Facilities Management

FACILITIES LAYOUT AND LOCATION

Outline

• The Problem• Objective of Facility Layout• Basic Types of Layout• Product versus Process Layout• Cellular Layouts• Proximity• Assignment Problem

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.

• The layout problem may also occur in other places such as grocery stores, hospitals, airports, etc.

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– waste or redundant movement

• 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

Objectives of Facility Layout

Basic Types of Layouts

• Process Layout– Used in a job shop for a low volume, customized products

• Product Layout– Used in a flow shop for a high volume, standard products

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– Suitable for producing a wide variety parts in moderate volume

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. Since a wide variety of products are produced, general purpose equipments and workers with varied skills are needed.


• 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• A process layout provides flexibility • A product layout provides efficiency.


• 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 forklifts are suitable for a process layout.


• 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 layout may need a new line balancing decision.

• 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

Cellular Layouts

Enter

Worker 1

Worker 2Worker

3

Exit

Key: Product routeWorker route

Machines

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 concentrated near the diagonal. This matrix provides the cells. For example, parts A, D and F require Machines 1, 2, 4, 8 and 10 which forms a cell.

Parts 1 2 3 4 5 6 7 8 9 10 11 12A x x x x xB x x x xC x x xD x x x x xE x x xF x x xG x x x xH x x x

Machines


Parts 1 2 4 3 5 6 7 8 9 10 11 12A x x x x xB x x x xC x x xD x x x x xE x x xF x x xG x x x xH x x x

Machines


Parts 1 2 4 3 5 6 7 8 9 10 11 12A x x x x xD x x x x xB x x x xC x x xE x x xF x x xG x x x xH x x x

Machines


Parts 1 2 4 8 3 5 6 7 9 10 11 12A x x x x xD x x x x xB x x x xC x x xE x x xF x x xG x x x xH x x x

Machines


Parts 1 2 4 8 3 5 6 7 9 10 11 12A x x x x xD x x x x xF x x xB x x x xC x x xE x x xG x x x xH x x x

Machines


Parts 1 2 4 8 10 3 5 6 7 9 11 12A x x x x xD x x x x xF x x xB x x x xC x x xE x x xG x x x xH x x x

Machines


Parts 1 2 4 8 10 3 6 9 5 7 11 12A x x x x xD x x x x xF x x xC x x xG x x x xB x x x xE x x xH x x x

Machines


12

12 3

4

5

6

7

8 910

11

A BCRaw materials

Cell1 Cell 2 Cell 3

Assembly


Each of A, B, C now visits only one area, minimizing jumping.

Advantages of Cellular Layouts

• Reduced material handling and transit time

• Reduced setup time

• Reduced work-in-process inventory

• Better use of human resources

• Better scheduling, easier to control and automate

Disadvantages of Cellular Layouts

• Sometimes cells may not be formed because of inadequate part families.

• Some cells may have a high volume of production and others very low. This results in poorly balanced cells.

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.

• Sometimes, machines are duplicated in different cells. This increases capital investment.

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 necessaryE- especially important I - importantO - ordinarily importantU- unimportantX- undesirable

Production area

Office rooms

Storage

Dock area

Locker room

Tool room

A A

A O

O

UO

O

U

U U

U

EX

I


Example: It’s ordinarily important to locate office rooms near loading/unloading area


• 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 area.

From-To Chart

PunchSaws Milling Press Drills

Saws 18 40 30Milling 18 38 75Punch Press 40 38 22Drills 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 period.

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 1 10 7 6 11 2 6 4 7 9 3 8 6 5 6 4 9 5 3 12

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

Machine A B C D 1 10 7 6 11 2 6 4 7 9 3 8 6 5 6 4 9 5 3 12

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

Machine A B C D 1 10 7 6 11 2 6 4 7 9 3 8 6 5 6 4 9 5 3 12

Assignment Method

1. Perform row reductions– Subtract minimum value in each row from all other row

values2. 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– Subtract minimum uncrossed value from all uncrossed

values & add it to all cells where two lines intersect. Go to Step 3.

Assignment Example

Row reduction Column reduction Line Test4 1 0 5 2 1 0 4 2 1 0 42 0 3 5 0 0 3 4 0 0 3 43 1 0 1 1 1 0 0 1 1 0 06 2 0 9 4 2 0 8 4 2 0 8

Number lines = 3 <> 4 = number of rows. So, modify matrix

Location

Machine A B C D 1 10 7 6 11 2 6 4 7 9 3 8 6 5 6 4 9 5 3 12

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.

