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# hmwk10_11 - ISyE 3232 H Ayhan Stochastic Manufacturing and...

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Unformatted text preview: ISyE 3232 H. Ayhan Stochastic Manufacturing and Service Systems Fall 2011 Homework 10 November 16, 2011 (For your practice only) 1. Suppose there are two tellers taking customers in a bank. Service times at a teller are independent, exponentially distributed random variables, but the ﬁrst teller has a mean service time of 3 minutes while the second teller has a mean of 6 minutes. There is a single queue for customers awaiting service. Suppose at noon, 3 customers enter the system. Customer A goes to the ﬁrst teller, B to the second teller, and C queues. To standardize the answers, let us assume that TA is the length of time in minutes starting from noon until Customer A departs, and similarly deﬁne TB and TC . (a) What is the probability that Customer A will still be in service at time 12:05? (b) What is the expected length of time that A is in the system? (c) What is the expected length of time that A is in the system if A is still in the system at 12:05? (d) How likely is A to ﬁnish before B? (e) What is the mean time from noon until a customer leaves the bank? (f) What is the average time until C starts service? (g) What is the average time that C is in the system? (h) What is the average time until the system is empty? (i) What is the probability that C leaves before A given that B leaves before A? (j) What are the probabilities that A leaves last, B leaves last, and C leaves last? (k) Suppose D enters the system at 12:10 and A, B, and C are still there. Let WD be the time that D spends in the system. What is the mean time that D is in the system? 2. Suppose passengers arrive at a MARTA station between 10am -5pm following a Poisson process with rate λ = 680 per hour. For notation, let N (t) be the number of passengers arrived in the ﬁrst t hours, S0 = 0, Sn be the arrival time of the nth passenger, Xn be the interarrival time between the (n − 1)st and nth passenger. (a) What is the expected number of passengers arrived in the ﬁrst 2 hours? (b) What is the probability that no passengers arrive in the ﬁrst one hour? (c) What is the mean number of arrivals between noon and 1pm? (d) What is the probability that exactly two passengers arrive between 2pm and 4pm? (e) What is the probability that ten passengers arrive between 2pm and 4pm given that no customer arrive in the ﬁrst half hour? (f) What is the average time of the ﬁrst arrival? (g) What is the expected time of the thirtieth arrival? (h) What is the expected time of the ﬁrst arrival given that no passenger arrives in the ﬁrst hour? (i) What is the expected arrival time of the ﬁrst passenger who arrives after 2pm? ISyE 3232 Stochastic Manufacturing and Service Systems Fall 2011 H. Ayhan Solutions to Homework 10 1. Let S1 and S2 be independent and exponentially distributed random variables with means 1 1 µ1 = 3 and µ2 = 6. They corresponds to the service time of the ﬁrst and the second teller. 1 (a) Since A starts service immediately after arrival, P(TA ≥ 6) = P(S1 ≥ 6) = e− 2 ×6 = 0.0498. (b) By the same reason, E(TA ) = E(S1 ) = 3. (c) This amount, in mathematical expression, will be E(TA | TA > 6) E(S1 | S1 > 6) = 6 + E(S1 ) // by memoryless property of exponential random variables = = 9. (d) P(TA < TB ) = P(S1 < S2 ) = µ1 = 2/3. µ1 + µ2 (e) The time from noon till a customer leaves is min{TA , TB }. Its expectation will be the same as 1 E(min{S1 , S2 }) = = 2. µ1 + µ2 (f) If a customer leaves, one of the tellers will be able to serve C . So it is the same as time till one departure, which is computed in (e). (g) The total time C spends in the system will be the summation of the waiting time min{TA , TB } and its service time. The service time will depend on