


If y dollar is available n years from now under a constant
annual compounding interest rate of r , present value of y
is
PV ( y)  y(1  r ) n  Discounted cash
2. Compute the current worth (the Net Present Value) of
the following cash-flow
x
t=0
x
t=1
x
x
t=2
t=3
NPV ( stream)  x  x(1  r )1  x(1  r ) 2  ...x(1  r )( n 1)
NFV ( stream)

(1  r )( n 1)
3. Continuous compounding/discounting
Annual interest rate = r
If q units of time in the year, effective interest rate per unit
r
time =
q
Since there are q periods in a year, x invested now has a
Future Value after 1 year
q

r
FV ( x)  x 1  
q

when q   , FV ( x)  xer
r
r
Proof: Let z  (1  )q . Then ln z  q ln(1  )
q
q

r r
r
Since  0 as q   , ln  1   
q q
q

ln z  r  z  er
Thus,
and ,
FV ( x)  xer for continuous compounding
PV ( y )  xe r for continuous discounting
For a horizon of n years,
 Continuous compounding
FVn ( x)  xenr
PVn ( x)  xe nr  Continuous discounting
4. For a continuous stream of cash-flow with continuous
discounting:
L
NPV ( stream)   x(t )e  rt dt
Return
s
0
t=L
t=0
Time
Similarly for future values.
5. Internal rate of return for an investment.
 = IRR (Internal rate of Return) if NPV of a cash-flow
using  interest rate is zero.
X
y1
t=0
t=1
y3
y2
t=2
t=3
NPV ( stream)   x  y1 (1   ) 1  y2 (1   )  2  ...  0
This means,
X
n 1
 yk (1   ) k
k 1
Given two investments, we should
■ choose the one that gives higher
rate of return 
■    * , some threshold parameter
6. Equivalent average rate of return
NPV ( x, )  Net Present Value of a policy x for some

Define R 

1
NPV ( x, )
R is the equivalent average rate of return.
An infinitely long cash-flow with return R after every t has
same NPV as that of the policy x.
R
R
R
R
NPV(Equivalent Stream) =
R  R(1   )1  R(1   ) 2  ...    R
= NPV ( x, )
1

7. NPV of an infinite stream (continuous discounting
model)
After every L units of time, return is Rt
NPV(infinite stream) =
 Rt et
t
If all Rt are same, say, R ,
NPV =
R
1  e L
8. A typical model. An equipment replacement policy, or
variations of this!
R(t ) revenue generated at time t
E (t ) operating expense on machine at time t
S(L) salvage value of machine at time L
 cost of a new machine
r continuous rate of discount
A. Single replacement model. NPV of replacing the
machine at time L.
Salvage
Value
Time
Operating
cost
Time
NPVsin ( L) 
L
 rL
 rt
 R(t )  E (t )  e dt + S ( L)   e
0
L must be such that NPVsin ( L) is maximum at L* , i.e.
d
NPVsin ( L)  0
dL
L*
B. Infinite chain replacement policy. Assume no
technological change.
Operating
Expense
Age
L
2L
3L
Assume R constant. Then
L
NPV ( L) 
 Re
 rt
0
1 e
t
dt
 rL
 rt
 rL
 E (t )e dt  (  S ( L))e
0
1  e rL
The second term C (L) is the cost term, and the first term
R
reduces to .
r
Approximation. Assume, E (t )  E0  t , both E0 and 
positive.
Then
E0    S ( L)e rL  Le  rL
C ( L) 
 

 rL
r r2
r 1  e rL
1 e
If S (L) is a constant, optimality condition is
(  S ) 

r
L 
*

r
2
(1  e
 rL*
)0
For rL*  1
1
 S ) 2
 2(
L*  




New machine at time t  0 , and then after every L units of
time.
C. Technological change over time
In this case, we might take the cost function to be like
E (t )  E0  t  L
This is Terborgh model. {G. Terborgh, “Dynamic
Equipment policy, McGraw Hill, 1949}
Optimality condition for S (L)  constant
(  S ) 
(   ) * (   )
 rL*
L 
(
1

e
)0
2
r
r
For rL*  1,
1
 S ) 2
 2(
L*  






9. Model for keeping a lossy capacitor charged. Dynamic
memory recharging system.
A normal capacitor loses its charge exponentially; so does a
memory cell or a pixel on a monitor-screen. Single
capacitor discharge:
q  q0 e  kt
Voltage
across
Capacitor
Time
How about injecting charge after every T units of time so
that the level is maintained above some threshold?
Charge
Time
Suppose q0 is injected after every T units. Therefore,
q (T )  q0  q0 e  kT
q(2T )  q0  q (T )e  kT = q0  q0 e  kT  q0 e  2 kT
…
q(nT )  q0 {1  e  kT  e  2 kT  ...  e nkT }
At t   , the amount of charge on the cell, is
q c  q0
want to maintain.
1
1  e  kT
.
q c is the desired level we