Embed
Email

predator-prey-systems

Document Sample

Shared by: xiaohuicaicai
Categories
Tags
Stats
views:
1
posted:
10/25/2011
language:
English
pages:
62
Predator Prey System

with a stable periodic orbit

1st Session - Simple Analysis



Systems Dynamics Study Group



Ellis S. Nolley

11/7/2001







12/20/2001 Systems Dynamics Study Group 1

Topics

• Overview

– Simple Analysis – 1st session, 11/7/2001

– Rigorous Analysis – 2nd session, 11/27/2001

– Simulation Results – 3rd session, 12/11/2001

• Mathematical Model

• Fixed Points

• Stable Periodic Orbit



Reference: McGehee & Armstrong, Journal of Differential Equations,

23, 30-52 (1977)







12/20/2001 Systems Dynamics Study Group 2

Model

x = amount of prey, y = amount of predator

g(x)

dx/dt = xg(x) – yp(x)

dy/dt = y[-s + cp(x)] g(x)





x

k

g(x) is a growth function,

g(x), monotonic non-increasing, dg(x)/dx 0

p(x) is predation function

p(x), monotonic increasing, dp(x)/dx >0 , p(0)=0



12/20/2001 Systems Dynamics Study Group 3

dx/dt = xg(x) – yp(x)

dy/dt = y[-s + cp(x)]

Fixed Points

3 Fixed points: (x*,y*), (0,0), (k,0)



(x*,y*)

dy/dt = 0, dx/dt = 0 for (x*,y*)

At dy/dt=0, y>0, then p(x*) = s/c, y*=x*g(x*)/p(x*)



Assume Lim p(x) = a, as x-> inf+

1) x* > s/c, otherwise there is no fixed point

2) y* > 0, in order to have a system

3) If there is a k, g(k)=0, then x* 0, y*>0







12/20/2001 Systems Dynamics Study Group 4

dx/dt = xg(x) – yp(x)



Fixed Points (Cont’d) dy/dt = y[-s + cp(x)]









Let’s look at the slope on x=k

y

dy/dx = (dy/dt)/(dx/dt)

At x=k, g(k)=0

Recall: -s+cp(x*)=0, p’(x)>0, x*0

-p(k) denominator 0, then delta y>0 (numerator)

So, the vectors are coming in.





12/20/2001 Systems Dynamics Study Group 5

dx/dt = xg(x) – yp(x)

Analysis at Fixed Points dy/dt = y[-s + cp(x)]









(0,0)



What happens at x=0 (y axis)?

dy/dt= y(-s) 0



So, (0,0) is a saddle point.

x

(0,0)



12/20/2001 Systems Dynamics Study Group 6

Analysis at dx/dt = xg(x) – yp(x)

dy/dt = y[-s + cp(x)]

Carrying Capacity

(k,0)

At (k,0), g(k) = 0

dx/dt = xg(x) – yp(x)

= xg(x)

g(x) is monotonic non-increasing.

For x0 y

For x>k, g(x) 0

So dy/dt = y[-s + cp(x)] > 0 for y>0, x=k

x

(k,0) is a saddle point k





12/20/2001 Systems Dynamics Study Group 7

dx/dt = xg(x) – yp(x)

dy/dt = y[-s + cp(x)]

Prey Isocline

y

At the prey isocline, dx/dt = 0 y(0)=g(0)/p’(0)





y= xg(x)/p(x)

and goes through (k,0) and (x*,y*).

To find y(0): by L’Hospital’s Rule, (x*,y*)





y(0) = [x g’(0) + g(0)]/p’(0) = g(0)/p’(0) > 0 x

k







Recall: L’Hospitals Rule: if f(x) & g(x) both go to either 0 or infinity as x->a,

Then lim f(x)/g(x)] = lim [df(x)/dx]/[dg(x)/dx], as x-> a



12/20/2001 Systems Dynamics Study Group 8

The Vector Space dx/dt = xg(x) – yp(x)

dy/dt = y[-s + cp(x)]



Since delta x0 near x=k,

then vector is up near x=k

y

Vectors can only turn around at the critical pt.



At x=x* above y*, dx/dt0

Left of x*, dy/dt 0

x

(x*,y*) is unstable if tangent is positive. k

Pick a line tangent to dy/dx at (k,x**)

All vectors cross it inward



12/20/2001 Systems Dynamics Study Group 9

dx/dt = xg(x) – yp(x)

Periodic Orbit dy/dt = y[-s + cp(x)]









y

• Fixed points are unstable.

