2852 spring 2010

MIT 2.852

Manufacturing Systems Analysis

Lectures 1516: 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.

Assembly-Disassembly Systems Assembly System

2.852 Manufacturing Systems Analysis 2/41 Copyright c2010 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 buer upstream.

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

Each buer 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 buer and inserts one part into each downstream buer.

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

A machine is starved if any of its upstream buers is empty. It is blocked if any of its downstream buers 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

i = 1 and (nb > 0 b U(i)) and

(nb < Nb b D(i))

s(0) .

The average level of Buer b is

nb(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 Examples

Alternate Assembly Line Designs

A product is made of three

subassemblies (blue, yellow,

and red). Each subassembly

can be assembled

independently of the others.

We consider four possible

production system

structures.

Machine 6 (the rst machine

of the yellow process) is the

bottleneck the slowest

operation 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.

2

1N = 2

1

3

N = 3 2

N = 2 N = 31 2

1 2 3


Equivalence Assembly System State Space

2

1N = 2

1

3

N = 3 2

3

3

3

2

2

2

2

2

2

1

1 1

1

1

1

3

3

3

00 10

01 11

1202

03 13 23

22

21

20 1 1

3

3

3

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

Equivalence Transfer Line State Space

3

3

3

2

2

2

2

2

2

1

1 1

1

1

1

3

3

3

03 13

02 12

1101

00 10 20

21

22

23 1 1

3N = 2 N = 31 2

3

1 2 3

3


Equivalence Unlabeled State Space

The transition graphs of the two systems

are the same except for the labels of the

states.

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,

3

3

3

2

2

2

2

2

2

1

1 1

1

1

1

3

3

3

1 1

prob(n1 A , n2

A) = prob(n1 T ,N2 n2

T )


Equivalence Assembly System Production Rate

Production rate = rate of ow of material into M1 1 3

= 1 p(n1, n2) n1=0 n2=0

2

1N = 2

1

3

N = 3 2

3

3

3

3

3

3

2

2

2

2

2

2

1

1 1

1

1

1

3

3

3

00 10

01 11

1202

03 13 23

22

21

20 1 1


Equivalence Assembly System Production RateTransfer Line Production Rate

Production rate = rate of ow of material into M1

= 1

1 n1=0

3 n2=0

p(n1, n2)

1N = 2 2N = 3 3

1 2 3

3

3

3

3

3

2

2

2

2

2

2

1

1 1

1

1

1

3

3

3

03 13

02 12

1101

00 10 20

21

22

23 1 1


Equivalence Assembly Systtem Production RatEqual Produc ion Rates

Therefore

PA PT =


Equivalence Assembly System n1

2 3 2 3

n1 = n1p(n1, n2) = n1 p(n1, n2)

1 1 n1=0 n2=0 n1=0 n2=0

2

1N = 2

1

3

N = 3 2

3

3

3

3

3

3

2

2

2

2

2

2

1

1 1

1

1

1

3

3

3

00 10

01 11

1202

03 13 23

22

21

20


Equivalence Assembly System n1Transfer Line n1

2 3 2 3 n1 = n1p(n1, n2) = n1 p(n1, n2) 1 1

n1=0 n2=0 n1=0 n2=0

N = 2 N = 31 2

1 2 3

3

3

3

3

3

3

2

2

2

2

2

2

1

1 1

1

1

1

3

3

3

03 13

02 12

1101

00 10 20

21

22

23


Equivalence Assembly System Production RateEqual n1

Therefore

n1 A = n1

T


Equivalence Assembly System n2

2 3 3 2

n2 = n2p(n1, n2) = n2 p(n1, n2) 1 1 n1=0 n2=0 n2=0 n1=0

2

1N = 2

1

3

N = 3 2

3

3

3

3

3

3

2

2

2

2

2

2

1

1 1

1

1

1

3

3

3

00 10

01 11

1202

03 13 23

22

21

20


Equivalence Assembly System n2Transfer Line n2

2 3 3 2

n2 = n2p(n1, n2) = n2 p(n1, n2) 1 1 n1=0 n2=0 n2=0 n1=0

3 N = 2 N = 31 2

3 1 2 3

3

3

3

3

2

2

2

2

2

2

1

1 1

1

1

1

3

3

3

03 13

02 12

1101

00 10 20

21

22

23


Equivalence Assembly System Production RateComplementary n1

Therefore

n2 A = N2 n2

T


Equivalence Theorem

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

Theorem:

Assume

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

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

Nb = Nb , b = 1, ..., kB .

There is a subset of buers such that for j 6 , u (j) = u(j) and d (j) = d(j); and for j , u (j) = d(j) and d (j) = u(j). That is,

there is a set of buers such that the direction of ow is reversed in the two networks.

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

Z , except that the buer levels in are replaced by the amounts of space in those buers.


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 .


Equivalence Theorem

Corollary:

Assume: The initial states s(0) and s (0) are related as follows: nj

(0) = nj (0) for j 6 , and nj (0) = Nj nj (0) for j .

Then

P (n (0)) = P(n(0))

nb (n (0)) = nb(n(0)), for j 6

nb (n (0)) = Nb nb(n(0)), for j


Equivalence Theorem

Corollary: That is,

the production rates of the two systems are the same,

the average levels of all the buers in the systems whose direction of ow has not been changed are the same,

the average levels of all the buers in the systems whose direction of ow 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

1

2N

2

3

1

1N

2 N

2B

1B

2 3

2

1

1

2

3

N

N

1N

2B

1B

3 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

= (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 innite buers


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2.852 Manufacturing Systems Analysis Spring 2010

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MIT 2.852


Lectures 1921 Scheduling: Real-Time Control of Manufacturing Systems

Stanley B. Gershwin

Spring, 2007

Copyright c2007 Stanley B. Gershwin.

Definitions

Events may be controllable or not, and predictable or not.

controllable uncontrollable predictable loading a part lunch unpredictable ??? machine failure

Copyright 2007 Stanley B. Gershwin. c 2

Definitions

Scheduling is the selection of times for future controllable events.

Ideally, scheduling systems should deal with all

controllable events, and not just production.

That is, they should select times for operations,

set-up changes, preventive maintenance, etc.

They should at least be aware of set-up changes, preventive maintenance, etc.when they select times for operations.


Definitions

Because of recurring random events, scheduling is an on-going process, and not a one-time calculation.

Scheduling, or shop floor control, is the bottom of the scheduling/planning hierarchy. It translates plans into events.

Copyright c 42007 Stanley B. Gershwin.

Issues in

Factory Control

Problems are dynamic ; current decisions influence future behavior and requirements.

There are large numbers of parameters, time-varying quantities, and possible decisions.

Some time-varying quantities are stochastic . Some relevant information (MTTR, MTTF, amount of inventory available, etc.) is not known.

Some possible control policies are unstable .


Dynamic Example

Programming Problem

Discrete Time, Discrete State, Deterministic F

68

A

1

B L

5

6

2 2

G 5

10

6

4M

9

2 D 7 72 4 N

5 9

4 E

5 I 8

6

13

6 2 C H

J

1

4 Z

K 3 O

Problem: nd the least expensive path from A to Z. Copyright c 62007 Stanley B. Gershwin.

