scheduling in flexible robotic manufacturing cells hakan gÜltekİn

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SCHEDULING IN FLEXIBLE ROBOTIC MANUFACTURING CELLS HAKAN GÜLTEKİN

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SCHEDULING IN FLEXIBLE ROBOTIC MANUFACTURING CELLS

HAKAN GÜLTEKİN

HAKAN GULTEKIN

Mc 1 Mc 2 Mc m

...Input Buffer Output Buffer

Robot Linear Tracks

Parts

The System:

HAKAN GULTEKIN

• Robot never loads an already loaded machine and never unloads an empty machine.

General Assumptions:

• No buffers between the workstations.

• Robot loads and unloads the machines and transports the parts between the machines.

HAKAN GULTEKIN

General Assumptions:

• Maximize production rate = minimize cycle time

• Cycle time: Long run average time to produce one part

• Loading and unloading times, i or

• Robot travel times are ij or.

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Mc 1 Mc 2

Robot move sequence:

Alternative 1:

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Mc 1 Mc 2

Alternative 1:

Robot move sequence:

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Mc 1 Mc 2

Alternative 1:

Robot move sequence:

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Mc 1 Mc 2

Alternative 1:

Robot move sequence:

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Mc 1 Mc 2

Alternative 2:

Robot move sequence:

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Mc 1 Mc 2

Alternative 2:

Robot move sequence:

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Mc 1 Mc 2

Alternative 2:

Robot move sequence:

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• Multiple or identical parts to be processed.

Mc 1 Mc 2

Multiple Parts Case:

Different sequences of parts result in different production rates.

Part input sequence:

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Mc 1 Mc 2

Identical Parts Case:

Different sequences of parts result in identical production rates.

• Multiple or identical parts to be processed.

Part input sequence:

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Robot move cycles:

• Cyclic production: Production by repeating a fixed sequence of robot moves. Easy to implement and control.

• n-unit cycle: Sequence of robot moves that load and unload each machine exactly n times in which the initialand final states of the cell are the same.

S1seminer.exe S2seminer.exe

S1:A0A1A2 S2:A0A2A1 S12S21:A0A1A0A2A1A2

S12s21.exe

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Previous results (identical parts):

• 1-unit cycles are optimal for two and three machine cells.

• If P1+P2 < 2, S1 is optimal. If P1+P2 > 2 S2 is optimal. If P1+P2 = 2both perform equally well.

• Six 1-unit cycles in three machine cells.

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Previous results (identical parts):

• 1-unit cycles need not be optimal for cells with four or greater number of machines.

• m! 1-unit cycles

• Algorithm to find the best 1-unit cycle.

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Previous results (multiple parts):

• 1-unit cycles need not be optimal even in two machine cells

• Algorithm that simultaneously finds the optimal part input sequence and robot move sequence.

• Two Machine Case:

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Previous results (multiple parts):

• Fix the robot move cycle to one of the six 1-unit robot move cycles and find the part input sequence.

• NP-Hard for 2 out of 6 1-unit cycles.

• Three Machine Case:

• Suboptimal! Worst case?

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Other special cell configurations:

• Parallel machines

• More than 1 robots

• Dual gripper robots

• Different assumptions on , and Pi’s

• Robotic cells with buffers in front of the machines

• Different objective functions

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• Each part has a number of operations to be performed on the machines.

Assumptions pertinent to this study:

• CNC machines.

• Operational flexibility: ability to interchange the ordering of operations.

• Process flexibility: ability to perform multiple operations on the same machine.

Cncasil.exe

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The problem: Find the allocation of the operations to the machines and the corresponding robot move sequence that jointly minimize the cycle time.

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Definitions:

k-allocation type : Part operations (o1,o2,o3,o4,o5)

(o1,o2)

k

M1 M2

(o3,o4,o5)(o2,o3,o5) (o1,o4)

(o4) (o1,o2,o3,o5)

(o1,o2)

k

(o3,o4,o5)(o2,o3,o5) (o1,o4)

(o4) (o1,o2,o3,o5)

HAKAN GULTEKIN

Example:

• Let =1, =2, o1=13, o2=17, o3=10, o4=5, o5=5 P = 50

• Assume a 2-m/c robotic cell and consider robot move cycle S2.

1-allocation S2: P1=o1+o3=23, P2=o2+o4+o5=27

TS2(1)= 4+4+max{2+4, P1, P2} = 39

2-allocation S2: P11=o1+o3=23, P12=o2+o4+o5 = 27, P21=o2+o4+o5=27, P22=o1+o3 = 23.

