Growth-decay models. = kp. = population density at time t. Initial condition: at t = 0, dp. 1. Constant decay: k. Solution: p = 2. Exponential decay:

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1 Growth-decay models. Let P (t) = population density at time t. Initial condition: at t = 0, p = p0 dp = p p 1. Constant decay: k Solution: = 0 kt p 0 2. Exponential decay: dp = kp Solution: p = kt p 0 e t p 0 t

2 3. Movement towards a stable plateau dp = a kp At equilibrium, rate is zero. Therefore, at equilibrium, a p = p = k Solution comes out as p = p p p ) e ( 0 kt p p 0 t Modeling approach is often based on Taylor s series expansion. Taylor s series: If value at time is a continuous function in time, its would be given by (1)

3 Euler model. A linear model. This means we are satisfied with (1a) Example. Currently, the number of jobs in CPU queue is 240. The CPU queue is currently changing at a linear rate of 4 jobs/sec. What would be the queue volume 30 secs from now? This method is the basis for doing many dynamic modeling techniques particularly if the various differential of changes are constant. Coupled Euler models Consider an ecological niche comprising Hawks (H), Rabbits (R) and Grassland (GL). They depend on each other, and these dependencies are indicated by directed edges on appropriate graph with appropriate labels indicating intensities of dependencies H R GL +0.11

4 Meaning of the legends: Each node is a specific specie: {Hawks, Rabbits, GrassLand} An arrow from a node to itself indicates the rate at which it grows. An arrow from a node to another node indicates how the latter node is affected by the current state of the first node. Hawks dine on Rabbits. More hawks less rabbits, and the rate of rabbit consumption is More rabbits mean more meat for hawks, and the rate of hawks growth is on Rabbit-Hawks interaction. Rabbits multiply at a rate 0.09 More rabbits, less grassland. The effect of reduction is 0.2 on Rabbit-GrassLand interaction. Data defining the eco-niche: Suppose maximum supportable value: 500 hawks per square mile. Starting value: 150 hawks. Normalized starting value = 0.3 Maximum supportable value for rabbits: 1000 rabbits per square mile. Starting value: 250 rabbits. Normalized starting value = 0.4 Maximum sustainable value for grasslands: 10 units per square mile. Starting value: 7 units. Normalized starting value = 0.7 Note that each normalized value must be contained between 0 and 1.0. Any value outside this range implies fractured ecosystem.

5 Model: The first few output is: initial; value: initial value: initial value: t H R GL The plot of Hawks (Red), Rabbits (Blue) and Grassland (Magenta) are shown below:

6 HAWKS RABBITS GRASSLAND 0.6 Species Population TIME t[year] Deduce your conclusions. Another way to approach this specific model would be via cross-impact matrix analysis mode. In this case, our impact matrix is just what we had before H(0.3) R(0.4) G(0.7) Hawks, H Rabbits, R GrassLand, G Alternative model: The value of a variable at time given by is (2) Where is the cross-impact ratio given as

7 (2a) Total negative impact = 1 + impacts on Total positive impact = 1 + on sum of all negative sum of all positive impacts Portion of output from this alternative model is shown below. t H R GL Obviously, the model is not linear like Euler model. Thus it might diverge somewhat from the simple model.

8 A graphical output outlines the profiles of the species next. SPECIES IN ALTERNATIVE MODEL HAWKS RABBITS GRASSLAND TIME t[year] Suggest a way to rationalize this model. Signed Digraphs as structural models. Aggregate structural models based on causal loops may provide important qualitative understanding of systems. Particularly on stability issues. Stable Sys Unstable

9 Both have an equilibrium point, but in one case any slight perturbation at an equilibrium point returns the system there (stable case); in the other case the system leaves the equilibrium point (unstable case). Some structural cases Job Availability (JA) attracts Migrants (M) More (M), more is the Labor population (L) More (L) less JA. Graphically the system is a stable lone as the product of the signs on the edges is negative. If it were a positive, it would lead to instability. JA M Positive Negative Positive A stable system L What if we have a system like this? Unstable system. Positive Wages Positive Positive Price Pressure Wage demands Positive Prices

10 Asymptotic stability. As in a thermostat controlled temperature. This could be depicted as a converging cycle spiraling down to the equilibrium temperature. Output Time Cobweb Model and stability issues. Suppose, we have a price-quantity model of a commodity C given by supply-demand equations of C. (3a) (3b) (3c)

11 Supply of the commodity C is determined by the last year s price, the demand for C is determined by the current price now. We want to know what would be the equilibrium price that would satisfy (3). Strategy. Start with some in (3b). Get From (3c) get From (3a) get the corresponding. Continue the loop for steps until the convergence takes place when at the limit of large or the system blows up! Graphically, D S Price Quantity What happens if the slope of the Supply line (curve) is made flatter relative to Demand line (curve) as shown below? D S Price Quantity

12 Now, a spiraling out phenomenon (unstable equilibrium). The equilibrium price cannot be reached as the system becomes unstable. Explosive situation. This happens whenever the absolute value of the slope of the Supply line (curve) is less than that of the Demand line (curve). What if the system equations are nonlinear but contains a lagged variable? Same issue. Demand (D) jointly determine the Price of the commodity (P).

13 At the intersection is the equilibrium price where the demand equals supply. At the unstable case, the trajectory spirals out from as shown below. If then the price is stable. What if the absolute values of slopes are equal? Oscillation about the equilibrium point. Limit cycle. Other simple stable-unstable dynamics. A. Romantic model by J. C. Sprott ( R = Romeo s love (or hate) for Juliet J = Juliet s love (or hate) for Romeo. dr dj = ar + bj = cr + dj with a b, c, d, as constants.

14 Some romantic nuances: a = 0 (Out of touch with own feelings ) b = 0 (Ignorant of others feelings) a > 0, b > 0 (Very romantic) a > 0, b < 0 (Narcissistic nerd) a < 0, b > 0 (Cautious lover) a < 0, b < 0 (As a hermit) Therefore, 6 6 possible combinations of the variables leading to a rich tapestry of dynamics! Examples: 1. With a=0.1 b =-0.1, c=0.1 and d=0.2 (Red:R, Blue: J)

15 2. With a=0.1, b=0.1, c=0.1, d= 0.2 (Red: R, Blue: J) 3. With a=0.1, b=0.1, c=0.1 and d=-0.2 (Red: R, Blue: J)

16 Some special cases. dr dj = bj = cr b > 0, c > 0 (Explosive love) b > 0, c < 0 (Never-ending dance?) With b=0.1 and c= -b, we get the never ending dance as shown below over 100 points.

17 b < 0, c > 0 (Never-ending dance?) b < 0, c < 0 (Unfinished Nerd-symphony) (As Red comes closer, Green moves away from it which makes the Red to fly in the opposite direction, etc.) B. Sales with advertising When advertising stops, sales decay exponentially. There is some t > τ when this occurs Sales has a saturation point M. i.e. sales cannot increase beyond M

18 Advertising level is A. Model: ds M S = A λs, A ( t ) = A when t < τ M = 0, otherwise If X = total advertising expenditure = A τ Then, ds X = bs + C, where b = + λ, τm C = X τ bt C bt Solution: S( t ) = S0e + ( 1 e ) b Advertising τ Time S( t ) = S( τ )e λ( τ t ) bt C bt S( t ) = S0e + ( 1 e ) b Typical policy: Advertising policy ought to be such that would push τ toward right but at the same time reaches

19 S( τ ) 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 = αmn dm = βn n m dn αm This yields = β dn = α dm β mdm α 2 i.e. β ( n N ) = ( m 2 M ) 2 N M

20 There is a likely draw if 2 2β N = αm or 2βN M = α α Vietnam data: < Guerrillas can win even if 2β vastly outnumbered provided both sides are subdivided into small pockets of engagements and guerrillas attack with local numerical superiority. Question: Do we need satellite surveillance for land warfare? ( Lancaster War model. A coupled growth-decay system. Two warring populations. n ( t ) : number of blue forces at time t m ( t ) : number of red forces at time t dn = αm dn αm =. dm dm βn = βn Initial populations N and M, respectively. 2 Solution: α ( M m ) = β ( N n ) 2 2 2

21 2 Forces fight to draw if α M = βn 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. 2 Also, d dn dm = α = αβ n d 2 n 2 = αβ n Its general solution would be n ( t ) = N cosh αβ t - similarly, m ( t ) = M cosh αβ t - β α α β N sinh αβ t M sinh αβ t A MATLAB run shows the profile of two functions against time for a given set of initial values. Another coupled system. Lotka-Volterra equations. Prey-Predator model. dm = λm( 1 αn )m Prey (host)

22 dn = λ ( 1 βm )n Predator (parasite) 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 - Parasite Host + This is structurally a stable arrangement. Similar, structural approach: Wages Wage demands + Price pressure Price A positive feedback loop. Unstable.

