FACILITIES LAYOUT AND LOCATION
Outline
• The Problem
• Objective of Facility Layout • Basic Types of Layout
• Product versus Process Layout • Cellular Layouts
• Proximity
The Problem
• In this lesson, we shall discuss how a plant or workplace should be laid out.
• Consider the problem of finding suitable locations for machines, workstations, storage areas and aisles within a plant.
• How to find suitable locations for departments, lounges and mail rooms and labs within a building that houses a faculty.
Objectives of Facility Layout
• A facility layout problem may have many objectives. In the context of manufacturing plants, minimizing material handling costs is the most common one.
• Other objectives include efficient utilization of – space
– labor • Eliminate
– bottlenecks
• Facilitate
– organization structure
– communication and interaction between workers – manufacturing process
– visual control • Minimize
– manufacturing cycle time or customer flow time – investment
• Provide
– convenience, safety and comfort of the employees – flexibility to adapt to changing conditions
Basic Types of Layouts
• Process Layout
– Used in a job shop for a low volume, customized products
• Product Layout
Basic Types of Layouts
• Fixed Position Layout
– Used in projects for large products e.g., airplanes, ships and rockets
• Cellular layouts
– A cell contains a group of machines dedicated for a group of similar parts
Product vs. Process Layouts
• A process layout is a functional grouping of machines. For example, a group of lathe machines are arranged in one area, drill machines in another area, grinding machines in another area and so on. Different job jumps from one area to another differently. Hence, the flow of jobs is difficult to perceive. This type of layout is suitable for a make-to-order or an assemble-to-order production environment, as in a job shop where customization is high, demand fluctuates, and volume of production low.
Product vs. Process Layouts
• A product layout arrangement of machines. Every job visits the machines in the same order. This type of layout is suitable for a make-to-stock or an assemble-to-stock production environment, as in a flow shop
where products are standard, demand stable, and volume of production high. Since variety is low, special purpose equipments and workers with a limited skill are needed.
• Advantage
2
Product vs. Process Layouts
• Inventory
• A product layout has a low work-in-process
inventory and high finished goods inventory because production is initiated by demand forecast.
• On the other hand, a process layout has a high work-in-process inventory and low finished goods inventory.
• Material handling
• A product layout can use automatic guided vehicles which travels in a fixed path. But, variable path
Product vs. Process Layouts
• Scheduling/line balancing
• In case of a process layout, jobs arrive throughout the planning period. A process layout requires
dynamic scheduling where a new scheduling decisions is made whenever a new job arrives.
• In case of a product layout, sequencing and timing of product flow are standard and set when the line is designed. With a change in demand, a product
• Every cell contains a group of machines which are dedicated to the production of a family of parts.
• One of the problems is to identify a family parts that require the same group of machines.
• These layouts are also called as group technology layouts.
Cellular Layouts
Enter
Worker 1
Worker 2
Worker 3
Exit
Key: Product route Worker route
12
1
2
3 4
5
6 7
8
9
10
11
A B C Raw materials Assembly
Cellular Layouts Example
• The previous slide shows a facility in which three parts A, B, C flow through the machines.
• The next slide provides the information in a matrix form which includes some other parts D, E, F, G, H.
• The rows correspond to the parts and columns to the machines.
