H & Q CH 12 Optimization over time.ppt

optimization over time Henderson , Quandt , CH 12 1
Introduction
So far we have optimization for a single period of time .the consumer
would spend all his money during the current period of time. in other
words the consumer would maximizes his utility index defined only
for goods consumed during the current period of time .
Time is introduced in both discrete and continuous terms . In this way
multi-period utility and production functions are defined .
We should have some simplified assumption as follows ;
1- the periods are equal in length.
2- market transaction is limited to the first day of each period.
3- during the remaining days ; consumers supply the factors they have
hold and consume the commodities they have purchased , and
entrepreneurs apply the inputs they have purchased and produced
the commodities for sale on the next marketing date.
In this way the current consumer expenditure is no longer bounded by
a single period budget constraint. He may spend more or less than
his current income and borrow or lend the difference .

Basic concepts
Assumptions;
1- consumers and entrepreneurs are free in borrowing and lending
which takes place on the first day of each period .
2- only one type of credit instrument , bonds , with a one period
duration is available.
3- borrowers sell bonds to lenders in exchange for specified amounts of
current purchasing power. If bt is the bond position (amount of
saving ) of individual at the end of trading on the t th marketing date,
then
if∆bt = 0 ( bt = bt-1 , saving =0 ), the individual is neither a borrower nor
a lender
if ∆bt <0 (bt < bt-1 negative saving), the individual is a borrower. He
would borrow ∆ bt =bt and he must pay bt plus the borrowing fee in
the next period (t+1). bt+bti=bt(1+i)
if ∆bt >0 (bt > bt-1 positive saving), the individual is a lender. He would
lend ∆ bt = bt and he must receive bt plus the borrowing fee in the
next period (t+1). bt+bti=bt(1+i)

Basic concepts
Market rate of interest.
Suppose that a consumer receives bt on the tth marketing date and
continues to reinvest both principal and the interest until the Ť
marketing date.
t=0→→→→→t=t→→→→→→t=Ť→→→→t=T
bt= value of consumer’s investment in the beginning of tth marketing
date.
b t+bt it =bt (1+it) = value of consumer’s investment in the beginning of
( t+1) marketing date
bt (1+it) +[ bt (1+it) ] it+1 = bt(1+it)(1+it+1)= value of investment in the
beginning of (t+2) marketing date.
bt(1+it)(1+it+1)……(1+iŤ-1)= value of investment in the beginning of the Ť
marketing date.
J= bt(1+it)(1+it+1)……(1+iŤ-1) – bt = total return on investment .
εtŤ=(J/bt)=(dJ/dbt)= [ (1+it)(1+it+1)……(1+iŤ-1) – 1 ] = average rate of
return= marginal rate of return.
εtt = 1 -1 =0 , εtt+1 = ( 1 + it ) -1= it
If it = it+1 = it+2 = it+3 =….. =iŤ-1 = i , then εtŤ = (1+ i)Ť-t – 1 ;
as it is seen the level of interest rates and not the order of their
sequence affects the rate of return

Basic concepts
Discount rate and present value
Rational consumer will not consider one dollar payable on the current
marketing date equivalent to one dollar payable on some future
marketing date. Because , consumption at present time has more
utility than consumption in future. Delaying consumption will lower
the utility (assuming constant price and no inflation) .
Discount rate payable on the tth ( t=1→→→t=t ) marketing date
(beginning from t=1) is equal to ;
[ (1+i1)(1+i2)…….(1+it-1) ] -1 = ( 1+ ε1t)-1
Investing one dollar in first marketing date (t=1) will result in ( 1+ ε1t)
dollars in the tth marketing date. in other words ( 1+ ε1t)-1 dollars in
the tth marketing date worth one dollar on the first marketing date
In the same way the present value of the income stream (y1,y2,y3,,,yŤ)
would be equal to Y= y1+[y2/(1+ε12)]+[y3/(1+ε13)]+….+[yŤ/(1+ε1Ť)]
If all the interest rates are positive then denominator will increase and
the discounted values would decrease .

Multi-period consumption
Multi-period utility function
U=U(q11,,,qn1 , q12,,,,qn2 , q13,,,,,qn3 ,,,,,, q1T,,,,,qnT)
qjt= quantity of qj consumed during the tth period .
Actual and expected commodity prices are fixed in values and
remained unchanged.
Ct= Σj=1
n pjtqjt = consumer total expenditure for commodities consumed
on the tth marketing date. t = 1,2,3,4,,,T.
Redefine the utility function as follows ;
V=V(c1, c2 , c3 ,,,,cT) indirect utility function
dv= [∂v/∂cŤ]dcŤ + [∂v/∂ct]dct= vŤdcŤ + vtdct =0 (indifference locus )
(vt/vŤ)=-(dcŤ/dct) = the rate of return at which consumption expenditure
on the Ť th marketing date must be increased to compensate for a
reduction of consumption expenditure on the tth marketing date to
leave the consumer satisfaction level unchanged during the time
horizon. This is called time substitution rate

For example if (dcŤ/dct)=-1.06 , it means that if consumption reduces by
one unit in tth marketing date, the consumption should increase by
0.06 in Ťth marketing date to keep satisfaction level unchanged. One
dollar worth of consumption in the tth period worth 1.06 dollar worth
of consumption in Ťth period. Or, the rate of time preference
between tth and Ťth marketing date is 0.06. This is the minimum
premium which is needed for the satisfaction level to remain
unchanged.
ηtŤ = (vt/vŤ) -1=-(dcŤ/dct) -1 =1.06 -1=0.06 =rate ot time preference ,
Ť>t , t,Ť=1,2,3,,,T
ηtŤ is usually positive except for some unusual cases in which the
expected worth (utility) of consumption expenditure is much larger in
Ťth marketing date comparing to tth marketing date (Ťth > tth ).
The consumer subjective rate of time preference are derived from his
utility function and depend upon his indirect utility function and the
level of his consumption expenditures. This is independent of the
market rate of interest and his borrowing and lending opportunities .

