lecture slides - assembly/disassembly systems

MIT 2.852Manufacturing Systems Analysis

Lectures 15–16: Assembly/Disassembly Systems

Stanley B. Gershwin

http://web.mit.edu/manuf-sys

Massachusetts Institute of Technology

Spring, 2010

c2.852 Manufacturing Systems Analysis 1/41 Copyright �2010 Stanley B. Gershwin.

http://web.mit.edu/manuf-sys

Assembly-Disassembly Systems Assembly System

2.852 Manufacturing Systems Analysis 2/41 Copyright c�2010 Stanley B. Gershwin.

Assembly-Disassembly Systems Assembly-Disassembly System with a Loop


Assembly-Disassembly Systems A-D System without Loops


Assembly-Disassembly Systems Disruption Propagation in an A-D System without Loops

F

E

E

E E

E

E

F

F

F

FF

F

F


Assembly-Disassembly Systems Models and Analysis

An assembly/disassembly system is a generalization of a transfer line:

◮ Each machine may have 0, 1, or more than one buffer upstream.

◮ Each machine may have 0, 1, or more than one buffer downstream.

◮ Each buffer has exactly one machine upstream and one machine downstream.

◮ Discrete material systems: when a machine does an operation, it removes one part from each upstream buffer and inserts one part into each downstream buffer.

◮ Continuous material systems: when machine Mi operates during [t, t + δt], it removes µi δt from each upstream buffer and inserts µi δt into each downstream buffer.

◮ A machine is starved if any of its upstream buffers is empty. It is blocked if any of its downstream buffers is full.



◮ A/D systems can be modeled similarly to lines: ◮ discrete material, discrete time, deterministic processing time,

geometric repair and failure times; ◮ discrete material, continuous time, exponential processing, repair, and

failure times; ◮ continuous continuous time, deterministic processing rate, exponential

repair and failure times; ◮ other models not yet discussed in class.

◮ A/D systems without loops can be analyzed similarly to lines by decomposition.

◮ A/D systems with loops can be analyzed by decomposition, but there are additional complexities.



◮ Systems with loops are not ergodic. That is, the steady-state distribution is a function of the initial conditions.

◮ Example: if the system below has K pallets at time 0, it will have K pallets for all t ≥ 0. Therefore, the probability distribution is a function of K .

Empty Pallet Buffer

Raw Part Input

Finished Part Output

◮ This applies to more general systems with loops, such the example on Slide 3.



◮ In general,

p(s|s(0)) = lim t→∞

prob { state of the system at time t = s|

state of the system at time 0 = s(0)}.

◮ Consequently, the performance measures depend on the initial state of the system:

◮ The production rate of Machine Mi , in parts per time unit, is

Ei (s(0)) = prob

�

αi = 1 and (nb > 0 ∀ b ∈ U(i)) and

(nb < Nb ∀ b ∈ D(i))

�

�

�

�

s(0)

�

.

◮ The average level of Buffer b is

n̄b(s(0)) = �

s

nb prob (s|s(0)).


Assembly-Disassembly Systems Decomposition

Mj

B (j, i)

B (n, i)

Mn

Mi

B (i, m)

Mm

B (i, q)

Mq

•

•

•

•

•

•

Part of Original Network


Assembly-Disassembly Systems Decomposition

Mu (j, i) B (j, i) Md (j, i)

Mu (n, i) B (n, i) Md (n, i)

•

• •

Mu (i, m) B (i, m) Md (i, m)

Mu (i, q) B (i, q) Md (i, q) •

• •

Part of Decomposition


Numerical examples Eight-Machine Systems

Deterministic processing time model



7.3351

7.3351

5.6516

5.6516

2.6449

7.3351

7.3351

Case 1:

ri = .1, pi = .1, i = 1, ..., 8;

Ni = 10, i = 1, ..., 7.



7.9444

7.9444

7.0529

7.0529

2.0555

7.9444

7.9444

Case 2:

Same as Case 1 except p7 = .2



7.7077

7.7077

4.1089

6.5276

2.2923

4.5640

7.7077

Case 3:




7.9017

7.9017

3.4593

6.9609

2.0983

7.9017

7.9017

Case 4:



Numerical ExamplesAlternate Assembly Line Designs

A product is made of threesubassemblies (blue, yellow,and red). Each subassemblycan be assembledindependently of the others.We consider four possibleproduction systemstructures.

