Transcript Document 7437731

Flexible Manufacturing Systems
(FMS)
by Ed Red
“…an automated, mid-volume, mid-variety, central computer-controlled
manufacturing system” Nanua Singh, Computer-Integrated Design and
Manufacturing, John Wiley & Sons, 1996
References:
1. Nanua Singh, Computer-Integrated Design and Manufacturing, John Wiley &
Sons, 1996
2. Mikell Groover, Automated Production Systems and Computer-Integrated
Manufacturing, Prentice-Hall, 3rd edition, 2008
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Objectives
• To review modern flexible manufacturing systems (FMS):
- Group technology (GT)
- Manufacturing cells
- Automated part handling equipment (AGV’s, etc.)
- Control software
- Analysis models
• To consider application conditions (student presentations)
• To test understanding of the material presented
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FMS characteristics
• A manufacturing cell used to implement group technology
(GT)
robots…machine tools…
• Independent machines performing multiple operations and
having automated tool interchange capabilities
• Automated material-handling between stations (move parts
between machines and fixturing stations)
• Hierarchical computer control architectures
• Often include CMM, inspection and part washing devices
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GT requirement:
Parts can be grouped into part families!
Similar manufacturing process
requirements (manufacturing
attributes), but with different
design attributes
Turned, drilled, milled…..
Cylindrical, hole, thread, chamfer,
tolerance, dimension…..
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GT requirement: Production machines
can be arranged into cells!
Group technology layout
Process type plant layout …dashed
lines indicates departments!
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GT part classification and coding
 Parts distinguished (classified) by design attributes and
manufacturing attributes.
 Part differentiated by coding methods for
• design retrieval
• automated process planning
• machine cell design
Basic structure
of Opitz coding
system
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GT Opitz form code
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GT example
For the part shown determine the form
code in the Opitz parts classification and
coding system..
Solution:
With reference to Figure 15.6, the five-digit code is developed as follows:
• Length-to-diameter ratio, L/D = 1.5
• External shape: stepped on both ends with screw thread on one end
• Internal shape: part contains a through-hole
• Plane surface machining: none
• Auxiliary holes, gear teeth, etc.: none
The parts’s form code in the Opitz system is
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Digit 1 = 1
Digit 2 = 5
Digit 3 = 1
Digit 4 = O
Digit 5 = O
15100
FMS
Highly automated GT manufacturing cell, consisting of a group of
processing workstations, interconnected by an automated material
handling and storage system, and controlled by a distributed
computer system (Groover defn)
What does flexible mean?
1. Can identify and operate on different part/product styles
2. Quick changeover of process/operating instructions
3. Quick changeover of physical setup
FMS operations:
1. Processing operations, or
2. Assembly operations
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FMS – automated part handling
Conveyor
AGV
AS/RS
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Single machine
FMS type - Distinguish by number of machines
1.
Single machine cell* can operate in batch mode (sequentially process
parts of a single style in defined lot sizes) or flexible mode (process
different part styles and adapt to different production schedules)
* No error recovery if machine breaks down since production will stop
2.
Flexible machine cell (FMC) consists of 2-3 machines plus part handling
Flexible Manufacturing Cell
equipment and limited part storage….simultaneous production of different
parts and error recovery.
3.
Flexible manufacturing system consists of 4 or more workstations connected
by common part handling system and distributed computer system. Other
stations may support the activities, such as a coordinate measuring machine
(CMM) or washing station. ….simultaneous production of different parts and
error recovery.
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Loop layout
Ladder
Open field
layout
layout
FMS layouts
In-line layout
1. In-line layout
2. Loop layout (secondary part handling systems)
3. Ladder layout
4. Open field layout
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FMS computer control system
1.
Workstation control
2.
Supervisory control among workstations (workstation coordination)
3.
Production control (part rate and mix)
4.
Traffic control (manage part delivery systems)
5.
Shuttle control (part handling between machine and primary handling
system)
6.