• Just by interchanging rows and columns, eventually a matrix is obtained where the “X” marks are all
Parts 1 2 3 4 5 6 7 8 9 10 11 12
A
x x
x
x
x
B
x
x
x x
C
x
x
x
D
x x
x
x
x
E
x x
x
F
x
x
x
G
x
x
x
x
H
x
x x
Machines
Parts 1 2 4 3 5 6 7 8 9 10 11 12
A
x x x
x
x
B
x
x
x x
C
x
x
x
D
x x x
x
x
E
x x
x
F
x
x
x
G
x
x
x
x
H
x
x x
Machines
Parts 1 2 4 3 5 6 7 8 9 10 11 12
A
x x x
x
x
D
x x x
x
x
B
x
x
x x
C
x
x
x
E
x x
x
F
x
x
x
G
x
x
x
x
H
x
x x
Machines
Parts 1 2 4 8 3 5 6 7 9 10 11 12
A
x x x x
x
D
x x x x
x
B
x
x
x x
C
x
x
x
E
x x
x
F
x
x x
G
x
x
x
x
H
x
x x
Machines
Parts 1 2 4 8 3 5 6 7 9 10 11 12
A
x x x x
x
D
x x x x
x
F
x
x x
B
x
x
x x
C
x
x
x
E
x x
x
G
x
x
x
x
H
x
x x
Machines
Parts 1 2 4 8 10 3 5 6 7 9 11 12
A
x x x x x
D
x x x x x
F
x
x x
B
x
x
x x
C
x
x
x
E
x x
x
G
x
x
x
x
H
x
x x
Machines
Parts 1 2 4 8 10 3 6 9 5 7 11 12
A
x x x x x
D
x x x x x
F
x
x x
C
x x x
G
x x x
x
B
x x x x
E
x
x
x
H
x x x
Machines
12
1
2 3
4
5 6
7
8 10 9
11
A C B
Raw materials
Cell1 Cell 2 Cell 3 Assembly
Cellular Layouts Example
Advantages of Cellular Layouts
• Reduced material handling and transit time
• Reduced setup time
• Reduced work-in-process inventory
• Better use of human resources
Disadvantages of Cellular Layouts
• Sometimes cells may not be formed because of inadequate part families.
• Some cells may have a high volume of
Disadvantages of Cellular Layouts
• When volume of production changes, number of workers are adjusted and workers are
reassigned to various cells. To cope with this type of reassignments, workers must be multi-skilled and cross-trained.
Activity Relationship Chart
• An activity relationship chart is a graphical tool used to represent importance of locating pairs of operations near each other.
• Importance is described using letter codes defined below:
A- absolutely necessary E- especially important I - important
O - ordinarily important U- unimportant
Production area
Activity Relationship Chart
Example: It’s ordinarily important to locate
office rooms near
Activity Relationship Chart
• Sample interpretation of the diagram on the previous slide:
• To find how important it is to locate office rooms near loading/unloading area, find the diamond shaped block at the intersection of office rooms and loading/unloading area. The block contains “O” meaning ordinarily
important. Therefore, it’s ordinarily important to locate office rooms near loading/unloading
Analisa hubungan aktivitas
Activity Relationship Diagram
Contoh
Perusahaan alat rumah tangga membuat rencana tata
letak fasilitas yang baru dengan kebutuhan ruang
sbb:
• Departemen A = 850 m
2• Departemen B = 1750 m
2• Departemen C = 850 m
2• Departemen D = 850 m
2• Departemen E = 500 m
2• Departemen F = 850 m
2• Departemen G = 650 m
2penyelesaian
Dept C Dept G
Dept F Dept B Dept E
Dept H Dept A Dept D
From-To Chart
Punch
Saws Milling Press Drills
Saws 18 40 30
Milling 18 38 75 Punch Press 40 38 22 Drills 30 75 22
• A from-to chart is used to analyze flow of
materials between departments. The example below shows distances in feet. So, the distance between Saws and Drills is 30 feet. The chart may also show material handling trips or cost per
Assignment Method
• Many methods can be used to solve the facility layout
problem. Here we discuss assignment method to minimize material handling costs.
• Suppose that some machines 1, 2, 3, 4 are required to be located in A, B, C, D. The cost of locating machines to
locations are known and shown below. For example, if Machine 2 is located to location C, the cost is 7 (say, hundred dollars per month).
Location
Machine A B C D
Assignment Method
• The problem is to locate the machines to minimize total material handling costs.
• One solution can be (not necessarily and optimal solution) to assign 1, 2, 3, 4 to respectively C, B, A, D. In such a
case total cost is 6+4+8+12=30 hundred dollars per month.
Location
Assignment Method
• Notice in this solution that every machine is assigned to one location and every location is assigned to one
machine. So, there is a single box in each row and each column. Every solution will must this property.
• If there are more locations than machines, dummy machines must be added with the same cost for all
locations. Assignment method finds an optimal solution.
Location
Assignment Method
1. Perform row reductions
– Subtract minimum value in each row from all other row values
2. Perform column reductions
– Subtract minimum value in each column from all other column values
3. Line Test
– Cross out all zeros in matrix using minimum number of
horizontal & vertical lines. If number of lines equals number of rows in matrix, optimum solution has been found, stop. 4. Matrix Modification
Assignment Example
Row reduction Column reduction Line Test
4 1 0 5 2 1 0 4 2 1 0 4 2 0 3 5 0 0 3 4 0 0 3 4 3 1 0 1 1 1 0 0 1 1 0 0 6 2 0 9 4 2 0 8 4 2 0 8
Number lines = 3 <> 4 = number of rows. So, modify matrix Location
Assignment Example
• More on the previous slide:
• From rows 1, 2, 3, 4 subtract respectively 6, 4, 5, 3 which are minimum numbers on the rows. The results are shown under row reduction.
