1. 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

2. 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

3. FMS characteristics
A manufacturing cell used to implement group technology (GT)
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
robots…machine tools…

4. 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…..

5. GT requirement: Production machines can be arranged into cells!
Group technology layout
Process type plant layout …dashed lines indicates departments!

6. 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

7. GT Opitz form code

8. 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 = Digit 1 = 1
External shape: stepped on both ends with screw thread on one end Digit 2 = 5
Internal shape: part contains a through-hole Digit 3 = 1
Plane surface machining: none Digit 4 = O
Auxiliary holes, gear teeth, etc.: none Digit 5 = O
The parts’s form code in the Opitz system is

9. 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

10. FMS – automated part handling
Conveyor
AGV
AS/RS

11. FMS type - Distinguish by number of machines
Single machine
FMS type - Distinguish by number of machines
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
Flexible machine cell (FMC) consists of 2-3 machines plus part handling equipment and limited part storage….simultaneous production of different parts and error recovery.
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.
Flexible Manufacturing Cell

12. FMS layouts In-line layout
Loop layout
Open field layout
Ladder layout
FMS layouts
In-line layout
Loop layout (secondary part handling systems)
Ladder layout
Open field layout

13. FMS computer control system
Workstation control
Supervisory control among workstations (workstation coordination)
Production control (part rate and mix)
Traffic control (manage part delivery systems)
Shuttle control (part handling between machine and primary handling system)
Workpiece monitoring (status of various systems)
Tool control (location and tool life)
Performance monitoring and reporting (report operational data)
Diagnostics (identify sources of error, preventive maintenance)

14. FMS design issues Workstation types
Variations in process routings and FMS layout (increasing product variety move you from in-line layouts to open field layouts)
Material handling system
Work in process (WIP) and storage capacity (FMS storage capacity must be compatible with WIP)
Tooling (numbers and types of tools at each station, tool duplication)
Workpiece monitoring (status of various systems)
Pallet fixtures (numbers in system, flexibility)

15. FMS operational issues
Scheduling (master production schedule) and dispatching (launching of parts into the system)
Machine loading
Part routing
Part grouping
Tool management
Pallet and fixture allocation

16. FMS quantitative analysis
Discrete event simulation – Used to model manufacturing cell or material handling system, as events occur at discrete moments in time and affect the status and performance of the system, e.g., parts arriving at the machine.
FMS quantitative analysis
Four models:
Deterministic models (don’t include operating characteristics, including queues, that may degrade performance, thus are a little optimistic)
Queueing models
Discrete event simulation (simulation)
Heuristic approaches

17. 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

18. 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
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)

19. 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

20. 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

21. 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.

22. 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
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 difficult interpretation!
Workload of handling system is Ln+1 = nt tn+1
i = station i j = part/product j (process routing)
k = operation in routing sequence

23. 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
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) = 2
“load” “machine at 2” “unload”
Load
Unload

24. 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

25. 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

26. 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:
FMS max production rate
Production rate of each part
Each station utilization
Overall FMS utilization

27. FMS example solution
First, determine bottleneck station by calculating workloads:
L1 = (4+2)(1.0)( ) = 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 = ( )(0.1) + ( )(0.2) + (3.5 -1)(0.3) + ( )(0.4) = 2.783
L5 = (2.873)(3.5) = 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 = pc/min. ( pc/hr)
L4/s4 = 4.0/1 = L5/s5 = 10.06/2 = 5.03

28. FMS example solution Production rate for each part:
RA = 8.333(0.1) = pc/hr.
RB = 8.333(0.2) = pc/hr.
RC = 8.333(0.3) = pc/hr.
RD = 8.333(0.4) = pc/hr.
Station utilization:
U1 = (6.0/1)(0.1389) = (83.33%)
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
Overall FMS utilization (exclude part handling):
U1 = [1(0.833) + 3(0.879) + 2(1.0) + 1(0.555)]/7= (86.1%)

29. 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.

30. FMS follow-on example
At 100% utilization the workload for part D increases to 21.6 – 7.0 = 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 = pc/hr.
Note that increasing the throughput for part D will change the part mix ratios previously presented.

31. 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)

32. 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 Î 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

33. 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 ³ dk " k
Can not exceed machine capacity: Sklj akljm tkjm Xkl £ bm " m
Positive number of units produced: Xkl ³ 0 " k, l

34. 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 ³ dk " k
Can not exceed machine capacity: Sklj akljm tkjm Xkl £ bm " m
Positive number of units produced: Xkl ³ 0 " k, l

35. 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 ³ " m
Demand for parts must be met: Sl Xkl ³ dk " k
Can not exceed machine capacity: Sklj akljm tkjm Xkl £ bm " m
Positive number of units produced: Xkl ³ 0 " k, l

36. Linear programming - example
Consider the manufacture of 5 part types on 4 machine types, each part requiring several operations. Table list the pertinent data. Develop a production plan for: 1) min cost model; 2) max throughput (min processing time); and 3) workload balancing.

37. Linear programming - example
The 3 models were solved using LINDO, a linear programming package, with the results shown in Table 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.

38. 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.
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

39. FMS
What have we learned?