Layout and Design

1
Layout and Design

• Introduction
• Fixed-position layout
• Job shop production I

2
Introduction

• Strategic decisions (location problems)
• Concretion and realization on tactical level
  (Layout and design, configuration)
• Operational decisions (operational planning)

3
Introduction

• An efficient layout facilitates and reduces costs
  of material flow, people, and information between
  areas.
• 4 concepts will be introduced here
• Fixed-position layout
• Large, bulky workpieces (ships,)
• Job shop production (Process/Function-oriented
  production)
• High variety, low volume
• Cellular manufacturing systems
• Individual products
• Flow shop production(Object-oriented production)
• Low variety, high volume
• cf. Heizer, J., Render, B., Operations
  Management, Prentice Hall, 2006, Chapter 9
• cf. Francis, R., McGinnis, L., White, J.,
  Facility Layout and Location An Analytical
  Approach, Prentice Hall, 1992

4
Introduction

• Example Production and assembly
• of 4 parts (A, B, C, D)
• A saw -gt turn -gt mill -gt drill
• B saw -gt mill -gt drill -gt paint
• C grind -gt mill -gt drill -gt paint
• D weld -gt grind -gt turn -gt drill
• Minimum equipment
• 1 weld
• 1 grind
• 1 saw
• 1 turn
• 2 mills
• 2 drills
• 1 paint

  Equipment requirements Equipment requirements Equipment requirements Equipment requirements Equipment requirements Equipment requirements Equipment requirements
Part Weld Grind Saw Turn Mill Drill Paint
A - - 0.5 0.5 0.3 0.2 -
B - - 0.4 - 0.5 0.3 0.2
C - 0.4 - - 0.3 0.5 0.3
D 0.3 0.5 - 0.3 - 0.2 -
cf. Heizer, J., Render, B., Operations
Management, Prentice Hall, 2006, Chapter 9 cf.
Francis, R., McGinnis, L., White, J., Facility
Layout and Location An Analytical Approach,
Prentice Hall, 1992
5
Introduction

• 1. Fixed-postion layout

cf. Heizer, J., Render, B., Operations
Management, Prentice Hall, 2006, Chapter 9 cf.
Francis, R., McGinnis, L., White, J., Facility
Layout and Location An Analytical Approach,
Prentice Hall, 1992
6
Introduction

• 2. Job shop layout

cf. Heizer, J., Render, B., Operations
Management, Prentice Hall, 2006, Chapter 9 cf.
Francis, R., McGinnis, L., White, J., Facility
Layout and Location An Analytical Approach,
Prentice Hall, 1992
7
Introduction

• 3. Cellular manufacturing system

cf. Heizer, J., Render, B., Operations
Management, Prentice Hall, 2006, Chapter 9 cf.
Francis, R., McGinnis, L., White, J., Facility
Layout and Location An Analytical Approach,
Prentice Hall, 1992
8
Introduction

• 4. Flow shop layout

cf. Heizer, J., Render, B., Operations
Management, Prentice Hall, 2006, Chapter 9 cf.
Francis, R., McGinnis, L., White, J., Facility
Layout and Location An Analytical Approach,
Prentice Hall, 1992
9
Introduction

• Selection of layout
• Characteristic of workpiece
• Variety of production
• Volume of production
• Combinations
• Components job shop and/or Celluars sytems
  assembly flow shop system

cf. Heizer, J., Render, B., Operations
Management, Prentice Hall, 2006, Chapter 9 cf.
Francis, R., McGinnis, L., White, J., Facility
Layout and Location An Analytical Approach,
Prentice Hall, 1992
10
Fixed-position layout

• Workpiece too large or cumbersome to be moved
  trough its processing steps -gt processes are
  brought to the product rather than otherwise.
• Processes are arranged in the right sequence
  around the workpiece
• Right processes at the workpiece at the right
  location in the right time

cf. Heizer, J., Render, B., Operations
Management, Prentice Hall, 2006, Chapter 9 cf.
Francis, R., McGinnis, L., White, J., Facility
Layout and Location An Analytical Approach,
Prentice Hall, 1992
11
Fixed-position layout

• Advantages
• Material movement is reduced.
• Promotes job enlargement by allowing individuals
  or teams to perform the whole job.
• Highly flexible can accommodate changes in
  product design, product mix, and production
  volume.
• Independence of production centres allows
  scheduling to achieve minimum total production
  time.

cf. Heizer, J., Render, B., Operations
Management, Prentice Hall, 2006, Chapter 9 cf.
Francis, R., McGinnis, L., White, J., Facility
Layout and Location An Analytical Approach,
Prentice Hall, 1992
12
Fixed-position layout

• Limitations
• Increased movement of personnel and equipment.
• Equipment duplication may occur.
• Higher skill requirements for personnel.
• General supervision required.
• Cumbersome and costly positioning of material and
  machinery.
• Low equipment utilization.

