Unformatted text preview: 2 IEOR 4404, Assignment #5 Solutions 3. The variables needed are: (1) the entering time of each customer, denoted by E (i) for the ith customer. (2) the tolerance time length of each customer, denoted by R(i) for the ith customer. (3) the leaving time of each customer, denoted by L(i) for the ith customer. Some other variables could be (1) time t. (2) service time S . The events are: (1) new customer joins this system. (2) customer exits the system since the tolerance time is exceeded. (3) the server ﬁnishes the service for one customer. The updating procedures are: (1) when new customer joins this system, the number of customers increases by 1. (2) when the smallest tolerance length is less than the current service length, some customer exits the system and the number of customers decreases by 1. (3) when the tolerance length of the next customer is larger than the current service length, the server ﬁnishes the service for the current customer and the number of customers in the system decreases by 1. Since we are interested in estimating the average number of customers by time T , we record the time that each customer spends in the system. Remark: It is possible to understand this question as (a) the average number of customers that have been served by time T , or (b) the average number of customers staying in the system during the interval [0, T ]. Here we deduce the algorithm for (b), while the algorithm for (a) can be obtained by simplifying several steps in the algorithm for (b). Step 1: I = 0, t = 0, the value of T . Step 2: According to the arrival process, generate the nonhomogeneous Poisson process. The procedure is like that in Problem 2. Now we have the total number of customers N that could be possibly involved before time T , and the entering time of each customer E (i), for i = 1, . . . , N . Step 3: According to distribution F , generate the tolerance time of each customer, R(i), for i = 1, . . . , N . Step 4: Initialize the leaving time of each customer in the case of not being served, L(i) = min(E (i) + R(i), T ), for i = 1, . . . , N . Once some customer is served, L(·) will be updated. Step 5: Record the arrival time set E = {E (i), for i = 1, . . . , N }. service starting time as S ∗ = min E . Step 6: According to G, generate the service time S . This service starts at S ∗ , the person served corresponds to min E , and service ends at S ∗ + S . If S ∗ + S > T , let L(arg min E ) = T stop; else, Set L(arg min E ) = S ∗ + S , go to step 7. IEOR 4404, Assignment #5 Solutions 3 Step 7: Update the waiting list E = E \{min E }, and also the customers whose tolerance time is exceeded, E = E \{e : e ∈ E and e < S ∗ + S }. If E = φ, stop; else go to step 8. Step 8: Set the next service time at S ∗ = max{S ∗ + S, min E }. Go to step 6. The average number of customers by time T is the average of
N i=1 [L(i) − E (i)] T 4. The arrival process is a Poisson process with rate 5. F is the uniform distribution on (0, 5). G is an exponential random variable with rate 4. T is 100. With 500 simulation runs, the expected average number in [0, T ] is estimated to be 5.80910 (the standard error is 0.03619), and the expected number of lost customers is estimated to be 118.66200 (the standard error is 0.92945). One possible MATLAB programming code is attached. 5. With the setting slightly changed, the expected average number in [0, T ] is estimated to be 10.14732 (the standard error is 0.06535), and the expected number of lost customers is estimated to be 93.87000 (the standard error is 1.21710). This result implies that the average number that depart before entering service will decrease. One possible MATLAB programming code is attached. 6. Extra Credit: For any time t ≥ 0, P (M (t) = m, W (t) = w) = P (M (t) = m, W (t) = wM (t) + W (t) = m + w) · P (M (t) + W (t) = m + w) = = m+w m pm (1 − p)w · e−λt (λt)m+w (m + w)! e−λtp (λtp)m e−λt(1−p) (λt(1 − p))w · m! w! The answer means both the processes of male arrivals M (t) and female arrivals W (t) are independent Poisson processes with rates λp and λ(1 − p), respectively. ...
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 Spring '10
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 Exponential distribution, average number

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