Dynamic Example

Programming Problem

Let g(i, j) be the cost of traversing the link from i to j. Let i(t) be the tth node on a path from A to Z. Then the path cost is

T

g(i(t 1), i(t)) t=1

where T is the number of nodes on the path, i(0) = A, and i(T ) = Z.

T is not specied; it is part of the solution.


Dynamic Example Programming Solution A possible approach would be to enumerate all possible paths (possible solutions). However, there can be a lot of possible solutions.

Dynamic programming reduces the number of possiblesolutions that must be considered. Good news: it often greatly reduces the number of possible solutions.

Bad news: it often does not reduce it enough to give an exact optimalsolution practically (ie, with limited time and memory). This is the curse of dimensionality .

Good news: we can learn something by characterizing the optimal

solution, and that sometimes helps in getting an analytical optimal

solution or an approximation.

Good news: it tells us something about stochastic problems.


Dynamic Example

Programming Solution

Instead of solving the problem only for A as the initial point, we solve it for all possible initial points.

For every node i, dene J(i) to be the optimal cost to go from Node i to Node Z (the cost of the optimal path from i to Z). We can write

T

J(i) = g(i(t 1), i(t))

t=1

where i(0) = i; i(T ) =Z; (i(t 1), i(t)) is a link for every t.


Dynamic Example


Then J(i) satises

J(Z) = 0

and, if the optimal path from i to Z traverses link (i, j),

J(i) = g(i, j) + J(j).

i j

Z


Dynamic Example Programming Solution Suppose that several links go out of Node i.

4j 3j

2j

5j

1j

6j

i Z

Suppose that for each node j for which a link exists from i to j, the optimal path and optimal cost J(j) from j to Z is known. Copyright 2007 Stanley B. Gershwin.c 11

Dynamic Example


Then the optimal path from i to Z is the one that minimizes the sum of the costs from i to j and from j to Z. That is,

J(i) = min [g(i, j) + J(j)]

j

where the minimization is performed over all j such that a link

from i to j exists. This is the Bellman equation .

This is a recursion or recursive equation because J() appears

on both sides, although with different arguments.

J(i) can be calculated from this if J(j) is known for every node j

such that (i, j) is a link.


Dynamic Example


Bellmans Principle of Optimality: if i and j are nodes on an optimal path from A to Z, then the portion of that path from A to Z between i and j is an optimal path from i to j.

A

Z

j

i


{ }

{ }

Dynamic Example


Example: Assume that we have determined that J(O) = 6 and J(J) = 11. To calculate J(K),

g(K, O) + J(O)J(K) = min

g(K, J) + J(J)

3 + 6 = min = 9.

9 + 11

Copyright 2007 Stanley B. Gershwin.c 14

Dynamic Example


Algorithm 1. Set J(Z) = 0. 2. Find some node i such that

J(i) has not yet been found, and for each node j in which link (i, j) exists, J(j) is already calculated.

Assign J(i) according to

J(i) = min [g(i, j) + J(j)]

j

3. Repeat Step 2 until all nodes, including A, have costs calculated.


KDynamic Example


F

A

B 11 L

5

6 H

C

14

D

E

I

O

J

9

G

M

N 4

8

14

13

11

17

12

13

6

11

Z


Dynamic Example


The important features of a dynamic programming problem are the state (i) ; the decision (to go to j after i); ( ) the objective function

Tt=1 g(i(t 1), i(t))

the cost-to-go function (J(i)) ; the one-step recursion equation that determines J(i)

(J(i) = minj [g(i, j) + J(j)]);

that the solution is determined for every i, not just A and not just nodes on the optimal path;

that J(i) depends on the nodes to be visited after i, not those between A and i. The only thing that matters is the present state and the future;

that J(i) is obtained by working backwards.


Dynamic Example


This problem was discrete time, discrete state, deterministic. Other versions: discrete time, discrete state, stochastic continuous time, discrete state, deterministic continuous time, discrete state, stochastic continuous time, mixed state, deterministic continuous time, mixed state, stochastic

in stochastic systems, we optimize the expected cost. Copyright c 182007 Stanley B. Gershwin.

( )

Programming Discrete time, discrete state

Stochastic Dynamic

Suppose g(i, j) is a random variable; or if you are at i and you choose j, you actually go to k with probability p(i, j, k).

Then the cost of a sequence of choices is random. The objective function is

T

E g(i(t 1), i(t))

t=1

and we can dene J(i) = E min [g(i, j) + J(j)]

j


Programming Continuous Time, Mixed State

Stochastic Example Dynamic

Context: The planning/scheduling hierarchy Long term: factory design, capital expansion, etc. Medium term: demand planning, stafng, etc. Short term: response to short term events part release and dispatch In this problem, we deal with the response to short term events. The factory and the demand are given to us; we must calculate short term production rates; these rates are the targets that release and dispatch must achieve.


DynamicProgramming

Continuous Time, Mixed State

Stochastic Example

x1

x2

d1

d2

u (t)1

u (t)2 Type 2

Type 1

r, p Perfectly exible machine, two part types. i time units required to make Type i parts, i = 1, 2.

Exponential failures and repairs with rates p and r. Constant demand rates d1, d2. Instantaneous production rates ui(t), i = 1, 2 control variables .

Downstream surpluses xi(t).




Objective: Minimize the difference between cumulative production and cumulative demand.

The surplus satises xi(t) = Pi(t) Di(t)

t

Cumulative Production and Demand production P (t)

surplus x (t)

i

i

idemand D (t) = d ti

Copyright c2007 Stanley B. Gershwin. 22



Feasibility: For the problem to be feasible, it must be possible to make approximately diT Type i parts in a long time period of length T, i = 1, 2. (Why approximately?)

The time required to make diT parts is idiT . During this period, the total up time of the machine ie, the time available for production is approximately r/(r + p)T .

Therefore, we must have 1d1T + 2d2T r/(r + p)T , or 2 r

idi r + p

i=1




If this condition is not satised, the demand cannot be met. What will happen to the surplus? The feasibility condition is also written

2 di r

i=1 i r + p

where i = 1/i. If there were only one part type, this would be

r

d

r + p

Look familiar?




The surplus satises xi(t) = Pi(t) Di(t)

where t Pi(t) = ui(s)ds; Di(t) = dit

0

Therefore dxi(t)

= ui(t) di dt




To define the objective more precisely, let there be a function g(x1, x2) such that

g is convex g(0, 0) = 0

lim g(x1, x2) = ; lim g(x1, x2) = . x1 x1

lim g(x1, x2) = ; lim g(x1, x2) = . x2 x2




Examples:

g(x1, x2) = A1x12 + A2x2

2

g(x1, x2) = A1|x1| + A2|x2|

g(x1, x2) = g1(x1) + g2(x2) where + gi(xi) = g(i+)xi + g(i)xi ,

xi + = max(xi, 0), xi = min(xi, 0),

g(i+) > 0, g(i) > 0.