• Next from columns A, B, C, D subtract respectively 2, 0, 0, 1 which are minimum numbers on the columns. The results are shown under column reduction.
Assignment Example
• A zero assignment is an assignment solution with
exactly one zero from each row and exactly one zero from each column. After we decide to stop computation, we find a zero assignment.
• As long as the minimum number of lines is less than the number of rows, it’s not possible to find a zero
assignment. If the minimum number of lines equals the number of rows, then there exists a zero assignment. • If we mistakenly decided that the minimum number of
Assignment Example
• See the matrix under line test on Slide 34. The minimum uncovered number is 1. There are three types of
numbers and these numbers are modified in three different ways:
– uncovered numbers: subtract minimum uncovered number 1 from all uncovered numbers.
– numbers covered by one line: do nothing
– numbers covered by two lines: add minimum
uncovered number 1 to all numbers covered by two lines.
Location
Machines AB CD 1 10 04
so at optimal solution
Location
Total material handling costs = 22
Assignment Example
• Explanation on the previous slide:
• Another line test is done on the modified matrix. It’s observed that the minimum number of lines = 4 = number of rows. So, the process stops.
• Next, a zero assignment is found. See one box on each row and one box one each column. The boxes denote the optimal assignment.
• So, locate machine 1 to B, machine 2 to A, machine 3 to D and machine 4 to C. To find the corresponding we
READING AND EXERCISES
Lesson 20
Reading:
– Section 10.1-10.4 pp. 557-573 (4th Ed.), pp. 535-552
(5th Ed.)
Exercises:
– 10.1 p. 568, 10.7, p. 573 (4th ED.)
LESSON 21: LOCATING A SINGLE FACILITY
THE RECTILINEAR DISTANCE PROBLEM
Outline
• Locating New Facilities
Locating New Facilities
• In Lessons 15-16, we consider the problem of locating new facilities. For example, consider locating
1. a facility used by many people: a hospital, a
gymnasium, computer center, student center, etc.
Locating New Facilities
• Distance is an important consideration in each of these location problems. It’s desirable to locate a facility that’s not too far from the users. Following are the issues:
• How the distance will be measured
• What importance (weight) will be assigned to various users
• Whether the location will be selected on the basis of total (weighted) distance or the maximum
Locating New Facilities
• First, there are two important types of distance
measures. People walks or drives along the streets, possibly by changing directions several times but airplanes and radio signals travel along a straight line without any change of directions. Utility cable are often laid out without changing directions.
• For the first type of cases, when directions are
changed more often, rectilinear distance measure is more appropriate. On the other hand for the latter
Locating New Facilities
• Suppose L(5,5) is the location of the new facility and A(2,1) is location of an user. • Between A and L, the
• Notations:
• X-coordinate of user location i = ai
• Y-coordinate of user location i = bi
• Weight assigned to user location i = wi
• Location of the proposed facility = (x,y)
• For the picture shown on the previous slide, there is only one user location. So, index i can be omitted. We have,
a=2, b=1, x=5, y=5
• Rectilinear distance
= distance along the x-axis +distance along the y-axis = |x-a|+|y-b|
=
• Euclidean distance
Locating New Facilities
Locating New Facilities
• Often different user locations may be assigned a different weight to reflect their relative importance. For the problem of locating a computer center, the weights may be the number potential users. For the problem of locating an airline hub, the weights may be the number of flights per week.
• The weights are important because the new facility may be located by minimizing total weighted
Locating New Facilities
• Sometimes, weights are not required because it may be more important to minimize the maximum
Locating New Facilities
• In summary, there are 4 important types of problems • Rectilinear distance
• Minimize weighted total (current lesson) • Minimize maximum (current lesson)
• Euclidean distance
• Minimize weighted total (next lesson) • Minimize maximum (not discussed) • Slide 2 gives application of each type of
• The new health-watch facility is targeted to serve five census tracts in Erie, Pennsylvania. Coordinates for the center of each census tract, along with the
projected populations, measured in thousands are shown next. Customers will travel from the five
census tract centers to the new facility when they need health care.