Machine 6 (the first machineof the yellow process) is thebottleneck — the slowestoperation of all.

2.852 Manufacturing Systems Analysis 17/41 cCopyright �2010 Stanley B. Gershwin.

Numerical Examples Alternate Assembly Line Designs



Now the bottleneck is Machine 5, the last operation of the blue process.


Equivalence Simple models

Consider a three-machine transfer line and a three-machine assembly system. Both are perfectly reliable (pi = 0) exponentially processing time systems.

µ2

1N = 2

µ1

µ3

N = 3 2

N = 2 N = 31 2

µ µ µ1 2 3


Equivalence Assembly System State Space

µ2

1N = 2

µ1

µ3

N = 3 2

µ3

µ3

µ3

µ

µ

µ

µ 2

µ 2

µ 2

µ 2

µ 2

µ 2

µ1

µ1 µ1

µ1

µ1

µ1

µ3

µ3

µ3

00 10

01 11

1202

03 13 23

22

21

20

µ1 µ1

3

3

3

2.852 cManufacturing Systems Analysis 22/41 Copyright �2010 Stanley B. Gershwin.

Equivalence Transfer Line State Space

µ3

µ3

µ3

µ

µ

µ

µ 2

µ 2

µ 2

µ 2

µ 2

µ 2

µ1

µ1 µ1

µ1

µ1

µ1

µ3

µ3

µ3

03 13

02 12

1101

00 10 20

21

22

23

µ1 µ1

3N = 2 N = 3

1 2

3µ µ µ

1 2 3

3


Equivalence Unlabeled State Space

◮ The transition graphs of the two systemsare the same except for the labels of thestates.

◮ Therefore, the steady-state probability 3

distributions of the two systems are the same, except for the labels of the states.

◮ The relationship between the labels of the 3

states is:

(n1 A , n2

A) ⇐⇒ (n1 T ,N2 − n2

T ) 3

◮ Therefore, in steady state,

µ3

µ3

µ3

µ

µ

µ

µ 2

µ 2

µ 2

µ 2

µ 2

µ 2

µ1

µ1 µ1

µ1

µ1

µ1

µ3

µ3

µ3

µ1 µ1

prob(n1 A , n2

A) = prob(n1 T ,N2 − n2

T )


� �

Equivalence Assembly System Production Rate

Production rate = rate of flow of material into M1

1 3

= µ1 p(n1, n2) n1=0 n2=0

µ2

1N = 2

µ1

µ3

N = 3 2

µ3

µ3

µ3

µ3

µ3

µ3

µ 2

µ 2

µ 2

µ 2

µ 2

µ 2

µ1

µ1 µ1

µ1

µ1

µ1

µ3

µ3

µ3

00 10

01 11

1202

03 13 23

22

21

20

µ1 µ1


Equivalence Assembly System Production RateTransfer Line Production Rate

Production rate = rate of flow of material into M1

= µ1

1 �

n1=0

3 �

n2=0

p(n1, n2)

1N = 2

2N = 3

µ3

µ1

µ2

µ3

µ3

µ3

µ3

µ3

µ3

µ 2

µ 2

µ 2

µ 2

µ 2

µ 2

µ1

µ1 µ1

µ1

µ1

µ1

µ3

µ3

µ3

03 13

02 12

1101

00 10 20

21

22

23

µ1 µ1


Equivalence Assembly Systtem Production RatEqual Produc ion Rates

Therefore

PA PT =


� � � �

� �

Equivalence Assembly System n̄1

2 3 2 3

n̄1 = n1p(n1, n2) = n1 p(n1, n2)

