Homework No. 2: Capacity Analysis. Little's Law. Partial Solution.

Question 8.8.1 Estimate the average number of back-logged cases and the average number of cases thatthe judge resolves per month by¯L =∑ 12i=1 L i12and ¯λ =∑ 12i=1 λ i12respectively.Then the average sojourn time of a case in the court per judge is (in months),W = ¯L ∑ 12¯λ = ∑ 12i=1 L ii=1 λ i.8.2 Leti = index of month;j = index of judge.The arrival rate to judge j is equal toThe overall arrival rate to the system:λ j = 1 12 λ ij .λ = ∑ jλ j .According to 8.1, we can estimate the average sojourn time at judge j byW j =∑ 12i=1 L ij∑ 12i=1 λ ij.Now, generalizing 6.1, we getW= ∑ jλ jλ W j .Alternative way.W =∑ ∑ 12j i=1 L ∑ij/12∑ ∑ 12j i=1 λ ij/12 = j L jλ=∑j λ jW jλ.2

Question 10. The ER (emergency room) can be divided into 4 subsystems:1. Queue for registration.2. Registration.3. Queue for doctors.4. Doctors.We know that the arrival rate λ to all subsystems is equal to 50 per hour = 5/6 per minute.Denote by W i and L i the waiting/processing time and average number of customers in subsystemi. We know:L 1 = 40; W 2 = 3 min; L 3 = 30; W 4 = 5 · 90% + 30 · 10% = 7.5 min.Now using Little formula for every subsystem we getW 1 = 405/6 = 48 min; L 2 = 5 6 · 3 = 156 ; W 3 = 305/6 = 36 min; L 4 = 5 · 7.5 = 6.25.6If we denote by L and W the average number of customers and the average waiting time in ERrespectively, thenL = L 1 + L 2 + L 3 + L 4 = 78.75; W = W 1 + W 2 + W 3 + W 4 = 94.5 min.Answer. A patience stays 94.5 minutes in the ER, on average. On average, 6.25 patients arebeing examined by doctors. There are 78.75 patients in the ER, on average.Remark. The ratiois very small.service timewaiting time = 3 + 7.548 + 36 = 0.1253