Multi period budget constraint
The consumer expect to receive the earned income stream (y1,,,y Ť ) on
the marketing dates within his panning horizon.
The consumer’s total income receipts on the tth marketing date are
some of his earned income and his interest income from bond held
during the preceding period and is equal to (yt+it-1bt-1)
Anticipated saving on the tth marketing date=St=(yt+it-1bt-1) - ct , t=1,,,,T
At the beginning of his earning life assume that, b0=inherited wealth =0
At the same manner bt = bt-1 + st , t=1,2,3,4,,,,T.
At early years of earnings ,the individual is indebt,( rising family, buying
home, low income earnings)
In the middle years , he saves to retire his debts. He establish a
positive bond position.
Finally in the old ages he spend the saving and liquidate his bonds

b1=y1 – c1
b2 = (y1 – c1)(1+i1) +( y2 - c2 )
b3 = ( y1 – c1)(1+i1)(1+i2) + (y2 – c2 )(1+i2) + y3 – c3
b4=(y1– c1)(1+i1)(1+i2)(1+i3)+(y2–c2 )(1+i2)(1+i3)+(y3–c3)(1+i3)+(y4 – c4)
……………………………………………………………………………….
bŤ=Σt=1
Ť (yt – ct)(1+εtŤ) Ť=1,2,3,,,,,,,T
εtŤ = (1+it)(1+it+1)(1+it+2)……..(1+iŤ-1) -1 = (1+i)Ť-t -1 , if it=it+!=….iŤ-1=i
Assuming that the individual is planning to leave his heir neither asset
nor debt . The budget constraint could be written as follows
bT = Σt=1
T (yt –ct) (1+εtŤ)=0,
Divide both sides by (1+ε 1Ť) and moving the consumption term to the
right , since (1+εtŤ)/(1+ ε1Ť)=(1+ε1t)-1, we get
Σt=1
Tyt(1+ε1t)-1 = Σt=1
T ct(1+ε1t)-1 ,
Discounted income stream = Discounted consumption stream

The consumption plan
V* = v(c1,,,,,cT) + µΣt=1
T(yt –ct)(1+ε1t)-1
F O C
∂v*/∂ct = vt - µ(1+ε1t)-1 =0 t=1,2,3…..T
∂v*/∂µ = Σt=1
T(yt –ct)(1+ε1t)-1 =0
(vt/vŤ)= -∂cŤ/∂ct = (1+ε1t)-1/ (1+ε1Ť)-1 = 1+εtŤ t,Ť= 1,2,3,4…T , Ť>t
ηtŤ = - (∂cŤ/∂ct) -1 → - (∂cŤ/∂ct)= 1+ ηtŤ
ηtŤ = εtŤ
rate of time preference between every pair of period (ηtŤ )= market
rate of return between the same pair of period (εtŤ )
ηtŤ < εtŤ → he will buy bonds
ηtŤ > εtŤ → he will sell bonds
The second order condition confirms that the utility function v(c1,,,,,cT)
should be regularly strictly quasi-concave . In other words the rate of
time preference should be decreasing .

Suppose that there are two period horizon; t=1,2
y0=y1 + y2 (1+i1)-1 present value of income stream
y0=c1 + c2 (1+i1)-1 constraint
u = time indifference curve
A = initial position B=consumption point
slope of u = -(1+η12)=∂c2/∂c1=
rate of time preference
c1
c2
y0=y1 + y2 (1+i1) -1
Slope=-(1+i1)
u1
B
A
C
consumer will buy AC worth of bond
on the first marketing date and will
receive and spend CB( principal and
interest )for the consumption good on
the second day .
initial position

Substitution and Income effect
V* = v(c1,c2) + µ[(y1 – c1) +(y2 – c2)(11+i1)-1]
F .O .C.
∂v*/∂c1 = v1 - µ = 0
∂v*/∂c2 = v2 - µ(1+i1)-1=0
∂v*/∂µ = (y1 – c1) +(y2 – c2)(11+i1)-1 = 0
Differentiate totally from the first order condition
v11 v12 -1 dc1 0
v21 v22 -(1+i1)-1 dc2 = - µ (1+i1)-2 di
-1 -(1+i1)-1 0 dµ -dy1 – (1+i1)-1dy2+(y2-c2)(1+i1)-2di1
Using Cramer’s rule ;
dc1=(0)(D11/D)-µ(1+i)-2di1(D21/D)+[-dy1–(1+i1)-1dy2+(y2-c2)(1+i1)-2di1]D31/D (I)
Where Dij is the cofactor of element in ith row and jth column
If di1≠0 , and dy1 = dy2 = =0 , then ;
∂c1/∂i1 = total effect = -µ(1+i1)-2D21/D + (y2 – c2 )(1+ i1)-2D31/D (II)
Y = y1 +y2(1+i1)-1 , present value of the consumer’s earned income .
An increase in y1 by one or y2 by (1+i1) would increase Y by one . So;