Workpiece monitoring (status of various systems)
7.
Tool control (location and tool life)
8.
Performance monitoring and reporting (report operational data)
9.
Diagnostics (identify sources of error, preventive maintenance)
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FMS design issues
1. Workstation types
2. Variations in process routings and FMS layout (increasing
product variety move you from in-line layouts to open field layouts)
3. Material handling system
4. Work in process (WIP) and storage capacity (FMS storage
capacity must be compatible with WIP)
5. Tooling (numbers and types of tools at each station, tool duplication)
6. Workpiece monitoring (status of various systems)
7. Pallet fixtures (numbers in system, flexibility)
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FMS operational issues
1. Scheduling (master production schedule) and dispatching
(launching of parts into the system)
2. Machine loading
3. Part routing
4. Part grouping
5. Tool management
6. Pallet and fixture allocation
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FMS quantitative analysis
Four models:
1. Deterministic models (don’t include operating characteristics,
including queues, that may degrade performance, thus are a little optimistic)
2. Queueing models
3. Discrete event Discrete
simulation
(simulation)
event simulation
– Used to model manufacturing cell
4. Heuristic
or material handling system, as events occur at discrete
approaches
moments in time and affect the status and performance of the
system, e.g., parts arriving at the machine.
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FMS bottleneck model
Bottleneck – output of a production system has an upper
limit, given an upper bounds on the product mix flowing
through the system
 Introduce the bottleneck model to provide initial FMS
parameter estimates
 Introduce terminology and symbols
 Demonstrate on examples
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FMS terminology and symbols
Part mix
pj = fraction of system output that is of style j
P = total number of part styles made in FMS in given
time period
P
 p j 1.0
j 1
Workstations and servers (workstation that can duplicate process
capabilities of another workstation )
n = number of workstations
si = number of servers at each station i (i = 1,2,…n, and we
include the load/unload station as an FMS workstation)
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FMS terminology and symbols
Process routing – for each part or product, defines
operational sequence, assigned workstations, and
associated process times, including loading and
unloading times
tijk = processing time for a part/product in a given
server, not including waiting time, where
i = station i
j = part/product j
k = particular operation in process routing sequence of operations
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FMS terminology and symbols
Work handling system – material handling system is
considered a special workstation and designate it as
station n + 1; then
sn+1 = number of carriers (servers) in handling system
(conveyors, carts, AGV’s, etc.)
Transport time
tn+1 = mean transport time required to move a part from
one workstation to the next station in the process
routing
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FMS terminology and symbols
Operation frequency – expected number of times a given
operation in the process routing is performed for each
work unit, e.g, an inspection of a dimension
fijk = operation frequency for operation k for part j at
station i
This parameter (fijk) is usually one since each operation is
usually performed once on a different workstation!
Exceptions might exist for part inspection stations. Note that
there are many zero values since not all parts and operations
go through every machine.
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FMS quantitative models
Average workload (Li) – mean operational time of station i per part,
calculated as (units are in min.)
Li = Sj Sk tijk fijk pj
i = station i
j = part/product j (process routing)
k = operation in routing sequence
Workload of the handling system is the mean transport time (tn+1)
multiplied by the average number of transports to complete part
process.
Average number of transports (nt) is the mean number of operations in
the process routing minus 1:
nt = Si Sj Sk fijkpj – 1
Workload of handling system is Ln+1 = nt tn+1
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difficult interpretation!
FMS example – determine nt
Simple system has machining station and load/unload station.
If system processes single part, determine nt.
One part (j = 1) so
p1 = 1.0
Load
Unload
fi1k = 1.0
3 routing operations: load part at 1-> route to station 2 for machining->
return to station 1 for unloading
Then
nt = 1(1.0) + 1(1.0) + 1(1.0) - 1 = 2
“load”
“machine at 2”
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“unload”
FMS quantitative models
FMS production is usually constrained by a bottleneck
station (consider the handling station also), which is the
station i with the highest workload per server as measured
by Li/si. Designate i = b the bottleneck station and calculate
the maximum production rate from
Rmax = sb/Lb
(number of parts per time for station b)
Note: This is valid even for parts not passing through the bottleneck station
because the part mix ratios are fixed and limited by the bottleneck station.
