Advanced Layout Algorithms Chapter 11

Layout Algorithms Optimal Heuristic

Optimal Algorithms Branch and bound Decomposition – Benders’ decomposition Cutting Plane Algorithms

B&B Objective Function Value

Branch and bound Algorithm

Step 1: Computer lower bound (LB*) by solving a linear assignment problem (LAP) with a matrix [w ij ]. Matrix [w ij ] is obtained by taking dot product of two vectors [f i ] and [d j ]. Vector [f i ], [d j ] are obtained by removing f ii, d jj 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={s 1,s 2,...,s q } 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 [w ij ]. – Matrix [w ij ] 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 [f i ] and [d j ]. Vector [f i ], [d j ] 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

Explain Branch and Bound Algorithm with Example 1 OfficeSite O f S1-112 [f ij ]=f [d ij ]=i21-21 i312 -4t312-1 c41144-e4211- e Figure 7.2Flow 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 Minimizecx Subject toAx+By>b x > 0 y = 0 or 1 Now, consider a feasible y solution vector to MIP - say y i. Then, MIP becomes the following linear model. LP i Minimizecx Subject toAx>b - By i x> 0

Dual of Linear Program LP i Minimizecx Subject toAx>b - By i x> 0 The dual of LP i is the following model DLP i Maximizeu(b - By i ) Subject touA < c u > 0

Dual of Linear Program Let u i be the optimal solution to DLP i From duality theory, u i (b - By i ) is equal to the optimal OFV of LPi (because LP i and DLP i are both feasible) Hence, u i (b - By i ) is equal to the OFV of some feasible solution to MIP (the one in which y = y i ). Because each variable y ij 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 y ij variables, then the number of y vectors is equal to 2 n. Of course, not all of these may be feasible to MIP. We assume that there are s feasible y solution vectors to MIP - {y 1, y 2,..., y i,..., y s }, arranged in any order

Dual of Linear Program Let DLP 1, DLP 2,..., DLP i,..., DLP s be the duals obtained by substituting y 1, y 2,..., y i,..., y s for y i in DLP i. Let u 1, u 2,..., u i,..., u s 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{u i (b - By)}1 < i < s Subject to y = 0 or 1 and feasible to MIP

Master Problem Minimize{u i (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>u i (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 > u i (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, y i = {0,0,...,0}, lower bound LB=0 and upper bound UB=infinity. Step 1: Solve DLP i. Let u i be the optimal solution to DLP i. If u i (b - By i ) < UB, set UB = u i (b - Byi) Step 2: Update MP by adding the constraint z > u i (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, y i = y* and return to step 1.

Explain Benders’ decomposition algorithm with Example 2 Machine Dimension Horizontal Clearance Matrix Flow Matrix x x x x Figure 7.6 Flow and clearance matrices and dimensions for four machines

LMIP 1 for Example 2 Minimize Subject to

Example 2 MIP MIN 25 XP XP XP XP XP XP XN XN XN XN XN XN34 SUBJECT TO C1) 999 Y12 + X1 - X2 >= 33.5 C2) 999 Y12 + X1 - X2 <= C3) 999 Y13 + X1 - X3 >= 32.5 C4) 999 Y13 + X1 - X3 <= C5) 999 Y14 + X1 - X4 >= 37.5 C6) 999 Y14 + X1 - X4 <= C7) 999 Y23 + X2 - X3 >= 37.5 C8) 999 Y23 + X2 - X3 <= C9) 999 Y24 + X2 - X4 >= 40.5 C10) 999 Y24 + X2 - X4 <= 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 XP XP XP XP XP XP XN XN XN XN XN XN34 SUBJECT TO C1) X1 - X Y12 >= 33.5 C2) X1 - X Y12 <= C3) X1 - X Y13 >= 32.5 C4) X1 - X Y13 <= C5) X1 - X Y14 >= 37.5 C6) X1 - X Y14 <= C7) X2 - X Y23 >= 37.5 C8) X2 - X Y23 <= C9) X2 - X Y24 >= 40.5 C10) X2 - X Y24 <= C11) X3 - X Y34 >= 40 C12) X3 - X 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) VARIABLE VALUE REDUCED COST XP XP XP XP XP XP X X X

Example 2 (Cont) DLP 1 MAX 33.5 U V U V U V U V U V U 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) VARIABLE VALUE REDUCED COST U U U WN WN WN WN WN WN

Feasibility Constraints If an upper bound on z is U, then we can write U as 1 > y ij + y jk - y ik > 0 i<n-1, i<j<n, j<k<n

Example 2 (Cont) MP 1 MIN Z1 + 2 Z2 + 4 Z3 + 8 Z Z Z Z Z Z Z Z Z Z Z 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 Z Z Z Z Z Z Z Z Z Z Z Z Y Y Y34 >= END INTE 21 NEW INTEGER SOLUTION OF AT BRANCH 1 PIVOT 3 OBJECTIVE FUNCTION VALUE 1) VARIABLE VALUE REDUCED COST Y LAST INTEGER SOLUTION IS THE BEST FOUND

Solution Table

Dual for LMIP 1

Modified Benders’ decomposition algorithm Step 0: Set i=1, y i = {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 u i be the solution to DLP i. If u i (b - By i ) < UB, set UB = u i (b - By i ) Step 2: Update MP by adding the constraint z > u i (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, y i =y* and return to step 1.

Simulated Annealing Algorithm nnumber of departments in the layout problem Tinitial temperature rcooling factor ITEMPnumber of times temperature T is decreased NOVER maximum number of solutions evaluated at each temp NLIMITmax number of new solutions to be accepted at each temp δdifference 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 of departments permitted in any row, column respectively. Step 1:Repeat step 2 NOVER times or until the number of successful new solutions is equal to NLIMIT. Step 2: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. Step 3:Set T=rT and ITEMP=ITEMP+1. If ITEMP is < 100, go to Step 1; otherwise STOP.

Simulated Annealing Algorithm

Modified Penalty Algorithm Minimize c 11 x 11 + c 12 x c 3n x 3n Subject to a 11 x 11 + a 12 x a 1n x 1n > b 1 a 21 x 21 + a 22 x a 2n x 2n < b 2 a 31 x 31 + a 32 x a 3n x 3n = b 3 x 21, x 22,..., x 3n > 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 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. Step 3: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. 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 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-e δ/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 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. Step 3: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 LTM ij =LTM ij +1. Step 4: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: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 N gen =1. Step 1:Determine the fitness of each solution in the current generation’s population 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 N gen = N gen + 1. If N gen < G, go to step 1. Otherwise, STOP.

Fitness functions & Population Generation 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 {x 1, x 2, …, x n } and {y 1, y 2, …, y n }, 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: {x 1, x 2, …, x r-1, y r,, y r+1, …, y s, x s+1, x s+2, …. x n } {y 1, y 2, …, y r-1, x r,, x r+1, …, x s, y s+1, y s+2, …. y n }

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

Multicriteria Layout

Model for CMS Design Parameters: i, j, kpart, machine, cell indices, respectively c i intercellular movement cost per unit for part i v i number of units of part i u ij cost of part i not utilizing machine j o ij number of times each part i requires operation on machine j M max maximum number of machines permitted in a cell M min minimum number of machines permitted in a cell C u maximum number of cells permitted S 1 sets of machine pairs that cannot be located in the same cell S 2 sets of machine pairs that must be located in the same cell nptotal number of part types nmtotal number of machines

Model for CMS Design Decision Variables

Model 1 for CMS Design Minimize Subject to

Model 2 for CMS Design Minimize Subject to

Model P (Primal problem) Minimize Subject to

Model D (Dual problem) Minimize Subject to

Model M (Master problem) Minimize Z Subject to Z > 0

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 Prese nt 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 for current period Layout for next period Generate candidate layoutsEstimate performance measures of layouts Determine Layout to be usedRefine selected layout

Need for Stochastic Analysis

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 MPA Results

63 Example Problem Data

64 Current Cellular Layout (L0) Cell Cell Cell 2 x y Legend x Machine type y Machine label (number)

65 Functional Layout (L1)

66 Cell. Layout with reorientation and reshaping (L2) Cell Cell Cell 2

67 Cell. Layout with reorientation and reshaping (L3) Cell Cell Cell 3

68 Virtual Cellular Layout (L4) Cell 1 Cell 2 Cell 3

69 Cellular Layout with Remainder Cell (L5) Cell Cell Cell 3

70 Distance Matrix for Layout L1

71 WIP for 6 layouts

72 MH Cost and Lead Times

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

74 WIP and MH Cost - Efficient Frontier

75 Final Layout Cell Cell Cell 3