from (I)→ dc1/dy = dc1/dy1 = (1+i1)-1 dc1/dy2 = -D31/D (if di1=0)
consider those changes by i which are accompanied by changes in c1 and c2 in such a
way that the level of consumer’s utility index remain unchanged, that is ;
→ du = v1dc1 + v2dc2 =0
from the first order condition we have → v2/v1 = (1+i1)-1 = - dc1/dc2 so;
→-dc1 - dc2 (1+i1) -1=o
Taking total differential from the third equation of first order condition
dy1 –dc 1+dy2(1+i1)-1 -dc2(1+i1)-1 - (1+i1)-2di1y2 + (1+i1)-2 di1c2 = 0
dy1 +dy2(1+i1)-1 - (1+i1)-2di1y2 + (1+i1)-2 di1c2 = 0
dy1 + dy2(1+i1)-1 - (1+i1) -2 di1 (y2 – c2)=0, or
- dy1 - dy2(1+i1)-1 + (1+i1) -2 di1 (y2 – c2)=0, substitute this in (I)
dc1= -µ(1+i1 )-2di1(D21/D) ( when utility does not change )
Substitution effect = (∂c1/∂i1)u=cons.= -µ(1+i)-2(D21/D) <0
Multiply the budget constraint by (1+i1 ) -1 , we get
-(y1 – c1) (11+i1)-1=(y2 – c2)(11+i1)-2 and substitute in (II) we get
total effect = ∂c1/∂i1 = - µ(1+i1)-2D21/D + (y1– c1 )(1+ i1)-1 (∂c1 /∂y1)i=cons. =
(∂c1/∂i1)u=cons +(y1– c1 )(1+ i1)-1 (∂c1 /∂y1)i=cons =
Substitution effect + income effect
Substitution effect = the effect of change in interest rate (price) on c1 holding utility
constant
Income effect = the income effect of change in interest rate on c1 . Purchasing power
effect of change in interest rate which cause chance in utility level.

(∂c2/∂i1)u=cons.= -µ(1+i1)-2D22/D >0 , µ>0, D22=-1<0 , D>0
i1↑→c2↑(c1 ↓ ) , more expensive consumption has not been chosen.
As it seen the substitution effect is negative , but the sign if income
effect is not clear beforehand ;
If (∂c1 /∂y1)i=cons >0 , except for extraordinary cases , the direction of
income effect is determined by the sign of consumer bond position
(y1 – c1);
If (y1 – c1)>0 , then ; income effect = (y1– c1 )(1+ i1)-1 (∂c1 /∂y1)i=cons >0,
and if │income effect│< │substitution effect│ → total effect =
∂c1/∂i1 <0 (when i↑ then c1↓ , saving↑ ), other wise ∂c1/∂i1 >0,
If (y1 – c1)<0 , then ; income effect = (y1– c1 )(1+ i1)-1 (∂c1 /∂y1)i=cons <0,
and total effect = ∂c1/∂i1 <0 (when i↑ then c1↓ , borrowing↓ ), ,

c1
c2
y0=y1 + y2 (1+i1) -1
Slope=-(1+i1)
B
A
C
initial position
Substitution and income effect

Investment theory of the firm
Generally time must elapse between the application of inputs and
securing the inputs. There are few assumptions ;
1- the entrepreneurs buys input and sells output only on marketing
dates within his horizon.
2- he performs the technical operation of his production process in the
time between marketing dates.
3- during the period t , he applies the inputs he has bought on the t th
marketing dates .
4- on the (t+1) th marketing date he sells the output which he secured
during the t th period .
5- input and output prices are fixed , so investment expenditures and
revenues from sale on each of the marketing dates acts as decision
variables.

Different cases can be formulized;
1- point –input point-output . All inputs are purchased in one
marketing date and all outputs are sold on a subsequent period .
Like winger aging or tree growing.
2- multipoint-input point-output . The production of an output which
requires application of inputs during a number of successive
periods. Like shipbuilding.
3- point-input multipoint-output. It is like investment in durable goods
which is purchased on one marketing date and is used for the
production of outputs during a number of successive periods.
4- multipoint-input multipoint-output . It is like investment for
replacement of durable goods during a time horizon.

The multi-period production function
Entrepreneurs plan his production process for the horizon of L
complete periods and L+1 marketing dates ;
F( q12 , ….qs L+1 , x11 ,,,,xnL)=0 i=1,2,3,…n , j=1,2,3….s
qjt = quantity of jth output secured during the (t-1) period and sold on
the t th marketing date.
Xit = quantity of i th input purchased on the t th marketing date and
applied to the production process during the t th period .
On the (L+1) marketing date , the entrepreneur plans to sell the outputs
secured during the L th period , but does not plan to purchase input.
The inputs applied during each period contribute to the production of
outputs during all periods.

The investment opportunity function
present and future prices are known .
input expenditure and output revenues are composite variables which
are related by an implicit investment opportunity function ;
H(I1 , I2 ,,,,,IL, R2 , R3 ,,,,RL+1 )=0
It = Σi=1
n rit xit investment on tth marketing date
Rt = Σj=1
s pjt qjt Revenue on tth marketing date
There are two kinds of investment opportunities for entrepreneurs ;
1- External investment opportunities → bonds
2- internal investment opportunities→ reinvesting
Each revenue depends upon all investment on any particular
marketing date, it is not possible to attribute the entire revenue on
the Ť marketing date to the investment on any particular marketing
date and in this manner average rate of return can not be calculated
. But marginal rate of return can be calculated.
Marginal rate of return from investment on the t th marketing date with
respect to revenue on the Ť th marketing date = ρtŤ =(∂RŤ/∂It) -1 =
-(∂H/∂It)/(∂H/∂Rt) -1
ρtŤ depends upon the level of all the planed revenues and investment
expenditures and it is independent of the market rates of interest.

The investment plan
Max Π* = Σt=2
L+1 Rt(1+ ε1t)-1 - Σt=1
L It(1+ ε1t)-1 +µH(I1, ,,,,,RL+1)
∂Π*/∂RŤ = (1 + ε1Ť)-1 + µ ∂H/∂RŤ = 0 Ť =2,3,4,,,,,L+1
∂Π*/∂It = - (1 + ε1t)-1 + µ ∂H/∂It = 0 t=1,2,3……L
∂Π*/∂µ = H(I1, ,,,,,RL+1) = 0
(∂H/∂It )/(∂H/∂RŤ )=[(1+i1)….(1+it-1)-1+1]-1 / [(1+i1)….(1+iŤ-1)-1+1]-1 =
1/ [(1+it)….(1+iŤ-1)-1+1]-1= [(1+it)….(1+iŤ-1)] = - (1+ εtŤ)
(∂H/∂It )/(∂H/∂RŤ)=-(1+ρtŤ)=-(1+ εtŤ) →ρtŤ=εtŤ optimum level of investment
Marginal internal rate of return (internal investment )=ρtŤ = εtŤ = market rate of
return (external investment )
t=1,2,3..L , Ť= 2,3,4,,,,L+1
ρtŤ < εtŤ , entrepreneur buy bonds , contract investment
ρtŤ > εtŤ , entrepreneur sell bonds , expand investment
S . O. C.
H11 H12 H1 H11 H12 H13 H1
H21 H22 H2 < 0 → 2H1H2H12 – H22H1
2 – H11H2
2 <0 H21 H22 H23 H2 > 0
H1 H2 0 H31 H32 H33 H3
H1 H2 H3 0

Calculating the ∂ρtŤ/∂It we will find that it is equal to ;
-1/(H2
3)(-2H1H2H12 + H22H1
2 + H11H2
2 ) = ∂ρtŤ/∂It
for having 2H1H2H12 – H22H1
2 – H11H2
2 <0 we should have ;
∂ρtŤ/∂It <0 , since H2>0 , that is ;
Marginal rate of return should be decreasing with respect to investment
as the result of second order condition .
Point-input point-output investment .
Investment on one marketing date receives the resultant revenue on
the next. In other words the effective planning horizon includes only
one full period ; R2 = h(I1) I1 = investment expenditure
Average internal rate of return = (R2 – I1)/I1
Π = R2(1+ i1 )-1 - I 1 = h(I1 )(1+ i1 )-1 - I 1

∂Π/∂I1 = h’(I1)(1+i1)-1 -1 =0
h’(I1) = (1+i1)
Revenue in the second period is equal to the amount of investment
plus the return to investment.
R2 = I1 + I1i1 = I1(1+i1)
(∂R2/∂I1) = (1+i1) = h’(I1)
ρ12 = ∂R2/∂I1 – 1 = h’(I1) – 1 → ρ12 + 1 = h’(I1)→ ρ12 + 1 =(1+i1)→
marginal rate of return between period 2 and 1= ρ12 = i1 =interest rate= ε12
Second order condition ; (∂2Π/∂I1
2)=h”(I1)(1+i1)-1 <0 → if i1> 0 → h”(I1)<0
h”(I1) = Marginal Internal Rate of Return is decreasing .
∂Π/∂I1 = h’(I1)(1+i1)-1 -1 =0 → h’(I1) = (1+i1) → h”(I1)dI1 – di1 =0 →
dI1/di1 =1/ h”(I1)<0 ,
If second order condition is satisfied , increase in interest rate will cause
investment expenditure to decrease .

Considering the following figure ;
I1 I1
$ $
A.R.R
M.I.R.R.
َََA.R.R.
M.I.R.R.
A.R.R. = (h(I1)-I1)/I1 = (R2 -I1)/I1
M.I.R.R. = h’(I1) -1
At point A → MIRR=i1
0 → Π is
max
I1
0
A
B
C
i1
i1
0
I1
0
I0 = Level of investment = I0
Interest
rate
S(0I0BC)=total return
S(oi1AIo)=total (opportunity) cost
S(i1ABC)=net return
In perfect competition
S(i1 ABC)=net return=0

Interest rate determination
Use of loan able fund rather than bonds could be treated as the
commodity for sale.
Demand for bonds is equivalent to a supply of loan able fund , and
supply of bonds is equivalent to demand for loan able fund . An
interest rate is the price of using loan able fund for specific period of
time .
An equilibrium current interest rate is the one for which the excess
demand for current loan able funds equals to zero. The equilibrium
interest rate reflects time preferences and productivity of investment.
In equilibrium the rate of time preference for each consumer and
marginal internal rate of return for each producer equals the interest
rate.
Excess demand for loan able funds by each consumer and
entrepreneur can be expressed as a function of the current and
expected interest rates. It is convenient to use excess demand
function , since entrepreneurs may demand loan able funds at one
interest rate and supply it at another one . If all the interest rates are
equal to i, then the excess demand would be a function of interest
rate

Investment theory and the role
of the firm .
Interest rate theory is characterized by the fact that time elapse between the
application of inputs and the attainment of the resultant output.
Continuous compounding and discounting.
We should note that time is continuous and transaction may take place at any
point in time.
If Wt is the interest of one dollar investment at the end of t years period ;
$1 at t=0→t=1[W1=(1+i)] →t=2 [W2= (1+i)2 ]→t=3 [W3=(1+i)3 ]→…t=t [Wt=(1+i)t ]
Wt = (1+ i) t
,, if interest rate compound once a year.
Wt = (1+ i/2) 2t
,, if interest rate compound twice a year , one half of i would
compound every six months .(1/2 of a year )
Wt = (1+ i/n)nt
,, if interest rate compound n times a year , 1/n of i would
compound every 1/n of a month.
Z=(1+i/n)nt ,→ Ln Z = nt Ln (1+i/n) → Ln Z =[Ln(1+i/n)] / (1/nt)
if n → ∞(continuous time)→Lim Ln Zn→∞ = 0/0
Using Hopital rule → Lim Ln Zn→∞ =(i)(t) → Lim Z n→∞ = eit
So the present value of u dollars payable at time t is equal to ue-it

of the firm .
Point and flow variables
Transaction may take place at any point in time and their values may
be the function of the time at which they occur.
R(T) =revenue (dollar value) , realized at time t =T
R(T)e-iT = present value of the revenue at time T
d[R(T)e-iT ]/dT = [R’(T) –iR(T)]e-iT = marginal discounted revenue with
respect to time.
We should note that inputs , outputs , costs, and revenues may be
realized as flow variables over time . Flow variables may occur at
constant rate over time or their rate may be a function of the time .
R=R(t) = rate of flow of income at instant t measured in dollars per
year.
R=R(T) denotes point value at time T .

