Transcript Slayt 1

OMSAN LOJİSTİK
Top Management Program in Logistics
& Supply Chain Management (TMPLSM)
Production and Operations Management
3: Process Analysis
Process Flow Analysis
Objectives:
• Increase Throughput Rate
• Reduce Throughput Time (and Inventories)
• Reduce Waste (and, more generally,
operating costs)
3
Throughput Rate
is determined by the
Bottleneck(s)
C
A
B
E
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D
But don’t be surprised if throughput rate is less
than the “capacity” of the bottleneck!
C
A
D
B
E
There is a difference between
“rated capacity” & “utilizable capacity”
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Why the Difference?
The Curse of Variability
“Rated
Capacity”
“Utilizable
Capacity”
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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.
Similar reasons apply also in
service operations
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
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Consider a powerful insight from
Queuing Models
Inventory or Queue
Throughput time rises rapidly
at higher levels of
capacity utilization
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10
8
6
4
2
0
20%
40%
60%
Capacity Utilization
80%
100%
Scheduling bottleneck operations very
close to their rated capacity can
create high WIP and long lead times
C
A
D
B
(or lead time,or waiting time)
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Inventory
E
10
8
6
4
2
0
0.0
0.2
0.4
0.6
Capacity Utilization
0.8
1.0
More variability
means less “utilizable capacity”
for same level of inventory (or waiting time)
(or waiting time)
Inventory
10
8
6
4
2
0
0.0
10
0.2
0.4
0.6
Capacity Utilization
0.8
1.0
The Triangle of Basic Choices
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High Inventory
(or long lead time)
(or lead time)
Inventory
8
6
4
2
0
0.0
0.2
0.4
0.6
0.8
Capacity Utilization
11
1.0
12
What if Arrivals & Service Times are not
uniform?
Arrivals
at Average Rate l
(jobs/hr)
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Service
at Average Rate m
(jobs/hr)
How do we measure variability?
We know standard deviation (“s”) is a
measure of variability
A (larger s)
Pr (x)
B (smaller s)
x
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A is clearly more variable than B
But which of these demand
patterns is more variable?
A
s=10
B
Pr (x)
s=12
20
m=50
80
150
m=120
Number of customers per hour (“x”)
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We
need something better than “s” alone
Exponential distribution:
Probability density,
f(t) = l exp (-l t)
Arrivals and service times in many situations follow
an exponential distribution
l
0
0
Mean
Time
Which makes life easy because
s=m in exponential distribution
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i.e.,
c=1
Variability in Arrivals, ca


When there is no variability ( s = 0)
ca = s / m = 0
When arrivals are “Random”
ca = 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)

And ca can be any other number
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Same story for variability of
service, cs
 When there is no variability in processing time ( s = 0)
cs = s / m = 0
 When processing times are “Random”
cs = s / m = 1
(I.e., processing times are exponentially distributed)
 And
cs can be any other number
(reflecting other distributions)
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How do find waiting time when
there is variability?
W = Wq + 1/m
Time in Queue
Wq (hrs/job)
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Service time:
1/m (hrs/job)
Equations!
Queue
3
2
Server (resource)
1
Arrivals
Arrival rate = l
Mean interarrival time = ma= 1 / l
Standard deviation in interarrival time = sa
Coefficient of variation in interarrival time
= ca = sa / ma
Average waiting time in queue = Wq
Average number in queue = Lq
Services
Processing rate = m
Mean service (processing) time = ms = 1 / m
Standard deviation in processing time = ss
Coefficient of variation in processing
time
= cs = ss / ms
Utilization =  = l / m
Average number in service = 
Average total waiting time in the system (in the queue and in service) = W = Wq + 1 / m
Average total number in the system (in the queue and in service) = L = Lq + 
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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?
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PK Formula
Wq = (1 /m) [ / (1 –)] [(ca2 + cs2) / 2]
Process
Time
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Capacity
Utilization
Variability
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
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Let’s apply the PK Formula
W q = (1 / m ) [  / (1 –  )] [(ca2 + cs2) / 2]
Arrival Rate
l
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Capacity.
Utilization
Waiting
Time
Customers/hr.

/(1 - )
Wq
Hours
2
4
6
8
9.9
.2
.4
.6
.8
.99
0.25
0.67
1.5
4
99
0.025
0.067
0.15
0.40
9.9
Waiting time
2.0
1.8
Waiting time, hours
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
-
0.2
0.4
0.6
Capacity utilization
0.8
1.0
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Remember the triangle
10
High Inventory
(or long lead time)
(or lead time)
Inventory
8
6
4
2
0
0.0
0.2
0.4
0.6
0.8
Capacity Utilization
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1.0
Basic Principle
 As the variability in an operating system becomes more
severe, it becomes increasingly difficult to simultaneously
achieve high utilization and low inventory.
 Generally 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
 Immediate consequence: Excess Capacity will be
required in systems containing variability.
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