• Next, find minimum number of lines to cross out all the zeros. Since the minimum number of lines = 3 < 4 = number of rows, more computation is necessary.

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 lines to cover all the zeros is 4, then we would stop, and attempted to find a zero assignment with no success. Thus, the mistake would be detected.

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.

• The modified matrix is shown next.

Location

Machines A B C D 1 1 0 0 4 2 0 0 4 5 3 0 0 0 0 4 3 1 0 8

Modify matrix Line Test

1 0 0 4 1 0 0 40 0 4 5 0 0 4 50 0 0 0 0 0 0 03 1 0 8 3 1 0 8

# lines = # rows so at optimal solution

Location

Machines A B C D 1 10 7 6 11 2 6 4 7 9 3 8 6 5 6 4 9 5 3 12

Total material handling costs = 22

Assignment Example

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 have to check with the original cost matrix. The total cost is 7+6+6+3 = 22 hundred dollars per month.

LOCATING A SINGLE FACILITYTHE RECTILINEAR DISTANCE PROBLEM

Outline

• Locating New Facilities• Minimize Weighted Sum of the Rectilinear Distances• Minimize Maximum Rectilinear Distance

Locating New Facilities

• 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.

2. emergency facilities: a fire station, police station, etc.

3. airline hub, utility cables such as phone cables etc.

4. a radio tower, etc.


• 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 distance


• 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 case, when the direction is not not changed so often, Euclidean distance measure is more appropriate.


• 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 rectilinear distance is given by total length of two broken lines and Euclidean distance by the length of the solid line.

A(2,1)

L(5,5)

3

45

R ectilineardis tance

Euc lideandis tance

• 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


22 byax A(2,1)

L(5,5)

3

45

R ectilineardis tance

Euc lideandis tance


• 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 distance which is a criteria used in locating facilities used by many users, a hospital, an airline hub, etc.


• Sometimes, weights are not required because it may be more important to minimize the maximum distance which is a criterion used in locating an emergency facility, a radio tower, etc.


• In summary, there are 4 important types of problems• Rectilinear distance

• Minimize weighted total • Minimize maximum

• Euclidean distance• Minimize weighted total • Minimize maximum

• 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 Locating New Facilities

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 Locating New Facilities

• 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 theRectilinear Distances


• Find the optimal value of x as follows:– Arrange the x-coordinates in ascending order– Compute the cumulative weights– The optimal value of x is obtained by dividing the

total weight by 2 and finding the first location at which the cumulative weight equals or exceeds this value.


• Find the optimal value of y similarly:– Arrange the y-coordinates in ascending order– Compute the cumulative weights– The optimal value of y is obtained by dividing the

total weight by 2 and finding the first location at which the cumulative weight equals or exceeds this value.

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)

Example: Minimize Weighted Sum of Rectilinear Distances

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)



• 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 of the rectilinear distances from (5.5, 4.0) to each location. This is done on the next two slides.


iiii

baw 04555

1

..

locations 5 to (5.5,4.0) location optimal From

distances rrectilinea the of sum Weighted


Continues from the previous slide

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 illustrated with an example.


• Compute five numbers c1, c2, c3, c4, c5:

)(min1

1 iini

bac

)(max1

2 iini

bac

)(min1

3 iini

bac

)(max1

4 iini

bac

),max( 34125 ccccc

• Define:

• All points along the line connecting (x1,y1) and (x2,y2) are optimal


2/)(,2/)( 5311311 cccyccx

2/)(,2/)( 5422422 cccyccx

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( 34125 ccccc

• All points along the line connecting ____________

and ____________ are optimal.

2

2

5311

311

/)(

/)(

cccy

ccx

2

2

5422

422

/)(

/)(

cccy

ccx



• 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).

• To get the optimal value, compute the maximum rectilinear distances from any optimal location, say (4.25, 5.0). This is done on the next two slides.

iii ba 05254 ..max

(4.25,5.0) location optimal From

distance rrectilinea Maximum




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.)– 10.25 p. 578, 10.32, 10.38, p. 585 (5th Ed.)

LESSON 22: LOCATING A SINGLE FACILITYTHE EUCLIDEAN DISTANCE PROBLEM

Outline

• Minimize Weighted Sum of the Squares of the Euclidean Distances (Gravity Problem)

• Minimize Weighted Sum of the Euclidean Distances

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)

– Minimize weighted sum of the Euclidean distances

• 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 =

22 )()( ii byax

The Euclidean Distance Problem

Example 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 centers to the new facility when they need health care.

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 The Euclidean Distance Problem

Gravity ProblemMinimize Weighted Sum of the Squares of the

Euclidean Distances

n

ii

n

iii

w

awx

1

1*

n

ii

n

iii

w

bwy

1

1*

• 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 squares of the Euclidean distances. The solution of the gravity problem is obtained using the following formula:

Example: Gravity Problem

Census Weight

Tract (i) (ai,bi) wi wiai wibi

1 (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

n

ii

n

iii

w

awx

1

1*

n

ii

n

iii

w

bwy

1

1*

Conclusion: The location ___________________ minimizes the weighted sum of the squares of the Euclidean distances.



Euclidean distance =

Square of the Euclidean distance =

22 )()( ii byax 22 )()( ii byax

5

locations 5 to )(5.72,3.76

location optimal from distances Euclidean

the of squares the of sum Weighted

1

22 763725i

iii baw ..



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

22 )()( ii

ii

byax

wg

Minimize Weighted Sum of the Euclidean Distances

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.

n

ii

n

iii

g

gax

1

1

n

ii

n

iii

g

gby

1

1

Minimize Weighted Sum of the Euclidean Distances

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 0.613 2.5 5 1.675 2 7 3.68

5.5 4 10 30.718 5 12 4.62

Trial location

Example: Minimize Weighted Sum of the Euclidean Distances


Sample computation for the previous slide:

21

21

1

21

21

11

763725 ba

w

byax

wg

..

22

22

2

22

22

22

763725 ba

w

byax

wg

..

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 2.723 2.5 5 1.67 5.00 4.175 2 7 3.68 18.41 7.36

5.5 4 10 30.71 168.93 122.868 5 12 4.62 36.99 23.12

41.29 230.84 160.23

Trial location


Total

Sample computation for the previous slide:


ii

ii

gb

ga

ga

ga

22

11

n

ii

n

iii

g

gax

1

1

n

ii

n

iii

g

gby

1

1= =

• Repeat the process with (x,y) = (5.59, 3.88). The result is (x,y) = (5.54, 3.94).






• Stop.



Euclidean distance = 22 )()( ii byax

5

1

22 993505i

iii baw ..

locations 5 to )(5.50,3.99 location from

distances Euclidean the of sum Weighted



Application

1. Minimize weighted sum of the rectilinear distances (done)– Facilities used by many people e.g., computer centre,

gymnasium2. Minimize maximum rectilinear distance (done)

– Emergency facilities e.g., police, fire 3. Minimize weighted sum of the Euclidean distances (done)

– Utilities, e.g., phone cable4. Minimize maximum Euclidean distance (not in book)

– Transmission towers e.g., radio towers5. Minimize weighted sum of the squares of the Euclidean

distances (done)– Approximation for 3.

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

– A Layout Construction Procedure, ALDEP

• 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.

Computerized Layout Technique

CRAFT - Computerized Relative Allocation of Facilities Technique

Following are some examples of questions addressed by CRAFT:

• Is this a good layout?

• If not, can it be improved?

10 20 30 40 50 60 70 80 90 100

1020

3040

5060

7080

9010

0

A

B

C

D

• 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

11, yx

2,

22121 yyxx

22 , yx

• 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)

9035758030 CACA yyxx

BABA yyxx

AA yx ,

BB yx ,


Centroid of A

=

Centroid of C

=

Distance (A,C)

=

10 20 30 40 50 60 70 80 90 100

1020

3040

5060

7080

9010

0

A

B

C

D

(80,85)

(30,25)


• 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.

CRAFT: Total Distance Traveled

A B C DA 2 7 4B 3 5 7C 6 7 3D 7 7 3

FromTo


(a) Material handling trips (given)

(a)

A B C DA 2 7 4B 3 5 7C 6 7 3D 7 7 3

FromTo


(a)

(b)(b) Distances (given)

A B C DA 50 90 60B 50 60 110C 90 60 50D 60 110 50

FromTo


A B C DA 2 7 4B 3 5 7C 6 7 3D 7 7 3

FromTo


(a)

(b)

(c)

(b) Distances (given)

Total distance traveled = 100+630+240+…. = 4640

A B C DA 50 90 60B 50 60 110C 90 60 50D 60 110 50

FromTo

A B C DA 100 630 240B 150 300 770C 540 420 150D 420 770 150

FromTo

(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.

