ch8-fms

ROBOTICS An IntroductionReferences:
1. Nanua Singh, Computer-Integrated Design and Manufacturing, John Wiley & Sons, 1996
2. Mikell Groover, Automated Production Systems and Computer-Integrated
Manufacturing, Prentice-Hall, 2nd edition, 2001
ME 482 - Manufacturing Systems
- Group technology (GT)
- Control software
- Analysis models
To test understanding of the material presented
Hierarchical computer control architectures
robots…machine tools…
Similar manufacturing process requirements (manufacturing attributes), but with different design attributes
Turned, drilled, milled…..
Process type plant layout …dashed lines indicates departments!
Group technology layout
Parts distinguished (classified) by design attributes and manufacturing attributes.
Part differentiated by coding methods for
design retrieval
GT Opitz form code
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 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 15100
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?
2. Quick changeover of process/operating instructions
3. Quick changeover of physical setup
FMS operations:
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.
Single machine
Ladder layout
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)
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)
Scheduling (master production schedule) and dispatching (launching of parts into the system)
Machine loading
Part routing
Part grouping
Tool management
Queueing models
Heuristic approaches
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 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
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)
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
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
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.
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
FMS example – determine nt
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
“load” “machine at 2” “unload”
Load
Unload
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
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
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:
Each station utilization
Overall FMS utilization
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)
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.
U1 = [1(0.833) + 3(0.879) + 2(1.0) + 1(0.555)]/7= 0.861 (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.
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.
Two criteria:
- production of parts at max production rate
Define:
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)
Sound familiar?
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
Optimizing operations allocation in an FMS with negligible setup
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
Constraints:
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 0 " 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.
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.
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 12.18.
What have we learned?

ch8-fms

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