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 )

time

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 

Time

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

2 5

2 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

Time

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 