Part A Julian Argo is a computer technician in a large insurance company. He responds to a variety of complaints from agents regarding their computers’ performance. He receives an average of one computer per hour to repair, according to a Poisson distribution. It takes Julian an average of 50 minutes to repair any agent’s computer. Service times are exponentially distributed. What percent of the time is Julian busy fixing computers? What is the average number of computers waiting for repair or being repaired? What is the probability that at any time there are at most 3 computers waiting to be repaired? What percent of the time will a computer have to wait for service? On average how long will a computer have to wait before Julian can begin working on it? Part B Consider the situation in Part A. The number of computers that need to be repaired has increased to 2 per hour, Poisson distributed. To deal with this situation two solutions have been proposed. (i) Julian can hire a helper to help him with the repair. The time for a repair goes down to 25 minutes, exponentially distributed. Since Julian and his helper will work on one computer, this is still an M/M/1 system. (ii) Another technician can be hired who will work at the same pace as Julian (50 minutes to repair a computer). However, arriving computers will be assigned at random to either Julian or the new technician. Thus the two technicians will each have their own line. Work rules prevent one technician from repairing computers in the other technician’s line. If you were (the owner of) a computer, which solution would you prefer – (1) or (ii)? As the manager of the repair shop which solution would you prefer – (i) or (ii)? Clearly state your reasons. Base your answers on queue characteristics for each solution.