• All vectors enter the region

and move away from the boundary.

(k,x**)

• Stable periodic orbit exists around

(x*,y*)

the unstable fixed point.

x

k









12/20/2001 Systems Dynamics Study Group 10

dx/dt = xg(x) – yp(x)

dy/dt = y[-s + cp(x)]

Next Session

Simple Mathematics – 1st session

Rigorous Mathematics – 2nd session, 11/27

Simulation Results – 3rd session









12/20/2001 Systems Dynamics Study Group 11

Thank you!



12/20/2001 Systems Dynamics Study Group 12

Predator Prey System

with a stable periodic orbit

2nd Session - Rigorous Analysis



Systems Dynamics Study Group



Ellis S. Nolley

11/27/2001





12/20/2001 Systems Dynamics Study Group 13

Topics

• Overview

– Simple Analysis – 1st session, 11/7/2001

– Rigorous Analysis – 2nd session, 11/27/2001

– Simulation Results – 3rd session, 12/11/2001

• Mathematical Model

• Fixed Points & Eigenvalues

• Poincare-Bendixon Theorem







4 key slides: #23 – 26





12/20/2001 Systems Dynamics Study Group 14

References

• McGehee & Armstrong, Journal of Differential Equations, 23, 30-52 (1977)

• Morris Hirsch & Stephen Smale, Differential Equations, Dynamical Systems

and Linear Algebra, 1974, Academic Press

Ch 3-5, Linear Systems, Eigenvalues & Exponentials of Operators

Ch 9-12, Stability, Differential Equations on Electrical Systems, Poincare-Bendixon

Theorem, Ecology

• Michael Spivak, Calculus on Manifolds, 1965, W.A Benjamin

• Raghavan Narasimhan, Analysis on Real & Complex Manifolds, 1968, North-

Holland Publishing Company









12/20/2001 Systems Dynamics Study Group 15

Where to find

these References

Mathematics Library, Vincent Hall,

Vincent Hall, 206 Church Street,

3rd Floor, Mpls, MN 55455

University of MN



http://onestop.umn.edu/Maps/VinH/VinH-map.html









12/20/2001 Systems Dynamics Study Group 16

Model

x = amount of prey, y = amount of predator



g(x)

dx/dt = xg(x) – yp(x)

dy/dt = y[-s + cp(x)]

g(x)





x

g(x) is a growth function, k

g(x), monotonic non-increasing, dg(x)/dx 0

p(x) is predation function

p(x), monotonic increasing, dp(x)/dx >0 , p(0)=0



12/20/2001 Systems Dynamics Study Group 17

dx/dt = xg(x) – yp(x)

Jacobian & Eigenvalue Review dy/dt = y[-s + cp(x)]

dx/dt = xg(x) – yp(x)

dy/dt = y[-s + cp(x)]

z’(t) = f(z) = [f1(z1,…zn), …, fn(z1…,zn)]

Note above: z1=x, z2=y, f1(z)=xg(x)-yp(x), f2(z)=y[-s+cp(x)]



dF(z,t)/dt = f(z); F(n)(z)=dnF(z)/dzn,n=0, … ∞; F(0)(z)=F(z)



If z є B(z0,ε) ={z|z-z0| 0





12/20/2001 Systems Dynamics Study Group 20

Why Re λ dx/dt = xg(x) – yp(x)

dy/dt = y[-s + cp(x)]

determines stability dz/dt = f(z)







z’ = f(z); f(z0)=0, z0 fixed point, λk eigenvalues.

Suppose λj has Re λj >0. Pick z close to z0



f(z) = k=0∞Σ f(k)(z0)(z-z0)k/k! = f(z0) + f(1)(z0)(z-z0) + … Taylor Series

~ f(1)(z0) (z-z0) = (z-z0)Σckλk ; d f(z)/z ~ Σckλk dt

ln f(z) ~ Σckλkt; f(z) ~ c*eΣλkt

|f(z)| ~ |c*| |eλjt| |eΣλkt|; λ = Re λ + i Im λ ; |ei w|= |Cos(Im w) + i Sin(Im w)| = 1



lim |f(z)| ~ lim(|c*| |eRe(λj)t | |eΣλkt|) as t-> ∞

Then, |eRe(λj)t | -> large because Re λj >0 So, z0 is an unstable fixed point.