of the firm .
R0T = ∫0
T R(T)e-it dt = present value of revenue stream R(T) from
t=0 to t=T
dR0T/dT = R(T) e-iT , marginal revenue of an income stream with
respect to time
Consider a income stream R(t) from t=0 to t=T , and a point value at T
with equal present value , RT e-iT = ∫0
TR(t)e-it dt , then ;
RT= ∫0
TR(t)e-i(T-t) dt , as it is seen a flow variable , R(t) , could be
converted into an equivalent point variable , R(T).
Consider a constant income flow , α , with present value equal to that of
a point value T , RT e-iT = ∫0
T αe-it dt = α ∫0
T e-it dt = α δ in which δ
equal to δ= [(1-e-iT)/i ] = ∫0
T e-it dt = present value of one dollar
income stream for T years .
Finding α from the RT e-iT = ∫0
T αe-it dt , and substituting from
[(1-e-iT)/i ] = ∫0
T e-it dt , we will find α = [i / (e-iT -1)] RT ,, which
provides a mean for converting a point value into an equivalent
constant flow.

of the firm .
Point –input Point-output
All inputs are applied at one point in time and all outputs are sold at a later
point in time. Winger aging .
I0 = cost for buying cask of grape juice . Frgmantation and aging is costless .
The only other cost is interest paid for I0
R(T) = sales value of winger at point T . T is the aging period .
Profit max→Π = R(T)e-iT – I0 , dΠ/dT = [R’(T)– i R(T)]e-it=0 (I)→ [R’(T)/R(T)] = i
[R’(T)/R(T)]= Proportionate rate of return with respect to time= i =
proportionate marginal rate of return with respect to time
S. O. C.→ d2Π/dT2 = [R”(T)-2iR’(T)+i2R’(T)]e-iT<0
Substituting from FOC →[ R”(T)R(T)–[R’(T)]2 ] / [R(T)]2 <0 →d[R’(T)/R(T)]/dT <0
Solving the first order equation we will get T=T0
If investment period = T0 , marginal earning from winger aging = earning from
investing R(T) in bond market.
If investment period < T0 , marginal earning from winger aging > earning from
If investment period > T0 , marginal earning from winger aging < earning from

of the firm .
Totally differentiate the first order condition we get;
R”(T)dT – iR’(T)dT – R(T)di =0
dT/di = R(T) / [R”(T) – i R’(T)] < 0 [R”(T) – i R’(T)] < 0
if i goes up it will force the entrepreneur to shorten his aging period
Continuous-Input Point-Output
Example ; tree growing , ship building .
Seedling cost = I0 (initial fixed cost )
Cultivation cost = G(t) per year , (variable cost during the investment
period)
Selling price of the tree at time t=T R=R(T)
Π=R(T)e-it – I0 - ∫0
TG(t)e-itdt =present value of profit
dΠ/dT = [R’(T) – iR(T) – G(T)]e-iT = 0
[R’(T) – G(T)] / R(T) = i →
proportionate rate of return net of cultivation cost=interest rate

of the firm .
Point-Input Continuous-Output
Investment in durable equipment which yields a revenue stream over
time . (swing machinery)
Suppose that the equipment yields revenue at a constant rate of R
dollars per year during its life.
I0 = I(T) = investment cost T =life time of machine .
Π = ∫0
T Re-itdt – I(T)
dΠ/dT = Re-iT – I’(T) = 0 → Re-iT = I’(T)
Present value of additional revenue from increased durability =
marginal cost of durability
S. O. C. → d2Π/dT2 = -iRe-iT – I”(T) < 0 → S.O.C. is satisfied if the
marginal cost of durability is increasing over time → I”(T)>0
Differentiating the first order condition →dT/di =[TRe-iT]/ [-iRe-iT-I”(T)] <0
if interest rate goes up (i↑)→ life time of machine should shorten (T)↓.

of the firm
Continuous–Input Continuous-output
In order to find out the mechanism of this type of investment we have to
illustrate some fundamental points.
1- retirement and replacement of durable equipment
A machine is used for the production ofa single output q which is sold
for the competitive price of P which is fixed .
I0 = purchased value of machine at time t=t
Ct = c(qt) , input cost is a function of production .
Mt = M(qt , t ) = maintenance cost , t=0,1,2,3,,,,,T
ST = S(T)=scrap value of machine at time T ,
S’(T)<0 = loss of market value from continuing to use machine.
Entrepreneur's optimization problem could be formulized into two parts;
First - determination of optimum input and output levels for each point
in time while machines are in operation .
Second - determination of optimal lives for one or more machine.

Investment theory and the role of the firm
2- quasi-rent function ,Z(t) function,
Entrepreneur decides to operate a machine from t=t0 to tT.
Optimization behavior at t=t is equal to maximize the present value of quasi
rent at t=t which is equal to present value of income at t=t minus cost at t=t. The
initial cost and scrap value is ignored .
Max Zt e-it = (pqt) e-it - c(qt) e-it - M(qt , t ) e-it
e-it can be cancelled from both sides, since optimization behavior at any point
in time is independent from the time which optimization take place.
∂Zt/∂qt = P – dc(qt)/d(qt) – dMt/dqt = 0 → P = dc(qt)/d(qt) + dMt/dqt
fixed rate of MR = P = dc(qt)/d(qt) + dMt/dqt = rate of increase in the flow of MC.
Solving the above relation for optimum qt as a function of time (t) , and
substitute it in the quasi-rent function , we will get the following relation;
Zt = Z(t) = maximum quasi-rent obtainable at each point in time from the
operation of machine .it is based upon the underlying optimal combination of
inputs and outputs .Zt holds for all values of t and its form is unaffected by the
selection of a particular value for machine life. Thus Zt can be used for The
analysis of machine life time without explicit introduction of revenues and cost
function . Since Zt would give the maximum level of profit as a function of the
time which they occurs.

