OMSAN LOJİSTİK

Top Management Program in Logistics & Supply Chain Management (TMPLSM) Production and Operations Management 3: Process Analysis

3 Process Flow Analysis Objectives: Increase Throughput Rate Reduce Throughput Time (and Inventories) Reduce Waste (and, more generally, operating costs) Objectives: Increase Throughput Rate Reduce Throughput Time (and Inventories) Reduce Waste (and, more generally, operating costs)

4 Throughput Rate is determined by the Bottleneck(s) AB CD E

5 AB CD E But don’t be surprised if throughput rate is less than the “capacity” of the bottleneck! There is a difference between “rated capacity” & “utilizable capacity” There is a difference between “rated capacity” & “utilizable capacity”

6 “Rated Capacity” “Utilizable Capacity” In Manufacturing due to Breakdowns and repairs Operators skill and experience Product & material variations Availability (arrival) of inputs Changes in schedule (setups) Rate of Rejects and Off-Specs Etc. Why the Difference? The Curse of Variability Similar reasons apply also in service operations

7 Focus on narrowing this difference especially in bottleneck operations But be aware of “side effects” of pushing for very high levels of capacity utilization, especially on throughput time and waste

8 Capacity Utilization Inventory or Queue 100% 80%60%20%40% Throughput time rises rapidly at higher levels of capacity utilization Consider a powerful insight from Queuing Models

9 Scheduling bottleneck operations very close to their rated capacity can create high WIP and long lead times AB CD E Capacity Utilization Inventory (or lead time,or waiting time)

10 More Variability More variability means less “utilizable capacity” for same level of inventory (or waiting time) Capacity Utilization Inventory (or waiting time)

11 The Triangle of Basic Choices Capacity Utilization Low Variability “Extra” Capacity Inventory (or lead time) High Inventory (or long lead time)

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13 What if Arrivals & Service Times are not uniform? Arrivals at Average Rate (jobs/hr) Service at Average Rate  (jobs/hr)

14 How do we measure variability? We know standard deviation (“s”) is a measure of variability x Pr (x) A (larger s) B (smaller s) A is clearly more variable than B

15 Pr (x) Number of customers per hour (“x”) m=50m=120 But which of these demand patterns is more variable? A s=10 B s= We need something better than “s” alone

16 Arrivals and service times in many situations follow an exponential distribution 0 0 Time Exponential distribution: Probability density, f(t) = exp (- t) Mean Which makes life easy because s=m in exponential distribution i.e., c=1

17 Variability in Arrivals, c a  When there is no variability ( s = 0) c a = s / m = 0 u When arrivals are “Random” c a = s / m = 1 Time between arrivals is exponentially distributed (which is the same as saying number arrivals in a given time interval is Poisson distributed) u And c a can be any other number

18 Same story for variability of service, c s u When there is no variability in processing time ( s = 0) c s = s / m = 0 u When processing times are “Random” c s = s / m = 1 (I.e., processing times are exponentially distributed) u And c s can be any other number (reflecting other distributions)

19 How do find waiting time when there is variability? Service time: 1/  (hrs/job) Time in Queue W q (hrs/job) W = Wq + 1/ 

QueueServer (resource) Arrivals Arrival rate = Mean interarrival time = m a = 1 / Standard deviation in interarrival time = s a Coefficient of variation in interarrival time = c a = s a / m a Average waiting time in queue = W q Average number in queue = L q Services Processing rate =  Mean service (processing) time = m s = 1 /  Standard deviation in processing time = s s Coefficient of variation in processing time = c s = s s / m s Utilization =  = /  Average number in service =  Average total waiting time in the system (in the queue and in service) = W = W q + 1 /  Average total number in the system (in the queue and in service ) = L = L q +  Equations!

21 Exercise … trying the Equations! Consider a store where customers arrive randomly at an average rate of 10 per hour. Service times have a mean of 5 minutes and a standard deviation of 2 minutes. Calculate: –What is the avg total time a customer spends in the system? –What fraction of time is the resource occupied? –What is the avg number of customers in the system? –What is the avg number of customers in the queue? –What would your answers be if all services took exactly 5 minutes?

22 PK Formula W q = (1 /  ) [  / (1 –  Process Time Capacity Utilization Variability )] [(c a 2 + c s 2 ) / 2]

23 The bartender Example Assume a bartender can serve on average 10 drinks per hour (but some drinks take more than 6 minutes and some less; Assume processing times are “random”, Cs=1) Assume Customers arrive “randomly” ( i.e., Ca=1) Let’s see what happens when different number of customers arrive per hour

24 Arrival Rate Customers/hr Capacity. Utilization   /(1 -  ) Waiting Time W q Hours q  W = (1 /) [ / (1 –  )] [(c a 2 + c s 2 ) / 2] Let’s apply the PK Formula

25 Waiting time Capacity utilization Waiting time, hours

26 Remember the triangle Capacity Utilization Low Variability “Extra” Capacity Inventory (or lead time) High Inventory (or long lead time)

27 Basic Principle uAs the variability in an operating system becomes more severe, it becomes increasingly difficult to simultaneously achieve high utilization and low inventory. arrival processservice process uGenerally speaking, all else equal, the more variable the arrival process and the service process, the more likely jobs are going to clump up, average service will be longer, and therefore the more waiting time there will be on average Excess Capacity uImmediate consequence: Excess Capacity will be required in systems containing variability.