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
n 1
n
n 1
n

i 1 j i 1 k 1 l  k 1
fij dkl xik x jl
Branch and bound Algorithm
LB*=71.5
LB*11
LB*14
LB*12
First Level
nodes
LB*13
LB*22
LB*23
LB*34 + LB*43 = 76
LB*24
Second
Level
nodes
Branch and bound Algorithm


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 nondecreasing 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
Branch and bound Algorithm


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
q
Matrix [w’ij] is obtained as follows:
f ik d jsk where

k 1
k  S , i  S , sk  L, and j  L.
Explain Branch and Bound Algorithm
with Example 1
Office
O
[fij]=
Site
1
2
3
4
f
1
-
17
12
11
f
2
17
-
12
4
i
3
12
12
-
c
4
11
4
•4
1
2
3
4
S
1
-
1
1
2
i
2
1
-
2
1
4
t
3
1
2
-
1
-
e
4
2
1
1
-
[dij]=
e
Figure 7.2
Flow and distance matrices for the LonBank layout problem
Branch and bound Algorithm


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 MIP
Minimize
cx
Subject to
Ax
x
y
+
By
>
>
=
b
0
0 or 1
Now, consider a feasible y solution vector to MIP - say yi.
Then, MIP becomes the following linear model.
LP i
Minimize
cx
Subject to
Ax
>
b - Byi
x
>
0
Dual of Linear Program
LP i
Minimize
Subject to
cx
Ax
x
>
>
b - Byi
0
The dual of LP i is the following model
DLP i
Maximize
u(b - Byi)
Subject to
uA
<
c
u
>
0
Dual of Linear Program
•
•
•
•
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
Dual of Linear Program
•
•
•
•
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 MIP
The above model can be restated as:
MP
Minimize
z
Subject to
z
>
ui(b - By) i = 1, 2, ..., s
y
=
0 or 1
and feasible to MIP
z
=
0 or 1
MP requires us to generate all the feasible y solution vectors and the
corresponding s dual problems - DLP 1, DLP 2, ..., DLP s
Not computationally feasible because the number of dual problems, though
finite, may be very large
The 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
1
2
3
4
Flow Matrix
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
n 1
Minimize


c
f
x

x
  ij ij  ij ij 
n
i 1 j i 1
Subject to
xi  x j  My ij  0.5(li  l j )  d ij , i  1,2,..., n  1; j  i  1,..., n
 xi  x j  M (1  yij )  0.5(li  l j )  d ij , i  1,2,..., n  1; j  i  1,..., n
xi  x j  xij  xij i  1, 2,..., n -1; j  i  1,...n
xij , xij  0 i  1,2,..., n  1; j  i  1,..., n
y ij  0 or 1, i  1,2,..., n  1; j  i  1,..., n
xi  0
i=1, 2,..., n
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
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
z  2 0 z 0  21 z1  2 2 z 2  ...  2 k z k 2 k 1  1  U
•
1 > yij + yjk - yik > 0
i<n-1, i<j<n, j<k<n
Table 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
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
All equal to 0
y34=1
y12=1
u, v, w vector
u12= u23=110;
u34=115
u12= u24=110;
v34=95
u13= 110;
u34=115; v12=50
Upper Bound
12,410
Lower Bound
0
11,940
0
9,850
0
Dual for LMIP 1
DLP i
Maximize
Subject to
n 1

 uij 0.5 li  l j   Myij   vij 0.5 li  l j   M  Myij 
i 1 j i 1
n
 u
j:i  j
ij
 vij  wij     uij  vij  wij   0 i  1, 2,..., n
 wij  cij fij
wij  cij f ij
(5)
(6)
j:i  j
i  1, 2,..., n  1; j  i  1,..., n
i  1, 2,..., n  1; j  i  1,..., n
li  l j   if facilities i and j are adjacent