Dynamic

Programming

Objective:

Continuous Time, Mixed State

Stochastic Example

T min E g(x1(t), x2(t))dt

0

x1

g(x ,x )1 2

x2




Constraints: u1(t) 0; u2(t) 0

Short-term capacity:

If the machine is down at time t, u1(t) = u2(t) = 0


u1


Stochastic Example Assume the machine is up for a short period [t, t + t]. Let t

Dynamic

be small enough so that ui is constant; that is ui(s) = ui(t), s [t, t + t]

The machine makes ui(t)t parts of type i. The time required to make that number of Type i parts is iui(t)t. Therefore

iui(t)t t 1/

i

or iui(t) 1 1/ 1

2

u2

0 i




Machine state dynamics: Dene (t) to be the repair state of the machine at time t. (t) = 1 means the machine is up; (t) = 0 means the machine is down.

prob((t + t) = 0|(t) = 1) = pt + o(t)

prob((t + t) = 1|(t) = 0) = rt + o(t)

The constraints may be written iui(t) (t); ui(t) 0

i




Dynamic programming problem formulation: T min E g(x1(t), x2(t))dt

subject to: 0

dxi(t) = ui(t) di

dt

prob((t + t) = 0|(t) = 1) = pt + o(t)

prob((t + t) = 1|(t) = 0) = rt + o(t)

iui(t) (t); ui(t) 0 i

x(0), (0) specified


Dynamic Elements of a DP Problem Programming state: x all the information that is available to determine the future evolution of the system.

control: u the actions taken by the decision-maker. objective function: J the quantity that must be minimized; dynamics: the evolution of the state as a function of the control variables and random events.

constraints: the limitations on the set of allowable controls initial conditions: the values of the state variables at the start of the time interval over which the problem is described. There are also sometimes terminal conditions such as in the network example.


Dynamic Elements of a DP Solution Programming

control policy: u(x(t), t). A stationary or time-invariant policy is of the form u(x(t)).

value function: (also called the cost-to-go function) the value J(x, t) of the objective function when the optimal control policy is applied starting at time t, when the initial state is x(t) = x.


Bellmans Continuous x, t

Equation Deterministic T

Problem: min g(x(t), u(t))dt + F (x(T ))

u(t),0tT 0

such that dx(t)

= f(x(t), u(t), t)dt

x(0) specied

h(x(t), u(t)) 0

x Rn, u Rm, f Rn, h Rk, and g and F are scalars. Data: T, x(0), and the functions f, g, h, and F . Copyright c 352007 Stanley B. Gershwin.

Bellmans Continuous x, t Equation Deterministic The cost-to-go function is

T J(x, t) = min g(x(s), u(s))ds + F (x(T ))

t T J(x(0), 0) = min g(x(s), u(s))ds + F (x(T ))

0 { } t1 T = min g(x(t), u(t))dt + g(x(t), u(t))dt + F (x(T )) .

u(t), 0 t1

0tT



Equation Deterministic

[ ] t1 T = min g(x(t), u(t))dt + min g(x(t), u(t))dt + F (x(T ))

u(t), 0 u(t), t1 0tt1 t1tT{ t1 }

= min g(x(t), u(t))dt + J(x(t1), t1) . u(t), 0

0tt1




where

T J(x(t1), t1) = min g(x(t), u(t))dt + F (x(T ))

u(t),t1tT t1

such that dx(t)

= f(x(t), u(t), t)dt

x(t1) specied

h(x(t), u(t)) 0


{ }

Bellmans Continuous x, t Equation Deterministic Break up [t1, T ] into [t1, t1 + t] [t1 + t, T ] : { t1+t

J(x(t1), t1) = min g(x(t), u(t))dt u(t1) t1

+J(x(t1 + t), t1 + t)}

where t is small enough so that we can approximate x(t) and u(t) with constant x(t1) and u(t1), during the interval. Then, approximately,

J(x(t1), t1) = min g(x(t1), u(t1))t + J(x(t1 + t), t1 + t) u(t1)


{ }



Or,

J(x(t1), t1) = min g(x(t1), u(t1))t + J(x(t1), t1)+ u(t1)

J J (x(t1), t1)(x(t1 + t) x(t1)) + (x(t1), t1)t

x t

Note that dx

x(t1 + t) = x(t1) + t = x(t1) + f(x(t1), u(t1), t1)t dt


{ }

{ }



Therefore

J(x, t1) = J(x, t1)

J J + min g(x, u)t + (x, t1)f(x, u, t1)t + (x, t1)t

u x t where x = x(t1); u = u(t1) = u(x(t1), t1). Then (dropping the t subscript)

J J (x, t) = min g(x, u) + (x, t)f(x, u, t)

t u x


( )



This is the Bellman equation . It is the counterpart of the recursion equation for the network example.

If we had a guess of J(x, t) (for all x and t) we could confirm it by

performing the minimization.

If we knew J(x, t) for all x and t, we could determine u by performing the

minimization. U could then be written

J u = U x, , t .

x

This would be a feedback law .

The Bellman equation is usually impossible to solve analytically or numerically. There are some important special cases that can be solved analytically.



Equation Example

Bang-Bang Control

min |x|dt 0

subject to

dx = u

dt

x(0) specied

1 u 1


{ }

{ }


Equation Example

The Bellman equation is

J J (x, t) = min |x| + (x, t)u .

t xu, 1u1

J(x, t) = J(x) is a solution because the time horizon is infinite and t does not appear explicitly in the problem data (ie, g(x) = |x| is not a function of t. Therefore

dJ 0 = min |x| + (x)u .

dxu, 1u1

J(0) = 0 because if x(0) = 0 we can choose u(t) = 0 for all t. Then x(t) = 0 for all t and the integral is 0. There is no possible choice of u(t) that will make the integral less than 0, so this is the minimum.



Equation Example

The minimum is achieved when

u =

Why?

1 if dJ

(x) > 0 dx

1 if dJ

(x) < 0 dx

undetermined if dJ

(x) = 0 dx



Equation Example

Consider the set of x where dJ/dx(x) < 0. For x in that set, u = 1, so

dJ 0 = |x| + (x)

dxor dJ

(x) = |x|dx

Similarly, if x is such that dJ/dx(x) > 0 and u = 1,

dJ

(x) = |x|dx



Equation Example

To complete the solution, we must determine where dJ/dx > 0, < 0, and = 0. We already know that J(0) = 0. We must have J(x) > 0 for all

x 0 because |x| > 0 so the integral of |x(t)| must be positive.= Since J(x) > J(0) for all x 0, we must have =

dJ (x) < 0 for x < 0

dx

dJ

(x) > 0 for x > 0

dx



Equation Example

Therefore

dJ

(x) >= x

dx

so

1 J = x 2

2 and 1 if x < 0

u = 0 if x = 0 1 if x > 0


Continuous x, t,Discrete

Stochastic { T } Bellmans

Equation

J(x(0), (0), 0) = min E g(x(t), u(t))dt + F (x(T ))

u 0

such that dx(t)

= f(x, , u, t)dt

prob [(t + t) = = = ijt for all i, j, i i | (t) j] = j

x(0), (0) specied

h(x(t), (t), u(t)) 0


Bellmans

Equation


Stochastic

Getting the Bellman equation in this case is more complicated because changes by large amounts when it changes. Let H() be some function of . We need to calculate

EH((t + t)) = E {H((t + t)) | (t)}

=

H(j)prob {(t + t) = j | (t)}

j


( )

Bellmans

Equation


Stochastic

= H(j)j(t)t + H((t)) 1 j(t)t + o(t)

j j=(t)=(t)

= H(j)j(t)t + H((t)) 1 + (t)(t)t + o(t) j =(t)

E {H((t + t)) | (t)} = H((t)) + H(j)j(t) t + o(t)

j

We use this in the derivation of the Bellman equation.