If all Re λj 0 as t-> ∞ So, z0 is a stable fixed point.



12/20/2001 Systems Dynamics Study Group 21

dx/dt = xg(x) – yp(x) det [f(1)(z0)-λI] =

dy/dt = y[-s + cp(x)] (g(x0)+x0g’(x0)-y0p’(x0)-λ) (-s+cp(x0)-λ)

dz/dt = f(z) Fixed Points – cy0p’(x0)p(x0) = 0









dx/dt = xg(x) – yp(x) = 0

dy/dt = y[-s + cp(x)] = 0

y





1. (0,0), p(0) = 0

2. (k,0), g(k) = 0 (x*,y*)



x

k

3. (x*,y*), 00



12/20/2001 Systems Dynamics Study Group 22

dx/dt = xg(x) – yp(x) det [f(1)(z0)-λI] =

dy/dt = y[-s + cp(x)] (g(x0)+x0g’(x0)-y0p’(x0)-λ) (-s+cp(x0)-λ)

dz/dt = f(z) (0,0) – cy0p’(x0)p(x0) = 0









(g(x0)+x0g’(x0)-y0p’(x0)-λ) (-s+cp(x0)-λ) – cy0p’(x0)p(x0) = 0

x0=y0=p(x0)=0

(g(0)-λ)(-s-λ)=0

λ=g(0),-s; y



λ1= g(0) > 0, corresponds to x axis x

(0,0)

λ2= -s 0, p(x*) 0 y

λ1= kg’(k) 0, corresponds to y axis

(k,0) is unstable x

k

12/20/2001 Systems Dynamics Study Group 24

dx/dt = xg(x) – yp(x) det [f(1)(z0)-λI] =

(g(x0)+x0g’(x0)-y0p’(x0)-λ) (-s+cp(x0)-λ)

dy/dt = y[-s + cp(x)]

dz/dt = f(z)

(x*,y*) – cy0p’(x0)p(x0) = 0





(g(x0)+x0g’(x0)-y0p’(x0)-λ) (-s+cp(x0)-λ) – cy0p’(x0)p(x0) = 0

Note: -s+cp(x0)=0; x0=x*, y0=y*

(g(x*)+x*g’(x*)-y*p’(x*)-λ)(-λ) – cy*p’(x*)p(x*) = 0

λ2 - [g(x*)+x*g’(x*)-y*p’(x*)]λ – cy*p’(x*)p(x*) = 0

B C>0

λ = (B +/– sqrt(B2 + 4C))/2



Note: slope of prey isocline, (dy/dt) at (x*,y*) y

= d(dx/dt)dx = g(x)+xg’(x)-yp’(x) = B

If B > 0, (x*,y*) is unstable

λ1 = [B – sqrt(B2 + 4C)]/2 0

λ2 = [B + sqrt(B2 + 4C)]/2 > 0

(x*,y*)

If B 0



C1 – has a continuous first

(x*,y*)

derivative

x

k







12/20/2001 Systems Dynamics Study Group 26

dx/dt = xg(x) – yp(x)

dy/dt = y[-s + cp(x)]



Poincaré-Bendixon Rationale

F(z+,t1)=z+1=(x1,y1)

F(z+,t2)=z+2=(x2,y2) y

Z–1

Z–2

lim z+k -> z, as k->∞ Z+1

Z+2

Z









B>0

F(z–,t1 )=z–1=(x1,y1)

F(z–,t2)=z–2=(x2,y2)

lim F(z–k) -> z-, as k->∞ x



z- 0



dx/dt = xg(x) – yp(x) (x*,y*)



dy/dt = y[-s + cp(x)], s>0, c>0 x

k

g(x) is a growth function,

g(x), monotonic non-increasing, g’(x) 0, g(k)=0

p(x) is predation function

p(x), monotonic increasing, p’(x) >0 , p(0)=0

(x*,y*) fixed point, x*>0, y*,>0 => x*g(x*)-y*p(x*)=0; -s+cp(x*)=0

B = g(x*)+ x*g’(x*)-yp’(x*) > 0



Then, the dynamical system has a stable periodic orbit.





12/20/2001 Systems Dynamics Study Group 28

Thank you!



