Page 428 - A First Course In Stochastic Models
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EXERCISES                            423

                Laplace transform of the equilibrium excess density of the vacation time. Decomposition
                results of this type are discussed more generally in Fuhrmann and Cooper (1985).
                9.7 Consider the (R, S) inventory model with limited order sizes. In this model the inventory
                position is reviewed every R periods. At each review the inventory position is ordered up
                to the level S provided that the order size does not exceed the constant Q; otherwise, the
                replenishment order is of size Q. It is assumed that Q > µ R , where µ R is the mean demand
                between two reviews. The lead time of a replenishment order is negligible. The cumulative
                demands between successive reviews are independent random variables. Demand in excess
                of on-hand inventory is back ordered.
                  (a) Let the random variable ξ k denote the cumulative demand between the kth and (k+1)th
                review and let   i denote the difference between S, the order-up-to level, and the inventory
                position just after the ith review. Verify the Lindley equation
                                       i = max(0,   i−1 + ξ i−1 − Q).
                  (b) Use the results (9.5.5) and (9.5.17) for the D/G/1 queue to derive an explicit expres-
                sion for the long-run fraction of demand that is back ordered when the demand variables ξ k
                have a Coxian-2 distribution.
                9.8 A certain product is produced at a constant rate of r > 0. The product is temporarily
                stored in a finite buffer with capacity K. The production is stopped when the buffer is full.
                A stopped production is resumed as soon as the stock level falls below K by a customer
                demand. Customers asking for the product arrive according to a Poisson process with rate
                λ. The demand of each customer is for a constant amount of D. The customer is satisfied
                with the amount in the buffer when the stock level is below D. It is assumed that λD < r.
                One wishes to choose the buffer size K such that the long-run fraction of customers with
                partially unsatisfied demand is below a prespecified level α with α small. Use results from
                Section 9.4.1 to show that the required buffer K(α) is approximately given by
                                                1    γ δ  !
                                         K(α) ≈  ln     ,
                                                δ    λα
                where δ > 0 is the unique solution of e δD  = 1 + rδ/λ and the constant γ is given by
                γ = (1 − ρ)/(δD − (1 − ρ)) with ρ = λD/r.
                9.9 A finite buffer storing a liquid material is emptied at a constant rate of r > 0. Customers
                bringing in the liquid material arrive according to a Poisson process with rate λ. The buffer
                has a finite capacity of K > 0. If a customer brings in an amount of work that is larger than
                the remaining room in the buffer, the whole amount of work of the customer is rejected. The
                amounts of work brought in by the customers are independent and identically distributed
                positive random variables. This queueing model is known as the M/G/1 queue with bounded
                sojourn time. Let π rej (K) be defined as the long-run fraction of customers who are rejected.
                  (a) For the case that the amount of work brought in by a customer is a constant D, argue
                that π rej (K) equals the loss probability in the M/G/1 queue with impatient customers
                from Section 9.4.2, where the service time equals D/r and the impatience time τ equals
                (K − D)/r. In particular, conclude that
                                               2 δD
                                         (1 − ρ) e  −δK
                                 π rej (K) ∼       e     as K → ∞
                                         δD − (1 − ρ)
                where ρ = λD/r < 1 and δ > 0 is the unique solution of e δD  = 1 + rδ/λ. If the amount of
                work brought in by a customer has an exponential distribution with mean α, then it follows
                from results in Gavish and Schweitzer (1977) that
                                              −ρ −(1−ρ)K/α
                               π rej (K) ∼ (1 − ρ)e  e     as K → ∞
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