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. 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: Savings

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.


• 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. This will now be shown.


• The matrix on the left is the previous matrix, before interchange (see Slide 7). The matrix on the right is after.

• Dist (A,B) and (C,D) does not change. • New dist (A,C) = Previous dist (A,D)• New dist (A,D) = Previous dist (A,C)• New dist (B,C) = Previous dist (B,D)• New dist (B,D) = Previous dist (A,C)

A B C DA 50 60 90B 50 110 60C 60 110 50D 90 60 50

FromToA B C D

A 50 90 60B 50 60 110C 90 60 50D 60 110 50

FromTo

InterchangeC,D

CRAFT: A Sample Computation of

Savings

A B C DA 2 7 4B 3 5 7C 6 7 3D 7 7 3

FromTo

(a)



Savings

A B C DA 2 7 4B 3 5 7C 6 7 3D 7 7 3

FromTo

(a)

(b)

A B C DA 50 60 90B 50 110 60C 60 110 50D 90 60 50

FromTo


(b) Distances (rearranged)


Savings

A B C DA 2 7 4B 3 5 7C 6 7 3D 7 7 3

FromTo

(a)

(b)

(c)

A B C DA 50 60 90B 50 110 60C 60 110 50D 90 60 50

FromTo

A B C DA 100 420 360B 150 550 420C 360 770 150D 630 420 150

FromTo


(b) Distances (rearranged)

Total distance traveled = 100+420+360+… = 4480Savings =

(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

4. Repeat the above 3 steps as longs as Step 1 finds an interchange with some (positive) savings.

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 coordinate of the centroid (of a shape like A)?

10 20 30 40 50 60 70 80 90 100

1020

3040

5060

7080

9010

0

A

B

C

D


Find the centroid of A

10 20 30 40 50 60 70 80 90 100

5060

7080

9010

0

A A1

A2

A1of Centroid

A1of Centroid

A2Area

A1Area

Let

22

11

2

1

yx

yx

A

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


21

2211

AA

xAxA

Y-coordinate Multiply

Rectangle Area of centroid (2) and (3)

(1) (2) (3) (4)

A1

A2

Total

Y-coordinate of the centroid of A


21

2211

AA

yAyA


Exact coordinate of area A is

10 20 30 40 50 60 70 80 90 100

5060

7080

9010

0

A A1

A2

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

ALDEPAutomated 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 neighbor heuristic is a tour construction procedure which may be improved by eliminating intersections.

ALDEP

• Given – Size of the facility – The departments– Size of the departments– Proximity relationships (activity relationship chart) and– A sweep width (defined later)

ALDEP constructs a layout.

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– Office rooms 10– Storage area 8– Dock area 8– Locker room 4– Tool room 4

ALDEPA: absolutely necessary

E: especially important

I: important

O: ordinarily important

U: unimportant

X: undesirable Production area

Office rooms

Storage

Dock area

Locker room

Tool room

A A

A O

O

UO

O

U

U U

U

EX

I

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 width = 2. Then, dock area will be allocated 2 4 = 8 blocks.

D DD DD DD D

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.

D DD DD DD DS S S SS S S S

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.

D DD DD DD DS S S SS S S S

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.

D D P P P PD D P P P PD D P P P PD D P PS S S SS S S S

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.

D D P P P PD D P P P PD D P P P PD D P P T TS S S S T TS S S S

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 rating is selected. This will now be discussed.

D D P P P P O OD D P P P P O OD D P P P P O OD D P P T T O OS S S S T T O OS S S S L L L L

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• If two departments are adjacent in the layout then the

weight corresponding to the rating between the two departments is added to the score.

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 OD D P P P P O OD D P P P P O OD D P P T T O OS S S S T T O OS S S S L L L L

• Adjacent departments:

(D,S) (D,P)

(S,P) (S,T)

(S,L) (P,T)

(P,O) (T,L)

(T,O) (L,O)

ALDEP

Adjacent Proximity NumericDepartments 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

Total = -763

• 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.

ALDEP

Facilities Management

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