Practical Research I Senior High Learning Guide, Summaries of Geology

Download Practical Research I Senior High Learning Guide and more Summaries Geology in PDF only on Docsity! Decision Modeling (NPTEL Online Course) Tutorial 4 (Queuing Theory - Module 16 to Module 20) ------------------------------------------------------------------------------------------ 1. Which type of queuing system is likely to form in front of two counters in a small railway ticketing centre (one for enquiry and the other for ticket booking) with Poisson arrival and Exponential Service? i. One M/M/2 ii. Two M/M/2 iii. One M/M/1 iv. Two M/M/1 2. Under M/D/1 model, average number of customers in the system is given by i. ii. iii. iv. 3. In a small barber shop, only one customer can get hair cut while another customer can wait in a chair. Any other arriving customer has to wait outside as there is only one chair available. The customers arrive randomly at 6 per hour. The service is exponential and takes 6 minutes on the average. Find the probability that an arriving customer will have to wait outside. i. 36 % ii. 40 % iii. 60 % iv. 64 % 4. An overhead crane of ABC Ltd. moves jobs from one machine to another and must be used every time a machine requires loading or unloading. The demand for service is random. Data taken by recording the elapsed time between service calls followed an exponential distribution having a mean of a call every 24 minutes. In a similar manner, the actual service time of loading or unloading took an average of 8 minutes. If the machine time is valued at Rs. 8.50 per hour, how much does the downtime cost per day? [Assume 8 hour work per day] i. 30 ii. 32 iii. 34 iv. 36 5. Arrivals to single bank counter are poisson distributed with a rate of 20 per hour. The average time for a customer to get service is 2 minutes and this time is exponentially distributed. What would be the average waiting time of a customer in the system? i. 2 minutes ii. 4 minutes iii. 6 minutes iv. 8 minutes 6. Consider the previous problem once again. Arrivals to single bank counter are poisson distributed with a rate of 20 per hour. The time for a customer to get service, however, is constant at 2 minutes. What would be the average waiting time of a customer in the system? i. 2 minutes ii. 4 minutes iii. 6 minutes iv. 8 minutes 7. A small railway ticket booking office has two counters – Counter 1 for enquiry and Counter 2 for ticket booking. Customer arrival is Poisson at 5 per hour to the enquiry and 10 per hour to the ticket booking counter. Exponentially distributed service time in each counter is 4 minutes per customer. Find by how much the average waiting time of a customer in the system reduces at Counter 1 (original enquiry counter) when the office decides to go for pooling of resources – i.e. an arriving customer will get enquiry or ticket booking facility at any of the counters. i. 0.333 minute ii. 0.667 minute iii. 2 minutes iv. 6.667 minutes 8. In a restaurant, customer arrival is Poisson at 10 per hour. In this restaurant, the customers do self-service. Exponentially distributed service time 3 minutes per customer. Find the average waiting time of a customer in the restaurant. i. 3 minutes ii. 6 minutes iii. 9 minutes iv. 12 minutes So, the probability that an arrival will find the place free = p(0) = 1 – = 1 – (3/5) = 2/5 p (1) = p(0) = 6/25 The probability that an arriving customer will have to wait outside = 1-[2/5 + 6/25] = 9/25 = 36% 4) Arrival rate =  = 60 2.5 per hour 24  Service rate = 60 7.5 per hour 8    Average waiting time in the system = 1 1 0.20 7.5 2.5      Average number of units in the system = 1 * 0.5     Number of hour per day is 8 hour Cost per machine hour Rs. 8.50 Total downtime cost per day = 0.5*8*8.50 Rs.34 5) Arrival rate = 20 per hour  Service rate = 2 minute per customer = 30 customers per hour   System Utilization factor = 20 2 30 3       Length of the system, L = 2 3 2 11 3      Average waiting time in system = 2 6 minutes 1 3 L    6) Since the service time is constant, the system is M/D/1 System Utilization factor = 20 2 30 3       Length of the system = 21 8 2 1 6       Average waiting time in system = 8 6 4 minutes 1 3 L    7) We have two counters (i) Enquiry counter (ii) ticket booking counter The arrival rate at the enquiry counter, 5 per hour  The service rate at the enquiry counter, 15 per hour  The arrival rate at the ticket booking counter, 10 per hour  The service rate at the ticket booking counter, 15 per hour  Case 1: Without pooling the resources Enquiry counter System Utilization factor, 5 1 15 3    Average no. of customers in system, L = /(1 – ) = (1/3)/(1 – 1/3) = 1/2 Average waiting time in system W = L/λ = (1/2)/5 = 1/10 Hour = 6 mins Ticket booking counter System utilization factor, = λ/μ = 10/15 = 2/3 Average no. of customers in system, L = /(1 – ) = (2/3)/(1 – 2/3) = 2 Average waiting time in system W = L/λ = 2/10 = 1/5 Hour = 12 mins Case 2: With pooling the resources Arrival rate of customers at the counters, λ = 5+10 = 15 per hour Service rate at any one counter = 15 per hour System utilization factor, = λ/sμ = 15/(2*15) = ½ Using the M/M/2 queuing formula, we have, 1 1 0 0 ( / ) ( / ) 1 ! ! 1 ( / ) n ss n P n s s                   = [((15/15)0/0!)+(15/15)1/1!)+(15/15)2/2!)*(1/(1–1/2))]-1 = [1+1+(1/2)*2]-1 =1/3 Wq = Lq/λ =(1/3)/15 =1/45 W = Wq+(1/μ) =(1/45)+(1/15)=4/45 = 5.33 mins. how much the average waiting time of a customer in the system reduces at Counter 1 (original enquiry counter) when the office decides to go for pooling of resources? 6 mins – 5.33 mins = 0.667 minute 8) Arrival rate λ = 10/hour Service rate = 3 mins per customer = 20/hour System utilization factor, = λ/μ = 10/20 = 1/2 So, Hence, Busy Period = 1–0.606 = 0.394 = 39.4% Average No. of customers in the restaurant Average waiting time in the restaurant: hour = 3 minutes 9) Arrival rate λ = 4/hour Service rate = 5/hour System utilization factor, = λ/μ = 4/5 Average number of machines in repair, L = /(1 – ) = (4/5)/(1-4/5) = 4 Cost of non-productive machine is Rs. 200 per hour (A) Total cost of Non productive machine is 200* 4 = Rs. 800 (B) Repairman’s charge = Rs 100 02 ( / ) ( ) ... !(1 ) s q n n s L n s P P s             2 2 (15 /15) *(1/ 2) *(1/ 3) 2!(1 1/ 2)   1 3  1 2 0 0.606P e e        1 2 L a      1 1 20 L W     