Individual production rates are
Rj = pj sb/Lb
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FMS quantitative models
Mean workstation utilization is the proportion of time that stations are
active as determined from
Ui = Rmax Li/si
( Ub = 1)
The average station utilization is
U = Si Ui/(n+1)
The overall FMS utilization is weighted by the number of servers at
each station (not including handling stations)
Us = Si siUi/ Si si
Number of busy servers at other than the bottleneck station
determined from
Bi = Rmax Li
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FMS example (from Groover)
An FMS with 4 stations is designed so that station 1 is load/unload, station 2
performs milling operations with 3 servers, station 3 performs drilling operations
with 2 servers, while station 4 performs part inspection on part samples. The part
handling system has a mean transport time of 3.5 min and 2 carriers. The FMS
produces parts A, B, C, and D with part mix fractions and routings shown in the
table.
Determine:
1.FMS max production rate
2.Production rate of each part
3.Each station utilization
4.Overall FMS utilization
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FMS example solution
First, determine bottleneck station by calculating workloads:
L1 = (4+2)(1.0)(0.1 + 0.2 + 0.3 + 0.4) = 6.0 min.
L2 = (20)(1.0)(0.1) + 25(1.0)(0.2) + (30)(1.0)(0.4) = 19.0 min.
Similarly,
L3 = 14.4 min. ;
L4 = 4.0 min.
nt = (4.5 - 1)(0.1) + (5.2 - 1)(0.2) + (3.5 -1)(0.3) + (3.333 - 1)(0.4) = 2.783
L5 = (2.873)(3.5) = 10.06 min.
…part handling station!
Now calculate Li/si to identify bottleneck:
L1/s1 = 6.0/1 = 6.0
L2/s2 = 19.0/3 = 6.333
L3/s3 = 14.4/2 = 7.2 …the bottleneck! Rmax = 2/14.4 = 0.1389 pc/min. (8.333
pc/hr)
L4/s4 = 4.0/1 = 4.0
L5/s5 = 10.06/2 = 5.03
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FMS example solution
Production rate for each part:
RA = 8.333(0.1) = 0.8333 pc/hr.
RB = 8.333(0.2) = 1.667 pc/hr.
RC = 8.333(0.3) = 2.500 pc/hr.
RD = 8.333(0.4) = 3.333 pc/hr.
Station utilization:
U1 = (6.0/1)(0.1389) = 0.8333
U2 = (19.0/3)(0.1389) = 0.879
U3 = (14.4/2)(0.1389) = 1.0
U4 = (4.0/1)(0.1389) = 0.555
U5 = (10.06/2)(0.1389) = 0.699
(83.33%)
Overall FMS utilization (exclude part handling):
U1 = [1(0.833) + 3(0.879) + 2(1.0) + 1(0.555)]/7= 0.861
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(86.1%)
FMS follow-on example (from Groover)
Determine the production rate of part D that will increase the utilization
of station 2 to 100%. Note that this is possible since part D does not go
through station 3, the bottleneck station, and station 2 is under utilized.
Solution:
Set U2 = 100% and solve U2 = 1.0 = L2(0.1389)/3 to get L2 = 21.6 min.
as compared to 19.0 min. previously.
Parts A, B and D are processed by station 2. Parts A and B are
constrained in their production rate by the other stations, but not part D
which is only processed by station 2.
We first determine the portion of the station 2 workload taken up by A
and B:
L2(by A+B) = 20(0.1)(1.0) + 25(0.2)(1.0) = 7.0 min.