of the firm
Retirement of a single machine
Max Π1 = [ ∫0
T Z(t)e-iT dt ] – I0 + S(T)e-iT , T= life time of machine ,
Π1 = present value of the profit stream for the first machine
dΠ1/dT = [ Z(T) – i S(T) + S’(T) ] e-iT = 0
Z(T) – i S(T) + S’(T) =0 → Z(T) + S’(T) = i S(T) F.O.C.
Z(T)= marginal quasi rent
S’(T) <0 , depreciation flow or marginal loss of scrap value
i S(T)= interest from investing the scrap value ,
S.O.C. ; d[ Z(T) – i S(T) + S’(T) ]/dT <0
d[ Z(T) + S’(T) ]/dT < d [i S(T) ]/dT S’(T)<0
Quasi-rent less depreciation flow decrease more rapidly than the
alternative bond-market return.

Replacement of a chain of machine (Continuous-Input Continuous-Output)
Infinite horizon , infinite chain of machine succeeding each other. Quasi-rent
function , initial cost, planned life of the machine and scrap value are the
same for each machine except for the dates of obtaining them. Πi = present
value from the operation of ith machine.
Π1=∫0
T Z(t)e-it dt – I0 + S(T)e-iT
Π2=∫T
2T Z(t-T)e-it dt – I0e-iT + S(T)e –i2T = Π1e-iT
Π3=∫2T
3T Z(t-2T)e-it dt – I0e-i2T + S(T)e-i3T =Π1e-i2T
……………………………………………………..
Πk=Π1e-i(k-1)T = [∫0
T Z(t)e-it dt – I0 + S(T)e-iT ]e-i(k-1)T
Π= Σk=1
∞ Πk = total profit from the chain of the machine
Π = Σk=1
∞ Πk = Π1(1 + e-iT + e-i2T +…. +e-i(k-1)T) , k →∞ Π = Π1[1/(1-e-iT)]
dΠ/dT = {[Z(T)–iS(T)+S’(T)]e-iT(1-e-iT)-ie-iT[ ∫0
T Z(t)e-it dt–I0 + S(T)e-iT]}/(1-e-iT)2
Multiplying the both sides by e-iT(1-e-iT) and rearranging the terms ,
Z(T) + S’(T) = (1/δ ) [∫0
T [Z(t)e-it dt – I0 + S(T)] ,
δ=(1-e-iT)/i=∫0
Te-itdt = present value of one dollar income stream for T years.
∫0
T Z(t)e-it dt – I0 + S(T)= present value of the return of new machine
( with life time equal to T years )net of its investment cost plus the scrap value
of the old machine.

Income stream per year fot T years present value of the investment
after T years.
one dollar δ
X [∫0
T Z(t)e-it dt – I0 + S(T)]
(1/δ ) [∫0
T Z(t)e-it dt – I0 + S(T)] = present value of the average return per
year of new machine net of its investment cost plus the scrap value of
the old machine.
[(Z(T) + S’(T)] = marginal rate of quasi-rent flow net of depreciation
machine is replaced when its marginal rate of quasi-rent flow net of
depreciation equals the present value of the average return per year of
new machine net of its investment cost plus the scrap value of the old
machine.
The first order condition in this case and one machine case are different in
the sense that;
In the one machine case , entrepreneur is looking for continuing to operate
the machine or investing its scrap value in the bond market. While in the
infinite number of machine case the entrepreneur is looking for operating
an existing machine or operating a new one .

Exhaustible resource
For example ; coal mines , oil well ..
The horizon of the extraction is n discrete time periods. Exhaustible
extraction is limited to a fixed aggregate extraction cost=C=c(qt)
Max V= Σt=1
n [ptqt – c(qt)](1+i)-t + λ(q0- Σt=1
nqt)
∂V/∂qt = [pt – c’(qt)](1+i)-t - λ =0
∂V/∂λ = q0 - Σt=1
n qt =0
[pt – c’(qt)](1+i)-t = λ , λ is the measure of scarcity
Present value of the difference between price and marginal cost for all
periods should be the same .
If pt = fixed , when time (t)↑ →(1+i)-t } ↓ → so we should have c’(qt)↓ →
qt ↓ , (if marginal cost is increasing as the result of second order
condition) .

Human Capital
Cost of education ;
1- direct cost ; like teacher’s salaries, textbook expenditures ,..
2- opportunity cost of earning forgone during studying ;
There is three questions ;
1- yes or no decision for continuing higher education
2- cost of training workers
3- optimal investment in human capital over an earning cycle
Investment in education
The question is whether to continue the education or enter the labor
market. Two alternative income stream;
1-graduate from high school at t0 and enter the labor force immediately
and get an income stream equal to g(t) .
2-going to college and spend t0 to t1 in the college. He would spend
some of his income as college expenses, graduate at t1 from college
and earn an income stream equal to f(t).

Human Capital
We could see the above alternative income streams in following figure;
T = working life time period .
yt
t
g(t)
f(t)
t0 t1 Ť T
Investment cost for college education =∫0
Ť [g(t)-f(t)]dt=Sa+Sb=
opportunity cost forgone (b) + direct cost (a)
Return from investment = ∫Ť
T [ f(t) – g(t) ] dt
When present value of investment cost =present value of the return
from investment ,→ the equilibrium rate of return (r) will be found which
could equalize the cost and return from the investment .
a
b
negative
income
=
cost

Human Capital
If r* is the rate at which present value of net return is equal to present
value of investment cost, then ;
=∫0
Ť [g(t)-f(t)] e-r*t dt = ∫Ť
T[ f(t) – g(t) ] e-r*t dt → ∫0
T [g(t)-f(t)] e-r*t dt =0
g(t) and f(t) are function of rate of return of income after graduation (r* ).
If r* > i college education is desirable
If r* < i college education is not desirable
i=market interest rate ,
Investment in training
In a competitive market labor will be paid according to his value of
marginal product.
Suppose that the government requires that the firm should hires some
members of a disadvantaged group whose initial marginal product is
less than the current wage (w) , but should be paid the same current
wage rate.