Example
Census Tract (ai,bi) Population, wi
A (2.5,4.5) 2 B (3.0,2.5) 5 C (5.5,4.0) 10 D (5.0,2.0) 7 E (8.0,5.0) 12
Example
• Firs, we shall discuss the problem of finding a facility location by minimizing the weighted sum of the
rectilinear distances from the new facility to all users. • Application: locating facilities used by many people. • First, the procedure will be discussed. Then, the
procedure will be illustrated with an example.
Minimize Weighted Sum of the
Rectilinear Distances
• Find the optimal value of x as follows:
– Arrange the x-coordinates in ascending order – Compute the cumulative weights
Minimize Weighted Sum of the
Rectilinear Distances
• Find the optimal value of y similarly:
– Arrange the y-coordinates in ascending order – Compute the cumulative weights
Census Cumulative Tract (ai,bi) Weight, wi Weight
A (2.5,4.5)
B (3.0,2.5) D (5.0,2.0) C (5.5,4.0) E (8.0,5.0)
Census Cumulative Tract (ai,bi) Weight, wi Weight
D (5.0,2.0) B (3.0,2.5) C (5.5,4.0) A (2.5,4.5) E (8.0,5.0)
Example: Minimize Weighted Sum of
Rectilinear Distances
• The optimal location (x,y) is (5.5,4.0). In this case, both x and y are taken from the same location C. It may also happen that optimal x corresponds to one location and optimal y corresponds to some other. • To get the optimal value, compute the weighted sum
Example: Minimize Weighted Sum of
Rectilinear Distances
Minimize Maximum
Rectilinear Distance
• Next, we shall discuss the problem of finding a facility location by minimizing the maximum
rectilinear distance from the new facility to all users. • Application: locating emergency facilities.
• The procedure is mechanical. First, the procedure will be discussed. Then, the procedure will be
• Define:
• All points along the line connecting (x1,y1) and (x2,y2)
are optimal
Example: Minimize Maximum
Rectilinear Distance
First, compute c1, c2, c3, c4, c5:
Tract (ai,bi) ai+bi -ai+bi
A (2.5,4.5)
B (3.0,2.5) D (5.0,2.0) C (5.5,4.0)
E (8.0,5.0)
Min c1= c3= Max c2= c4=
max(
2 1,
4 3)
5
c
c
c
c
• All points along the line connecting ____________ and ____________ are optimal.
Example: Minimize Maximum
Rectilinear Distance
• Notice that instead of getting one optimal location we obtained infinite number of optimal locations I.e., any point along the lines connecting (4.25,5.0) and
(5.5,3.75).
maxi 4.25 ai 5.0 bi
(4.25,5.0) location
optimal From
distance r
rectilinea Maximum
Example: Minimize Maximum
Rectilinear Distance
READING AND EXERCISES
Lesson 21
Reading:
– Section 10.8-10.9 pp. 598-606 (4th Ed.), pp. 575-584
(5th Ed.)
Exercises:
– 10.25 p. 600, 10.32, 10.38, pp. 608-609 (4th Ed.)
LESSON 22: LOCATING A SINGLE FACILITY
THE EUCLIDEAN DISTANCE PROBLEM
Outline
• Minimize Weighted Sum of the Squares of the Euclidean Distances (Gravity Problem)
The Euclidean Distance Problem
• Lesson 15 discusses the rectilinear distance
problems. In this lesson, we shall discuss Euclidean distance problems with the following two objectives
– Minimize weighted sum of the squares of the Euclidean distances a.k.a the gravity problem (approximation to the other objective)
• Notations:
– X-coordinate of existing facility i = ai
– Y-coordinate of existing facility i = bi
– Weight assigned to user location i = wi
– Location of the proposed facility = (x,y)
– Euclidean distance =
(
x
a
i)
2
(
y
b
i)
2Example
The Euclidean Distance Problem
• The new health-watch facility is targeted to serve five census tracts in Erie, Pennsylvania. Coordinates for the center of each census tract, along with the
projected populations, measured in thousands are shown next. Customers will travel from the five
Census Tract (ai,bi) Population, wi
A (2.5,4.5) 2 B (3.0,2.5) 5 C (5.5,4.0) 10 D (5.0,2.0) 7 E (8.0,5.0) 12
Example
Gravity Problem
Minimize Weighted Sum of the Squares of the
Euclidean Distances
• The problem of minimizing weighted sum of the Euclidean distances is difficult and solved by an
iterative procedure. The initial solution of the iterative procedure is obtained from the solution of the gravity problem which minimizes weighted sum of the
Example: Gravity Problem
Census
Weight
Tract (
i
) (
a
i,
b
i)
w
iw
ia
iw
ib
i1 (2.5,4.5)
2
2 (3.0,2.5)
5
4 (5.5,4.0)
10
3 (5.0,2.0)
7
5 (8.0,5.0)
12
Conclusion: The location ___________________ minimizes the weighted sum of the squares of the Euclidean distances.
