University:
California State University, Northridge
Course:
SOM 686 | Operations Management
Academic year:

2024
Views:

435

Pages:

20
Author:

Nevaeh Bauer

Safety Capacity Exponential Model 20 In the exponential model, the interarrival and processing times are assumed to be independently and exponentially distributed with means 1/Ri and Tp. Independence of interarrival and processing times means that the two types of variability are completely unsynchronized. Complete randomness in interarrival and processing times. Exponentially distribution is Memoryless: regardless of how long it takes for a person to be processed we would expect that person to spend the mean time in the process before being released. Safety Capacity The Exponential Model 21 Poisson Arrivals – Infinite pool of potential arrivals, who arrive completely randomly, and independently of one another, at an average rate Ri , constant over time Exponential Processing Time – Completely random, unpredictable, i.e., during processing, the time remaining does not depend on the time elapsed, and has mean Tp Computations – Ci = Cp = 1 – K = ∞ , use Ii Formula – K < ∞ , use Performance.xls Safety Capacity Example 22 Interarrival time = 6 secs  Ri = 10/min Tp = 5 secs  Rp = 12/min for 1 server and 24 /min for 2 servers Rs = 12-10 = 2 c ρ Rs Ii Formula Ti= Ri / Ii T= Ti+ 5/60 I= Ii + c ρ 1 0.83 2 4.16 0.42 0.5 5 2 0.42 14 0.18 0.02 0.1 1 Safety Capacity t ≤ t in Exponential Distribution 23 Mean inter-arrival time = 1/Ri Probability that the time between two arrivals t is less than or equal to a specific vaule of t P(t≤ t) = 1 - e-Rit, where e = 2.718282, the base of the natural logarithm Example 8.5: If the processing time is exponentially distributed with a mean of 5 seconds, the probability that it will take no more than 3 seconds is 1- e-3/5 = 0.451188 If the time between consecutive passenger arrival is exponentially distributed with a mean of 6 seconds ( Ri =1/6 passenger per second) The probability that the time between two consecutive arrivals will exceed 10 seconds is e-10/6 = 0.1888 Safety Capacity Performance Improvement Levers 24 – Decrease variability in customer inter-arrival and processing times. – Decrease capacity utilization. – Synchronize available capacity with demand. Safety Capacity Variability Reduction Levers 25 Customers arrival are hard to control – Scheduling, reservations, appointments, etc…. Variability in processing time – Increased training and standardization processes – Lower employee turnover rate = more experienced work force – Limit product variety Safety Capacity Capacity Utilization Levers 26 If the capacity utilization can be decreased, there will also be a decrease in delays and queues. Since ρ=Ri/RP, to decrease capacity utilization there are two options: – Manage Arrivals: Decrease inflow rate Ri – Manage Capacity: Increase processing rate RP Managing Arrivals – Better scheduling, price differentials, alternative services Managing Capacity – Increase scale of the process (the number of servers) – Increase speed of the process (lower processing time) Safety Capacity Synchronizing Capacity with Demand 27 Capacity Adjustment Strategies – Personnel shifts, cross training, flexible resources – Workforce planning & season variability – Synchronizing of inputs and outputs Safety Capacity Effect of Pooling 28 Ri/2 Queue 1 Ri Ri/2 Queue 2 Server 1 Server 2 Server 1 Ri Queue Server 2 Safety Capacity Effect of Pooling 29 Under Design A, – We have Ri = 10/2 = 5 per