Bellmans

Equation


Stochastic

{ } T J(x(t), (t), t) = min E g(x(s), u(s))ds + F (x(T ))

u(s), t

ts

Bellmans

Equation

t+t = min E g(x(s), u(s))ds


Stochastic

u(s), t

0st+t

[ ] T + min E g(x(s), u(s))ds + F (x(T ))

u(s), t+t

t+tsT


Bellmans

Equation


Stochastic

= min u(s),

E

{ t+t t

g(x(s), u(s))ds

tst+t

+J(x(t + t), (t + t), t + t)

Next, we expand the second term in a Taylor series about x(t). We leave (t + t) alone, for now.


{ }

Bellmans

Equation


Stochastic

J(x(t), (t), t) =

min E g(x(t), u(t))t + J(x(t), (t + t), t) +

u(t)

J J (x(t), (t + t), t)x(t) + (x(t), (t + t), t)t + o(t).

x t where x(t) = x(t + t) x(t) = f(x(t), (t), u(t), t)t + o(t)


Bellmans

Equation


Stochastic

Using the expansion of EH((t + t)),

J(x(t), (t), t) = min g(x(t), u(t))t u(t)

+ J(x(t), (t), t) + J(x(t), j, t)j(t)t j J J

+ (x(t), (t), t)x(t) + (x(t), (t), t)t + o(t)x t

We can clean up notation by replacing x(t) with x, (t) with , and u(t) with u. Copyright 2007 Stanley B. Gershwin. c 56

Bellmans

Equation


Stochastic

J(x, , t) =

min u g(x, u)t + J(x, , t) +

j

J(x, j, t)jt J J

+ (x, , t)x + (x, , t)t + o(t)

x t

We can subtract J(x, , t) from both sides and use the expression for x to get ...


Bellmans

Equation


Stochastic

0 = min g(x, u)t + J(x, j, t)jt u

j J J

+ x

(x, , t)f(x, , u, t)t + t

(x, , t)t + o(t) or,


{ }

Bellmans

Equation


Stochastic

J (x, , t) = J(x, j, t)j+

t j

J min g(x, u) + (x, , t)f(x, , u, t)

u x

Bad news: usually impossible to solve; Good news: insight.


Bellmans

Equation


Stochastic

An approximation: when T is large and f is not a function of t, typical trajectories look like this:

x

t


{ }

Bellmans

Equation


Stochastic

That is, in the long run, x approaches a steady-state probability distribution. Let J be the expected value of g(x, u), where u is the optimal control. Suppose we started the problem with x(0) a random variable whose probability distribution is the steady-state distribution. Then, for large T , T

EJ = minu E 0 g(x(t), u(t))dt + F (x(T ))

JT


Bellmans

Equation


Stochastic

For x(0) and (0) specied

J(x(0), (0), 0) JT + W (x(0), (0))

or, more generally, for x(t) = x and (t) = specied, J(x, , t) J(T t) + W (x, )


Flexible

Manufacturing

System Control

Single machine, multiple part types. x, u, d are N -dimensional vectors. T

min E g(x(t))dt

subject to: 0

dxi(t) = ui(t) di, i = 1, ..., N

dt

prob((t + t) = 0|(t) = 1) = pt + o(t)

prob((t + t) = 1|(t) = 0) = rt + o(t)

iui(t) (t); ui(t) 0 i

x(0), (0) specified


{ }

Flexible

Manufacturing

System Control

Dene () = {u| i iui }. Then, for = 0, 1,

J (x, , t) = J(x, j, t)j+

t j

J min g(x) + (x, , t)(u d)

u() x


{ }

Flexible

Manufacturing

System Control

Approximating J with J(T t) + W (x, ) gives:

J = (J(T t) + W (x, j))j+ j

W min g(x) + (x, , t)(u d)

u() x Recall that

j = 0... j


{ }

Flexible

Manufacturing

System Control

so

J = W (x, j)j+ j

W min g(x) + (x, , t)(u d)

u() x

for = 0, 1


,[ ]

Flexible

Manufacturing

System Control

This is actually two equations, one for = 0, one for = 1.

W J = g(x) + W (x, 1)r W (x, 0)r (x, 0)d,

x for = 0,

W J = g(x) + W (x, 0)p W (x, 1)p + min (x, 1)(u d)

u(1) x for = 1.


,[ ]

Flexible Single-part-type case

Manufacturing

System Control Technically, not exible! Now, x and u are scalars, and

(1) = [0, 1/ ] = [0, ]

dW J = g(x) + W (x, 1)r W (x, 0)r (x, 0)d,

dxfor = 0,

dW J = g(x) + W (x, 0)p W (x, 1)p + min (x, 1)(u d)

0u dx for = 1.



Manufacturing

System Control

See book, Sections 2.6.2 and 9.3; see Probability slides #

91120.

When = 0, u = 0.

When = 1,

if dW < 0, u = ,dx

if dW = 0, u unspecied,dx

if dW > 0, u = 0.dx



Manufacturing

System Control

W (x, ) has been shown to be convex in x. If the minimum of W (x, 1) occurs at x = Z and W (x, 1) is differentiable for all x, then

dW < 0 x < Z dx

dW = 0 x = Zdx

dW > 0 x > Z dx

Therefore, if x < Z, u = , if x = Z, u unspecied, if x > Z, u = 0. Copyright c 702007 Stanley B. Gershwin.

Flexible Single-part-type case ManufacturingSystem Control Surplus, or inventory/backlog:

Production policy: Choose Z (the hedging point ) Then, if = 1,

if x < Z, u = ,

if x = Z, u = d,

if x > Z, u = 0;

if = 0,

u = 0.

How do we choose Z?

dx(t) = u(t) d

dt Cumulative

Production and Demand production

d t + Z

hedging point Z

surplus x(t)

demand dt

t



Manufacturing

System Control Determination of Z

Z J = Eg(x) = g(Z)P (Z, 1)+ g(x) [f(x, 0) + f(x, 1)] dx

in which P and f form the steady-state probability distribution of

x. We choose Z to minimize J . P and f are given by

f(x, 0) = Aebx

f(x, 1) = A d ebx d

P (Z, 1) = Apd ebZ


[ ]

Flexible Single-part-type case ManufacturingSystem Control Determination of Z where

r pb =

d d and A is chosen so that Z

[f(x, 0) + f(x, 1)] dx + P (Z, 1) = 1

After some manipulation, A =

bp( d) e bZ

db( d) + pand

db( d)P (Z, 1) =

db( d) + p



Manufacturing


Since g(x) = g+x+ + gx , ifZ 0, then Z

J = gZP (Z, 1) gx [f(x, 0) + f(x, 1)] dx;

ifZ > 0, 0 J = g+ZP (Z, 1) gx [f(x, 0) + f(x, 1)] dx

Z + g+x [f(x, 0) + f(x, 1)] dx.

