37 38 39 40 flexible manufacturing systems

8/12/2019 37 38 39 40 Flexible Manufacturing Systems

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Flexible Manufacturing Systems

Lecture 39


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FMS

Flexible Manufacturing Systems (FMS) is amanufacturing philosophy based on the conceptof effectively controlling material flow through a

network of versatile production stations using anefficient and versatile material handling andstorage systems.

FMS is a system that consists of numerousprogrammable machine tools connected by anautomated material handling system.


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shorter lead times, meeting demandfluctuations, handling volume and variety,reduction in space and people and obtaining

better control due to automation.

The main disadvantage is that the initial

installation cost and operational costs are highand it is necessary to have enough volumes tojustify the use of FMS.


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classified into

Flexible Manufacturing Cell (FMC),

Flexible manufacturing System (FMS) and

Flexible manufacturing Line (FML).

FMCs have high flexibility but handle less volume

while FML have less flexibility but can handle verylarge volumes.


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Types of flexibility

Operation flexibility is the ability to perform more than one operation on a

given part type.

Part flexibility is the ability to perform operations on more than one part

at a time.

Change over flexibility is the ability to change over from one part to

another in negligible change over times, in parallel when an operation is

being performed on another piece.

Routing flexibility means that a particular part can be delivered to any one

of the number of alternative stations

Path flexibility results from the existence of more than one possible path

from a specified origin to a specified destination.


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Production planning problems (Stecke)

Part Selection problemto choose the sub set of partsthat are to be made in the FMs.

Machine grouping problemto group machines suchthat any machine in a group can handle all jobs that

other machines in the group can process. Production ratio problemto decide the ratios in

which the parts chosen in 1 can be made.

Resource allocation problemto allocate limited

number of pallets and fixtures to the selected parttypes.

Allocate operations and tools to machine to meetstated objectives subject to technological constraints.


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Part Selection problem

Choose set of parts that maximizes profitsubject to time availability constraints

Binary Knapsack problem

Consider 4 parts that can be made in an FMS.The profits are 12, 10, 8 and 9 and the times

required are 20, 16, 15 and 12 respectively.The available time is 45. Find the parts thatare made in the FMS?


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The ratios of Cj/tjare 0.6, 0.625, 0.533 and 0.75.The rearranged order (using variable y), we havethe problem

Maximize 9Y1+ 10Y2+ 12Y3+ 8Y4

Subject to 12Y1+ 16Y2+ 20Y3+ 15Y4 45;Yj= 0,1

The optimum solution to this problem is Y1= Y2=Y4= 1 with Z = 27. Parts 1, 3 and 4 are made usingthe FMS.


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Heuristic

A heuristic solution after rearranging the

variable would be Y1= Y2= 1with Z = 21. We

observe that 17 time units are free and we can

insert Y4= 1 to get the same solution.


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Machine Loading

Balance load subject to

Each job to one machine

Tool slot requirement and capacity Binary integer programming


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Example

1 2 3 4 5 6 7 Ti

M1 1 4 7 8 6 7 9 10

M2 2 6 5 7 8 8 7 10

sj 2 3 3 4 2 2 4

The optimum solution to the problem is X11

= X22

= X32

= X41

= X51

=

X61= X72= 1 with Z = 22. This means that jobs {1, 4, 5, 6} are

assigned to M1 and jobs {2, 3, 7} are assigned to M2. The slot

requirements are 10 and 9 respectively. The loads are 22 and 18

respectively.


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Allocating the job to the machine with the

least processing times we assign jobs 1, 2, 5, 6

to M1 and jobs 3, 4, 7 to M2. The total

processing times are 18 and 19 respectively.

The tool slots required are 9 and 11. The tool

slot constraint is violated. Interchanging 2 and

7 we allot 1, 5, 6, 7, to M1 with time = 23 and2, 3, 4 to M2 with time = 18. The tool slots

required are 10 and 10.


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Tool slot saving

Jobs 3 and 7saving = 2

Jobs 1 and 3saving = 1

Jobs 6 and 7

saving = 2 Jobs 4 and 5saving = 1

Allocating the job to the machine with the least processing timeswe assign jobs 1, 2, 5, 6 to M1 and jobs 3, 4, 7 to M2. The total

processing times are 18 and 19 respectively. The tool slots required

are 9 and 9 (there is a saving of 2 slots if we assign jobs 3 and 7 to

the same machine). The present solution is feasible.


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Let us solve the problem for T = 8 in each machine. Theinitial allocation is {1, 2, 5, 6} and {3, 4, 7}. The tool slotrequirements are 9 and 9. Both are infeasible. We considerM1 and move job 1 to M2. The job allocations are {2, 5, 6}and {1, 3, 4, 7} with tool slots required = 7 and 10.

M2 is infeasible with respect to tool slots. We move job 5 toM1 to get the allocation {2, 4, 5, 6} and {1, 3, 7} withrequired tool slots = 10 and 6 respectively. Now M1 isinfeasible. We move job 6 to M2 to get the allocation {2, 4,5} and {1, 3, 6, 7} with tool slots required = 8 and 8. Wehave a feasible solution. The load on the two machines is18 and 23 respectively.