µ

µ

µ

µ1 µ1 n1=0 n2=0 n1=0 n2=0

µ2

1N = 2

µ1

µ3

N = 3 2

3

3

3

µ3

µ3

µ3

µ 2

µ 2

µ 2

µ 2

µ 2

µ 2

µ1

µ1 µ1

µ1

µ1

µ1

µ3

µ3

µ3

00 10

01 11

1202

03 13 23

22

21

20


� �

Equivalence Assembly System n̄1Transfer Line n̄1

2 3 2 3� � � �

n̄1 = n1p(n1, n2) = n1 p(n1, n2) µ1 µ1 n1=0 n2=0 n1=0 n2=0

N = 2 N = 31 2

µ µ µ1 2 3

3

3

3

µ3

µ3

µ3

µ 2

µ 2

µ 2

µ 2

µ 2

µ 2

µ1

µ1 µ1

µ1

µ1

µ1

µ3

µ3

µ3

03 13

02 12

1101

00 10 20

21

22

23


Equivalence Assembly System Production RateEqual n̄1

Therefore

n̄1 A = n̄1

T


� � � �

� �

Equivalence Assembly System n̄2

2 3 3 2

n̄2 = n2p(n1, n2) = n2 p(n1, n2) µ µ1 1 n1=0 n2=0 n2=0 n1=0

µ2

1N = 2

µ1

µ3

N = 3 2

µ µ3

µ µ3

µ µ3

3

3

3

µ 2

µ 2

µ 2

µ 2

µ 2

µ 2

µ1

µ1 µ1

µ1

µ1

µ1

µ3

µ3

µ3

00 10

01 11

1202

03 13 23

22

21

20


� � � �

� �

Equivalence Assembly System n̄2Transfer Line n̄2

2 3 3 2

n̄2 = n2p(n1, n2) = n2 p(n1, n2) µ µ1 1 n1=0 n2=0 n2=0 n1=0

µ µ3 N = 2 N = 3

1 2

µ µ3

µ µ µ1 2 3

µ µ3

3

3

3

µ 2

µ 2

µ 2

µ 2

µ 2

µ 2

µ1

µ1 µ1

µ1

µ1

µ1

µ3

µ3

µ3

03 13

02 12

1101

00 10 20

21

22

23


Equivalence Assembly System Production RateComplementary n̄1

Therefore

n̄2 A = N2 − n̄2

T


Equivalence Theorem

◮ Notation: Let j be a buffer. Then the machine upstream of the buffer is u(j) and the machine downstream of the buffer is d(j).

◮ Theorem:◮ Assume

′ ◮ Z and Z are two exponential A/D networks with the same number of

machines and buffers. Corresponding machines and buffers have the ′ same parameters; that is, µi = µi , i = 1, ..., kM and

′ Nb = Nb , b = 1, ..., kB . ′

◮ There is a subset of buffers Ω such that for j 6∈ Ω, u (j) = u(j) and ′ ′ ′ d (j) = d(j); and for j ∈ Ω, u (j) = d(j) and d (j) = u(j). That is,

there is a set of buffers such that the direction of flow is reversed in the two networks.

′◮ Then, the transition equations for network Z are the same as those of

Z , except that the buffer levels in Ω are replaced by the amounts of space in those buffers.


Equivalence Theorem

′ ◮ That is, the transition (or balance) equations of Z can be written by

transforming those of Z .

◮ In the Z equations, replace nj by Nj − nj for all j ∈ Ω.


Equivalence Theorem

Corollary: ◮ Assume:

◮ The initial states s(0) and s ′(0) are related as follows: nj ′(0) = nj (0)

′for j 6∈ Ω, and nj (0) = Nj − nj (0) for j ∈ Ω.

◮ Then

P ′ (n ′ (0)) = P(n(0))

n̄b′ (n ′ (0)) = n̄b(n(0)), for j 6∈ Ω

n̄b′ (n ′ (0)) = Nb − n̄b(n(0)), for j ∈ Ω


Equivalence Theorem

Corollary: That is,

◮ the production rates of the two systems are the same,

◮ the average levels of all the buffers in the systems whose direction of flow has not been changed are the same,

◮ the average levels of all the buffers in the systems whose direction of flow has been changed are complementary; the average number of parts in one is equal to the average amount of space in the other.


Equivalence Equivalence class of three-machine systems

N N1 2

µ1

2N

µ2

µ3

µ1

1N

2 N

2B

1B

µ2

µ3

2

1

1

2

3

N

N

µ

µ

µ

1N

2B

1B

µ µ µ3 2 1


Equivalence Equivalence classes of four-machine systems

Representative members


Equivalence Example of equivalent loops

2

2

1

1

4

4

3

3

4

3

3

2

2

1

4

1

Ω= (3, 4)

(a) A Fork/ Join Network (b) A Closed Network


Equivalence To come

◮ Loops and invariants

◮ Two-machine loops

◮ Instability of A/D systems with infinite buffers


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2.852 Manufacturing Systems Analysis Spring 2010

For information about citing these materials or our Terms of Use,visit:http://ocw.mit.edu/terms.

http://ocw.mit.edu

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