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solver.py
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'''
Created on Sep 3, 2018
@author: mohame11
'''
import math
import numpy as np
from scipy.optimize import *
def M_M_1_K_log_solve(x, K, PK):
print x
logp = math.log(1-x) + K*math.log(x) - math.log(1-(x**(K+1)))
return logp - math.log(PK)
def M_M_1_K_solve(x, K, PK):
p = (1.0-x) * (x**K) / (1.0 - (x**(K+1)))
return p - PK
def M_M_1_K(x, K):
p = (1.0-x) * (x**K) / (1.0 - (x**(K+1)))
return p
def M_M_m_K_log(x, m, K):
#print x
c1 = K * math.log(x)
c2 = math.log(math.factorial(m))
c3 = (K-m) * math.log(m)
logC_K = c1 - c2 - c3
part2 = 0.0
for n in range(1,m):
part2 += float(x**n) / float(math.factorial(n))
part3 = 0.0
for n in range(m,K+1):
part3 += float(x**n) / float(m**(n-m))
logP0 = -1 * math.log(1.0 + part2 + part3)
logP_K = logC_K + logP0
return logP_K
def M_M_m_K_log_solve(x, m, K, PK):
#print x
c1 = K * math.log(x)
c2 = math.log(math.factorial(m))
c3 = (K-m) * math.log(m)
logC_K = c1 - c2 - c3
part2 = 0.0
for n in range(1,m):
part2 += float(x**n) / float(math.factorial(n))
part3 = 0.0
for n in range(m,K+1):
part3 += float(x**n) / float(m**(n-m))
logP0 = -1 * math.log(1.0 + part2 + part3)
logP_K = logC_K + logP0
return logP_K - math.log(PK)
def f(x, m, K):
#print x
part1 = (x**K) / (math.factorial(m) * m**(K-m))
#part1 = K * math.log(x) - math.log((math.factorial(m) * m**(K-m)))
#part1 = math.exp(part1)
part2 = 0.0
for n in range(1,m):
part2 += float(x**n) / float(math.factorial(n))
part3 = 0.0
for n in range(m,K+1):
part3 += float(x**n) / float(m**(n-m))
part3 *= 1.0/math.factorial(m)
tot = part1 * (1.0/(1.0+part2+part3))
#tot = part1 + math.log(1.0/(1.0+part2+part3))
#tot = math.exp(tot)
return tot
#return tot - PK
def M_M_m_K(x, m, K):
#print x
part1 = (x**K) / (math.factorial(m) * m**(K-m))
#part1 = K * math.log(x) - math.log((math.factorial(m) * m**(K-m)))
#part1 = math.exp(part1)
part2 = 0.0
for n in range(1,m):
part2 += float(x**n) / float(math.factorial(n))
part3 = 0.0
for n in range(m,K+1):
part3 += float(x**n) / float(m**(n-m))
part3 *= 1.0/math.factorial(m)
tot = part1 * (1.0/(1.0+part2+part3))
#tot = part1 + math.log(1.0/(1.0+part2+part3))
#tot = math.exp(tot)
#return tot
return tot
def M_M_m_K_solve(x, m, K, PK):
#print x
part1 = (x**K) / (math.factorial(m) * m**(K-m))
#part1 = K * math.log(x) - math.log((math.factorial(m) * m**(K-m)))
#part1 = math.exp(part1)
part2 = 0.0
for n in range(1,m):
part2 += float(x**n) / float(math.factorial(n))
part3 = 0.0
for n in range(m,K+1):
part3 += float(x**n) / float(m**(n-m))
part3 *= 1.0/math.factorial(m)
tot = part1 * (1.0/(1.0+part2+part3))
#tot = part1 + math.log(1.0/(1.0+part2+part3))
#tot = math.exp(tot)
#return tot
return tot - PK
def main():
m = 1 # #workers
K = 972 # queue size in terms of #request
my_lambda = 8400 #arrival per unit time
PK = 1775.0/124259 #prob of failure
#PK = 0.001
rho_0 = 0.1 #lambda/mu (initial point to start)
#mu Amit's intuition is 8400 packet/sec
print 'PK=', PK
rho = fsolve(M_M_m_K_log_solve, rho_0, (m, K, PK)) #solve for rho
diff = M_M_m_K_log_solve(rho, m, K, PK)
pkk = f(rho, m, K, PK)
mu = my_lambda / rho
print 'using M/M/m/K \nrho_0=%.5f, rho_final=%.5f, PK\'=%.5f, PK\'-PK=%.5f, mu=%.5f, lambda=%.5f' % (rho_0, rho, pkk, diff, mu, my_lambda)
'''
rho = fsolve(M_M_1_K_log_solve, rho_0, (K, PK)) #solve for rho
diff = M_M_1_K_log_solve(rho, K, PK)
pkk = M_M_1_K(rho, K)
mu = my_lambda / rho
print 'using M/M/1/K\nrho_0=%.5f, rho_final=%.5f, PK\'=%.5f, PK\'-PK=%.5f, mu=%.5f' % (rho_0, rho, pkk, diff, mu)
'''
#sol = newton(f, x, (m, K, PK))
#diff = f(sol, m, K, PK)
#print 'using newton\nrho_0=%.5f, rho_final=%.5f, PK\'-PK=%.5f' % (x, sol, diff)
if __name__ == "__main__":
main()
print('DONE!')