Example: Gravity Problem
Euclidean distance =
Square of the Euclidean distance =
Example: Gravity Problem
Use the following procedure to find a location that
minimizes weighted sum of the Euclidean distances. The procedure is iterative, starts with a trial solution and converges to an optimal solution.
Step 1: Consider a trial location (x,y). The solution to the gravity problem is a good trial solution.
Step 2: For each location (ai,bi) compute
Step 3: Modify the x and y values as follows:
Step 4: If one or both of (x,y) changes, repeat the process with the modified (x,y). Go to step 2 with modified (x,y). If none of (x,y) changes, stop.
Note: This procedure is better done by spreadsheet. It’s easy to set up one.One is also available from the course web site.
Step 1: Consider the trial solution (x,y) = (5.72, 3.76) that is obtained from the gravity problem.
Step 2:
x i 5.72
y i 3.76
a i b i wi g i
2.5 4.5 2 0.61
3 2.5 5 1.67
5 2 7 3.68
5.5 4 10 30.71
8 5 12 4.62
Trial location
Example: Minimize Weighted Sum of the
Euclidean Distances
Sample computation for the previous slide:
Step 3:
Modified x = (230.84/41.29) = 5.59
y = (160.23/41.29) = 3.88
x i 5.72
y i 3.76
a i b i wi g i a i g i b i g i
2.5 4.5 2 0.61 1.51 2.72
3 2.5 5 1.67 5.00 4.17
5 2 7 3.68 18.41 7.36
5.5 4 10 30.71 168.93 122.86
8 5 12 4.62 36.99 23.12
41.29 230.84 160.23 Trial location
Example: Minimize Weighted Sum of the
Euclidean Distances
Sample computation for the previous slide:
• Repeat the process with (x,y) = (5.59, 3.88). The result is (x,y) = (5.54, 3.94).
• Repeat the process with (x,y) = (5.54, 3.94). The result is (x,y) = (5.52, 3.97).
• Repeat the process with (x,y) = (5.52, 3.97). The result is (x,y) = (5.51, 3.98).
• Repeat the process with (x,y) = (5.51, 3.98). The result is (x,y) = (5.51, 3.99).
• Repeat the process with (x,y) = (5.51, 3.99). The result is (x,y) = (5.50, 3.99).
• Repeat the process with (x,y) = (5.50, 3.99). The result is (x,y) = (5.50, 3.99).
• Stop.
Example: Minimize Weighted Sum of the
Euclidean Distances
Application
1. Minimize weighted sum of the rectilinear distances (done) – Facilities used by many people e.g., computer centre,
gymnasium
2. Minimize maximum rectilinear distance (done) – Emergency facilities e.g., police, fire
3. Minimize weighted sum of the Euclidean distances (done) – Utilities, e.g., phone cable
4. Minimize maximum Euclidean distance (not in book) – Transmission towers e.g., radio towers
5. Minimize weighted sum of the squares of the Euclidean distances (done)
READING AND EXERCISES
Lesson 22
Reading:
– Section 10.10 pp. 609-612 (4th Ed.), pp. 586-588 (5th
Ed.)
Exercises:
LESSON 23/24:
COMPUTERIZED LAYOUT TECHNIQUE
Outline
• Computerized Layout Technique
– A Layout Improvement Procedure, CRAFT • Distance Between Two Departments • Total Distance Traveled
• Savings and a Sample Computation • Improvement Procedure
• Exact Centroids
• Suppose that we are given some space for some
departments. How shall we arrange the departments within the given space?