23 Actual Room temp - Temperature difference Desired Room temp + + Heating A negative feedback system. Product of the signs on the arcs is negative. This system is stable. Typical thermostat System. Actual Room temp Desired Room temp Time For our predator-prey problem, dm dn System is in equilibrium when = 0 and = 0 Let equilibrium populations be m e and n e, respectively. Then α n e = 1 and β m e = 1. General system profile in terms of equilibrium volumes is

24 dm n dn m = λ m( 1 )m and = λ n( 1 ) n n m e e from these two, we get dm dn λm = λ n 1 1 n m n e m n m e from which we get contour solutions on the m-n plane 1 λ m lnm m m e 1 + λ n lnn n n e = const Logistic growth model

25 dp p = kp(1 ). Note. Two equilibrium points: p = 0 and m p = m. The equation is clearly separable. Also note that 1 p( m p) = 1 m 1 p + 1 m p Therefore, the solution comes out to be i.e. p kt mce = with the constant kt m + Ce p = p 0 mp 0 + ( m p0) e kt C = mp0 m p 0 At t, p = p m. From such a reference point, p = p 0 + p p ( p p0 ) e 0 kt A meta-stable population model. More complex than logistic model. dp = ap( b p)( p c) with a, b, c constants. This is often demonstrated in tipping behavior in a restaurant.

26 Boom/Bust dynamics Consider a population existing on a non-renewable resource. dx dy = caxy dx ( x is population) = axy ( y is resource) Dynamics of this model appears to be x (t) What if we allow the resource to be renewable? We now have Sustained oscillations with BB economy. Our equations might appear as dx = caxy dx and t

27 dy = axy + by We will repeat boom/bust phenomenon cyclically over time. Cf Lotka-Volterra model of predator/prey. 4. Geopolitics of war. A focal state is the state of our investigation. Suppose it has an area A. More the area more revenue R via taxation, more recruitment for military to wage wars W which would earn more area. Only three variables: A, W, R. A + R da +? W = c1w R = c2a and W = c3r c4 There may be some adversaries and peaceniks always wanting to thwart war effort; hence a c4with a negative sign. If earned revenue is below c 4 / c 3 war would be negatively appearing with revenue. Then

28 da = c1 c2c3a c1c 4 = ca a a The equilibrium point is A 0 =. If A starts below A 0 it c will disappear in time. If it starts above A 0, it will increase exponentially. A 0 Not much interesting solution. How about introducing the notion of distance into it? Larger the area more difficult it is to control its periphery. Assume area to be a circle with radius r. Therefore when A increases it enforces increase of its logical distance from center as A. Assume increasing logical distance corresponds to decay in population exponentially. Then a reasonable model could be asserted as: t

29 da = cae A h a = ca( 1 A h +...) a yielding additional equilibrium points A 1 and A 2. da A 0 A 1 A 2 If A is below A 1 it will disappear in time; if it is above A2 in the neighborhood of A2 it d be declining, but below A2 it d be increasing. A2 is thus a stable equilibrium point. What if the area of the focal state is a linear strip having different countries at the side? Since the space one dimensional, the degree of influence away from center is inversely proportional to distance leading to an equation like

30 da = cae Model A/ h a = ca(1 A h + A 2 2h 2...) a Marchland Collective solidarity inclusion. Assume S is solidarity factor varying between 0 and 1. If 0, the society lives as bunch of islands with little communication between them. At the other extreme, society lives as one national unit. Then a reasonable structure would be to suggest da A = c1 SA(1 ) h a Bust how does S change if it s not fixed? Let us assume a model for S Collective solidarity = living with neighbors in empathy, helping each other, standing out for each other is minimal near the center of power, maximum at the periphery.

31 ds A Then = c 2 1 S(1 S ) A with 0 S 1 0 If A < A0 (too near to Washington, S declines) If A > A0 S increases. S( 1 S) maximizes at 0.5 In this case, our model is a pair of equations: da ds A = c1 SA(1 ) a h A = c 1 S(1 A 0 2 S (From Historical Dynamics by Peter Turchin.) )