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FMS follow-on example
At 100% utilization the workload for part D increases to 21.6 – 7.0 = 14.6
min., where it was 19.0 – 7.0 = 12.0 min. at 87.9% utilization. The production
rate for part D is now increased to 14.6(3.333)/12.0 = 4.055 pc/hr.
Note that increasing the throughput for part D will change the part mix ratios
previously presented.
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Optimizing operations allocation in an
FMS with negligible setup
Sound familiar?
Two criteria:
- production of parts with minimum cost
- production of parts at max production rate
Define:
K part types having demand dk
(k = 1,......K)
M machine types each having capacity bm
(m = 1,.....M)
Jk operations performed on part type k
(j = 1,.......Jk)
ckjm = unit processing cost to perform jth operation on kth part
on mth machine; else, set the cost to infinity (set high)
tkjm = unit processing time to perform jth operation on kth part
on mth machine; else, set the time to infinity (set high)
ME 482 - Manufacturing Systems
Optimizing operations allocation in
an FMS with negligible setup
Define flexibility factor, akljm :
Assume operations can be performed on alternative machines. Part can
be manufactured along a number of routes. For example, if a part has
three operations and if the first, second, and third operations can be
performed as:
- operation 1 on two machines
- operation 2 on three machines
- operation 3 on two machines
then a set of alternative process plans (l  L, where L is the total
number of alternative plans) would include 2 x 3 x 2 = 12 possible
processing routes. Define
akljm = 1 if in plan l the jth operation on the kth part is performed on
the mth machine; else, set the factor to 0
ME 482 - Manufacturing Systems
Optimizing operations allocation in
an FMS with negligible setup
Minimum cost to manufacture all parts:
Minimize Z1 = Skljm akljm ckjm Xkl
“Linear programming”
where Z1 is the objective function and Xkl is a decision variable representing
the number of units of part k to be processed using plan l.
Constraints:
Demand for parts must be met:
Sl Xkl
Can not exceed machine capacity:
Sklj akljm tkjm Xkl 
 dk
Positive number of units produced: Xkl  0
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"k
bm
" k, l
"m
Optimizing operations allocation in an
FMS with negligible setup
Maximize throughput (minimize total process time for parts):
Minimize objective function Z2 = Skljm akljm tkjm Xkl
Constraints:
Demand for parts must be met:
Sl Xkl
Can not exceed machine capacity:
Sklj akljm tkjm Xkl 
 dk
Positive number of units produced: Xkl  0
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"k
bm
" k, l
"m
Optimizing operations allocation in
an FMS with negligible setup
Balance workload on machines (minimize the maximum of the process times):
Minimize objective function Z3 = maximum{ Skljm akljm tkjm Xkl }
Constraints:
Minimized max > other workloads: Z3 - Skljm akljm tkjm Xkl  0
Demand for parts must be met:
Sl Xkl
Can not exceed machine capacity:
Sklj akljm tkjm Xkl 
 dk
Positive number of units produced: Xkl  0
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"k
bm
" k, l
"m
"m
Linear programming - example
Consider the manufacture of 5 part types on 4 machine types, each part
requiring several operations. Table 12.18 list the pertinent data. Develop a
production plan for: 1) min cost model; 2) max throughput (min processing
time); and 3) workload balancing.
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Linear programming - example
The 3 models were solved using LINDO, a linear programming package,
with the results shown in Table 12.19. The table shows that parts can be
produced through a number of alternative process plans. Another table (next
slide) can be generated to show the machine loading for various operations
allocation strategies.
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Linear programming - example
Note that all three models result in 100% utilization of machines m2 and m3,
making these bottleneck machines. Consider machine m1. Its resource
utilization for the 3 models are 2400, 2400, and 2045 units of time,
respectively. This information is useful for production scheduling and also
for preventive maintenance.
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To calculate these
values simply multiply
all the operations on
each machine (each
part through the
machine is an
operation) by the time
required for each
operation as given in
Table 12.18.
FMS
What have we learned?
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