0


[ ]


Manufacturing


To minimize J: ( )

ln Kb(1 + g )

if g+ Kb(g+ + g) < 0, Z = g+

. b

if g+ Kb(g+ + g) 0, Z = 0

where K = p p 1 p

= = b(bd d2b + p) b(r + p)( d) b db( d) + p

Z is a function of d, , r, p, g+, g.


{ ( )}

{ ( )}

Flexible Single-part-type case ManufacturingSystem Control Determination of Z That is, we choose Z such that

e bZ = min 1,Kb g+ + g

g+ or

bZ 1 g+ e = max 1, Kb g+ + g


( ) [ ]

[ ]

Flexible Single-part-type case ManufacturingSystem Control Determination of Z

0 prob(x 0) = (f(x, 0) + f(x, 1))dx

( ) 0d = A 1 + e bxdx

d d 1

= A 1 + = A d b b( d)

bp( d) bZ = e db( d) + p b( d)

p bZ = e db( d) + p


{ ( )} { }


Manufacturing


Or, [ ] { ( )} prob(x 0) =

p max 1,

1 g+

db( d) + p Kb g+ + g

It can be shown that p

Kb = p + bd( d)

Therefore prob(x 0) = Kb max 1,

1 g+

Kb g+ + g p g+

= max , p + bd( d) g+ + g


Flexible Single-part-type case ManufacturingSystem Control Determination of Z That is, if

p >

g+ , then Z = 0 and p + bd( d) g+ + g

prob(x 0) = p

;

p + bd( d)

if p

0 and

p + bd( d) g+ + g

prob(x 0) = g+

. g+ + g

This looks a lot like the solution of the newsboy problem. Copyright 2007 Stanley B. Gershwin. c 79

Flexible Single-part-type case Manufacturing

Z vs. dSystem Control Base values: g+ = 1, g = 10 d = .7, = 1., r = .09, p = .01.

100

90

80

70

60

Z 50

40

30

20

10

0

d 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9



System Control Manufacturing

Z vs. g+

Base values: g+ = 1, g = 10 d = .7, = 1., r = .09, p = .01.

70

60

50

40

0 0.5 1 1.5 2 2.5 3 g+

Z 30

20

10

0

3.5




Z vs. g


14

12

10

8 Z

6

4

2

0 0 1 2 3 4 5 6 7 8 9 10 11

g




Z vs. p


1400

1200

1000

800

Z

600

400

200

0

0.005 0.01 0.015 0.02 0.025 0.03 0.035

p 0 0.04


Flexible Two-part-type case


x1

x2

d1

d2

u (t)1

u (t)2 Type 2

Type 1

r, p

u11/ 1

1/ 2

u2

0

Capacity set (1) when machine is up.


{ }


Manufacturing

System Control

We must nd u(x, ) to satisfy

W min (x, , t) u

u() x Partial solution of LP:

If W/x1 > 0 and W/x2 > 0, u1 = u2 = 0. If W/x1 < W/x2 < 0, u1 = 1, u2 = 0. If W/x2 < W/x1 < 0, u2 = 2, u1 = 0.

Problem: no complete analytical solution available.


1

0

2

Flexible Two-part-type case ManufacturingSystem Control Case: Exact solution if Z = (Z1, Z2) = 0

x2

x1

1 2u = u = 0

dx dt

2

2u = 0

u = 01

1u = 1

2

u =

u11/

1/ 2

u2

0

u11/ 1

1/ 2

u2

1/ 1

1/

u2

0 u1


1

2

0

Flexible Two-part-type case ManufacturingSystem Control Case: Approximate solution if Z > 0

x2 1 2u = u = 0

dx dt

2

2u = 0

u = 01

1u = 1

2

u =

u11/

1/ 2

u2

0

1/ 1

1/

u2

0

u11/ 1

1/ 2

u2

x1

u1


45



Two parts, multiple machines without buffers: e 12

4

61e

e

34e

23e

3

12

x2

Z

6 x1

5

56e

6

1 2

3

4

5

u 2

e34

12e

u1

I

e23

e45

e56

d 61e



Manufacturing

System Control

Proposed approximate solution for multiple-part, single machine system: Rank order the part types, and bring them to their hedging points in that order.




Surplus and tokens

Operating Machine M according to the hedging point policy is equivalent to operating this assembly system according to a nite

B buffer policy. D

M

S

FG




Surplus and tokens

D is a demand generator . Whenever a demand arrives, D

sends a token to B.

S is a synchronization machine. S is perfectly reliable and in

nitely fast.

M

D

S

FG

B

FG is a nite nished goods buffer. B is an innite backlog buffer.



Manufacturing

System Control Material/token policies

Operator An operation cannot take Machine place unless there is a

Part Part token available. OperationConsumable Waste

Tokens authorize Token Token

production.

These policies can often be implemented either with nite

buffer space, or a nite number of tokens. Mixtures are also

possible.

Buffer space could be shelf space, or oor space indicated with paint or tape.


Multi-stage Proposed policy systems

To control M B M B M1 1 2 2 3

add an information flow system:

B1 M B M2 2 3M1

S S2 3

D

S1 BB1

SB2

BB2

SB3

BB3

SB1


Multi-stage Proposed policy systems

B1 M B M2 2 3M1

S S2 3

D

S1 BB1

SB2

BB2

SB3

BB3

SB1

Bi are material buffers and are nite. SBi are surplus buffers and are nite. BBi are backlog buffers and are innite. The sizes of Bi and SBi are control parameters. Problem: predicting the performance of this system.


Multi-stage

systems

Three Views of Scheduling

Three kinds of scheduling policies, which are sometimes exactly the same.

Surplus-based: make decisions based on how much production exceed demand.

Time-based: make decisions based on how early or late a product is.

Token-based: make decisions based on presence or absence of tokens.


Multi-stage Objective of Scheduling systems Surplus and time

and Demand

earliness

production P(t)

demand D(t)

surplus/backlog x(t)

Objective is to keep cumulative production close to cumulative demand.

Cumulative

Production

Surplus-based policies look at vertical differences between the graphs.

Time-based policies look at the horizontal t differences.


Multi-stage Other policies systems CONWIP, kanban, and hybrid

CONWIP: finite population, infinite buffers kanban: infinite population, finite buffers hybrid: finite population, finite buffers


Multi-stage Other policies systems CONWIP, kanban, and hybrid CONWIP

Supply Demand

Token flow

Demand is less than capacity. How does the number of tokens affect performance (production rate, inventory)?


0.85

0.855

0.86

P

Multi-stage Other policies systems CONWIP, kanban, and hybrid

0.835

0.84

0.845

0.865

0.87

0.875

0 20 40 60 80 100 120 0

5

10

20

25

30

Aver

age

Buffe

r Lev

el

n1 n2 n3

15

0 20 40 60

Population Population

80 100 120

cCopyright 2007 Stanley B. Gershwin. 99

Multi-stage Other policies

systems Basestock

Demand


Multi-stage Other policies systems FIFO First-In, First Out. Simple conceptually, but you have to keep track of arrival times.

Leaves out much important information: due date, value of part, current surplus/backlog state, etc.


Multi-stage Other policies systems EDD Earliest due date. Easy to implement. Does not consider work remaining on the item, value of the item, etc..