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The solution obtained in Illustration 12.3 for T = 8 has jobs2, 4, 5 assigned to M1 and jobs 1, 3, 6, 7assigned to M2.This is infeasible for T = 6. It is also obvious that we wouldrequire more than 1 batch on each M1 and M2 to completethe jobs allocated to the machines. We could use the same

allocation and use the tool saving advantage as well as duedates (if available) to schedule the jobs on the machines.

Alternately we could ignore the tool constraint and assignjobs to machines to balance workload. Assigning the jobs tothe machine with minimum processing times gives us {1,2,5, 6} and {3, 4, 7} with load = 18 and 19. We can solve theloading and scheduling problem individually on themachines.


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Flexible Manufacturing Systems

Lecture 39


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Batching and sequencing

Consider three jobs (J1 to J3) with processing times 5, 7 and 7 to beproduced in an FMS with a single machine. The machine has 6 toolslots and the tool slot requirements are 3, 4 and 4. The due datesfor the jobs are 12, 18 and 10 respectively. Also consider thesituation where we can have a tool slot if jobs 1 and 2 are assignedtogether. Also consider that changeover time between batches is 3units?

if we ignore the tool requirement constraint and the batchingconstraint, the optimum solution that minimizes total tardiness isJ3-J1-J2. The completion times are 7, 12, 19 and the total tardiness

is 1. If we ignore the tool saving constraint and consider 6 tool slots,we require three batches one each for J3, J1 and J2. The actualcompletion times including the batch changeover times are 7, 15,25. The due dates are 10, 12 and 18 and the total tardiness is 10.


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If we include the tool saving constraint for the

sequence J3-J1-J2, we will have 2 batches with

J3 in the first batch and J1-J2 in the second

batch (due to tool savings). The completiontimes are 7, 15, 22. The total tardiness is 7.


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Loading considering different tools for

each machine Here we assume that the number of tools required can be

different for different machines. We also assume that theprocessing times are different for different machines. Weformulate the problem for n jobs and 2 machines (M1 andM2). Let the tool slots available be Tj. Let there be Ajtools

available for machine j (Aj Tj). Each job i requires Sij toolson machine j. These are specific, for example, job 1 wouldrequire tool S11and tool S21on machine 1 and wouldrequire tool S32on machine M2. We would allocate jobs tomachines such that the sum of processing times is

minimized. Let Xk= 1 if tool k is chosen for use in M1 andlet Yk= 1 if tool k is chosen for use in M2. Let Zij= 1 if job i isallotted to machine j.

Th bj ti i t


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2

1 1

n

ij iji j

p Z

2

1

1ijj

Z

1

K

k k

k

X T

1

K

k k

k

Y T

The objective is toMinimize

Subject to

Ziij

Sij, ; Zij, Xk, Yk= 0,1.


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Consider 7 jobs that are to be processed. Twomachines, each with 4 slots is available. A job

is to be assigned to only one machine. The

data is given in table 12.4

1 2 3 4 5 6 7 Ti

M1 1 4 7 8 6 7 9 4

M2 2 6 5 7 8 8 7 4A1 A2

B1 B3

A2

B4

A1 A3

B2 B5

A2 A4

B3 B4

A3 A4 A5

B2 B3

A4 A5

B4

A3 A5

B1 B2


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The binary Integer programming formulation has 24 binary variables and35 constraints. The optimum solution is X2= X3= X4= X5= 1; Y1= Y2= Y3=Y5= 1; Z12= Z21= Z32= Z41= Z51= Z61= Z72= 1 with objective value = 39.Tools A2, A3, A4 and A5 are chosen for M1 and jobs 2, 4, 5, and 6 areallotted to M1. Tools B1, B2, B3 and B5 are chosen for M2 and jobs 1, 3and 7 are allotted to M2. The loads are 25 and 14 adding to a total of 39.

Based on minimum processing time we allot J1 to M1. M1 is assigned

tools A1 and A2. J2 goes to M1 because it has all the tools required. Basedon minimum processing time J3 goes to M2. M2 has tools B2 and B5. J5 isallotted to M2 and tool B3 is added to M2. J4 also goes to M2 and tool B4is added to M2. J6 goes to M2 because the tools are already available. Job

7 goes to m1 and tools A3 and A5 are added. Jobs {1, 2, 7} and tools A1,A2, A3, A5 go to M1. The total processing time is 14. Jobs {3, 4, 5, 6} andtools B2, B3, B4, B5 go to M2. The total processing time is 26.


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Sequencing jobs on machines

Four jobs have to be processed on M2. The processingtimes are 6, 5, 7 and 8. The due dates are 10, 25, 15 and 20.Sequence the jobs to minimize flow time, tardiness andnumber of tardy jobs?

SPT sequence minimizes flow time. The sequence is J2-J1-J3-J4. The completion times are 5, 11, 18, 26. The minimumtotal flow time possible is 60. The mean flow time is 15.This sequence has 3 tardy jobs and the total tardiness is 10.

The EDD sequence is J1-J3-j4-J2. The completion times are

6, 13, 21 and 26. The mean flow time is 66/4 = 16.5. Thissequence has 2 tardy jobs with total tardiness = 2.

37 38 39 40 flexible manufacturing systems

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