• We shall assume that the given space is rectangular shaped and every department is either rectangular shaped or composed of rectangular pieces.
• We shall discuss
– a layout improvement procedure, CRAFT, that attempts to find a better layout by pair-wise interchanges when a layout is given and
– a layout construction procedure, ALDEP, that
constructs a layout when there is no layout given.
CRAFT - Computerized Relative Allocation of
Facilities Technique
Following are some
examples of questions
• 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.
• Centroid of a rectangle is the point where two diagonals meet. So, if a rectangle has two opposite corners and then the centroid is
• See Slide 7 for an example of finding centroid.
CRAFT: Distance Between Two Departments
• Finding centroid of a shape composed of rectangular pieces involves more computation and discussed on Slides 21-25.
• 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
• Let
– Centroid of Department A = – Centroid of Department B =
• Then, the distance between departments A and B, Dist(A,B)
• The distance formula is illustrated with an example on the next slide. The distance between departments A and C is the rectilinear distance between their
centroids (30,75) and (80,35). Distance (A,C)
90
Centroid of A =
Centroid of C =
• 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.
• The next 3 slides illustrates finding total distance traveled.
A B C D
A 2 7 4
B 3 5 7
C 6 7 3
D 7 7 3
F r o m T o
CRAFT: Total
Distance Traveled
(a) Material handling trips (given)
A B C D
(a) Material handling trips (given)
(a)
(b) (b) Distances (given)
A B C D
A B C D
(a) Material handling trips (given)
(a)
(b)
(c) (b) Distances (given)
Total distance traveled = 100+630+240+….
= 4640
(c) Sample computation: distance traveled (A,B) = trips (A,B) dist (A,B)
=
• As stated before, given a layout CRAFT first finds the total distance traveled as illustrated on the previous 3 slides. 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.
• 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. For the layout shown on Slide 7:
– feasible pairs are {A,B}, {A,C}, {A,D}, {B,C}, {C,D} – and an infeasible pair is {B,D}
• For the layout shown on Slide 7, 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. See Slides 21-25.
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
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.
CRAFT: A Sample Computation of Savings
from a Feasible Pairwise Interchange
• The matrix on the left is the previous matrix, before
CRAFT: A Sample
Computation of
Savings
A B C D
A 2 7 4
B 3 5 7
C 6 7 3
D 7 7 3
F r o m T o
(a)
CRAFT: A Sample
Computation of
Savings
(a) Material handling trips (given)
CRAFT: A Sample
Computation of
Savings
(a) Material handling trips (given)
(b) Distances (rearranged)
Total distance traveled = 100+420+360+…
= 4480
Savings =
(c) Sample computation: distance traveled (A,B) = trips (A,B) dist (A,B)
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
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 on Slide 7 and interchange
departments A and D. The
resulting picture is shown on the right.
• How to compute the exact
CRAFT: Exact Coordinates of Centroids
Find the centroid of A
X-coordinate Multiply Rectangle Area of centroid (2) and (3) (1) (2) (3) (4)
A1 A2 Total
X-coordinate of the centroid of A
CRAFT: Exact Coordinates of Centroids
Y-coordinate Multiply Rectangle Area of centroid (2) and (3) (1) (2) (3) (4)
A1 A2 Total
Y-coordinate of the centroid of A
CRAFT: Exact Coordinates of Centroids
CRAFT: Exact Coordinates of Centroids
Exact coordinate of area A is
10 20 30 40 50 60 70 80 90 100
50
60
70
80
90
10
0
A A1
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)
ALDEP
Automated Layout Design Program
• While CRAFT is an improvement procedure, ALDEP is a construction procedure.
• CRAFT requires an initial layout, which is improved by CRAFT. ALDEP does not need any initial layout. ALDEP constructs a layout when there is none.
• Previously, in Lesson 13, we have discussed a
construction procedure and an improvement procedure in the context of vehicle scheduling. The nearest
ALDEP
• Given
– Size of the facility – The departments
– Size of the departments
– Proximity relationships (activity relationship chart) and – A sweep width (defined later)
ALDEP
• The size of the facility and the size of the departments are expressed in terms of blocks.
• The procedure will be explained with an example. Suppose that the facility is 8 blocks (horizontal) 6 block (vertical).