Multi-stage Other policies systems SRPT Shortest Remaining Processing Time Whenever there is a choice of parts, load the one with least remaining work before it is finished.

Variations: include waiting time with the work time. Use expected time if it is random.


Multi-stage Other policies systems Critical ratio Widely used, but many variations. One version:

Processing time remaining until completion Dene CR =

Due date - Current time Choose the job with the highest ratio (provided it is positive). If a job is late, the ratio will be negative, or the denominator will be zero, and that job should be given highest priority.

If there is more than one late job, schedule the late jobs in SRPT order.


Multi-stage Other policies systems Least Slack This policy considers a parts due date. Define slack = due date - remaining work time When there is a choice, select the part with the least slack.

Variations involve different ways of estimating remaining time.


Multi-stage Other policies systems Drum-Buffer-Rope Due to Eli Goldratt. Based on the idea that every system has a bottleneck. Drum: the common production rate that the system operates at, which is the rate of ow of the bottleneck.

Buffer: DBR establishes a CONWIP policy between the entrance of the system and the bottleneck. The buffer is the CONWIP population.

Rope: the limit on the difference in production between different stages in the system.

But: What if bottleneck is not well-dened?


Conclusions

Many policies and approaches. No simple statement telling which is better. Policies are not all well-dened in the literature or in practice. My opinion: This is because policies are not derived from rst principles. Instead, they are tested and compared. Currently, we have little intuition to guide policy development and choice.


MIT OpenCourseWarehttp://ocw.mit.edu



MIT 2.852


Lecture 1: Overview

Stanley B. Gershwin

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

Massachusetts Institute of Technology

Spring, 2010


Books

Required Manufacturing Systems Engineering (MSE) by Stanley B. Gershwin ... obtainable from author.

Optional Factory Physics by Hopp and Spearman The Goal by Goldratt Stochastic Models of Manufacturing Systems by Buzacott and

Shanthikumar Production Systems Engineering by Li and Meerkov


Course Overview Goals

To explain important measures of system performance.

To show the importance of random, potentially disruptive events in factories.

To give some intuition about behavior of these systems.

To describe some current tools and methods.


Problems

Manufacturing systems engineering is not as well-developed as most other elds of engineering.

Practitioners are encouraged to rely on gurus, slogans, and black boxes.

There is a gap between theoreticians and practitioners.


Problems

The research literature does not always focus on real-world problems ... but practitioners are often unaware of what does exist.

Terminology, notation, basic assumptions are not standardized.

There is a separation of product, process, and system design.


Problems

Confusion about objectives: maximize capacity? minimize capacity variability? maximize capacity utilization? minimize lead time? minimize lead time variability? maximize prot?

Systems issues are often studied last, if at all.


Problems

Manufacturing gets no respect. Systems not designed with engineering methods. Product designers and sales sta are not informed of manufacturing

costs and constraints.

Black box thinking. Factories not treated as systems to be analyzed and engineered. Simplistic ideas often used for management and design.


Problems

Reliable systems intuition is lacking. As a consequence, there is ...

Management by software Managers buy software to make production decisions, rather than to

aid in making decisions.

Management by slogan Gurus provide simple solutions which sometimes work. Sometimes.


Observation

When a system is not well understood, rules proliferate.

This is because rules are developed to regulate behavior.

But the rules lead to unexpected, undesirable behavior. (Why?)

New rules are developed to regulate the new behavior.

Et cetera.


Observation Example

A factory starts with one rule: do the latest jobs rst .

Over time, more and more jobs are later and later.

A new rule is added: treat the highest priority customers orders as though their due dates are two weeks earlier than they are.

The low priority customers nd other suppliers, but the factory is still late.


Observation Example

Why?

There are signicant setup times from part family to part family. If setup times are not considered, changeovers will occur too often, and waste capacity.

Any rules that that do not consider setup times in this factory will perform poorly.


Denitions

Manufacturing: the transformation of material into something useful and portable.

Manufacturing System: A manufacturing system is a set of machines, transportation elements, computers, storage buers, people, and other items that are used together for manufacturing. These items are resources.


Denitions

Manufacturing System:

Alternate terms: Factory Production system Fabrication facility

Subsets of manufacturing systems, which are themselves systems, are sometimes called cells, work centers, or work stations .


Basic Issues

Increasingly, there are ... frequent new product introductions, and short product lifetimes, and short process lifetimes.

Consequently, ...

factories are built and rebuilt frequently, and there is not much time to tinker with a factory. It must be operational

quickly.


Basic Issues Consequent Needs

Tools to predict performance of proposed factory design.

Tools for optimal real-time management (control) of factories.

Manufacturing Systems Engineering professionals who understand factories as complex systems.


Basic Issues Quantity, Quality and Variability

Quantity how much and when.

Quality how well.

In this course, we emphasize quantity.

General Statement: Variability is the enemy of manufacturing.

General Statement: Know your enemy!


Basic Issues More Denitions

Make to Stock (O the Shelf): items available when a customer arrives appropriate for large volumes, limited product variety, cheap raw

materials

Make to Order: production started only after order arrives appropriate for custom products, low volumes, expensive raw materials


Basic Issues Conicting Objectives

Make to Stock: large nished goods inventories needed to prevent stockouts small nished goods inventories needed to keep costs low

Make to Order: excess production capacity (low utilization) needed to allow early,

reliable delivery promises minimal production capacity (high utilization) needed to to keep costs

low


Basic Issues Concepts

Complexity: collections of things have properties that are non-obvious functions of the properties of the things collected.

Non-synchronism (especially randomness) and its consequences: Factories do not run like clockwork.


Basic Issues What is an Operation?

Operation Part Part

Waste

Machine Operator

Consumable

Nothing happens until everything is present.


Basic Issues Waiting

Whatever does not arrive last must wait.

Inventory: parts waiting.

Underutilization: machines waiting.

Idle work force: operators waiting.


Basic Issues Causes of Poor Performance

Operation Part Part

Waste

Machine Operator

Consumable

Reductions in the availability, or ...

Variability in the availability ...

... of any one of these items causes waiting in the rest of them and reduces

performance of the system.


Kinds of Systems Flow shop

... or Flow line , Transfer line , or Production line.

Machine Buffer

Traditionally used for high volume, low variety production.

What are the buers for?


Kinds of Systems Assembly system

Assembly systems are trees, and may involve thousands of parts.


Loops Closed loop (1a)

Limited number of pallets or xtures:

Empty Pallet Buffer

Raw Part Input


Pallets or xtures travel in a closed loop. Routes are determined. The number of pallets in the loop is constant.

Pallets or xtures take up space and may be expensive.


Loops Closed loop (1b)

Limited number of tokens:

Empty Token Buffer

Raw Part Input


Tokens travel in a closed loop. Routes are determined. The number of pallets in the loop is constant.

Tokens take up no space and cost nothing.

What are the tokens for?


Loops Reentrant (2)

System with reentrant ow and two part types

M 1

B11

B21

M 2 M 3

M 4

B51 B12 B22

B61

B32 B B31

B41

71

Type 2 Type 1Type 1 Type 2

Routes are determined. The number of parts in the loop varies.

Semiconductor fabrication is highly reentrant.


