Jo 13 M, =30 = My 3. Let's look at a linked series of M/M/1 computers: (i) 10 jobs a minute are being sent to computer 1 which takes 2 seconds for each job. 50% of the completed jobs are sent to computer 2 (the rest are done and go out of the sys (ii) Computer 2 also takes 2 seconds for each job. All of the finished jobs are sent to computer 3. (iii) 10 jobs a minute are being sent directly to computer 3. Computer 3 takes 1 second per job. 50% of the finished jobs are sent to computer 1 and 30% are sent to computer 2 (the rest are done). a) (4 pts) Set up the system of equations and solve for A1, 12, and 13. M3 = 60 * , = 10 + .5 x3 1 , = 10 + . 5 ( 107 12 ) 2 : 12 = . 5 * , +, 343 12 = 1 51, +13 (10+ dz ) 3 : to = 10 + 2 2 d , - 15 1 2 = 15 12 = 23, 33 = 703 + 2 ( - 15x,. 7. 7 /2 = 3 ) * , = 15 + 15 12 = 30 1 9 /2 = 21 13 = 100 / 3 099 b) (3 pts) Find E(N1), E(N2), and E(N3). . 173 E ( N ) = M - 1 1 ' Po = T PI=$1 - E ( N. ) = 8 1. 2 , P1 9 9 Cor PI = 9 , p2 = / 9 E ( N2.) = 3.5 2: Po = 2 5.4 P. = 9 9 p3 = $ ... ] E ( N ) = 11250 3: Po = . 247 c) (3 pts) Find the probability that there is a total of 1 job in the system. Let ( a , b, c ) = ( # jobs in ), # in 2, # in J p ( total = 1 ) = p ( ), ,, 0 ) + 1 / 0 , 1 , 0 ) + P ( 0 , 0 , , ) 0 PR = ( 1 -p )p - po = Ip pic ( 1- p ) p 5 0244 d) (3 pts) Find E(T) (you can use Little's Theorem with A being the rate that jobs are entering the system, you will also need the sum of the mean number of jobs at each computer). E ( N ) = E ( N. ) + E ( NZ ) + E ( NB ) = 12.75 0 1 = 20 @ - E ( T ) = 30 ( 12 . 75 ) = 3 = . 6375 min2. For the M/ M/ 3 system, where 2.5 jobs arrive each minute on average and it takes 1 minute to nish each: a) (3 pts) nd the the steady state probabilities 130,131,302, 333. b) (2 pts) nd P(W