<p>Some exponential growth-decay model</p><p>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</p><p> n0 n( t )</p><p> time</p><p>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</p><p>B. Sales with advertising</p><p>► 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. </p><p> dS  M  S  Model:  A   S , A( t )  A when t   dt  M   0, otherwise</p><p>If X = total advertising expenditure = A</p><p> dS  X  X Then,  bS  C , where b      , C  dt M  </p><p>C Solution: S( t )  S ebt  ( 1  ebt ) 0 b Advertising </p><p>Time</p><p>S( t )  S( )e( t )</p><p>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.</p><p>Time Guerrilla warfare. Two different groups. </p><p>Defenders: m( t ) Return fire blindly</p><p>Guerrillas: n( t ) Fire defenders having the later in full view</p><p>A typical model may be:</p><p> 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</p><p>There is a likely draw if 2N  M 2 or</p><p>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.</p><p>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. </p><p>Solution: ( M 2  m2 )   ( N 2  n2 )</p><p>Forces fight to draw if M 2  N 2</p><p>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. </p><p> d  dn  dm d 2n Also,      n   n dt  dt  dt dt 2</p><p>Its general solution would be  n( t )  N cosh  t - N sinh  t  similarly,  m( t )  M cosh  t - M sinh  t </p><p>A MATLAB run shows the profile of two functions against time for a given set of initial values.</p><p>L a n c a s t e r m o d e l 3 0</p><p>2 5</p><p>2 0</p><p>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.</p><p>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. </p><p>Structurally, they are related in this manner</p><p>-</p><p>ParasiteParasite HostHost</p><p>+ This is structurally a stable arrangement. Similar, structural approach: </p><p>WagesWages PricePrice pressure pressure +</p><p>WageWage demands demands PricePrice</p><p>A positive feedback loop. Unstable. ActualActual - TemperatureTemperature DesiredDesired RoomRoom temp temp differencedifference RoomRoom temp temp + +</p><p>HeatingHeating A negative feedback system. Product of the signs on the arcs is negative. This system is stable. Typical thermostat System.</p><p>ActualActual DesiredDesired RoomRoom temp temp RoomRoom temp temp</p><p>Time</p><p>For our predator-prey problem, </p><p> dm dn System is in equilibrium when  0 and  0 dt dt Let equilibrium populations be me and ne , respectively.</p><p>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 </p><p> n   1  m dm   ne    m dn n  m   1  n  me  from which we get contour solutions on the m-n plane</p><p>1  m  1  n   lnm     lnn    const m  me  n  ne </p>