Loops Rework loop (3)

rework

rejectinspection

Routes are random. The number of parts in the loop varies.


Kinds of Systems Job shop

Machines not organized according to process ow.

Often, machines grouped by department: mill department lathe department etc.

Great variety of products.

Dierent products follow dierent paths.

Complex management.


Two Issues

Ecient design of systems;

Ecient operation of systems after they are built.


Time

Most factory performance measures are about time.

production rate: how much is made in a given time. lead time: how much time before delivery. cycle time: how much time a part spends in the factory. delivery reliability: how often a factory delivers on time. capital pay-back period: the time before the company get its

investment back.


Time

Time appears in two forms:

delay capacity utilization

Every action has impact on both.


Time Delay

An operation that takes 10 minutes adds 10 minutes to the delay that

a workpiece experiences while undergoing that operation; every other workpiece experiences that is waiting while the rst is being

processed.


Time Capacity Utilization

An operation that takes 10 minutes takes up 10 minutes of the available time of a machine, an operator, or other resources.

Since there are a limited number of minutes of each resource available, there are a limited number of operations that can be done.


Time More Denitions

Operation Time: the time that a machine takes to do an operation.

Production Rate: the average number of parts produced in a time unit. (Also called throughput. )

If nothing interesting ever happens (no failures, etc.),

Production rate = 1

operation time

... but something interesting always happens.


Time More Denitions

Capacity: the maximum possible production rate of a manufacturing system, for systems that are making only one part type. Short term capacity: determined by the resources available right now. Long term capacity: determined by the average resource availability.

Capacity is harder to dene for systems making more than one part type. Since it is hard to dene, it is very hard to calculate.


Randomness, Variability, Uncertainty More Denitions

Uncertainty: Incomplete knowledge.

Variability: Change over time.

Randomness: A specic kind of incomplete knowledge that can be quantied and for which there is a mathematical theory.


Randomness, Variability, Uncertainty

Factories are full of random events: machine failures changes in orders quality failures human variability

The economic environment is uncertain demand variations supplier unreliability changes in costs and prices



Therefore, factories should be

designed as reliably as possible, to minimize the creation of variability;

designed with shock absorbers, to minimize the propagation of variability;

operated in a way that minimizes the creation of variability;

operated in a way that minimizes the propagation of variability.



Therefore, all engineers should know probability... especially manufacturing systems engineers .

Probability is an important prerequisite for this course.


The Course Mechanics

Reading: Mainly Chapters 29 of MSE . (Chapter 9 up to 9.3.)

Grading: project and class participation.

Homework optional.

2.852 Manufacturing Systems Analysis 42/44 Copyright c2010 Stanley B. Gershwin. .

The Course Topics

Probability Basics, Markov processes, queues, other examples.

Transfer lines Models, exact analysis of small systems, approximations of large

systems.

Extensions of transfer line models Assembly/disassembly, loops, system optimization

Real-time scheduling

Quality/Quantity interactions

New material


The Course

Emphasis on mathematical modeling and analysis.

Emphasis on intuition.

Comparison with 2.854: Narrower and deeper.



MIT OpenCourseWare http://ocw.mit.edu


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

MIT 2.852


Lecture 14-16 Line Optimization

Stanley B. Gershwin

Spring, 2007

Copyright c2007 Stanley B. Gershwin.

Line Design

Given a process, find the best set of machines and buffers on which it can be implemented.

Best: least capital cost; least operating cost; least average inventory; greatest profit, etc.

Constraints: minimal production rate, maximal stockout probability, maximal floor space, maximal inventory, etc..

To be practical, computation time must be limited. Exact optimality is not necessary, especially since the parameters are not known perfectly.


Optimization

Optimization may be performed in two ways: Analytical solution of optimality conditions; or Searching

For most problems, searching is the only realistic possibility.

For some problems, optimality cannot be achieved in a reasonable amount of time.


Search Optimization

Dn

past designs and performance.Modify design as a function of

D = D(D , D , ..., D , P , P , ..., P )n+1 n0 01 n 0

n = 0

n = P( )Pn

Is performance satisfactory?

yes

no

n n+1 1

Propose Design D

Evaluate Design D

Quit

Typically, many designs are tested.


Issues Optimization For this to be practical, total computation time must be limited. Therefore,we must control both computation time per iteration and the number of iterations .

Computation time per iteration includes evaluation time and the time to determine the next design to be evaluated.

The technical literature is generally focused on limiting the number of iterations by proposing designs efficiently.

The number of iterations is also limited by choosing a reasonabletermination criterion (ie, required accuracy).

Reducing computation time per iteration is accomplished by using analytical models rather than simulations using coarser approximations in early iterations and more accurate

evaluations later.


Problem

Statement

X is a set of possible choices. J is a scalar function defined on

X. h and g are vector functions defined on X.

Problem: Find x X that satisfies J(x) is maximized (or minimized) the objective subject to h(x) = 0 equality constraints

g(x) 0 inequality constraints


Taxonomy

static/dynamic

deterministic/stochastic

X set: continuous/discrete/mixed

(Extensions: multi-criteria optimization, in which the set of all

good compromises between different objectives are sought; games, in which there are multiple optimizers, each preferring different xs but none having complete control; etc.) Copyright c 72007 Stanley B. Gershwin.

Continuous

Variables and

Objective

X = Rn. J is a scalar function defined on Rn. h( Rm) and g( Rk) are vector functions defined on Rn.

Problem: Find x Rn that satisfies J(x) is maximized (or minimized) subject to h(x) = 0

g(x) 0


Continuous Unconstrained Variables and Objective One-dimensional search

Find t such that f(t) = 0.

This is equivalent to Find t to maximize (or minimize) F (t)

when F (t) is differentiable, and f(t) = dF (t)/dt is continuous.

If f(t) is differentiable, maximization or minimization depends on the sign of d2F (t)/dt2.


Continuous

Variables and

Objective

Assume f(t) is decreasing. Binary search: Guess t0 and f(t0 ) t1 such that f(t0) > 0 and f(t1) < 0. Let t2 = (t0 + t1)/2. f(t2 )

If f(t2) < 0, then repeat f(t1 )

with t = t0 and t 0 1 =t2.

If f(t2) > 0, then repeat with t0 t2 and t= =t1.

Unconstrained

One-dimensional search

f(t)

t0 t1t2

t0 t1

t

1



t0 t2 t1

Example:

f(t) = 4 t2

0 1.5

1.5

1.875

1.875

1.96875

1.96875

1.9921875

1.9921875

1.998046875

1.998046875

1.99951171875

1.99951171875

1.9998779296875

1.9998779296875

1.99996948242188

1.99996948242188

1.99999237060547

1.99999237060547

1.99999809265137

1.5

2.25

1.875

2.0625

1.96875

2.015625

1.9921875

2.00390625

1.998046875

2.0009765625

1.99951171875

2.000244140625

1.9998779296875

2.00006103515625

1.99996948242188

2.00001525878906

1.99999237060547

2.00000381469727

1.99999809265137

2.00000095367432

3 3 2.25

2.25

2.0625

2.0625

2.015625

2.015625

2.00390625

2.00390625

2.0009765625

2.0009765625

2.000244140625

2.000244140625

2.00006103515625

2.00006103515625

2.00001525878906

2.00001525878906

2.00000381469727

2.00000381469727



f(t)

Newton search, exact tangent: f(t0 )

Guess t0. Calculate

df(t0)/dt.

