How to solve this question using pyhton's code For Checkpoint C you will extend Checkpoint B to do the following:1. Track and then output a summary report that features0 the average kelp population across the simulated timescale,o the average urchin population across the simulated timescale,o the minimum kelp population observed during the simulation,o the maximum urchin population observed during the simulation.Hint: your program will employ the min-finding and max-finding patterns we learned in class with for-loops.Sample OutputSample input/output behavior for the checkpoint are provided below. Your program's spacing, spelling, capitalization, and punctuation willneed to match the sample output EXACTLY for this project.Ex 1 Sample Input/OutputGiven inputs xx, 6, 7, 8, ko, uo, n as:1.5-001.052.5100210The program outputs==> Bull Kelp and Purple Urchin Population Simulator <==---Model ParametersKelp growth rate:Kelp death rate:Urchin birth rate:Urchin death rate:Initial PopulationKelp population (in thousands) at t = 0:Urchin population (in thousands) at t = 0:--- Simulation ---Timescale:Time t = 0: 100.000k kelp, 2.000k urchinsTime t = 1: 249.800k kelp, 7.000k urchinsTime t = 2: 622.751k kelp, 76.930k urchinsTime t = 3:Time t = 4:Time t = 5:Time t = 6:Time t = 7:1508.970k kelp, 2280.018k urchins331.946k kelp, 168603.957k urchins0.000k kelp, 2545463.659k urchins0.000k kelp, 0.000k urchins0.000k kelp, 0.000k urchinsTime t = 8: 0.000k kelp, 0.000k urchinsTime t = 9: 0.000k kelp, 0.000k urchinsTime t = 10: 0.000k kelp, 0.000k urchins--- Simulation Statistics ---Average kelp population: 255.770kAverage urchin population: 246948.506kMin kelp population was 0.000k at t=5.000Max urchin population was 2545463.659k at t=5.000 1 print('==> Bull Kelp and Purple Urchin Population Simulator <==\n')2 print('- Model Parameters ---')a float(input ("Kelp growth rate: \n"))if a<0:print("Error: cannot have a negative growth rate")exit()b =float(input("Kelp death rate: \n"))8 if b<0:91011 c =float12 if c<0:131445616171819 ko293031print("Error: cannot have a negative death rate")exit()ko = max(0, float(input()))20 u0= max(0, float(input()))21k =k033343522u =u023 print( \n--- Initial Population ---')24 print (f"Kelp population (in thousands) at t = 0: ")25 print (f"Urchin population (in thousands) at t = 0: \n")2627 print('--- Simulation ---')m m mprint("Error:exit()36(input("Urchin birth rate: \n"))print("Error: cannot have a negative birth rate")exit()d =float (input ("Urchin death rate: \n"))if d<0:37cannot have a negative death rate")n = int(input("Timescale: \n"))if n < 0:print("Error: cannot have a negative timescale ")exit()for t in range (n+1): # 0, 1print (f"Time t = {t}: {k:.3f}k kelp, {u:.3f}k urchins")k_next = max(0, k + a*k - b*k*u)u_nextmax (0, u + c*k*u - d*u)k = k_nextu = u next

Program Approach : Take input from the user for the model parameters: kelp growth rate, kelp…

For Checkpoint C you will extend Checkpoint B to do the following: 1. Track and then output a summary report that features o the average kelp population across the simulated timescale, o the average urchin population across the simulated timescale, o the minimum kelp population observed during the simulation, o the maximum urchin population observed during the simulation. Hint: your program will employ the min-finding and max-finding patterns we learned in class with for-loops. Sample Output Sample input/output behavior for the checkpoint are provided below. Your program's spacing, spelling, capitalization, and punctuation will need to match the sample output EXACTLY for this project. Ex 1 Sample Input/Output Given inputs xx, 8, 7, 8, ko, uo, n as: 1.5 .001 .05 100 2 10 The program outputs =>Bull Kelp and Purple Urchin Population Simulator <== Model Parameters --- Kelp growth rate: Kelp death rate: Urchin birth rate: Urchin death rate: ---Initial Population --- Kelp population (in thousands) at t = 0: Urchin population (in thousands) at t = 0: --- Simulation --- Timescale: Time t= 0: 100.000k kelp, 2.000k urchins Time 1: 249.800k kelp, 7.000k urchins Time = 2: 622.751k kelp, 76.930k urchins Time = 3: 1508.970k kelp, 2280.018k urchins Time 4: 331.946k kelp, 168603.957k urchins Time = 5: Time = 6: Time = 7: 0.000k kelp, 2545463.659k urchins 0.000k kelp, 0.000k kelp, 0.000k urchins 0.000k urchins Time 8: 0.000k kelp, 0.000k urchins Time = 9: 0.000k kelp, 0.000k urchins Time 10: 0.000k kelp, 0.000k urchins

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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How to solve this question using pyhton's code

Expert Solution

Step 1

Program Approach :

Take input from the user for the model parameters: kelp growth rate, kelp death rate, urchin birth rate, and urchin death rate.

Initialize variables to store the population of kelp and urchin at each time step.

Use a for loop to simulate the population of kelp and urchin over the specified timescale. At each time step, update the population of kelp and urchin using the given differential equations, and store the new values in the respective variables.

Calculate the average population of kelp and urchin over the simulated timescale by taking the mean of the population values stored in the respective variables.

Find the minimum kelp population and the maximum urchin population observed during the simulation by iterating through the population values stored in the respective variables and keeping track of the minimum and maximum values.

Print the simulation results, including the model parameters, initial population, simulated population over time, and simulation statistics (average population, minimum kelp population, and maximum urchin population).
End the program.