TS2(2)= 4+4+1/2max{2+4, P11, P22} +1/2max{2+4, P21, P12} = 37

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2-machine case:

• Limited tool magazine capacity

• In most cases number of required tools exceeds the tool magazine capacity

Then for each part we have 3 sets of operations:1. Operations that can only be processed on the first

machine with total processing time P1.2. Operations that can only be processed on the second

machine with total processing time P2.3. Operations that can be processed on both machines

with total processing time P.

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2-machine case:

The problem: Find the allocation of the operations that are in the third set to the machines and the corresponding robot move sequence that jointly minimize the cycle time.

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1. If P1 + P2 ≥ 2then S2 gives the minimum cycle time,2. Else, 2.1. If 2P + P1 + P2 ≤ 2, then S1 gives the minimum cycle time 2.2. Else, 2.2.1. If 2P1 + P2 + P ≤ 2 + 6then S12S21 gives the minimum cycle time, 2.2.2. Else, depending on the allocations of the operations for S2 either S2 or S12S21 gives the minimum cycle time.

Theorem:

2-machine case:

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P2 = P1

2-machine case:

Grafik.exe

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• Note that whenever P = 0, then there is no allocation problem and the result reduces to:

S1 is optimal if P1+P2 < 2and S2 is optimal otherwise.

The problem as well as the result becomes identical to two machine identical parts case.

2-machine case:

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Duplicate all cutting tools: P1 = P2 = 0.

2-machine case:

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A new robot move cycle:

3_mc_proposed.exe

• Tproposed= 4+2(m+1)+1/m(max{0, P-2(m-1)-(m-1)(m+2)})

Lower bound: max{2(m+1)(+)+min{P, }, 4+4+P/m}

- Load and unload all machines once: 2(m+1)

- Travel from input buffer to output buffer and return back: 2(m+1)

- For all machines either wait or do some other activity: ∑min{Pi,} ≥ min{P,}.

First component:

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- Either wait or do some other activity: Pi,- Unload Mi: - Transport to M(i+1): - Load M(i+1): - Transport to M(i-1): 2- Unload M(i-1): - Transport to Mi: - Load Mi: 4+4+maxi{Pi} ≥ 4+4+P/m

Second component:

Lower bound:

max{2(m+1)(+)+min{P, }, 4+4+P/m}

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• The new robot move cycle gives the minimum cycle time.

• Changing the layout from in-line robotic cell to robot-centered cell reduces the cycle time of the new cycle even further.

2-machine case:

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Example:

o1 = 40, o2 = 45, o3 = 50, o4 = 60, o5 = 50, o6 = 55= 2, = 10

Allocation: M1 M2 M3 o1,o4 P1=100 o2,o6 P2=100 o3,o5 P3=100

TS6 = 4+4+max{4+8, P1, P2, P3} = 148

Tproposed = 1/3(max{12+24, P+8+14}) = 152

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Theorem: The proposed cycle gives the minimum cycle time if:• (m-2) ≤ 2 or • P ≤ 2(m2-1)+(m2+2m-2)

Theorem: For (m-2) > 2 and P > 2(m2-1)+(m2+2m-2), using the proposed cycle instead of the optimal robot move cycle has the following worst case performance bound:

26

231

2

2

mm

mmm=3 1.08m=4 1.157m=5 1.226m=6 1.384

m-machine case:

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Optimal number of machines:

Theorem: The optimal number of machines, m*, for the proposed cycle is either or )2(2/1

Theorem: Proposed cycle with optimal number of machines dominates all traditional robot move cycles with optimal number of machines.

P 417204 22 where .

1 )2(2/1

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Bicriteria Robotic Cell Scheduling:

• No existing studies considering cost objectives in robotic cell scheduling literature.

• Minimize manufacturing cost vs Minimize cycle time.

• Manufacturing Cost = Machining Cost+Tooling Cost

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Cut

ting

spee

d,

Cut

ting

spee

d, vv

Bicriteria Robotic Cell Scheduling:

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Cost

Processing time

Total cost

Machining cost

Tooling cost

PuPl

Bicriteria Robotic Cell Scheduling:

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• Increasing the processing times reduces the cost but may increase the cycle time.

• Reducing the processing times increases the cost but may reduce the cycle time.

• Find the set of efficient solutions.

Bicriteria Robotic Cell Scheduling:

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Cycle timeTl Tu

Manufacturingcost

Cl

A

B

C

Dominated Solution

Nondominated Solutions

Bicriteria Robotic Cell Scheduling:

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Cycle time

Manufacturingcost

S1

S2

Bicriteria Robotic Cell Scheduling:

T

C

HAKAN GULTEKIN

www.ie.bilkent.edu.tr/~robot

QUESTIONS