Some Exponential Growth-Decay Model

Some exponential growth-decay model

A. Fishery model. n( t ) : Fish population at time t. dn  n   attrition rate for fish dt      ,  = natural mortality rate   fishing mortality rate At t  0,n( t )  n0 t Then n( t )  n0e

n0 n( t )

Total catch up to time t,  C( t )  ( n  n )  0    When more than one variety of fish active  j j And, the volume of jth catch would be  C ( t )  j ( n  n ) j  0

B. Sales with advertising

► When advertising stops, sales decay exponentially. There is some t   when this occurs ► Sales has a saturation point M. i.e. sales cannot increase beyond M ► Advertising level is A.

dS  M  S  Model:  A   S , A( t )  A when t   dt  M   0, otherwise

If X = total advertising expenditure = A

dS  X  X Then,  bS  C , where b      , C  dt M  

C Solution: S( t )  S ebt  ( 1  ebt ) 0 b Advertising 

S( t )  S( )e( t )

C S( t )  S ebt  ( 1  ebt ) 0 b Typical policy: Advertising policy ought to be such that would push  toward right but at the same time reaches S( ) quickly. Probably, a big initial push and then followed by a steady advertising pulses with a certain periodicity. e.g.

Time Guerrilla warfare. Two different groups.

Defenders: m( t ) Return fire blindly

Guerrillas: n( t ) Fire defenders having the later in full view

A typical model may be:

dn  mn dt dm  n dt dn m n m This yields    dn    mdm dm  N M  i.e.  ( n  N )  ( m2  M 2 ) 2

There is a likely draw if 2N  M 2 or

2N M    Vietnam data:  0.001. Guerrillas can win even if 2 vastly outnumbered provided both sides are subdivided into small pockets of engagements and guerrillas attack with local numerical superiority. Lancaster War model. A coupled system.

Two warring populations. n( t ) : number of blue forces at time t m( t ) : number of red forces at time t dn  m dt dn m   . dm dm n  n dt Initial populations N and M , respectively.

Solution: ( M 2  m2 )   ( N 2  n2 )

Forces fight to draw if M 2  N 2

This suggests that if the Blue system is 4 times as effective as the red (weapons, environment, doctrine, etc.), Red will need twice the initial force to seize a draw.

d  dn  dm d 2n Also,      n   n dt  dt  dt dt 2

Its general solution would be  n( t )  N cosh  t - N sinh  t  similarly,  m( t )  M cosh  t - M sinh  t 

A MATLAB run shows the profile of two functions against time for a given set of initial values.

L a n c a s t e r m o d e l 3 0

1 5 0 0 . 5 1 1 . 5 2 2 . 5 3 3 . 5 4 4 . 5 5 t i m e Another coupled system. Lotka-Volterra equations.

Prey-Predator model. dm   ( 1  n )m Prey (host) dt m dn   ( 1  m )n Predator (parasite) dt n Higher the parasite population, lower is the host growth rate. Higher the host population, higher is the parasite growth rate.

Structurally, they are related in this manner

ParasiteParasite HostHost

+ This is structurally a stable arrangement. Similar, structural approach:

WagesWages PricePrice pressure pressure +

WageWage demands demands PricePrice

A positive feedback loop. Unstable. ActualActual - TemperatureTemperature DesiredDesired RoomRoom temp temp differencedifference RoomRoom temp temp + +

HeatingHeating A negative feedback system. Product of the signs on the arcs is negative. This system is stable. Typical thermostat System.

ActualActual DesiredDesired RoomRoom temp temp RoomRoom temp temp

For our predator-prey problem,

dm dn System is in equilibrium when  0 and  0 dt dt Let equilibrium populations be me and ne , respectively.

Then ne  1 and me  1. General system profile in terms of equilibrium volumes is dm n dn m  m ( 1  )m and  n( 1  )n dt ne dt me from these two, we get

 n   1  m dm   ne    m dn n  m   1  n  me  from which we get contour solutions on the m-n plane

1  m  1  n   lnm     lnn    const m  me  n  ne 