M ij   n
 li   otherwise i  1, 2,...n  1; j  i  1,..., n
 i 1
uij , vij  0, wij unrestricted
i  1, 2,..., n  1; j  i  1,..., n
(7)
(8)
(9)
(10)
Dual for LMIP 1
u12
+
-u12
+
-u23
+
u23
u34
-
u34
w12
+
w13
+w14
=0
w12
+
w23
-
w23
=0
+
w34
=0
-
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 problem
T
initial temperature
r
cooling factor
ITEMP number of times temperature T is decreased
NOVER maximum number of solutions evaluated at each temp
NLIMIT max number of new solutions to be accepted at each temp
δ
difference in OFVs of previous (best) & current solutions
Simulated Annealing Algorithm
Step 0:
Step 1:
Step 2:
Step 3:
Set: S = initial feasible solution; z = corresponding OFV; T=999.0; r=0.9;
ITEMP=0; NLIMIT=10n; NOVER=100n; p, q = maximum number of
departments permitted in any row, column respectively.
Repeat step 2 NOVER times or until the number of successful new
solutions is equal to NLIMIT.
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
departments to ensure there is no overlapping. If the resulting solution
S* has an OFV < z, set S=S* and z=corresponding OFV. Otherwise,
compute δ = difference between z and the OFV of solution S*, and set
S=S* with a probability e-δ/T.
Set T=rT and ITEMP=ITEMP+1. If ITEMP is < 100, go to Step 1; otherwise STOP.
Simulated Annealing Algorithm
7
8
4
9
5
1
2
6
3
Simulated Annealing Algorithm
7
4
1
8
5
9
6
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:
Step 2:
Step 3:
Step 4:
Step 5:
Step 6:
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;
Apply the MP algorithm to the initial feasible layout. If the departments overlap, modify the
coordinates 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.
If i < n-1, exchange the positions of departments i and j; otherwise go to step 4. If the exchange of
the 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 there
is no overlapping. If the resulting solution has an OFV z* < z, set S=S*; z=z*; i=1; j=i+1 and
repeat step 3. Otherwise, set j=j+1. If j > n, set i=i+1, j=i+1 and repeat step 3.
Repeat step 5 NOVER times or until the number of successful new solutions is equal to NLIMIT.
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 departments
to ensure there is no overlapping. If the resulting solution S* has an OFV < z, set S=S* and
z=corresponding OFV. Otherwise, compute δ = difference between z and the OFV of solution S*,
and set S=S* with a probability 1-eδ/T.
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:
Step 2:
Step 3:
Step 4:
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.
Construct an initial solution using any construction algorithm. Obtain values for the
following two short-term memory parameters - size of tabu list (t = 0.33n – 0.6n) and
maximumnumber of iterations (v = 7n-10n). Construct the zero tabu list (TL) vector
and set iteration counter k=1.
For iteration k, examine all possible pairwise exchanges to the current solution and
make the exchange {i,j} that leads to the greatest reduction in the OFV and satisfies
one 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 the
number of elements in TL is greater than t, drop the last element.
Update LTM matrix by setting LTMij=LTMij+1.
Set k=k+1. If k>v, invoke long term memory by replacing the original distance matrix
D with D+LTM and go to step 2. Otherwise STOP.
Genetic Algorithm
Step 0:
Step 1:
Step 2:
Step 3:
Obtain the maximum number of individuals in the population N
and the maximum number of generations G from the user,
generate N solutions for the first generation’s population randomly and
represent each solution as a string. Set generation counter Ngen=1.
Determine the fitness of each solution in the current generation’s
population and record the string with the best fitness.
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.
Update Ngen = Ngen + 1. If Ngen < G, go to step 1. Otherwise, STOP.
Fitness functions & Population
Generation
 n1 n

F  1/   cij fij xi  x j 
 i 1 j i 1


gij  xi  x j  yi  y j 
N N

F  k   wij  where wij 
ni n j
 i 1 j 1 
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}
Population Generation