12/20/2001 Systems Dynamics Study Group 29

Predator-Prey System

with a stable periodic orbit

Systems Dynamics Study Group

3rd Session – Simulation Results



Ellis S. Nolley

12/20/2001





12/20/2001 Systems Dynamics Study Group 30

References



• McGehee & Armstrong, Journal of Differential Equations, 23, 30-52 (1977)



• Vensim ® PLE software (free for educational use)

www.vensim.com/download.html



• Vensim Tutorial by Craig Kirkwood, Arizona State University

www.public.asu.edu/~kirkwood/sysdyn/SDRes.htm



• Vensim User Guide

www.vensim.com/ffiles/venple.pdf









12/20/2001 Systems Dynamics Study Group 31

Topics

• Overview

– Simple Analysis – 1st session, 11/7/2001

– Rigorous Analysis – 2nd session, 11/27/2001

– Simulation Results – 3rd session, 12/11/2001

• Model Parameters

• Simulation Results

• Vensim Techniques

• Bifurcation



• Extra: Mathematics of Parameter Selection



12/20/2001 Systems Dynamics Study Group 32

y

Model

B>0

dx/dt = xg(x) – yp(x)

(x*,y*)

dy/dt = y[-s + cp(x)], s>0, c>0

x

k

g(x) is a growth function,

g(x), monotonic non-increasing, g’(x) 0, g(k)=0

p(x) is predation function

p(x), monotonic increasing, p’(x) >0 , p(0)=0

(x*,y*) fixed point, x*>0, y*,>0 => x*g(x*)-y*p(x*)=0; -s+cp(x*)=0

B = g(x*)+ x*g’(x*)-yp’(x*) > 0



Then, the dynamial system has a stable periodic orbit.





12/20/2001 Systems Dynamics Study Group 33

dx/dt = xg(x) – yp(x) g(0)>0, g’(x)0, B>0

Model Parameters

g(x) = a0+a1x, x*~147.4, a0=54, a1= -0.15

p(x) = b ln(x+1), b=4, s=200, c=10



g(0) = 54>0, g’(x)= -0.150



B= g(x)+xg’(x)-xg(x)p’(x)/p(x), for x=x*, p(x*)=s/c

~ 54+2(-0.15)147.4+4(1)[54-0.15(147.4)]/(200/10)

since x/(x+1) ~ 1

~ 54 - 44.2 - 6.4 = 3.4 > 0







12/20/2001 Systems Dynamics Study Group 34

Inside Orbit

X & Y Log Inside: time 0 - 40



500

400

Y - Predator









300

200

100

0 Time Series of first 3 Periods

0 100 200 300 400

120









Amount X & Y

X - Prey 100

80

x

60

40 y

20

0

0 0.5 1 1.5 2 2.5

Time









12/20/2001 Systems Dynamics Study Group 35

Outside Orbit

X & Y Log Outside: time 0 - 40



500



400

Y - Predator









300

200



100



0

0 100 200 300 400

X - Prey Time Series of first 2 Periods



500









Amount X & Y

400

300 x

200 y

100

0









0

0.08

0.16

0.24

0.32

0.4

0.48

0.56

0.64

0.72

0.8

0.88

0.96

Time









12/20/2001 Systems Dynamics Study Group 36

Inside & Outside Orbits

X & Y Log Inside: time 0 - 40



500

X & Y Log Outside: time 0 - 40

400









Y - Predator

300

200

500

100



400 0

Y - Predator









0 100 200 300 400

X - Prey

300



200



100



0

0 100 200 300 400

X - Prey





12/20/2001 Systems Dynamics Study Group 37

Combined Inside & Outside Orbits



X & Y Log Inside: time

X & Y Log Outside: time00- -40

40



500

500

400

400

Y - Predator









300

300

200

200

100

100

0

0

0

0 100 200 300

300 400

400

X - Prey





12/20/2001 Systems Dynamics Study Group 38

dx/dt = xg(x) – yp(x)

Vensim Model Layout dy/dt = y[-s + cp(x)]









12/20/2001 Systems Dynamics Study Group 39

g(x)









12/20/2001 Systems Dynamics Study Group 40

p(x)









12/20/2001 Systems Dynamics Study Group 41

dx/dt









12/20/2001 Systems Dynamics Study Group 42

x









12/20/2001 Systems Dynamics Study Group 43

dy/dt









12/20/2001 Systems Dynamics Study Group 44

y









12/20/2001 Systems Dynamics Study Group 45

Other Vensim Techniques



• Select Runge Kutta Integration (RK4).

