MGT3303
Process Analysis Fundamentals
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1
Process Definition
Input
Process
Output

A process is a collection of operations connected by a flow of
transactions and information that transforms various inputs into
outputs using resources.

A process maybe deterministic or stochastic (probabilistic)
• Components are processed on an automated machine.
• People walk to a window, buy a ticket and leave with a ticket.
• People arrive at scheduled times and receive some service.
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2
Process Diagrams
 Process Diagrams are used to analyze various
processes.
 We first start by a precedence diagram, that includes
the following symbols
Operation
Precedence
 Example: Precedence diagram for product X, (times in
seconds)
1 : 20
4 : 40
3 : 10
2 : 50
6 : 30
7 : 20
5 : 30
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3
Process Diagrams
Based on the precedence diagram we draw a process
diagram, which includes the following symbols:
Resource at a
stage
Buffer / Storage
Material flow
Information flow
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Example: Trinidad’s “circored iron ore” plant
Fluidized bed based Circored® technology: The process
uses hydrogen from natural gas to reduce iron ore fines,
significantly decreasing the costs of the steelmaking
process due to the elimination of the agglomeration
step. The high quality, low cost product can be used as
pure scrap substitute in electric arc furnaces as well as
for increasing the capacity of existing blast furnaces
with corresponding coke savings.
- Trinidad with ample natural gas
- Port Lisas in the Claxton Bay, most
dependable port in the Caribbean
- Proximity to iron ore rich South America
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Example: Trinidad’s “circored iron ore” plant
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6
Process Flow Chart (Diagram)
CFB
Preheater
1st Stage
CFB Reactor
CFB: Circulating
Fluidized Bed
1st Stage
Reactor
~ 110 m
Inclined
Bucket Elevator
Briquetting
Plant
Process
Gas
Heat
Exchanger
Iron
Ore
Fines
Electrical
Substation&
Control
Room
Process Gas
Compressor
Fired Gas Heaters
HBI Product
~ 160 m
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7
Process Flow Chart (Diagram)
Flash heater
Pre-Heater
1st (cfb)
Reactor
2nd (fb) Reactor
Discharge
Briquetting
Pile of
Iron ore fines
Finished
goods
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8
Process Parameters

Input rate gives the entering rate of jobs; in [Q/T].

Output rate ( i.e. flow rate, production rate, throughput rate)
gives the exit rate of jobs ; in [Q/T].

Capacity is the maximum output rate possible; [Q/T].

Cycle time (CT) is the reciprocal of capacity; in [T/Q].

Minimum Manufacturing Lead Time (MLT) (min flow time) is the
fastest processing time for a transaction to go through the process,
when the process is empty; in [T].
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Process Parameters
When the process is running:
 Actual cycle time is the reciprocal of output rate; in
[T/Q].
 Average Manufacturing Lead Time (MLT) (flow time)
is the average time required for a transaction to go
through the process; in [T].
 Work-In-Process (WIP) is the average over time
number of transactions in the process; in [Q].
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10
Example 1
M1
X
Y
4
(a) A new component is scheduled every 10 minutes.
The Gantt chart : ???




input rate = ?
output rate = ?
actual cycle time = ?
The average MLT = ?
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Example 1
Solution for (a):
 Since each job enters the system every 10 minutes, the
input rate will be 1/10 [units / min].
 The output rate will be 1/10 [units / min].
 The actual cycle time is 10 [min / units].
 The average MLT is 4 [min].
 Notice in the Gantt chart that between time units 4 to
10, the machine is idle.
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Example 1
(b) A new component is scheduled every 4 minutes.
The Gantt chart : ???




input rate = ?
output rate = ?
actual cycle time = ?
The average MLT = ?
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Example 1
Solution for (b):
 Since each job enters the system every 4 minutes, the
input rate is 1 / 4 [unit / min].
 Also, the output rate will be same as the input rate, i.e.
Output rate = 1 / 4 [unit / min] = 15 [units / hour].
 The actual cycle time is 4 [min / units].
 The average MLT is 4 [min].
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Example 1
(c) A new component is scheduled every 3 minutes.
The Gantt chart : ???
 input rate = ?
 output rate = ?
 actual cycle time = ?
 The average MLT = ?
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Example 1

Since each job enters the system every three minutes, the input rate will
be 1/3 [unit / min] or 20 [units / hour].

M1 can process jobs every 4 minutes. So the output rate will be only 1/4
[unit / min] = 15 [units / hour].

You can see that every subsequent job stays one more minute in the buffer
that the previous job and we won’t be able draw all theses jobs in the
buffer.

The actual cycle time is 4 [min / units]. The minimum MLT is 4 [min].

The MLT tends to infinity as time goes to infinity. In fact, each successive
transaction stays 1 minute longer and the system never becomes stable.
The average MLT is undefined.

The WIP (in an instance of time) and the average WIP also go to infinity
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Important Properties
 Any process will not become stable if the input rate is
greater than the system capacity
 Output rate can never exceed capacity.
 For sequential stages, the capacity is determined by the
slowest stage called the bottleneck.
 When machines are in parallel (performing the same
operation), capacities add.
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Example 2
 We still consider a single stage process: (in seconds)
M1, 10
M2, 16
M3, 20
 Part (a): If you need to process 1125 units per hour, do
you need an extra machine?
 Part (b): If so, what should be the per-unit processing
time on this extra machine?
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Example 2

Part (a): We have three machines in parallel and should first find
the process capacity. When machines are in parallel, the capacity
of the process is equal to the sum of capacities of individual
machines.
Capacity of M1 = 1/10 [units / second] = 360 [units / hour],
Capacity of M2 = 225 [units / hour],
Capacity of M3 = 180 [units / hour].
System capacity = 765 [units / hour].
We therefore need an extra machine.