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
Population Generation

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
n
n
n
n
n
n
n
n
Minimize C   fik c jl xij xkl Maximize R   rijkl xij xkl
i 1 j 1 k 1 l 1
i  k j l
n
Subject to
x
ij
j 1
n
x
i 1
ij
 1 i  1, 2,..., n

n
Subject to
x
 1 i  1, 2,..., n
x
1
j 1
n
1
j  1, 2,..., n
xij  0 or 1 i, j  1, 2,..., n

i 1 j 1 k 1 l 1
i  k j l
Minimize w1C-w2R
Subject to above constraints
i 1
ij
ij
j  1, 2,..., n
xij  0 or 1 i, j  1, 2,..., n
tij if locations i and j are adjacent
rijkl  
0 otherwise
Multicriteria Layout
Ci
D
B
C
A
Ri
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 j is not required for part i
aij 
 1 otherwise
Model for CMS Design

Decision Variables
 0 if part i is not processed in cell k
xik 
 1 otherwise
 0 if machine j is not in cell k
y jk 
 1 otherwise
Model 1 for CMS Design
np nm Cu
Minimize
 c v o a x
i 1 j 1 k 1
Cu
Subject to
x
k 1
ik
i i ij ij ik
np nm Cu
(1  y jk )   uij (1  aij ) xik y jk
i 1 j 1 k 1
Cu
1
i =1,2,...,np
ysk  ytk  1
y
k 1
jk
1
j =1,2,...,nm
k =1,2,...,Cu , {s, t} S1
y sk  ytk  0 k  1,2,..., Cu , {s, t}  S 2
nm
M min   y jk  M max
j 1
0  xik  1
k =1,2,...,Cu
i =1,2,...,np, k =1,2,...,Cu
y jk  0 or 1 j  1,2,..., nm, k  1,2,..., Cu
Model 2 for CMS Design
np nm Cu
Minimize
np nm Cu
 c v o a x   u
i i ij ij ik
i 1 j 1 k 1
Cu
Subject to
x
k 1
ik
ij
i 1 j 1 kC1
y
u
1
i =1,2,...,np
k 1
jk
(1  aij )  ci vi oij aij  zijk
1
j =1,2,...,nm
ysk  ytk  1 k =1,2,...,Cu , {s, t} S1
y sk  ytk  0 k  1,2,..., Cu , {s, t}  S 2
nm
M min   y jk  M max
0  xik  1
j 1
k =1,2,...,Cu
y jk  0 or 1 j  1,2,..., nm, k  1,2,..., Cu
zijk  xik
i =1,2,...,np, j =1,2,...,nm, k =1,2,...,Cu
zijk  y jk
i =1,2,...,np, j =1,2,...,nm, k =1,2,...,Cu
xik  y jk  zijk  1 i =1,2,...,np, j =1,2,...,nm, k =1,2,...,Cu
i =1,2,...,np, k =1,2,...,Cu
0  zijk  1
i =1,2,...,np, j =1,2,...,nm, k =1,2,...,Cu
Model P (Primal problem)
np nm Cu
Minimize
 c v o a x   u
i 1 j 1 k 1
Cu
Subject to
np nm Cu
x
k 1
ik
i i ij ij ik
1
0  xik  1
zijk  xik
zijk  y jk
(1  aij )  ci vi oij aij  zijk
i =1,2,...,np
i =1,2,...,np, k =1,2,...,Cu
i =1,2,...,np, j =1,2,...,nm, k =1,2,...,Cu
i =1,2,...,np, j =1,2,...,nm, k =1,2,...,Cu
xik  y jk  zijk  1
0  zijk  1
i 1 j 1 k 1
ij
i =1,2,...,np, j =1,2,...,nm, k =1,2,...,Cu
i =1,2,...,np, j =1,2,...,nm, k =1,2,...,Cu
Model D (Dual problem)
np nm Cu
Minimize
 0  l
i 1 j 1 k 1
nm
Subject to
l
j 1
ijk
ijk
np nm Cu
np nm Cu
np
i 1 j 1 k 1
i 1 j 1 k 1
i 1
  ( y jk )  mijk   ( y jk  1)  nijk   pi
Cu
nm
 0  mijk  (1) nijk  pi   ci vi aij oij
k 1
i =1,2,...,np; k =1,2,...,np
j 1
lijk  mijk  nijk  0  pi   uij (1  aij )  ci vi aij oij i =1,2,...,np, j =1,2,...,nm, k =1,2,...,Cu
lijk , mijk , nijk  0 i =1,2,...,np, j =1,2,...,nm, k =1,2,...,Cu
pi free i =1,2,...,np
Model M (Master problem)
Minimize
Subject to
Z
Cu
y
k 1
jk
1
j =1,2,...,nm
ysk  ytk  1
k =1,2,...,Cu , {s, t} S1
y sk  ytk  0 k  1,2,..., Cu , {s, t}  S 2
nm
M min   y jk  M max
j 1
k =1,2,...,Cu
np
np nm Cu
 Cu