Investment in training
For the disadvantaged group suppose that;
MP=f(t) , marginal product of labor is a function of time
VMPL = P MP = f(t) < w = the current wage rate If P=1→ VMPL = MP
Suppose the firm provides on job training program for this disadvantage
group till at time t=T , their marginal productivity would increase and
be equal to w , the current wage rate .
∫0
T[w-f(t)]e-it dt = Money paid to the disadvantage group in excess of
their marginal productivity till they productivity increase to w
cost of training = direct cost of training plus money paid to the
disadvantage group in excess of their marginal productivity. The
distribution of this cost depends upon the institutional setting . In a
competitive market the entire cost will be born by disadvantage
group, since in these markets the labor will be paid only by their
value of marginal product .

Earning cycle investment
Human capital is subject to depreciation over time .it is possible to offset
depreciation and increase the stock of human capital through further
learning . The optimal human capital rate is an important question .
t=0 → t=T , earning cycle .
Kt = K1t + K2t 1
Kt = stock of human capital at time t
K1t = quantity of human capital used to generate income
K2t = quantity of human capital used to generate more human capital
respectively.
yt= αK1t = income at time t , α>0 2 income generated at time t
qt = a K2t
β , a>0 , 0<β<1 3 human capital generated at time t
dKt/dt =qt - δKt 4 human capital depreciation rate
Ct = A K2t 5 investment cost for the production of
human capital , which is equal to forgone earnings plus direct cost.
Optimal investment in human capital ;
Max V= ∫0
T yte-it dt , ( present value of the individual income stream )
S.T. 1 , 2 ,3 , 4, 5

Earning cycle investment
Some aspects of optimization process ;
1- marginal cost of producing a unit a human capital =MCt=dct/dkt
∂MCt/∂kt >0 , ∂MCt/∂ t = 0
2- marginal revenue from one unit of human capital =MR=dRt/dkt
∂MRt/∂kt = constant , and ∂MRt/∂ t < 0
The observations suggest that ;
1- during the early years of working age ; MR > MC . The entire stock of
human capital is used to produce more human capital , K2t=Kt
2- during the middle years, marginal revenue is decreasing . Stock of
capital is used to produce more human capital and also used to obtain
income . K1t >0 , K2t> 0 , and MR = MC
3 – during the ending years of working life , MR<MC , the additions to
human capital is not enough to compensate the depreciation.

Problems
12-1
Consider two alternative income streams ; y1 = 300 , and y2 =321 ,and
y1 =100 , and y2 =535 . For what rate of interest would the consumer
be indifferent between the two streams .
solution
300 + 321/(1+r) = 100 + 535/(1+r)
r = 0.07
12-2
A consumer consumption–utility function for a two period horizon is
U=c1c2
0.6 . His income stream is y1=1000 , y2=648 , and the market
rate of interest is 0.08. determine values for c1 and c2 that
maximizes his utility function . Is he a borrower or lender ?
Solution
Max U=c1c2
0.6
s.t. 1000 + 648/(1+0.08) = c1 + c2/(1+0.08)
L =c1c2
0.6 + λ [ 1000 + 648/(1+0.08) – c1 –c2 / (1+0.08) ]

Problems
∂L/∂c1 = c2
0.6 - λ =0
∂L/∂c2 = 0.6 c1c2
-0.4 - λ/(1+0.8) = 0
∂L/∂λ = 1000 + 648/(1+0.08) – c1 –c2 / (1+0.08) = 0
1600 – c1 – c1 / 2(1+ 0.08) = 0
C1 = 1093.6 > 1000 borrower
C2 = 546.8 < 648
12-3
An entrepreneur invest on one marketing date and receives the
resultant revenue on the next . The explicit form of the investment –
opportunities function is R2 = 24 (I1)1/2 , and the market rate of
interest is 0.20 . Find his optimum investment level .
Solution
R2 = 24 (I1)1/2 revenue from investing I1 inside the firm .
I1 +( I1) i = I1 + 0.2 I1 revenue from investing I1 in the bond
I1 + 0.2 I1 =24 (I1)1/2
I1 = 400 R2 = 480

Problems
12-4
Consider a bond market in which only the consumers borrow and lend. Assume
that all 150 consumers have the same two-period consumption –utility
function : U=c1c2 . Let each of the 100 consumers have the expected
income stream y1=10000 , y2=8400 , and let each of the remaining 50
consumers have the expected income stream y1= 8000 , y2 = 14000 . At
what rate of interest will the bond market be in equilibrium ?
Solution
The Lagrangian function for each consumer group is
V*= c1c2 +µ [ (y1 – c1 ) + (y2 – c2) (1+i)-1 ]
F. O. C.
∂V*/ ∂c1 = c2 - µ = 0
∂V*/ ∂c2 = c1 - µ (1+i)-1 = 0
∂V*/ ∂ µ = (y1 – c1 ) + (y2 – c2) (1+i)-1 = 0 , and solve for c1 ;
C1 = [ y1 + y2(1+i)-1 ] / 2 The consumer excess demand for bond is ;
y1 - c 1 = [y1 – y2(1+i)-1]/2 ,
Bond market equilibrium requires that aggregate excess demand by the two
groups of consumers equal zero ;
100[5000–4200(1+i)-1]+50[4000–7000(1+ i)-1]=700000–770000(1+i)-1=0→
i=0.10

Problems
12-5
An entrepreneur receives 100 dollars at t=5 , determine an equivalent
constant continuous income-stream from t=0 to t=5 if the interest
rate is 10 percent . Note that e0.5 = 1.64872 .
Solution
t=0→→→→→→t=5 . RT = 100
RT e-it = ∫0
t y e-it dt
y =[ i/(e-it - 1)]RT = 154.149
12-6
Consider an entrepreneur engaged in a point input point output vinegar
aging process .his initial cost is 20 , the sales value of the vinegar is
R(T)=100 T1/2 . And the rate of interest is 0.05 . How long is his
optimal investment period .