minute, and TP= 5 seconds, c =1 and K =50, we arrive at a total flow time of 8.58 seconds Under Design B, – We have Ri =10 per minute, TP= 5 seconds, c=2 and K=50, we arrive at a total flow time of 6.02 seconds So why is Design B better than A? – Design A the waiting time of customer is dependent on the processing time of those ahead in the queue – Design B, the waiting time of customer is only partially dependent on each preceding customer’s processing time – Combining queues reduces variability and leads to reduce waiting times Safety Capacity Effect of Buffer Capacity 30 Process Data – Ri = 20/hour, Tp = 2.5 mins, c = 1, K = # Lines – c Performance Measures K 4 5 6 Ii 1.23 1.52 1.79 Ti 4.10 4.94 5.72 Pb 0.1004 0.0771 0.0603 R 17.99 18.46 18.79 ρ 0.749 0.768 0.782 Safety Capacity Economics of Capacity Decisions Cost of Lost Business Cb – $ / customer – Increases with competition Cost of Buffer Capacity Ck – $/unit/unit time Cost of Waiting Cw – $ /customer/unit time – Increases with competition Cost of Processing Cs – $ /server/unit time – Increases with 1/ Tp Tradeoff: Choose c, Tp, K – Minimize Total Cost/unit time = Cb Ri Pb + Ck K + Cw I (or Ii) + c Cs 31 Safety Capacity Optimal Buffer Capacity Cost Data – Cost of telephone line = $5/hour, Cost of server = $20/hour, Margin lost = $100/call, Waiting cost = $2/customer/minute Effect of Buffer Capacity on Total Cost K $5(K + c) $20 c $100 Ri Pb $120 Ii TC ($/hr) 4 25 20 200.8 147.6 393.4 5 30 20 154.2 182.6 386.4 6 35 20 120.6 214.8 390.4 32 Safety Capacity Optimal Processing Capacity 33 c K=6–c Pb Ii TC ($/hr) = $20c + $5(K+c) + $100Ri Pb+ $120 Ii 1 5 0.0771 1.542 $386.6 2 4 0.0043 0.158 $97.8 3 3 0.0009 0.021 $94.2 4 2 0.0004 0.003 $110.8 Safety Capacity Performance Variability 34 Effect of Variability – Average versus Actual Flow time Time Guarantee – Promise Service Level – P(Actual Time ≤ Time Guarantee) Safety Time – Time Guarantee – Average Time Probability Distribution of Actual Flow Time – P(Actual Time ≤ t) = 1 – EXP(- t / T) Safety Capacity Effect of Blocking and Abandonment Blocking: the buffer is full = new arrivals are turned away Abandonment: the customers may leave the process before being served Proportion blocked Pb Proportion abandoning Pa 35 Safety Capacity Effect of Blocking and Abandonment Net Rate: Ri(1- Pb)(1- Pa) Throughput Rate: R=min[Ri(1- Pb)(1- Pa),Rp] 36 Safety Capacity Example 8.8 - DesiCom Call Center 37 Arrival Rate Ri= 20 per hour=0.33 per min Processing time Tp =2.5 minutes (24/hr) Number of servers c=1 Buffer capacity K=5 Probability of blocking Pb=0.0771 Average number of calls on hold Ii=1.52 Average waiting time in queue Ti=4.94 min Average total time in the system T=7.44 min Average total number of customers in the system I=2.29 Safety Capacity Example 8.8 - DesiCom Call Center Throughput Rate R=min[Ri(1- Pb),Rp]= min[20*(1-0.0771),24] R=18.46 calls/hour Server utilization: R/ Rp=18.46/24=0.769 38 Safety Capacity Example 8.8 - DesiCom Call Center 39 Number of lines 5 6 7 8 9 10 Number of servers c 1 1 1 1 1 1 Buffer Capacity K 4 5 6 7 8 9 Average number of calls in queue 1.23 1.52 1.79 2.04 2.27 2.47 Average wait in queue Ti (min) 4.10 4.94 5.72 6.43 7.08 7.67 Blocking Probability Pb (%) 10.04 7.71 6.03 4.78 3.83 3.09 Throughput R (units/hour) 17.99 18.46 18.79 19.04 19.23 19.38 Resource utilization .749 .769 .782 .793 .801 .807

Capacity Planning in Services Industry, Part 2

Related Documents

Recommended Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock