A group of n processors is arranged in an ordered list. When a job arrives, the firstprocessor in line attempts it; if it is unsuccessful, then the next in line tries it; if ittoo is unsuccessful, then the next in line tries it, and so on. When the job is suc-cessfully processed or after all processors have been unsuccessful, the job leavesthe system. At this point we are allowed to reorder the processors, and a new jobappears. Suppose that we use the one-closer reordering rule, which moves theprocessor that was successful one closer to the front of the line by interchangingits position with the one in front of it. If all processors were unsuccessful (or ifthe processor in the first position was successful), then the ordering remains thesame. Suppose that each time processoriattempts a job then, independently ofanything else, it is successful with probabilitypi(a) Define an appropriate Markov chain to analyze thismodel.
(b) Show that this Markov chain is time reversible.
(c) Find the long-run probabilities.

The solution from the solution manual is the following: Consider states x=(x1,...,xi−1,xi,xi+1,...,xn) and x1=(x1,...,xi−1,xi+1,xi,...,xn). Note that each time j is moved to the left we multiply by qj/pj and each time it moves to the right we multiply by (qj/pj)−1. Since xj, which is initially in position j, is to have a net move of j−xj positions to the left (so it will end up in position j−(j−xj)=xj) it follows from the above that π(x)=C∏j(qxj/pxj)j−xj. The value of C, which is equal to π(1,2,...,n), can be obtained by summing over all states x and equating to 1. Since the solution given by the above value of π(x) satisfies the time reversibility equations it follows that the chain is time reversible and these are the limiting probabilities.

( - 4 , -7 )

all the lines must parallel  

step-by-step explanation:

coordinates of c = ( - 1 , 2 )

3 units left = ( -4 , 2 )

9 units down = ( - 4 , 11 )

find the vertical and horizontal asymptotes of the function f x = 5 x − 1 . to find the vertical asymptote, equate the denominator to zero and solve for . since the degree of the polynomial in the numerator is less than that of the denominator, the horizontal asymptote is .

step-by-step explanation: