advanced layout algorithms chapter 8. layout algorithms optimal heuristic

Advanced Layout Algorithms

Chapter 8

Layout Algorithms

Optimal Heuristic

Optimal Algorithms

Branch and bound Decomposition

– Benders’ decomposition

Cutting Plane Algorithms

B&B Objective Function Value

1 1

1 1 1 1

n n n n

ij kl ik jli j i k l k

f d x x

Branch and bound Algorithm

LB*=71.5

LB*14

LB*13LB*12

LB*11

LB*22

LB*23 LB*24

LB*34 + LB*43 = 76

SecondLevelnodes

First Levelnodes


Step 1: Computer lower bound (LB*) by solving a linear assignment problem (LAP) with a matrix [wij]. Matrix [wij] is obtained by taking dot product of two vectors [fi] and [dj]. Vector [fi], [dj] are obtained by removing fii, djj and arranging the remaining flow and distance values in non-increasing and non-decreasing order, respectively

Step 2:Computer lower bound for other nodes

Branch and bound

Assignment: Matching a department with a specific location and vice‑versa

Partial assignment: An assignment in which a subset of n departments is matched with an equal-sized subset of locations and vice‑versa

Complete assignment: All the n departments are matched with n locations and vice‑versa

A complete assignment obtained from a partial assignment must not disturb the partial assignment but only grow from it

Lower bound calculation for partial assignment

Given a partial assignment in which a certain subset S={1,2,...,q} of n departments is assigned to a subset L={s1,s2,...,sq} of n locations, the optimal objective function for a complete assignment is equal to the sum of the products of flow and distance computed for these three categories of departments:

– Pairs of departments i, j such that i, j belongs to S;– Pairs of departments i, j such that i belongs to S, j does not

belong to S; and– Pairs of departments i, j such that i, j do not belong to S


Step 2a: Calculate cost of partial assignment Step 2b: Computer lower bound (LB*ij) for lower level

nodes by solving a LAP with a matrix [wij].– Matrix [wij] is obtained by adding two matrices [w’ij] and [w’’ij]– Matrix [w’’ij] is obtained by taking half the dot product of two

vectors [fi] and [dj]. Vector [fi], [dj] are obtained by arranging flow and distance values in non-increasing and non-decreasing order, respectively. Do this only for ‘available’ departments and locations

– Matrix [w’ij] is obtained as follows: where

q

kjsik k

df1

, , , and .kk S i S s L j L

Explain Branch and Bound Algorithm with Example 1

Office Site

O 1 2 3 4 1 2 3 4

f 1 - 17 12 11 S 1 - 1 1 2

[fij]= f 2 17 - 12 4 [dij]= i 2 1 - 2 1

i 3 12 12 - 4 t 3 1 2 - 1

c 4 11 4 •4 - e 4 2 1 1 -

e

Figure 7.2 Flow and distance matrices for the LonBank layout problem


Why is the specialized B&B algorithm more efficient than a general purpose B&B?

How to terminate algorithm for large problems– Terminate after preset CPU time limit has exceeded– Terminate after preset number of nodes have been

examined

Benders’ decomposition algorithm

Consider this MIPMinimize cxSubject to Ax + By > b

x > 0y = 0 or 1

Now, consider a feasible y solution vector to MIP - say yi.Then, MIP becomes the following linear model.LP iMinimize cxSubject to Ax > b - Byi

x > 0

Dual of Linear Program

LP iMinimize cxSubject to Ax > b - Byi

x > 0

The dual of LP i is the following modelDLP iMaximize u(b - Byi)Subject to uA < c

u > 0


• Let ui be the optimal solution to DLP i• From duality theory, ui(b - Byi) is equal to the optimal OFV of

LPi (because LP i and DLP i are both feasible)• Hence, ui(b - Byi) is equal to the OFV of some feasible

solution to MIP (the one in which y = yi). Because each variable yij in the vector y can take on a value of 0 or 1 only and because the number of such variables is finite, it is clear that the number of y vectors are also finite

• In fact, if there are n yij variables, then the number of y vectors is equal to 2n. Of course, not all of these may be feasible to MIP. We assume that there are s feasible y solution vectors to MIP - {y1, y2, ..., yi, ..., ys}, arranged in any order


• Let DLP 1, DLP 2, ..., DLP i, ..., DLP s be the duals obtained by substituting y1, y2, ..., yi, ..., ys for yi in DLP i. Let u1, u2, ... , ui, ..., us be the optimal solution vectors to DLP 1, DLP 2, ..., DLP i, ..., DLP s, respectively

• The optimal OFV of each corresponds to the OFV of some feasible y solution vector to MIP

• Because we have considered all feasible solution vectors, the dual with the least OFV among DLP 1, DLP 2, ..., DLP s, provides the optimal OFV to MIP

• Thus, the original problem MIP may be reduced to the following problem:

Minimize {ui(b - By)} 1 < i < s Subject to y = 0 or 1 and feasible to MIP

Master Problem

Minimize {ui(b - By)} 1 < i < s Subject to y = 0 or 1 and feasible to MIPThe above model can be restated as:

MPMinimize zSubject to

z > ui(b - By) i = 1, 2, ..., s y = 0 or 1 and feasible to MIP

z = 0 or 1MP requires us to generate all the feasible y solution vectors and the corresponding s dual problems - DLP 1, DLP 2, ..., DLP sNot computationally feasible because the number of dual problems, though finite, may be very largeThe dual associated with each of these has to be solved - a time consuming task

Solving the Master Problem

• However, we can overcome the computational problem by generating a subset of the constraints in MP and solving a restricted problem

• Because we are solving MP with only a small subset of constraints, its optimal solution will provide a lower bound on MIP

• Thus, beginning with few or no constraints, we solve MP, obtain a new y vector, setup DLP i corresponding to this y vector and obtain an upper bound

• Using the optimal solution to DLP i, we add the corresponding constraint [z > ui(b - By)] in the master problem MP and solve it

• If the resulting lower bound is greater than or equal to the upper bound, we stop because the last solution to MP provides the optimal solution to MIP. Otherwise, we repeat the procedure until the termination criterion is met

Benders’ decomposition algorithm

Step 0: Set i=1, yi = {0,0,...,0}, lower bound LB=0 and upper bound UB=infinity.Step 1: Solve DLP i. Let ui be the optimal solution to DLP i. If ui(b - Byi) < UB, set UB = ui(b - Byi)Step 2: Update MP by adding the constraint z > ui(b - By). Solve MP. Let y* be the optimal solution and z be the optimal OFV of MP. Set LB = z. If LB > UB, stop. Otherwise, set i = i+1, yi = y* and return to step 1.

Explain Benders’ decomposition algorithm with Example 2

Machine Dimension Horizontal Clearance Matrix Flow Matrix

1 2 3 4 1 2 3 4

1 25x20 1 - 3.5 5.0 5.0 1 - 25 35 50

2 35x20 2 3.5 - 5.0 3.0 2 25 - 10 15

3 30x30 3 5.0 5.0 - 5.0 3 35 10 - 50

4 40x20 4 5.0 3.0 5.0 - 4 50 15 50 -

Figure 7.6 Flow and clearance matrices and dimensions for four machines

LMIP 1 for Example 2

1

1 1

n n

ij ij ij iji j i

c f x x

Minimize

Subject to

1, 2,..., -1; 1,...i j ij ijx x x x i n j i n

0 i=1,2,...,ix n

nijnixx ijij ,...,1;1,...,2,10,

nijnidllMyxx ijjiijji ,...,1;1,...,2,1,)(5.0

nijnidllyMxx ijjiijji ,...,1;1,...,2,1,)(5.0)1(

nijnioryij ,...,1;1,...,2,1,10

Example 2

MIP MIN 25 XP12 + 35 XP13 + 50 XP14 + 10 XP23 + 15 XP24 + 50 XP34 + 25 XN12 + 35 XN13 + 50 XN14 + 10 XN23 + 15 XN24 + 50 XN34 SUBJECT TO C1) 999 Y12 + X1 - X2 >= 33.5 C2) 999 Y12 + X1 - X2 <= 965.5 C3) 999 Y13 + X1 - X3 >= 32.5 C4) 999 Y13 + X1 - X3 <= 966.5 C5) 999 Y14 + X1 - X4 >= 37.5 C6) 999 Y14 + X1 - X4 <= 961.5 C7) 999 Y23 + X2 - X3 >= 37.5 C8) 999 Y23 + X2 - X3 <= 961.5 C9) 999 Y24 + X2 - X4 >= 40.5 C10) 999 Y24 + X2 - X4 <= 958.5 C11) 999 Y34 + X3 - X4 >= 40 C12) 999 Y34 + X3 - X4 <= 959 C13) - XP12 + XN12 + X1 - X2 = 0 C14) - XP13 + XN13 + X1 - X3 = 0 C15) - XP14 + XN14 + X1 - X4 = 0 C16) - XP23 + XN23 + X2 - X3 = 0 C17) - XP24 + XN24 + X2 - X4 = 0 C18) - XP34 + XN34 + X3 - X4 = 0 END INTE 6

Example 2 (Cont)

LP 1 MIN 25 XP12 + 35 XP13 + 50 XP14 + 10 XP23 + 15 XP24 + 50

XP34 + 25 XN12 + 35 XN13 + 50 XN14 + 10 XN23 + 15 XN24 + 50 XN34

SUBJECT TO C1) X1 - X2 + 999 Y12 >= 33.5 C2) X1 - X2 + 999 Y12 <= 965.5 C3) X1 - X3 + 999 Y13 >= 32.5 C4) X1 - X3 + 999 Y13 <= 966.5 C5) X1 - X4 + 999 Y14 >= 37.5 C6) X1 - X4 + 999 Y14 <= 961.5 C7) X2 - X3 + 999 Y23 >= 37.5 C8) X2 - X3 + 999 Y23 <= 961.5 C9) X2 - X4 + 999 Y24 >= 40.5 C10) X2 - X4 + 999 Y24 <= 958.5 C11) X3 - X4 + 999 Y34 >= 40 C12) X3 - X4 + 999 Y34 <= 959 C13) - XP12 + XN12 + X1 - X2 = 0 C14) - XP13 + XN13 + X1 - X3 = 0 C15) - XP14 + XN14 + X1 - X4 = 0 C16) - XP23 + XN23 + X2 - X3 = 0 C17) - XP24 + XN24 + X2 - X4 = 0 C18) - XP34 + XN34 + X3 - X4 = 0 C19) Y12 = 0 C20) Y13 = 0 C21) Y14 = 0 C22) Y23 = 0 C23) Y24 = 0 C24) Y34 = 0 END TITLE ( MIN)

OBJECTIVE FUNCTION VALUE 1) 12410.000 VARIABLE VALUE REDUCED COST XP12 33.500000 .000000 XP13 71.000000 .000000 XP14 111.000000 .000000 XP23 37.500000 .000000 XP24 77.500000 .000000 XP34 40.000000 .000000 X1 111.000000 .000000 X2 77.500000 .000000 X3 40.000000 .000000

Example 2 (Cont)

DLP 1 MAX 33.5 U12 - 965.5 V12 + 32.5 U13 - 966.5 V13 + 37.5

U14 - 961.5 V14 + 37.5 U23 - 961.5 V23 + 40.5 U24 - 958.5 V24 + 40 U34 - 959 V34

SUBJECT TO C1) - WP12 + WN12 <= 25 C2) WP12 - WN12 <= 25 C3) - WP13 + WN13 <= 35 C4) WP13 - WN13 <= 35 C5) - WP14 + WN14 <= 50 C6) WP14 - WN14 <= 50 C7) - WP23 + WN23 <= 10 C10) WP23 - WN23 <= 10 C11) - WP24 + WN24 <= 15 C12) WP24 - WN24 <= 15 C13) - WP34 + WN34 <= 50 C14) WP34 - WN34 <= 50 C15) U12 - V12 + U13 - V13 + U14 - V14 + WP12 - WN12

+ WP13 -WN13 + WP14 - WN14 <= 0 C16) - U12 + V12 + U23 - V23 + U24 - V24 - WP12 + WN12

+ WP23 -WN23 + WP24 - WN24 <= 0 C17) - U13 + V13 - U23 + V23 + U34 - V34 - WP13 + WN13

- WP23 +WN23 + WP34 - WN34 <= 0 C18) - U14 + V14 - U24 + V24 - U34 + V34 - WP14 + WN14

- WP24 + WN24 - WP34 + WN34 <= 0 END TITLE ( MAX)

LP OPTIMUM FOUND AT STEP 10 OBJECTIVE FUNCTION VALUE 1) 12410.000 VARIABLE VALUE REDUCED COST U12 110.000000 .000000 U23 110.000000 .000000 U34 115.000000 .000000 WN12 25.000000 .000000 WN13 35.000000 .000000 WN14 50.000000 .000000 WN23 10.000000 .000000 WN24 15.000000 .000000 WN34 50.000000 .000000

Feasibility Constraints

• If an upper bound on z is U, then we can write U as

• 1 > yij + yjk - yik > 0 i<n-1, i<j<n, j<k<nTable 7.1. Feasible combinations of y variables for a triplet {i,j,k}

yij yik yjk Feasible?

0 0 0 Yes

0 0 1 Yes

0 1 1 Yes

1 0 0 Yes

1 1 0 Yes

1 1 1 Yes

1 0 1 No

0 1 0 No

Uzzzzz kk

k 122...222 12

21

10

0

Example 2 (Cont)

MP 1 MIN Z1 + 2 Z2 + 4 Z3 + 8 Z4 + 16 Z5 + 32 Z6 + 64 Z7 + 128 Z8 + 256 Z9 + 512 Z10 + 1024 Z11 + 2048 Z12 + 4096 Z13 + 8192 Z14 + 16384 Z15 SUBJECT TO C1) Y12 - Y13 + Y23 >= 0 C2) Y12 - Y13 + Y23 <= 1 C3) Y12 - Y14 + Y24 >= 0 C4) Y12 - Y14 + Y24 <= 1 C5) Y23 - Y24 + Y34 >= 0 C6) Y23 - Y24 + Y34 <= 1 C7) Z1 + 2 Z2 + 4 Z3 + 8 Z4 + 16 Z5 + 32 Z6 + 64 Z7 + 128 Z8 + 256 Z9 + 512 Z10 + 1024 Z11 + 2048 Z12 + 4096 Z13 + 8192 Z14 + 16384 Z15 + 109890 Y12 + 109890 Y23 + 114885 Y34 >= 12410 END INTE 21 NEW INTEGER SOLUTION OF .000000000 AT BRANCH 1 PIVOT 3 OBJECTIVE FUNCTION VALUE 1) .00000000 VARIABLE VALUE REDUCED COST Y34 1.000000 .000000 LAST INTEGER SOLUTION IS THE BEST FOUND

Solution Table

y vector u, v, w vector Upper Bound Lower Bound All equal to 0 u12= u23=110;

u34=115 12,410 0

y34=1 u12= u24=110; v34=95

11,940 0

y12=1 u13= 110; u34=115; v12=50

9,850 0

Dual for LMIP 1

DLP i

Maximize

1

1 1

0.5 0.5n n

ij i j ij ij i j iji j i

u l l My v l l M My

(5)

Subject to

: :

0 1,2,...,ij ij ij ij ij ijj i j j i j

u v w u v w i n

(6)

1,2,..., 1; 1,...,ij ij ijw c f i n j i n

(7)

1,2,..., 1; 1,...,ij ij ijw c f i n j i n

(8)

1

1,2,... 1; 1,...,

i j

nij

ii

l l if facilities i and j are adjacent

Ml otherwise i n j i n

(9)

, 0, 1,2,..., 1; 1,...,ij ij iju v w unrestricted i n j i n

(10)

Dual for LMIP 1

u12 + w12 + w13 +w14 = 0

-u12 + u23 - w12 + w23 = 0

-u23 + u34 - w23 + w34 = 0

- u34 - w34 = 0

Modified Benders’ decomposition algorithm

Step 0: Set i=1, yi = {0,0,...,0} and upper bound UB=infinity.

Step 1: Because DLP i has a unique solution, find this using the technique discussed above. Let ui be the solution to DLP i. If ui(b - Byi) < UB, set UB = ui(b - Byi)Step 2: Update MP by adding the constraint z > ui(b - By) and z > UB-epsilon. Solve MP. If the solution is infeasible, we have found an epsilon-optimal solution to MIP. Otherwise, let y* be the feasible solution. Set i=i+1, yi=y* and return to step 1.

Simulated Annealing Algorithm

n number of departments in the layout problemT initial temperaturer cooling factorITEMP number of times temperature T is decreasedNOVER maximum number of solutions evaluated at each tempNLIMIT max number of new solutions to be accepted at each

tempδ difference in OFVs of previous (best) & current

solutions


Step 0: Set: S = initial feasible solution; z = corresponding OFV; T=999.0; r=0.9;ITEMP=0; NLIMIT=10n; NOVER=100n; p, q = maximum number ofdepartments permitted in any row, column respectively.

Step 1: Repeat step 2 NOVER times or until the number of successful newsolutions is equal to NLIMIT.

Step 2: Pick a pair of departments randomly and exchange the position of thetwo departments. If the exchange of the positions of the two departmentsresults in the overlapping of some other pair(s) of departments,appropriately modify the coordinates of the centers of the concerneddepartments to ensure there is no overlapping. If the resulting solutionS* has an OFV < z, set S=S* and z=corresponding OFV. Otherwise,compute δ = difference between z and the OFV of solution S*, and setS=S* with a probability e-δ/T.

Step 3: Set T=rT and ITEMP=ITEMP+1. If ITEMP is < 100, go to Step 1; otherwise STOP.


7 8 9

4 5 6

1 2 3


7 8 5

4 9 6

1 2 3

Modified Penalty Algorithm

Minimize c11 x11 + c12 x12 + ... + c3n x3n

Subject to a11 x11 + a12 x12 + ... + a1n x1n > b1

a21 x21 + a22 x22 + ... + a2n x2n < b2

a31 x31 + a32 x32 + ... + a3n x3n = b3

x21, x22, ..., x3n > 0

Hybrid Simulated Annealing Algorithm

Step 0: Set: S = initial feasible solution; z = corresponding OFV; T=999.0; r=0.9; ITEMP=0;NOVER=100n; NLIMIT=10n; and p, q = maximum number of departments permitted in any row,column respectively;

Step 2: Apply the MP algorithm to the initial feasible layout. If the departments overlap, modify thecoordinates of the departments to eliminate overlapping. If z* (OFV of the resulting solution S*)is < z, set z=z*; S=S*. Set i=1; j=i+1.

Step 3: If i < n-1, exchange the positions of departments i and j; otherwise go to step 4. If the exchange ofthe positions of departments i, j results in the overlapping of some other pair(s) of departments,appropriately modify the coordinates of the centers of the concerned departments to ensure thereis no overlapping. If the resulting solution has an OFV z* < z, set S=S*; z=z*; i=1; j=i+1 andrepeat step 3. Otherwise, set j=j+1. If j > n, set i=i+1, j=i+1 and repeat step 3.

Step 4: Repeat step 5 NOVER times or until the number of successful new solutions is equal to NLIMIT.Step 5: Pick a pair of departments randomly and exchange the position of the two departments. If the

exchange of the positions of the two departments results in the overlapping of some other pair(s)of departments, appropriately modify the coordinates of the centers of the concerned departmentsto ensure there is no overlapping. If the resulting solution S* has an OFV < z, set S=S* andz=corresponding OFV. Otherwise, compute δ = difference between z and the OFV of solution S*,and set S=S* with a probability 1-eδ/T.

Step 6: Set T=rT and ITEMP=ITEMP+1. If ITEMP is < 100, go to Step 4; otherwise STOP.

SA and HSA Algorithms

Do Example 3, 4 and 5 using SINROW and MULROW

Tabu Search Algorithm

Step 1: Read the flow (F) and distance (D) matrices. Construct the zero long term memory(LTM) matrix of size nxn, where n is the number of departments in the problem.

Step 2: Construct an initial solution using any construction algorithm. Obtain values for thefollowing two short-term memory parameters - size of tabu list (t = 0.33n – 0.6n) andmaximumnumber of iterations (v = 7n-10n). Construct the zero tabu list (TL) vectorand set iteration counter k=1.

Step 3: For iteration k, examine all possible pairwise exchanges to the current solution andmake the exchange {i,j} that leads to the greatest reduction in the OFV and satisfiesone of the following two conditions.(i) Exchange {i,j} is not contained in the tabu list.(ii) If exchange{i,j} is in the tabu list, it satisfies the aspiration criteria.Update tabu list vector TL by including the pair {i,j} as the first element in TL. If thenumber of elements in TL is greater than t, drop the last element.Update LTM matrix by setting LTMij=LTMij+1.

Step 4: Set k=k+1. If k>v, invoke long term memory by replacing the original distance matrixD with D+LTM and go to step 2. Otherwise STOP.

Genetic Algorithm

Step 0: Obtain the maximum number of individuals in the population Nand the maximum number of generations G from the user,generate N solutions for the first generation’s population randomly

andrepresent each solution as a string. Set generation counter Ngen=1.

Step 1: Determine the fitness of each solution in the current generation’spopulation and record the string with the best fitness.

Step 2: Generate solutions for the next generation’s population as follows.(i) Retain 0.1N of the solutions with the best fitness in the previous population.(ii) Generate 0.89N solutions via mating.(iii) Select 0.01N solutions from the previous population randomly and mutate them.

Step 3: Update Ngen = Ngen + 1. If Ngen < G, go to step 1. Otherwise, STOP.

Fitness functions & Population Generation

1

1 1

1/n n

ij ij i ji j i

F c f x x

1 1

N Nij i j i j

ij iji j i j

g x x y yF k w where w

n n

Population Generation– Mating (70-90%)– Retain a small percentage (10-30%) of

individuals from the previous generation, and– Mutate, i.e. randomly alters a randomly selected

chromosome (or individual) from the previous population (0.1 to 1%)

Population Generation

Mating– Two-point crossover method,– Partially matched crossover method– In the two-point crossover method, given two parent

chromosomes {x1, x2, …, xn} and {y1, y2, …, yn}, two integers r, s, such that 1 < r < s < n are randomly selected and the genes in positions r to s of one parent are swapped (as one complete substring without disturbing the order) with that of the other to get two offspring as follows:

{x1, x2, …, xr-1, yr,, yr+1, …, ys, xs+1, xs+2, …. xn}{y1, y2, …, yr-1, xr,, xr+1, …, xs, ys+1, ys+2, …. yn}


Mating– Partially matched crossover method– Partially matched crossover method is just like

two-point, but genes are exchanged only if they lead to a feasible solution

Mutate– Take a solution and simply swap two genes


Mutate Reproduction Method in which a prespecified

percentage of individuals are retained based on probabilities that are inversely proportional to their OFVs

Clonal Propagation Method in which xN individuals with the best fitness are retained. (x is the prespecified proportion of individuals that are to be retained from the previous generation and N is the population size), and

Multicriteria Layout

Minimize w1C-w2R Subject to above constraints

1

1 1,2,...,n

ijj

Subject to x i n

1

1 1,2,...,n

iji

x j n

1 1 1 1

n n n n

ik jl ij kli j k l

i k j l

Minimize C f c x x

1

1 1,2,...,n

ijj

Subject to x i n

1

1 1,2,...,n

iji

x j n

0 1 , 1,2,...,ijx or i j n

1 1 1 1

n n n n

ijkl ij kli j k l

i k j l

Maximize R r x x

if locations and are adjacent

0 otherwise

ij

ijkl

t i jr

0 1 , 1,2,...,ijx or i j n

Multicriteria Layout

Ri

Ci

A

BC

D

Model for CMS Design

Parameters: i, j, k part, machine, cell

indices, respectively ci intercellular movement

cost per unit for part i vi number of units of part i uij cost of part i not utilizing

machine j oij number of times each

part i requires operation on machine j

Mmax maximum number of machines permitted in a cell

Mmin minimum number of machines permitted in a cell

Cu maximum number of cells permitted

S1 sets of machine pairs that cannot be located in the same cell

S2 sets of machine pairs that must be located in the same cell

np total number of part types nm total number of machines

0 if machine is not required for part

1 otherwise ij

j ia

Model for CMS Design

Decision Variables

0 if part is not processed in cell

1 otherwise ik

i kx

0 if machine is not in cell

1 otherwise jk

j ky

Model 1 for CMS Design

Minimize

Subject to

1 1 1 1 1 1

(1 ) (1 )u uC Cnp npnm nm

i i ij ij ik jk ij ij ik jki j k i j k

c v o a x y u a x y

1

1 =1,2,...,uC

ikk

x i np

1

1 =1,2,...,uC

jkk

y j nm

11 =1,2,..., , { , }sk tk uy y k C s t S

2},{,,...,2,10 StsCkyy utksk min max

1

=1,2,...,nm

jk uj

M y M k C

0 1 =1,2,..., , =1,2,...,ik ux i np k C

ujk Cknmjory ,...,2,1,,...,2,110

Model 2 for CMS Design

Minimize

Subject to 1

1 =1,2,...,uC

ikk

x i np

1

1 =1,2,...,uC

jkk

y j nm

11 =1,2,..., , { , }sk tk uy y k C s t S 2},{,,...,2,10 StsCkyy utksk

min max1

=1,2,...,nm

jk uj

M y M k C

0 1 =1,2,..., , =1,2,...,ik ux i np k C ujk Cknmjory ,...,2,1,,...,2,110

1 1 1 1 1 1

(1 )u uC Cnp npnm nm

i i ij ij ik ij ij i i ij ij ijki j k i j k

c v o a x u a c v o a z

=1,2,..., , =1,2,..., , =1,2,...,ijk ik uz x i np j nm k C =1,2,..., , =1,2,..., , =1,2,...,ijk jk uz y i np j nm k C

1 =1,2,..., , =1,2,..., , =1,2,...,ik jk ijk ux y z i np j nm k C

0 1 =1,2,..., , =1,2,..., , =1,2,...,ijk uz i np j nm k C

Model P (Primal problem)

Minimize

Subject to1

1 =1,2,...,uC

ikk

x i np

0 1 =1,2,..., , =1,2,...,ik ux i np k C

=1,2,..., , =1,2,..., , =1,2,...,ijk ik uz x i np j nm k C

=1,2,..., , =1,2,..., , =1,2,...,ijk jk uz y i np j nm k C

1 =1,2,..., , =1,2,..., , =1,2,...,ik jk ijk ux y z i np j nm k C

0 1 =1,2,..., , =1,2,..., , =1,2,...,ijk uz i np j nm k C

1 1 1 1 1 1

(1 )u uC Cnp npnm nm

i i ij ij ik ij ij i i ij ij ijki j k i j k

c v o a x u a c v o a z

Model D (Dual problem)

Minimize

Subject to

1 1 1 1 1 1 1 1 1 1

0 ( ) ( 1)u u uC C Cnp np np npnm nm nm

ijk jk ijk jk ijk ii j k i j k i j k i

l y m y n p

1 1 1

0 ( 1) =1,2,..., ; =1,2,...,uCnm nm

ijk ijk ijk i i i ij ijj k j

l m n p c v a o i np k np

0 (1 ) =1,2,..., , =1,2,..., , =1,2,...,ijk ijk ijk i ij ij i i ij ij ul m n p u a c v a o i np j nm k C

, , 0 =1,2,..., , =1,2,..., , =1,2,...,ijk ijk ijk ul m n i np j nm k C

free =1,2,...,ip i np

Model M (Master problem)

Minimize Z

Subject to

Z > 0

11 =1,2,..., , { , }sk tk uy y k C s t S

2},{,,...,2,10 StsCkyy utksk

min max1

=1,2,...,nm

jk uj

M y M k C

1 1 1 1 1 1 1

u uC Cnp np npnm nm

ijk ijk jk i ijki j k i i j k

Z m n y p n

1

1 =1,2,...,uC

jkk

y j nm

52

Next Generation Factory Layouts

A Re-configurable facility is one that can adapt efficiently and effectively to frequent changes in product mix and volume and aid in mass customization and lean manufacturing environments

53

Reconfigurable, dynamic and robust layout problems

Scenario

Demand Scenario 2

Demand Scenario 1

Reconfigurable layout problem only considers one deterministic layout context for current and the next available future planning periods. Dynamic layout considers layout contexts for multiple periods and robust layout considers layout context for multiple scenarios and multiple periods.

tFuture Planning Period 2

Present

Future Planning Period 1

Current Period

Robust Layout

Dynamic Layout

Reconfigurable Layout

…

…

54

Traditional View

Static Problem

Assumptions– Product range and composition fairly constant– Product Mix Changes known prior to the design

stage

55

Why a Re-configurable Facility?

Why not? Besides …..• Changes in

Manufacturing Environment

• Changes in Materials and Process Technology

56

What are the experts saying?

Visionary Manufacturing Challenges for 2020 Two of six challenges to remain productive

and profitable in 2020– To “achieve concurrency in all operations”– To “reconfigure manufacturing enterprises rapidly

in response to changing needs and opportunities” Two enabling technologies companies need

to overcome above challenges– Adaptable processes and equipment– Reconfiguration of manufacturing operations

57

Modifications to the Facility Layout Problem

Design for Relocation– Machine tools– Inherent features in layout– Material handling equipment– Support facilities

Re-configurable Factory Layout

58

Examples Supporting Re-configurable Layout Systems

Scalable Machines (NSF/UofM, TRIFLEX)

Portable Machines (Southwestern Industries, Climax)

Conveyor mounted cells (NT)

Modular Automated Parking System (Robotic Parking, Inc.)

59

Four phase Approach for Reconfigurable Layouts

Change in layout parameters?

Layout forcurrent period

Layout fornext period

Generate candidate layouts Estimate performance measures of layouts

Determine Layout to be usedRefine selected layout

Need for Stochastic Analysis

1 2 3

4 5 6

7 8 9

10 11 12

1 2 5

3 4 6

10 11 7

8 9 12

61

Following Assumed to be Known

First two moments of external arrival rate for each product

First two moments of service time for each processing operation

Set-up Times Batch Size - Process as well as Transfer Failures Empty Travel

62

L_q WqC M Whitt/MP

ASimu. Whitt

MPASimu. Whitt

MPASimu.

1 1 0.6668 0.639 3.3409 3.7602 3.9966 2.2273 2.5068 2.6521 2 0.7250 0.6959 3.6966 3.3888 1.7222 5.6009 5.1345 2.60431 3 0.3292 0.3116 1.1487 1.4647 1.8392 0.6564 0.8370 1.05181 4 0.1890 0.1797 0.2297 0.2338 0.4404 0.0753 0.0767 0.144521 5 0.2153 0.2044 0.3516 0.3608 0.4904 0.0879 0.0902 0.12292 1 0.5675 0.5605 1.6321 1.7626 3.9969 1.1658 1.2590 2.85432 2 0.7350 0.7353 3.9742 4.5636 2.9090 2.2710 2.6078 1.67112 3 0.3700 0.3679 0.6234 0.6605 1.6588 0.2653 0.2811 0.704822 4 0.6600 0.6577 5.8562 6.8056 10.159 2.1689 2.5206 3.75973 1 0.4405 0.4419 0.8267 0.9690 2.9143 0.2802 0.3285 0.98783 2 0.5270 0.5348 1.4018 1.6770 2.0923 0.6095 0.7291 0.91073 3 0.6200 0.6299 1.0329 1.0329 1.12 0.8263 0.8263 0.89923 4 0.1825 0.1823 0.0491 0.0519 0.2748 0.0135 0.0142 0.07557

MPA Results

63

Example Problem Data

Table 1 Operation Sequences of Products Produced in the Facility

Product#

Sequence Arrival Rate(per hour)

1 1 4 8 9 0.2

2 1 4 7 4 8 7 0.3

3 1 2 4 7 8 9 0.1

4 1 4 7 9 0.3

5 1 6 10 7 9 0.2

6 6 10 7 8 9 0.1

7 6 4 8 9 0.2

8 3 5 2 6 4 8 9 0.1

9 3 5 6 4 8 9 0.1

10 4 7 4 8 0.2

11 6 0.3

12 11 7 12 0.1

64

Current Cellular Layout (L0)

46

46

87

98

65

24

53

32

11

Cell 1

1116

717

717

1116

1116

1015

1214

Cell 3

1012

1012

913

711

711

610

19

Cell 2

xy

Legend

x Machine type

y Machine label

(number)

65

Functional Layout (L1)

1212

1111

1111

1111

1010

1010

1010

33

99

99

55

77

77

77

77

88

66

66

44

44

11

11

22

66

Cell. Layout with reorientation and reshaping (L2)

46

46

87

98

65

24

53

32

11

Cell 1

1115

710

710

1115

1115

1014

1213

Cell 3

912

11

1011

69

1011

710

710

Cell 2

67

Cell. Layout with reorientation and reshaping (L3)

65

26

57

38

11

44

44

83

92

Cell 1

11

92

69

1012

710

1012

710

Cell 2

1213

1012

1111

1111

1111

710

710

Cell 3

68

Virtual Cellular Layout (L4)

78

1111

78

1111

1212

1111

109

11

65

78

78

109

109

910

11

65

32

46

53

46

910

24

87

Cell 1

Cell 2

Cell 3

69

Cellular Layout with Remainder Cell (L5)

47

710

47

710

53

35

22

66

89

911

11

Cell 1

11

64

108

108

912

Cell 2

108

713

713

1115

1214

1115

1115

Cell 3

70

Distance Matrix for Layout L1

M 1 2 3 4 5 6 7 8 9 10 11 12

1 0 1.5 3.5 2 2.5 1 4.5 2.5 3 5.5 6.5 7.5

2 1.5 0 4 3.5 3 2.5 6 4 4.5 7 8 9

3 3.5 4 0 5.5 1 2.5 5 4 2.5 3 4 5

4 2 3.5 5.5 0 4.5 3 2.5 1.5 3 3.5 4.5 5.5

5 2.5 3 1 4.5 0 1.5 4 3 1.5 4 5 6

6 1 2.5 2.5 3 1.5 0 3.5 1.5 2 4.5 5.5 6.5

7 4.5 6 5 2.5 4 3.5 0 2 2.5 2 3 4

8 2.5 4 4 1.5 3 1.5 2 0 1.5 3 4 5

9 3 4.5 2.5 3 1.5 2 2.5 1.5 0 2.5 3.5 4.5

10 5.5 7 3 3.5 4 4.5 2 3 2.5 0 1 2

11 6.5 8 4 4.5 5 5.5 3 4 3.5 1 0 1

12 7.5 9 5 5.5 6 6.5 4 5 4.5 2 1 0

71

WIP for 6 layouts

WIP(Average queue length)

M L0 L1 L2 L3 L4 L5

1 2.42 3.00 3.00 3.00 3.00 3.002 0.90 0.90 0.90 1.36 0.90 0.903 0.52 0.90 0.52 2.09 0.52 0.524 0.90 1.15 0.90 1.15 0.90 1.485 4.24 0.52 4.24 4.24 2.66 0.906 1.14 2.66 1.15 0.90 1.15 4.247 2.09 0.90 2.09 0.52 2.09 1.638 1.08 2.09 1.08 0.90 0.90 2.619 3.87 1.36 1.48 1.48 2.64 2.0510 1.48 2.64 0.90 0.90 1.36 0.2211 0.32 0.74 3.76 0.74 0.74 0.5912 3.75 1.01 0.91 2.61 1.01 2.9513 0.90 1.01 1.01 2.8614 1.01 1.46 1.0115 1.46 0.73 0.7316 0.7317 2.85

Sum 29.67 17.87 24.16 20.91 17.88 25.71

72

MH Cost and Lead Times

Material Handling Cost (distance) Due Lead Time (hours)

P Lambda L0 L1 L2 L3 L4 L5 Date L0 L1 L2 L3 L4 L5

1 0.2 7 5 7.5 6.5 7 4.5 7.87 8.35 7.88 7.81 7.87 7.88 7.67

2 0.3 21 10.5 27.5 26.5 21 8 9.10 9.29 9.09 9.10 9.09 9.10 9.49

3 0.1 17 11 20.5 19.5 20 7.5 17.30 17.59 17.30 17.23 17.29 17.30 16.89

4 0.3 16 7 18.5 16.5 12 4.5 6.61 6.92 6.61 6.55 6.61 6.61 11.41

5 0.2 10 10 10.5 12.5 10 8.5 18.30 28.73 16.18 23.35 15.76 16.18 20.42

6 0.1 12 10 12 13 14 7 11.95 15.36 11.94 15.56 11.52 11.95 10.98

7 0.2 5 6 5 4 6 5 17.89 17.96 9.48 17.97 18.03 9.48 17.83

8 0.1 9 12.5 9 7 12.5 8 38.73 38.79 30.32 38.81 38.87 30.32 38.67

9 0.1 9 8.5 9 7 10.5 8 28.22 28.28 19.80 28.29 28.35 19.80 28.15

10 0.2 10.5 6.5 16.5 16.5 11.5 4.5 6.01 5.82 6.01 6.01 6.00 6.01 6.04

11 0.3 0 0 0 0 0 0 2.55 2.55 2.94 2.55 2.55 2.94 2.55

12 0.1 5.5 7 6.5 7 4 3 6.23 9.44 6.23 6.22 6.22 6.23 9.46

13 0.1 3.5 1 3.5 3 1 2 4.38 4.38 4.38 4.38 4.38 4.38 4.38

14 0.3 5.5 5 6.5 5 4 4 8.08 11.40 7.96 8.20 7.94 7.96 11.17

15 0.1 13 10.5 15.5 11.5 12.5 17.5 16.10 25.03 15.98 16.22 15.96 15.98 19.19

16 0.2 13 10.5 15.5 11.5 12.5 17.5 16.10 25.03 15.98 16.22 15.96 15.98 19.19

17 0.1 5.5 7 6.5 7 4 3 6.83 10.04 6.83 6.82 6.82 6.83 10.06

18 0.3 4.5 5.5 4.5 6 5.5 6 8.69 12.17 8.69 12.37 8.27 8.69 8.06

19 0.2 0 0 0 0 0 0 1.97 1.97 1.97 1.97 1.97 1.97 1.97

MHD_Cost 30.65 22.75 36.35 33.9 30 20.35 OverDue

8.09 0.12 2.58 0.05 0.12 4.49

73

Cost with vector {5,2,10,0.1}

Unit Layout

Criteria Cost L0 L1 L2 L3 L4 L5

WIP 5 29.67 17.87 24.16 20.91 17.88 25.71MaterialHandling

2 30.65 22.75 36.35 33.9 30 20.35Over Due 10 8.09 0.12 2.58 0.05 0.12 4.49Relocation 0.1 0 118 15 60 122 67

Overall Cost 290.55 147.85 220.8 178.85 162.6 220.85

Table 6. Overall cost with unit cost vector of {1, 10, 1, 0.1}

Unit Layout

Criteria Cost L0 L1 L2 L3 L4 L5

WIP 1 29.67 17.87 24.16 20.91 17.88 25.71MaterialHandling

10 30.65 22.75 36.35 33.9 30 20.35Over Due 1 8.09 0.12 2.58 0.05 0.12 4.49Relocation 0.1 0 118 15 60 122 67

Overall Cost 344.26 257.29 391.74 364.96 330.2 240.4

74

WIP and MH Cost - Efficient Frontier

C1 (5,2)

C2 (1,10)

138.29

134.85

149.40

WIP

MH

D.C

ost

10 15 20 25 30

1015

2025

3035

L0

L1

L2

L3

L4

L5

Figure 10. WIP and material handling cost as layout selection criteria

75

Final Layout

47

710

47

710

53

35

22

66

89

911

11

Cell 1

11

64

108

108

912

Cell 2

108

713

713

1115

1214

1115

1115

Cell 3

advanced layout algorithms chapter 8. layout algorithms optimal heuristic

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