Z     mijk  nijk   y jk    pi   nijk
i 1 j 1  k 1
i 1
i 1 j 1 k 1

np nm
Z>0
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
52
Reconfigurable, dynamic and
robust layout problems
Scenario
…
Demand
Scenario 2
Robust Layout
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.
Reconfigurable
Layout
Dynamic
Layout
…
Demand
Scenario 1
53
Prese
nt
Current Period
Future
Period 1
Planning
Future
Period 2
Planning
t
Traditional View
Static Problem
Assumptions
–
–
54
Product range and composition fairly constant
Product Mix Changes known prior to the design
stage
Why a Re-configurable Facility?


•
•
Why not?
Besides …..
Changes in
Manufacturing
Environment
Changes in Materials
and Process
Technology
55
What are the experts saying?


Visionary Manufacturing Challenges for 2020
Two of six challenges to remain productive
and profitable in 2020
–
–

Two enabling technologies companies need
to overcome above challenges
–
–
56
To “achieve concurrency in all operations”
To “reconfigure manufacturing enterprises rapidly
in response to changing needs and opportunities”
Adaptable processes and equipment
Reconfiguration of manufacturing operations
Modifications to the Facility
Layout Problem
 Design
–
–
–
–
for Relocation
Machine tools
Inherent features in layout
Material handling equipment
Support facilities
 Re-configurable
57
Factory Layout
Examples Supporting Reconfigurable Layout Systems
Scalable Machines (NSF/UofM, TRIFLEX)
Portable Machines (Southwestern Industries,
Climax)
Conveyor mounted cells (NT)
Modular Automated Parking System (Robotic
Parking, Inc.)
58
Four phase Approach for
Reconfigurable Layouts
Change in layout parameters?
Generate candidate layouts
Estimate performance measures
of layouts
Refine selected layout
Determine Layout to be used
Layout for
current period
Layout for
next period
59
Need for Stochastic Analysis
1
2
3
1
2
5
4
5
6
3
4
6
7
8
9
10
11
7
10
11
12
8
9
12
Following Assumed to be Known






