advanced layout algorithms chapter 8. layout algorithms optimal heuristic
TRANSCRIPT
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
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 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
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
– 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
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 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
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 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
Simulated Annealing Algorithm
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.
Simulated Annealing Algorithm
7 8 9
4 5 6
1 2 3
Simulated Annealing Algorithm
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}
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
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