Problems
12-6 solution
Π=R(T) e-iT - Io = 100 T1/2 e-iT - 20
dΠ/dT = [ R’(T) – iR(T) ]e-iT = [50T-1/2 – (0.05)(100)T1/2 ] e-iT = 0
i= R’(T)/R(T)→→→0.05 = 50T-1/2 / 100 T1/2 →→→→T =10 .
12-7
An entrepreneur is engaged in a repeated point input point output
investment process . He invests Io dollars and receives a revenue of
R(T) dollars T years later . At T he will again invest Io dollars and
receive another revenue of of R(T) dollars at 2T . Assume that he
repeats this process indefinitely . Interest is compound continuously
at constant rate of i .
What is the present value of the entrepreneurs profit from such an
infinite chain ? Formulate his first order condition for profit
maximization . Compare this result with the first order condition for
the unrepeated case.

Problems
12-7 solution
Π1 = R(T) e-iT – I0 ; unrepeated case
Π2 = R(T) e-2iT - Ioe-iT = Π1 e-iT
Π3 = Π1 e-2iT
……………….
Πn = Π1 e-(n-1)iT
Π= Σi=1
∞ Πi =Π1(1+e-iT +e-2iT+e-3iT+...∞)=Π1/(1 - e-iT)=
[R(T) e-iT- I0 ]/(1 - e-iT)
dΠ/dT ={ (1 - e-iT)[R’(T) – iR(T)] e-iT - i e-iT [R(T) e-iT – Io ]} / (1 - e-iT)2 =0
(1 - e-iT)[R’(T) – iR(T)] =i [R(T) e-iT – Io]
∫o
T e-iT = (1 - e-iT)/ i = γ present value of one dollar income stream
[R’(T) – iR(T)] = 1/γ [R(T) e-iT – Io] F .O . C . Repeated case
[R’(T) – iR(T)] =marginal present value of profit (or net revenue) of increasing
the life of first casting (T) by one year .
[R(T) e-iT – Io] = net present revenue of profit from a new casting process after
T years .
1/γ [R(T) e-iT – Io]= net present revenue of profit from a new casting process for
the first year ( during the T years life of investment ).
Unrepeated case:
d(Π1 )/dT=d[R(T) e-iT – I0 )] / dT =0 → [R’(T) – iR(T)]=0 → R’(T) = iR(T)

Problems
12-8
an entrepreneur is engaged in tree growing . He purchases a seedling
for 4 dollars , incurs a cultivation cost flow at a rate G(t) = 0.4t
dollars per year during the life of a tree and sells the tree at t=T for
R(T) = 4+ 8T – T2 dollars . The market rate of interest is 20 percent.
Determine an optimal length for his cultivation period , T . Apply the
appropriate second order condition to verify that your solution is a
maximum .
12-8 solution
Π = R(T) e-iT - Io - ∫o
T G(t) e-it dt
dΠ/dT = [ R’(T) – iR(T) – G(T) ] e-iT = 0
[R’(T) – G(T) ] / R(T) = i
(-2T + 8 – 0.4T )/( 4+ 8T – T 2 ) = 0.20
T2 – 20T +36 =0 , T =2 , T=18 .
d2Π/dT2 <0 →→→T=2

Problems
12-9
An entrepreneur is considering the variable revenues and cost from the
operation of a machine to produce the output Q which sells at a
fixed price p=52. his input cost flow be at the rate ct=5qt
2 dollars per
year, and his maintenance cost flow would be at the rate Mt=2qt+3t
dollars per year. Construct a quasi-rent function for machine .
12-9 solution
Zt = pqt – c(qt) – M(q, t)
∂Zt / ∂qt = p - ∂ct/∂qt - ∂Mt / ∂qt = 0
52 = 10 qt + 2 →→ qt = 5
Zt = pqt – c(qt) – M(q, t) = 52 (5) – 5(25) – 2(5) -3t = 135 - 3t .

Problems
12-10
An entrepreneur plans for a one-machine horizon. He purchases the
machine for 500 dollars . Its scrap value at time T is S(T)=500-40T.
The rate of interest is 0.05. The machine yields a quasi-rent flow at
the rate Zt = 85 – 4t dollars per year . When should the entrepreneur
retire this machine ?
12-10 solution
Π= ∫o
T Z(t) e-iT dt – I0 +S(T)e-iT
dΠ/dT = [Z(T) – iS(T) +S’(T)]e-iT = 0
Z(T) + S’(T) = iS(T)
85 – 4T -40 = 0.05(500 -40 T) , →→t=10
12-11
An entrepreneur with a two years horizon decides to extract 100 units
of output from an exhaustible resource . His extraction cost is
Ct=0.5qt
2 and the interest rate is 0.10 percent , and the constant
selling price for the output is 100 dollars . How much output should
he extract in each year ?

Problems
12-11 solution
Max Π= Σt=1
t=2 [ ptqt – c(qt) ] (1+i)-t + λ(q0 – Σt=1
t=2 qt)
dΠ/dqt = 0 →→→ [Pt - c’(qt) ] (1+i)-t = λ
[P1 - c’(q1 ) ] (1+i)-1 = λ
[P2 - c’(q2 ) ] (1+i)-2 = λ
[P1 - c’(q1 ) ] = [P2 - c’(q2 ) ] (1+i)-1
(100 –q1) = (100 – q2)(1.01)-1
q1+q2=100
q1 = 52.3 q2 = 47.7
THE END

H & Q CH 12 Optimization over time.ppt

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