Facility DesignWeek 10 cont Computerized Layout Planning By

Facility Design-Week 10 (cont) Computerized Layout Planning By Anastasia L. Maukar 1

CRAFT- Computerized Relative Allocation of Facilities Technique • • • Created in 1964 by Buffa and Armour Process layout approach A heuristic computer program – Compares process departments – CRAFT requires an initial layout, which is improved by CRAFT. 2

From-To Chart • Determines which of the two departments has a better from-to matrix, we can calculate the moment of the matrix as follows: 3

CRAFT • Input for CRAFT: – – initial spatial array/layout flow data cost data Number and location of fixed department • Secara umum, dapat ditambahkan dummy yg berfungsi untuk: – – Fill building irregularities. Represent obstacles or unusable areas in facility Represent extra space in the facility Aid in evaluating aisle locations in the final layout. 4

CRAFT Following are some examples of questions addressed by CRAFT: • Is this a good layout? • If not, can it be improved? 5

CRAFT The possible interchange: – – – only pairwise interchage only three-way interchange pairwise interchage followed by three-way interchange – three-way interchage followed by pairwise interchange – best of two-way and three-way interchage 6

CRAFT: Distance Between Two Departments • Consider the problem of finding the distance between two adjacent departments, separated by a line only. • People needs walking to move from one department to another, even when the departments are adjacent. • An estimate of average walking required is obtained from the distance between centroids of two departments. • The distance between two departments is taken from the distance between their centroids. • People walks along some rectilinear paths. An Euclidean distance between two centroids is not a true representative of the walking required. The rectilinear distance is a better approximation. • So, Distance (A, B) = rectilinear distance between centroids of departments A and B 7

CRAFT: Distance Between Two Departments • Let – Centroid of Department A = – Centroid of Department B = • Then, the distance between departments A and B, Dist(A, B) • Ex: the distance between departments A and C is the rectilinear distance between their centroids (30, 75) and (80, 35). Distance (A, C) 8

Centroid of A = (30, 75) Centroid of C = (80, 35) Distance (A, C) = 90 10 20 30 40 50 60 70 80 90 100 CRAFT: Distance Between Two Departments A D (80, 85) C B (30, 25) 9 10 20 30 40 50 60 70 80 90 100

CRAFT: Total Distance Traveled • If the number of trips between two departments are very high, then such departments should be placed near to each other in order to minimize the total distance traveled. • Distance traveled from department A to B = Distance (A, B) Number of trips from department A to B • Total distance traveled is obtained by computing distance traveled between every pair of departments, and then summing up the results. • Given a layout, CRAFT first finds the total distance traveled. 10

CRAFT: Total Distance Traveled (a) Material handling trips(given) (b) Distances (given) 11

CRAFT: Total Distance Traveled (a) Material handling trips (given) (b) Distances (given) (c) Sample computation: distance traveled (A, B) = trips (A, B) dist (A, B) =………. . Total distance traveled = 100+630+240+…. = 4640 (b) (c) 12

CRAFT: Savings • CRAFT then attempts to improve the layout by pair-wise interchanges. • If some interchange results some savings in the total distance traveled, the interchange that saves the most (total distance traveled) is selected. • While searching for the most savings, exact savings are not computed. At the search stage, savings are computed assuming when departments are interchanged, centroids are interchanged too. This assumption does not give the exact savings, but approximate savings only. • Exact centroids are computed later. 13

CRAFT: Savings • Savings are computed for all feasible pairwise interchanges. Savings are not computed for the infeasible interchanges. • An interchange between two departments is feasible only if the departments have the same area or they share a common boundary. – Feasible pairs are {A, B}, {A, C}, {A, D}, {B, C}, {C, D} – and an infeasible pair is {B, D} • In this example savings are not computed for interchanging B and D. Savings are computed for each of the 5 other pair-wise interchanges and the best one chosen. • After the departments are interchanged, every exact centroid is found. This may require more computation if one or more shape is composed of rectangular pieces. 14

CRAFT: A Sample Computation of Savings from a Feasible Pairwise Interchange • To illustrate the computation of savings, we shall compute the savings from interchanging Departments C and D • New centroids: A (30, 75) Unchanged B (30, 25) Unchanged C (80, 85) Previous centroid of Department D D (80, 35) Previous centroid of Department C • Note: If C and D are interchanged, exact centroids are C(80, 65) and D(80, 15). So, the centroids C(80, 85) and D(80, 35) are not exact, but approximate. 15

CRAFT: A Sample Computation of Savings from a Feasible Pairwise Interchange • The first job in the computation of savings is to reconstruct the distance matrix that would result if the interchange was made. • The purpose of using approximate centroids will be clearer now. • If the exact centroids were used, we would have to recompute distances between every pair of departments that would include one or both of C and D. • However, since we assume that centroids of C and D will be interchanged, the new distance matrix can be obtained just by rearranging some rows and columns of the original distance matrix. 16

CRAFT: A Sample Computation of Savings from a Feasible Pairwise Interchange • The matrix on the left is the previous matrix, before interchange. The matrix on the right is after. • Dist (A, B) and (C, D) does not change. • New dist (A, C) = Previous dist (A, D) Interchange • New dist (A, D) = Previous dist (A, C) C, D • New dist (B, C) = Previous dist (B, D) • New dist (B, D) = Previous dist (A, C) 17

CRAFT: A Sample Computation of Savings (a) Material handling trips (given) (b) Distances (rearranged) (c) Sample computation: distance traveled (A, B) = trips (A, B) dist (A, B) = Total distance traveled = 100+420+360+… = 4480 Savings = (b) (c) 18

CRAFT: Improvement Procedure • To complete the exercise 1. Compute savings from all the feasible interchanges. If there is no (positive) savings, stop. 2. If any interchange gives some (positive) savings, choose the interchange that gives the maximum savings 3. If an interchange is chosen, then for every department find an exact centroid after the interchange is implemented 4. Repeat the above 3 steps as longs as Step 1 finds an interchange with some (positive) savings. 19

CRAFT: Exact Coordinates of Centroids • Sometimes, an interchange may result in a peculiar shape of a department; a shape that is composed of some rectangular pieces • For example, consider the layout (from example) and interchange departments A and D. The resulting picture is shown on the right. • How to compute the exact coordinate of the centroid (of a shape like A)? 20

50 60 70 80 90 100 CRAFT: Exact Coordinates of Centroids A A 1 A 2 10 20 30 40 50 60 70 80 90 100 Find the centroid of A 21

CRAFT: Exact Coordinates of Centroids Rectangle (1) A 1 A 2 Total Area (2) X-coordinate Multiply of centroid (2) and (3) (4) X-coordinate of the centroid of A 22

CRAFT: Exact Coordinates of Centroids Rectangle (1) A 1 A 2 Total Area (2) Y-coordinate Multiply of centroid (2) and (3) (4) Y-coordinate of the centroid of A 23

50 60 70 80 90 100 CRAFT: Exact Coordinates of Centroids A A 1 A 2 10 20 30 40 50 60 70 80 90 100 Exact coordinate of area A is 24

CRAFT: Some Comments • An improvement procedure, not a construction procedure • At every stage some pairwise interchanges are considered and the best one is chosen • Interchanges are only feasible if departments have the same area; or they share a common boundary • Departments of unequal size that are not adjacent are not considered for interchange • Estimated cost reduction may not be obtained after interchange (because the savings are based on approximate centroids) • Strangely shaped departments may be formed 25

Computerized Layout Planning Graphical Representation “Points and lines” representation is not convenient for analysis 26

Layout Evaluation – An Algorithm needs to distinguish between “good” layouts and “bad” ones – Develop scoring model, s = g (X ) – Adjacency-based scoring (Komsuluk Bazli Skorlama) • Based on the relationship chart and diagram • Xi is the number of times an adjacency i is satisfied, i=A, E, I, O, • • U, X Aldep uses (wi values) A=64, E=16, I=4, O=1, U=0, and X=-1024 Scoring model has intuitive appeal; the ranking of layouts is sensitive to the weight values. Layout “B” may be preferred to “C” with certain weights but not with others. Therefore, the specification of the weights is very important. The weights wi can also be represented by the flow amounts 27 between the adjacent departments instead of scores assigned to A,

1 Exampl e. Receiving 1 2 E Milling U Press U Screw Machine Assembly I O E U U A Plating 4 I I 6 O I O U U 4 I E Shipping 3 5 6 U U U 7 1 2 3 5 7 2 1 2 3 4 5 6 7 3 4 5 I E O I 6 U O A E 7 4+1 =5 16+4+0 =20 U 1+0 ---64 16 Total Score U 3 Press O 7 U 2 E Milling Screw Machine E I I A 5 Assembly O =64 =16 106 4 Shipping 6 Plating =1 1 Receiving 28

1 Receiving Milling Press Screw Machine Assembly Plating 2 E U U I O E U U A 4 I 6 O I 5 U U 4 6 U I O I E Shipping 3 U U 2 7 1 Exercise: Find the score of the layout shown below. Use A=8, E=4, I=2, O=1, U=0 and X=-8. 3 5 7 3 Press 7 Shipping 1 Receiving 6 Plating 2 Milling 4 Screw Machine 5 Assembly 29

Layout Evaluation (cont’d) – Distance-based scoring (Mesafe Bazli Skorlama) • Approximate the cost of flow between activities • Requires explicit evaluation of the flow volumes and costs • cij covers both the i to j and the j to i material flows • Dij can be determined with any appropriate distance metric – Often the rectilinear distance between department centroids • Assumes that the material flow system has already been specified (cij=flow required* cost /flow-distance) • Assumes that the variable flow cost is proportional to distance • Distance often depends on the aisle layout and material handling equipment 30

CRAFT - Example 2 Initial Layout Ma Flow Data Total Cost Distance Data 31

CRAFT - Example 2 • • • A & D interchange Total cost = 1. 095 A (60, 10) dan D (25, 30) A & C interchange Total cost = 99 C & D interchange Total cost = 1. 040 B & D interchange Total cost = 945 estimated total cost • B & C tidak dapat dipertukarkan 32

CRAFT - Example 2 • Yang dipilih adalah pertukaran B & D menghasilkan layout baru dg department centroid, sesuai dengan luas yang diinginkan pada layout awal – – (XA, YA) = (25, 30) (XB, YB) = (55, 10) (XC, YC) = (20, 10) (XD, YD) = (67. 5, 25) 33

CRAFT - Example 2 To From A B C D 50 25 47. 5 35 27. 5 40’ A B 30’ 20’ 10’ 50 20’ C 25 35 D 47. 5 27. 5 40’ 60’ 80’ 62. 5 34

Layout Evaluation – Distance-Based Scoring – Distance-based scoring – Impact of aisle layout and direction of travel 35

Benefits and Problems • Benefit – A Computer Program – Flexibilty • Problems – Greedy Algorithm – Inefficient – End result may need to be modified 36

Summary It is beneficial to use CRAFT but you should also realize that the program is not flawless. The user must understand how the program works of the end product is not as efficient as you had hope. 37