Choose t1 so that df (t0)f(t0) + (t1 t0) dt = 0. f(t1 )

Repeat with t 0 = t1 until

|f(t 0)| is small enough.

t0 t1 t


Continuous Variables and Objective

Example:

f(t) = 4 t2

Unconstrained


t0 3 2.16666666666667 2.00641025641026 2.00001024002621 2.00000000002621 2



f(t) Newton search, approximate tangent:

f(t0 )

Guess t0 and t1. Calculate

approximate slope

f (t1)f (t0)s = t1t0

. f(t2 )

Choose t2 so that f(t1 )

f(t0) + (t2 t0)s = 0.

Repeat with t 0 = t1 and

t = t2 until |f(t )| is small1 0enough.

t0 t1t2 t


Continuous Variables and Objective

Example:

f(t) = 4 t2

Unconstrained


t0 0 3 1.33333333333333 1.84615384615385 2.03225806451613 1.99872040946897 1.99998976002621 2.0000000032768 1.99999999999999 2


Continuous Unconstrained Variables and Objective Multi-dimensional search

x2

J Optimum

SteepestAscent Directions Optimum often found

by steepest ascent or

hill-climbing methods.

x1


{ }

Continuous Unconstrained

Variables and

Objective Gradient search To maximize J(x), where x is a vector (and J is a scalar function that has nice properties): 0. Set n = 0. Guess x0. 1. Evaluate J (xn).x2. Let t be a scalar. Define Jn(t) = J xn + t

J (xn)

x

Find (by one-dimensional search ) t , the value of t thatnmaximizes Jn(t).

3. Set xn+1 = xn + t J (xn). nx4. Set n n + 1. Go to Step 1. also called steepest ascent or steepest descent .


Continuous Constrained Variables and Objective

Equality constrained: solution is on the constraint surface.

Problems are much easier when constraint is linear, ie, when the surface is a plane. In that case, replace J/x by its projection onto the constraint plane.

But first: find an initial feasible guess.

x1

x2

J

Constrained Optimum

h(x , x ) = 01 2



x2

>1 2g(x , x )

ConstrainedJ Optimum Inequality constrained:

solution is required to

be on one side of the

plane.

0

x1

Inequality constraints that are satisfied with equality are called effective or active constraints.

If we knew which constraints would be effective, the problem would reduce toan equality-constrained optimization.



6

8*(x+y).25*(x**4+y**4)4

20 16

2

0

2

4

6

12 8 4 0

4 8

12 16 20 24

2 0 2 4 6

Minimize 8(x + y) (x4 + y4)/4

subject to x + y 0

Solving a linearly-constrained problem is relatively easy. If the

solution is not in the interior, search within the boundary plane.



6

8*(x+y).25*(x**4+y**4)4

20 16

2

0

2

4

6

12 8 4 0

4 8

12 16 20 24

2 0 2 4 6

Minimize 8(x + y) (x4 + y4)/4

subject to x (x y)2 + 1 0

Solving a nonlinearly-constrained problem is not so easy. Searching within the boundary is numerically difficult. Copyright 2007 Stanley B. Gershwin. c 21

Continuous Nonlinear and Linear Programming Variables and Objective Optimization problems with continuous variables, objective, and constraints are called nonlinear programming problems, especially when at least one of J, h, g are not linear. When all of J, h, g are linear, the problem is a linear programming problem.


Continuous Multiple Optima Variables and Objective

x1

x2

J Local (or Relative)Maxima

Global Maximum

Danger: a search might find a local, rather than the global, optimum. Copyright 2007 Stanley B. Gershwin. c 23

Continuous Primals and Duals

Variables and

Objective

Consider the two problems:

min f(x) max j(x)

subject to j(x) J subject to f(x) F

f(x), F , j(x), and J are scalars. We will call these problems duals of one another. (However, this is not the conventional use of the term.) Under certain conditions when the last inequalities are effective, the same x satisfies both problems.

We will call one the primal problem and the other the dual problem .


Continuous Variables and Objective Generalization:

min f(x)

subject to h(x) = 0

g(x) 0

j(x) J

Primals and Duals

max j(x)

subject to h(x) = 0

g(x) 0

f(x) F


Buffer Space Problem statement Allocation

M B1 1 M B2 2 M B3 3 M B4 4 M B5 5 M6

Problem: Design the buffer space for a line. The machines have already been selected. Minimize the total buffer space needed to achieve a target production rate. Other problems: minimize total average inventory; maximize profit (revenue - inventory cost - buffer space cost); choose machines as well as buffer sizes; etc.


Buffer Space Problem statement

Allocation

Assume a deterministic processing time line with k machines with ri and pi known for all i = 1, ..., k. Assume minimum buffer size N MIN. Assume a target production rate P . Then the function P (N1, ..., Nk1) is known it can be evaluated using the decomposition method. The problem is:

Primal problem: k1

Minimize

Ni

i=1

subject to P (N1, ..., Nk1) P

Ni N MIN, i = 1, ..., k 1.

In the following, we treat the Nis like a set of continuous variables.


Properties of P (N1, ..., Nk1)Buffer Space

Allocation

P (, ..., ) = min ei i=1,...,k

P (N MIN, ..., N MIN) 1

Buffer Space Properties of P (N1, ..., Nk1) Allocation Example 3-machine line

200 Optimal curve

P=0.8800 P=0.8825 P=0.8850 P=0.8875 P=0.8900P P=0.8925

0.91 150 P=0.8950 P=0.8975 0.9 P=0.9000

0.89 0.88 0.87 N2 100 0.86 0.85 0.84 0.83

100 50 90

8070

6010 20 30 40 50 60 3040

50 N2

N1 70 80 90 1020 0

0 50 100 150 200100 N1

r1 = .35 r2 = .15 r3 = .4 p1 = .037 p2 = .015 p3 = .02 e1 = .904 e2 = .909 e3 = .952


Minimize k1

Ni

Buffer Space Solution Allocation Primal problem

i=1

subject to P (N1, ..., Nk1) P

Ni N MIN, i = 1, ..., k 1.

Difficulty: If all the buffers are larger than N MIN, the solution will satisfy P (N1, ..., Nk1) = P

. (Why?) But P (N1, ..., Nk1) is nonlinear and cannot be expressed in closed form. Therefore any solution method will have to search within this constraint and all steps will be small and there will bemany iterations.

It would be desirable to transform this problem into one with linear constraints.


Buffer Space Solution Allocation Dual problem

Maximize P (N1, ..., Nk1)

k1

subject to Ni N TOTAL specied i=1

Ni N MIN, i = 1, ..., k 1.

All the constraints are linear. The solution will satisfy the N TOTAL constraint with equality (assuming the problem is feasible). (1. Why? 2. When would the problem be infeasible?) This problem is consequently relatively easy to solve.


Buffer Space Solution

Allocation N2 Dual problem

N1 +N2 +N3 = NTOTAL

N3

Constraint set (if N MIN = 0). N1


Buffer Space Solution

Allocation Primal strategy

Solution

2852 spring 2010

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