61
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
MPA Results

C M Whitt/MP
A
1 1
0.6668
1 2
0.7250
1 3
0.3292
1 4
0.1890
1 5
0.2153
2 1
0.5675
2 2
0.7350
2 3
0.3700
2 4
0.6600
3 1
0.4405
3 2
0.5270
3 3
0.6200
3 4
0.1825
62
L_q
Simu .
Whitt
0.639
0.6959
0.3116
0.1797
0.2044
0.5605
0.7353
0.3679
0.6577
0.4419
0.5348
0.6299
0.1823
3.3409
3.6966
1.1487
0.2297
0.3516
1.6321
3.9742
0.6234
5.8562
0.8267
1.4018
1.0329
0.0491
MPA
3.7602
3.3888
1.4647
0.2338
0.3608
1.7626
4.5636
0.6605
6.8056
0.9690
1.6770
1.0329
0.0519
Wq
Simu .
Whitt
3.9966
1.7222
1.8392
0.4404
0.4904
3.9969
2.9090
1.6588
10.159
2.9143
2.0923
1.12
0.2748
2.2273
5.6009
0.6564
0.0753
0.0879
1.1658
2.2710
0.2653
2.1689
0.2802
0.6095
0.8263
0.0135
MPA
2.5068
5.1345
0.8370
0.0767
0.0902
1.2590
2.6078
0.2811
2.5206
0.3285
0.7291
0.8263
0.0142
Simu .
2.652
2.6043
1.0518
0.14452
0.1229
2.8543
1.6711
0.70482
3.7597
0.9878
0.9107
0.8992
0.07557
Example Problem Data
Table 1 Operation Sequence s of Produc ts Produced in the Facilit y
63
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
Current Cellular Layout (L0)
Cell 1
6
5
4
x
Legend
x Machine type
y Machine label
(number)
2
6
y
4
5
9
64
6
10
7
1
13
11
16
10
12
9
1
11
11
12
2
1
8
7
10
3
16
11
3
7
11
7
4
6
8
Cell 3
Cell 2
10
17
7
9
16
15
12
17
14
Functional Layout (L1)
11
10
11
12
11
12
7
10
10
11
11
7
10
10
11
7
7
7
9
10
4
8
1
6
5
1
6
1
6
5
4
8
9
3
4
7
9
9
3
4
7
7
1
6
2
2
65
Cell. Layout with reorientation and
reshaping (L2)
Cell 1
6
Cell 3
5
4
11
2
6
15
4
Cell 2
4
5
6
8
3
10
12
9
11
7
6
1
10
7
7
9
10
15
10
10
1
1
11
15
11
7
2
1
8
11
10
3
7
66
9
14
12
10
13
Cell. Layout with reorientation
and reshaping (L3)
Cell 1
6
4
5
2
4
4
6
5
4
8
7
3
12
3
9
8
1
10
13
9
2
1
1
Cell 3
10
2
6
1
Cell 2
67
11
10
11
12
7
7
10
11
11
12
7
9
11
11
12
10
10
7
10
Virtual Cellular Layout (L4)
Cell 1
1
1
1
Cell 2
6
6
5
7
7
7
8
11
7
8
11
11
11
9
10
11
12
Cell 3
12
68
10
11
9
9
10
4
8
6
6
9
9
8
4
4
10
2
3
2
5
8
5
3
1
10
7
Cellular Layout with Remainder
Cell (L5)
Cell 1
5
Cell 2
2
3
3
1
2
6
5
1
1
4
6
6
1
4
7
4
10
7
8
Cell 3
8
7
9
9
7
10
10
10
10
7
9
11
8
8
12
13
12
7
14
11
13
11
11
15
69
15
15
Distance Matrix for Layout L1
M
1
2
3
4
5
6
7
8
9
10
11
12
70
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
WIP for 6 layouts
71
M
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Sum
L0
2.42
0.90
0.52
0.90
4.24
1.14
2.09
1.08
3.87
1.48
0.32
3.75
0.90
1.01
1.46
0.73
2.85
29.67
L1
3.00
0.90
0.90
1.15
0.52
2.66
0.90
2.09
1.36
2.64
0.74
1.01
17.87
WIP(Ave rage queue length)
L2
L3
3.00
3.00
0.90
1.36
0.52
2.09
0.90
1.15
4.24
4.24
1.15
0.90
2.09
0.52
1.08
0.90
1.48
1.48
0.90
0.90
3.76
0.74
0.91