• The departments and the required number of blocks are: – Production area 14 blocks
ALDEP
A: absolutely necessary E: especially important I: important
O: ordinarily important U: unimportant
X: undesirable Production area
ALDEP
• The proximity relationships are shown on the previous slide.
• ALDEP starts to allocate the departments from the upper left corner of the facility. The first department is chosen at random. By starting with a different department, ALDEP can find a different layout for the same problem.
• Let’s start with dock rooms (D). On the upper left corner 8 blocks must be allocated for the dock area.
• The sweep width defines the width in number of blocks. Let sweep
ALDEP
• To find the next department to allocate, find the
department that has the highest proximity rating with the dock area as given on Slide 30. Storage area (S) has the highest proximity rating A with the dock area.
• So, the storage area will be allocated next. The storage area also needs 8 blocks.
• There are only 2 2 = 4 blocks,
remaining below dock area (D). After allocating 4 blocks, the down wall is hit after which further
allocation will be made on the adjacent 2 (=sweep width)
columns and moving upwards.
ALDEP
• See carefully that the allocation started from the upper left corner and started to move downward with an width of 2 (=sweep width) blocks.
• After the down wall is hit, the allocation continues on the adjacent 2 (=sweep width) columns on the right side and starts moving up.
• This zig-zag pattern will continue. • Next time, when the top wall will
be hit, the allocation will continue on the adjacent 2 (=sweep width) columns on the right side and
starts moving down.
ALDEP
• To find the next department to allocate, find the
department that has the highest proximity rating with storage area as given on Slide 30.
• Production area (P) has the highest proximity rating A with the storage area.
• The production area needs 14 blocks. • After allocating 8 blocks, the top
wall is hit and the remaining 6 blocks are allocated on the adjacent 2 (=sweep width) columns moving downward.
ALDEP
• To find the next department to allocate, find the
department that has the highest proximity rating with production area as given on Slide 30.
• Tool room (T) has the highest proximity rating A with the production area.
• The tool room needs 4 blocks. So, 4 blocks are allocated. • Next, there is a tie. See from Slide
30 that both locker room (L) and office room (O) have the proximity rating of U with the tool room.
• Ties are broken at random. So,
any of the locker room or the office room can be allocated next.
ALDEP
• Let’s choose locker room (L) room at random. Then, the last department must be office room (O). The resulting layout is shown below.
• Note that since the ALDEP chooses the first department at random and since the ties are broken at random,
ALDEP can give many solutions to the same problem. • Using the layout, the adjacency
relationships and the proximity ratings, we can find an overall rating of each layout. Then, the layout with the highest overall
ALDEP
• After a layout is obtained, a score for the layout is computed with the following conversion of proximity relationships:
A = 43 = 64, E = 42 = 16
I = 41 = 4, O = 40 = 1
U = 0, X = -45 = -1024
ALDEP
• Let’s compute the overall rating of the layout constructed. To do this, we shall list every pair of adjacent
departments. For each pair, a letter rating will be obtained from the activity relationship chart (a.k.a. rel chart) on
Slide 30 and then the score will be converted to a numeric score using the conversion scheme on the previous slide.
D D P P P P O O • Adjacent departments:
ALDEP
Adjacent Proximity Numeric
Departments Ratings (Slide 30) Scores (Slide 37)
(D,S) A 43=64
(D,P) I 41=4
(S,P) A 43=64
(S,T) O 40=1
(S,L) U 0
(P,T) A 43=64
(P,O) O 40=64
(T,L) U 0
(T,O) U 0
(L,O) X -45= -1024
• The process is repeated several times and the layout with the highest score is chosen.
• Notice the large negative weight associated with X ratings.
• If the departments which cannot be next to each other, are adjacent in a layout, then the layout score reduces significantly.
• This is important because ALDEP also uses a cut-off score (if not specified by the user this cut-off is zero) to eliminate any layout which has a layout score less than the cut-off score.
READING AND EXERCISES
Lesson 23/24 Reading:
– Section 10.6 pp. 575-581 (CRAFT), 581-582 (ALDEP) (4th Ed.), pp. 555-559 (CRAFT), 560-561 (ALDEP) (5th
Ed.)
– Appendix 10-A pp. 626-628 (4th Ed.), pp. 602-604 (5th
Ed.)
Exercises:
– 10.13, 10.15, 10.19, pp. 586-587 (4th Ed.), pp.