Part (b): Since we need the capacity of 1125, the extra machine
should have the capacity of (1125 – 765) = 360 [units / hour].
The processing time on the extra machine (say M4) should be 10
seconds.
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Little’s Law

Prof. John D. C. Little of MIT Sloan School of Management,
developed the proof for the following formula for a stable system:
L=λ*W
where L is the average number of customers in the system, λ is
the arrival rate and W is the average time in the system.

Let’s apply it in our case:
WIP = average MLT * output rate
= average MLT / actual CT
VERY IMPORTANT ….. !!!!
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Example 3
 A process involves two sequential tasks.
M1, 10
M2, 8
Part (a)






CT for station 1?
CT for station 2?
Bottleneck Station?
Process Capacity?
CT for the process?
Minimum MLT?
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Example 3
 Cycle time for Station 1 is 10 [min. / unit]
 Cycle time for Station 2 is 8 [min. / unit]
 Bottleneck is Station 1
 Process Capacity is 1/10 [units / min.]
 Cycle time for the process is 10 [min. / unit]
 Minimum MLT: It will take at least 10 [min.] + 8 [min.] =
18 minutes when system is empty
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Example 3
 Part (b): Consider the following schedule. Starting at
time 0, one new job enters the process at station 1
every 10 mins.
i. Calculate the actual cycle time and the flow rate (i.e.
output rate).
ii. Calculate the average MLT
iii. Calculate WIP contribution of each resource and the
total.
iv. Verify Little’s formula.
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Example 3
i. From the Gantt chart, a job will come off the process at times 18, 28,
38, 48, 58, ….; i.e. every 10 minutes. The actual CT = 10 [min. /
unit] and the flow rate = 1/10 [units / min].
ii. The first job enters the process at 0 and comes out at 18, the
second job enters the process at 10 and comes out (after 18
minutes) at 28. In fact every job will stay in the process of 18
minutes. Therefore average MLT = 18 [min.]
iii. Notice from the Gantt chart that station 1 is always occupied by a
job. So the contribution of station 1 to the WIP is 1.00 [units].
Therefore the WIP contribution of station 2 is 0.80 [units] and the
total WIP is 1.80 [units].
iv. You can now easily verify Little's formula: WIP = MLT * Flow rate =
MLT / CT
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Example 4
(a)
 CT for station 1?
 CT for station 2?
 Bottleneck?
 Process Capacity?
 CT for the process?
 Minimum MLT?
M1, 10
M3, 8
M2, 10
(b)
 Jobs are sent to M1 every 16 minutes starting at time 0
and to M2 every 16 minutes starting at time 8. Verify
Little’s formula?
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Example 4 : Gantt Chart
M1:
1
3
M2:
2
M3:
1
0
7
4
2
10
8
5
18
6
3
26
4
34
5
42
6
50
58
16
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Example 4
 WIP contributions:
For M1: 10 / (10 + 6) = 0.625 [units]
For M2: 10 / (10 + 6) = 0.625 [units]
For M3: 1.00 [units]
Total : 2.250 [units]

Each job stays in the system for 18 minutes, hence average MLT =
18 [min.]
From the Gantt chart, actual CT = 8 [min. / unit]
 WIP = MLT / CT
= 18 [min.] / 8 [min. / unit] = 2.250 [units]
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Example 5 : assembly operations

Product AA is processed at station 1 containing 2 machines in parallel.
Product BB is processed at Station 2 then at station 3. One unit of AA and
one unit of BB is picked from the buffer and assembled at station 4. The
assembled product is then processed at station 5 containing three
machines in parallel. All processing times are in minutes.
S1
R1, 16
M1, 9
S4
AA
M2, 6
P1, 4
R2, 18
R3, 18
BB
X1, 3
Y1, 5
S5
S2
S3
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Example 5

What is the process capacity?
First we need to determine the CT for each station.
CT values for station 2, 3 and 4 are 3 [min/unit], 5 [min./unit] and 4 [min./unit]
respectively.
CT calculations for parallel machines: Find a common multiple of processing
times for the station
Station 1: Common multiple = 18 minutes. In 18 minutes we can process 2 jobs
on M1 and 3 jobs on M2, or a total of 5. Therefore CT for station 2 = 18/5
[min/unit] = 3.6 [min./unit],
Station 5: Common multiple = 144 minutes. In this time we can process 9 jobs on
R1 and 8 jobs each on R2 and R3 or a total of 25. Therefore CT for station 5 =
144/25 [min/unit] = 5.76 [min/unit].
The Bottleneck is station 5 (longest CT). Therefore, the process capacity will be
[60 / 5.76] ≈ 10.4 [units / hour].
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Example 5

What is the minimum MLT?
One unit of AA will reach the buffer in 6 minutes (through M2) and
one unit of BB will reach the buffer in 8 minutes. Thus we need at
least 8 minutes before we can start the assembly at station 4. At
station 5, the smallest time is 16 min. Therefore:
minimum MLT = 8 + 4 + 16 = 28 [minutes]
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Some Key insights
 Capacity of a process is the maximum output rate. If
input rate becomes higher than the capacity, the
process will not become stable. MLT for incoming jobs
will become higher and higher.
 If we need to increase the capacity of any process, we
need to add resources at the bottleneck stage. This
may shift bottleneck to another stage.
 For two schedules it is possible to have the same cycle
time (output rate) but different WIP and MLT. The lower
MLT and WIP the more efficient is the process.
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