2.61
1.01
1.01
1.46
0.73
24.16
20.91
L4
3.00
0.90
0.52
0.90
2.66
1.15
2.09
0.90
2.64
1.36
0.74
1.01
L5
3.00
0.90
0.52
1.48
0.90
4.24
1.63
2.61
2.05
0.22
0.59
2.95
2.86
1.01
0.73
17.88
25.71
MH Cost and Lead Times
Material Handling Cost
P Lambda L0
L1
L2
L3
1
0.2
7
5
7.5
6.5
2
0.3
21
10.5 27.5 26.5
3
0.1
17
11
20.5 19.5
4
0.3
16
7
18.5 16.5
5
0.2
10
10
10.5 12.5
6
0.1
12
10
12
13
7
0.2
5
6
5
4
8
0.1
9
12.5
9
7
9
0.1
9
8.5
9
7
10
0.2
10.5 6.5
16.5 16.5
11
0.3
0
0
0
0
12
0.1
5.5
7
6.5
7
13
0.1
3.5
1
3.5
3
14
0.3
5.5
5
6.5
5
15
0.1
13
10.5 15.5 11.5
16
0.2
13
10.5 15.5 11.5
17
0.1
5.5
7
6.5
7
18
0.3
4.5
5.5
4.5
6
19
0.2
0
0
0
0
MHD_ Cost
72
30.65 22.75 36.35 33.9
(distance)
L4
L5
7
4.5
21
8
20
7.5
12
4.5
10
8.5
14
7
6
5
12.5
8
10.5
8
11.5 4.5
0
0
4
3
1
2
4
4
12.5 17.5
12.5 17.5
4
3
5.5
6
0
0
30
Due
Date
7.87
9.10
17.30
6.61
18.30
11.95
17.89
38.73
28.22
6.01
2.55
6.23
4.38
8.08
16.10
16.10
6.83
8.69
1.97
L0
8.35
9.29
17.59
6.92
28.73
15.36
17.96
38.79
28.28
5.82
2.55
9.44
4.38
11.40
25.03
25.03
10.04
12.17
1.97
Lead Tim e (hours)
L1
L2
L3
L4
7.88 7.81 7.87 7.88
9.09 9.10 9.09 9.10
17.30 17.23 17.29 17.30
6.61 6.55 6.61 6.61
16.18 23.35 15.76 16.18
11.94 15.56 11.52 11.95
9.48 17.97 18.03 9.48
30.32 38.81 38.87 30.32
19.80 28.29 28.35 19.80
6.01 6.01 6.00 6.01
2.94 2.55 2.55 2.94
6.23 6.22 6.22 6.23
4.38 4.38 4.38 4.38
7.96 8.20 7.94 7.96
15.98 16.22 15.96 15.98
15.98 16.22 15.96 15.98
6.83 6.82 6.82 6.83
8.69 12.37 8.27 8.69
1.97 1.97 1.97 1.97
20.35 Over
Due
8.09
0.12
2.58
0.05
0.12
L5
7.67
9.49
16.89
11.41
20.42
10.98
17.83
38.67
28.15
6.04
2.55
9.46
4.38
11.17
19.19
19.19
10.06
8.06
1.97
4.49
Cost with vector {5,2,10,0.1}
Unit
Criteria
Cost
WIP
5
Material
2
Handling
Over
Due
10
Relocation
0.1
Overall Cost
Layout
L0
L1
L2
L3
L4
L5
29.67
30.65
8.09
0
290.55
17.87
22.75
0.12
118
147.85
24.16
36.35
2.58
15
220.8
20.91
33.9
0.05
60
178.85
17.88
30
0.12
122
162.6
25.71
20.35
4.49
67
220.85
Table 6. Overall co st with unit cost vector of {1, 10, 1, 0.1}
Unit
Criteria
Cost
WIP
1
Material
10
Handling
Over Due
1
Relocation
0.1
Overall Cost
73
Layout
L0
L1
L2
L3
L4
L5
29.67
30.65
8.09
0
344.26
17.87
22.75
0.12
118
257.29
24.16
36.35
2.58
15
391.74
20.91
33.9
0.05
60
364.96
17.88
30
0.12
122
330.2
25.71
20.35
4.49
67
240.4
WIP and MH Cost - Efficient Frontier
35
149.40
L2
L3
134.85
L4
25
L0
C1 (5,2)
L1
L5
15
20
MHD.Cost
30
138.29
10
C2 (1,10)
10
15
20
25
WIP
Figur e 10. WIP and material hand li ng cost as layou t selection crit eria
74
30
Final Layout
Cell 1
5
Cell 2
2
3
1
2
1
1
6
1
4
Cell 3
3
6
5
4
6
8
4
7
7
9
9
10
7
7
10
8
10
10
8
9
11
12
10
8
7
12
13
7
11
13
11
15
11
15
75
14
15