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For a better understanding of (426), it is helpful to point out that qkj can be interpreted as the long-run average number of transitions per time unit to state j when averaging over the time the process is in state k A rigorous proof of the result (426) is given in Section 43 By (426), the long-run average number of transitions into state j per time unit =

(427)

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Together (424), (425) and (427) give the equilibrium equations (422) These equations may be abbreviated as rate out of state j = rate into state j (428)

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This principle is the ow rate equation method To formulate the equilibrium equations in speci c applications, it is convenient to use the transition rate diagram that was introduced in the previous section Putting the transition rate diagram in a physical context, one might think that particles with a total mass of 1 are distributed over the nodes according to the equilibrium distribution {pj } Particles move from one node to another node according to the transition rates qij In the equilibrium situation the rate at which particles leave any node must be equal to the rate at which particles enter that node The rate in = rate out principle (428) allows for a very useful generalization More generally, for any set A of states with A = I , rate out of the set A = rate into the set A (429)

j A k A /

1929 1945

k A /

The balance principle (429) enables us to write down a recursive equation for the pj when I = {0, 1, , N } and qij = 0 for i 1 and j i 2, where N Then, by taking A = {i, , N } with i = 0 and applying the balance principle (429), we get

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qkj ,

(4210)

This recursive relation is used quite often in queueing applications; see 5 In queueing applications it is often the case that direct transitions from any state i are either to higher states or to the state i 1 directly below state i A recursive computation of the state probabilities is usually much faster than a computation by any other method Also the recursion scheme (4210) is numerically stable since it involves no subtractions Next we apply the ow rate equation method to the two examples discussed in the previous section Example 411 (continued) Inventory control for an in ammable product In this example the equilibrium probability pj represents the long-run fraction of time that the stock in the tank equals j units Assumptions 412 and 421 are trivially satis ed (eg take state Q as regeneration state r) Using the transition rate diagram in Figure 411 and equating the rate at which the process leaves state i to the rate at which the process enters state i, it follows that p0 = p1 , pj = pj +1 , pQ = p0 These equilibrium equations together with the equation p0 + p1 + + pQ = 1 have a unique solution (in this special case an explicit solution can be given: p0 = (1 + Q / ) 1 and p1 = = pQ = ( / )p0 ) Next we can answer the questions posed earlier:

j = 1, 2, , Q 1,

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( 0 0 )k k!

n Tuesday, October 29, 1929, stockholders dumped a record 164 million shares on the trading floor, opening the floodgates for $50 billion in stock losses and sparking the Great Depression A few months later, workers finished the Empire State Building; but the world s tallest structure remained almost completely empty until after World War II because no one could pay the rent1 When a reporter asked Babe Ruth, who signed a contract for $80,000 in 1930, if he felt it was unseemly to make more money than President

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