• Select initial points (x,y)=(1,1) for an outside orbit

and (x,y)=(125,200) for an inside orbit.

• Select 0.005 for a step size in Model/Settings

• Select a custom graph/table to export to Excel

– Control Panel, Graphs, New, Name title, select

variables x & y, click on scale between them

– Click on As Table, click on “running down”

– Click on Ok, close





12/20/2001 Systems Dynamics Study Group 46

Run and Export Text File



• Click on Run Simulation

• Click on Control Panel

• Click on graph name, click on Display

• Click on File, then Save As









12/20/2001 Systems Dynamics Study Group 47

Import Text File into Excel

• Run Excel

• Click Open, select txt type, select file, click Open,

Finish.

• Click on Chart Wizard, XY (scatter), click on Data

Source icon (to right of data range), click and drag

over x & y data, click on Data Source icon, complete

the chart.

• Create a time series chart using t,x,y data the same

way as above, dragging over several periods of data.

• Then, alter step size and initial points in Vensim

• Create other charts for parameter changes by

edit/copy sheet, run new simulation & copy/paste

simulation data onto new sheet’s data region

12/20/2001 Systems Dynamics Study Group 48

Example

Excel

Result









12/20/2001 Systems Dynamics Study Group 49

dx/dt = xg(x) – yp(x) g(0)>0, g’(x)0, B>0









12/20/2001 Systems Dynamics Study Group 50

X & Y Log Outside: time 0 - 40 a0=54 X & Y Log Outside: time 0 - 40 a0=51

500 500

400 400

Y - Predator









Y - Predator

300 300

200 200

100 100



0 0

0 100 200 300 400 0 100 200 300 400

X - Prey X - Prey









X & Y Log Outside: time 0 - 40 a0=40 X & Y Log Outside: time 0 - 40 a0=49.7



250 400

350

200

Y - Predator









300







Y - Predator

150 250

200

100 150

50 100

50

0 0

0 100 200 300 0 100 200 300 400

X - Prey X - Prey







12/20/2001 Systems Dynamics Study Group 51

Projection of

Stable Attractor onto X Axis

x







Actual

~ 147.4 boundary shape

is not described







a0

~ 22.1 ~ 49.7





12/20/2001 Systems Dynamics Study Group 52

dx/dt = xg(x) – yp(x) g(0)>0, g’(x)0







• Generalized Lotka-Volterra Predator-Prey Model

• Internal Fixed Point (x*,y*)

– Stable when B0

• Existence Proof y

• Simulation Results

• Vensim techniques

• Bifurcation B>0





(x*,y*)



x

k



12/20/2001 Systems Dynamics Study Group 53

Another Reference

• C. Neuhauser, “Mathematical Challenges in Spatial Ecology,” Notices

of the American Mathematical Society, 48, 1304-1314 (Dec 2001)



http://www.ams.org/notices/200111/fea-neuhauser.pdf



University of Minnesota, EEB dept of CBS









12/20/2001 Systems Dynamics Study Group 54

Predator-Prey models Competition

Typical Competition Beliefs

• Survival of the fittest

• Competition develops excellence

• Diversity increases stability

• Complexity decreases stability

• One competitor per niche

• Good designs stabilize desirable behavior

and destabilize undesirable behavior

What are likely outcomes of well defined systems?

What systems produce specific outcomes?

12/20/2001 Systems Dynamics Study Group 55

Thank you!



12/20/2001 Systems Dynamics Study Group 56

Extra



Mathematics of Parameter Selection









12/20/2001 Systems Dynamics Study Group 57

dx/dt = xg(x) – yp(x) g(0)>0, g’(x)0, B>0

Recall that any continuous function within a closed bounded region can

be uniformly approximated by polynomials. (Stone-Weierstrauss)

Let g(x) ε R(m), p(x) ε R(n) real polynomials of degree m & n

g(x)=0Σmakxk, a0>0, a10, g’(x)0

p(x)= 1Σnbkxk, b0=0, b1>0, bn>0, since p(0)=0, p’(x)>0 for x>0

B= g(x)+xg’(x)-xg(x)p’(x)/p(x), for x=x*

+ - -

= 0Σmakxk+x(1Σmkakxk-1)-(0Σmakxk)(1Σnkbkxk)/(1Σnbkxk)

= 0Σmakxk+1Σmkakxk -(0Σmakxk)(1Σnkbkxk)/(1Σnbkxk)

= a0[1-(1Σnkbkxk)/(1Σnbkxk)]+ 1Σmkakxk -(1Σmakxk)(1Σnkbkxk)/(1Σnbkxk)

Note: if aj0 & bk>0 for all k>0, then (1Σnkbkxk)/(1Σnbkxk)>1,

then B0 for some j:1= 3

12/20/2001 Systems Dynamics Study Group 58

dx/dt = xg(x) – yp(x) g(0)>0, g’(x)0, B>0









g(x)= 0Σmakxk , a0>0, a10, a10, p(0)=0

Find x=x*, s/c = b ln(x +1), x*= e[s/(bc)] - 1

y*= (a0x+a1x2 )(c/s)









12/20/2001 Systems Dynamics Study Group 59

dx/dt = xg(x) – yp(x) g(0)>0, g’(x)0, B>0



B= g(x)+xg’(x)-xg(x)p’(x)/p(x), for x=x*

= a0 + a1x + a1x – x(a0 + a1x)b/[x+1] /(b [ln (x+1)])

= a0 (1-(cb/s)[x/(x+1)]) + 2a1x - [(a0 + a1x)b[x/(x+1)]/(s/c)], since p(x*)=s/c

= a0 (1-(cb/s)[x/(x+1)]) + a1x(2-(cb/s)[x/(x+1)]) > 0,

a0 > - a1x[(2s-cb)[x/(x+1)]/s][s/(s-cb[x/(x+1)])]

a0 > - a1x[(2s-cb[x/(x+1)])/(s-cb[x/(x+1)])]

2s>cb[x/(x+1)] and s>cb[x/(x+1)] => s>cb[x/(x+1)]

or 2s s cb>cb[x/(x+1)], since x - a1x[(2s-cb[x/(x+1)])/(s-cb[x/(x+1)])]

4) Verify y* = (a0x+a1x2 )(c/s) >0

5) Verify that B>0



12/20/2001 Systems Dynamics Study Group 60

dx/dt = xg(x) – yp(x)

Log (Cont’d)

g(0)>0, g’(x)0, B>0







Model Parameter Selections

1) b=4, c=10, s > bc=40, Let s=200

2) x* = e[200/(4*10)] –1 = e 5 –1 ~ 148.4 – 1 = 147.4

3) a0 > - a1x(2s-cb[x/(x+1)])/(s-cb[x/(x+1])

= - a1*147.4(2*200 - 40[1])/(200 - 40[1]) , since [x/(x+1)]~1

= - a1*147.4(360/160),

= - a1(332.7)

let a1= -0.15, a0 > 49.7 , Let a0 > 54

4) y* = x*g(x*)/p(x*) = 147.4(200-0.15*147.4)/(200/10) =

~ 235.0 > 0

5) B = g(x) + xg’(x) – xg(x)p’(x)/p(x) =

= (a0+a1x) + 2a1x – b[x/(x+1)] (a0+a1x)

= [54 – 0.15(147.4)] + 4(-0.15)[54 – 0.15(147.4)] , since [x/(x+1)]~1

= 54 – 44.2 – 6.4 = 3.4 > 0



12/20/2001 Systems Dynamics Study Group 61

Thank you!



12/20/2001 Systems Dynamics Study Group 62


Shared by: xiaohuicaicai
Other docs by xiaohuicaicai
LOGFRAMES_ MONITORING AND EVALUATION
Views: 0  |  Downloads: 0
JELSApndx3SophLanguage
Views: 0  |  Downloads: 0
1997TrumpetCompetitionNYTimes
Views: 0  |  Downloads: 0
Eng_wk52_31
Views: 0  |  Downloads: 0
ENVIRONMENTAL MONITORING PROGRAMME FOR
Views: 0  |  Downloads: 0
Marketing - Ulster Business School
Views: 0  |  Downloads: 0
speech-swallowing
Views: 1  |  Downloads: 0
May_FY11_Awards_Report_Web
Views: 0  |  Downloads: 0
Related docs
By registering with docstoc.com you agree to our
privacy policy

You are almost ready to download!

You are almost ready to download!