at which teller the customer gets service. So by conditioning, we have E(service time of C ) = = = E ( S 1 | T A < T B ) P ( T A < T B ) + E ( S 2 | T B < T A ) P( T B < T A ) 3 × 2/3 + 6 × 1/3 by problem (e) 4. Thus, the expected time C in the system is 2 + 4 = 6. (h) The expected time from noon till C start service is E(min{TA , TB }), which has already been studied in (e). Let’s now study the expected time from C enter the service till the system is empty. No matter which teller C goes to, both of the teller are busy upon this time. By memoryless property, the two tellers can be imagined just start service together at this time. The expected time left for both of them to ﬁnish is still E(max{S1 , S2 }). So E(time until system is empty) = E(min{TA , TB }) + E(max{S1 , S2 }) = = 2E(min{S1 , S2 }) + E(S1 ) + E(S2 ) − E(min{S1 , S2 }) E(S1 ) + E(S2 ) = 9. 1 (i) B leaves before A means that C start service at teller 2. P( T C < T A | T B < T A ) = = P(S2 < S1 ) // again by memoryless property µ2 = 1/3. µ1 + µ2 (j) The probability that A leaves last is P ( T B < T A , TC < T A ) = = P( T C < T A | T B < T A ) P( T B < T A ) 1/3 × 1/3 = 1/9. The probability that B leaves last is P ( T A < T B , TC < T B ) = = = P( T C < T B | T A < T B ) P( T A < T B ) P(TC < TB | TA < TB ) × 2/3 2/3 × 2/3 = 4/9. // computation of the ﬁrst 2/3 is similar as in (i) The probability that C leaves last is P ( T A < T C , TB < T C ) = P(TA < TC , TB < TA ) + P(TB < TC , TA < TB ) = P ( T A < T C | T B < T A ) P( T B < T A ) + P( T B < T C | T A < T B ) P( T A < T B ) = 2/3 × 1/3 + 1/3 × 2/3 = 4/9. (k) By memoryless property, we can imagine that everything just starts at 12:11. In order that D enter service, two customers must leave. The average time from 12:11 till one departure is same as in (e), which is 3. The average time from then to next departure will be the same, 3, due to memoryless property. Now, it is left to compute the expected service time of D. Using a similar conditioning in (g) E(service time of D ) = E ( S 1 | T A < T B , TC < T B ) P ( T A < T B , TC < T B ) + E ( S 1 | T B < T A , TA < T C ) P ( T B < T A , TA < T C ) + E ( S 2 | T B < T A , TC < T A ) P ( T B < T A , TC < T A ) = = + E ( S 2 | T A < T B , TB < T C ) P ( T A < T B , TB < T C ) 3 × 2/3 × 2/3 + 3 × 1/3 × 2/3 + 6 × 1/3 × 1/3 + 6 × 2/3 × 1/3 4. So the total time D spends in the system is 2 + 2 + 4 = 8. 2. (a) P(T ≤ 1) = 1 − e−10/7 ≈ 0.7603 and P(T ≤ 1) = 1 − e−20/7 ≈ 0.9426. (b) The tardiness will be T ￿ = max (T − 1), 0. ￿ E( T ￿ ) = = ∞ (t − 1)λe−λt dt 1 ￿∞ −λ e sλe−λs dt 0 = 2 e 10/7 0.7 ≈ 0.1678 (c) Let g1 (t), g2 (t), g3 (t) denote the proﬁt for the “\$1250”, “\$1000”, “\$750” contract with delivery time t respectively. Then they can be written as: 0 ≤ t ≤ 1/2 1250 2500(1 − t) 1/2 < t ≤ 1 g 1 ( t) = 0 t>1 0≤t≤1 1000 1000(2 − t) 1 < t ≤ 2 g 2 ( t) = 0 t>2 0≤t≤7 750 750(14 − t)/7 7 < t ≤ 14 g 3 ( t) = 0 t > 14 Let T be a random variable with parameter δ , let’s compute E[g1 (T )]: f1 (δ ) = E[g1 (T )] = = ￿ 0 .5 0 1 1250 d−t/δ dt − δ 1 ￿ 1 1 2500(1 − t) d−t/δ dt δ 0.5 1 1250 − 2500δ (e− 2δ − e− δ ) Similarly, f2 (δ ) = E[g2 (T )] = f3 (δ ) = E[g3 (T )] = 1 2 1000 − 1000δ (e− δ − e− δ ) 14 750 − 7 750 − δ ( e δ − e− δ ) 7 Plug-in 0.7 into each of the expected proﬁt function, f1 (0.7) = 812.69, f2 (0.7) = 872.45 and f3 (0.7) = 749.99. Hence, choosing the “\$1000” contract will give us the max expected proﬁt. (d) Using a software package like Maple or Mathematica, we can plot fi (δ ) for i = 1, 2, 3. f1 (·) achieves the largest value, which means that “\$1250 contract” is optimal, if 0 ≤ δ ≤ 0.5375; f2 (·) achieves the largest value, which means that “\$1000 contract” is optimal, if 0.5375 ≤ δ ≤ 1.055; f2 (·) achieves the largest value, which means that “\$750 contract” is optimal, if 1.055 ≤ δ . 3 ...
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