cf. Heizer, J., Render, B., Operations
Management, Prentice Hall, 2006, Chapter 9 cf.
Francis, R., McGinnis, L., White, J., Facility
Layout and Location An Analytical Approach,
Prentice Hall, 1992
13
Job shop production

• High variety low volume Rapid changes in mix
  or volume
• In-between conditions
• Collection of processing departments or cells
• Each containing a collection of machines
  processing similiar operations
• Each product (group) undergoes a different
  sequence of operations
• Different products have different material flows
  and are moved from one department to another in
  the appropriate sequence.
• High degree of interdepartmental flow

cf. Heizer, J., Render, B., Operations
Management, Prentice Hall, 2006, Chapter 9 cf.
Francis, R., McGinnis, L., White, J., Facility
Layout and Location An Analytical Approach,
Prentice Hall, 1992
14
Job shop production

• Advantages
• Better utilization of machines can result -gt
  fewer machines are required.
• A high degree of flexibility exists relative to
  equipment or manpower allocation for specific
  tasks (break down,)
• Comparatively low investment in machines
• The diversity of tasks offers a more interesting
  and satisfying occupation for the operator.
• Specialized supervision is possible.

cf. Heizer, J., Render, B., Operations
Management, Prentice Hall, 2006, Chapter 9 cf.
Francis, R., McGinnis, L., White, J., Facility
Layout and Location An Analytical Approach,
Prentice Hall, 1992
15
Job shop production

• Limitations
• Longer flow lines are needed -gt material handling
  is more expensive.
• Production planning and control systems are more
  involved than for other layouts.
• Usually, total production time is longer than for
  other layouts.
• Due to the fact that jobs have to queue before
  being processed in a machine job comparatively
  large amounts of in-process inventory occur.
• Comparatively high degree of (machine) idle time
  because machines have to wait until the
  subsequent job is finished with its foregoing
  process.
• Space and capital are tied up by work in process.
• Because of the diversity of the jobs in
  specialized departments, higher grades of skill
  are required.

cf. Heizer, J., Render, B., Operations
Management, Prentice Hall, 2006, Chapter 9 cf.
Francis, R., McGinnis, L., White, J., Facility
Layout and Location An Analytical Approach,
Prentice Hall, 1992
16
Job shop production

• Eliminiation of negative effects
• Production planning (operational level)
• Optimized machine allocation (tactical level)
• -gt Assignment problems
• LAP (linear assignment problem)
• QAP (quadratic assignment problem)

17
Linear Assignment Problem

• Simplest optimization problem in intra-company
  location planning
• Gegeben
• n machines (activities, workers)
• n potential locations (periods, projects)
• cij ... cost of running maschine i on location
  j
• Any machine can be assigned to any location
• It is required to use all locations by assigning
  exactly one machine to each location
• total cost of the assignments are to be
  minimized.

18
Linear Assignment Problem

• 3 machines, 4 locations and the following costs
  cij

location j
i \ j 1 2 3 4
machine 1 13 10 12 11
i 2 15 ? 13 20
3 5 7 10 6
Machine 2 cannot be assigned to location 2 -gt
cost ?
19
Linear Assignment Problem

• If number of locations ? number of machines
• Add dummymachines (-rows) or dummylocations
  (-columns) with costs 0 (this is always possible)

locations j
i \ j 1 2 3 4
machine 1 13 10 12 11
i 2 15 ? 13 20
3 5 7 10 6
dummy 4 0 0 0 0
Location with dummy machine -gt location stays
empty
20
Linear Assignment Problem

• Formulation as TP
• Each LAP can be interpreted as special case of a
  TP with each supplier (machine) having a
  capacity of 1 and each customer ( location)
  having a demand of 1.
• Since, for the TP it is guaranteed (due to the
  special problem structure) that all decision
  variables are integer, we end up with a feasible
  solution for the LAP (n variables with value 1
  and all others 0 (LAP mn basis variables TP
  mn-1 basis variables)
• TP

21
Linear Assignment Problem

• 1 if maschine i is assigned to location j
• 0 otherwise

xij
Cost
Constraints
1 für i 1,...,n ... each machine is
assigned exactly once 1 für j 1,...,n ...
each location is allocated with 1 machine 0
oder 1 für i 1,...,n und j 1,...,n
22
Linear Assignment Problem

• In order to solve LAPs exactly we are going to
  make use of the following important problem
  characteristic
• it is always possible to reduce (or increase) all
  entries of any column or row by a certain value
  without changing the optimal solution (only the
  absolute costs change, the relation stays the
  same). We use this characteristic to generate the
  maximum number of 0 entries.
• Example
• Optimal solution (column minimum method) A-I
  B-II C-III
• Cost 1236

I II III
A 1 8 15
B 6 2 10
C 7 9 3
23
Linear Assignment Problem

• Cost reduction (Columns -gt Rows)
• The relation of assignment costs for each
  machine/locations does not change.
• Optimal solution (column minimum method) A-I
  B-II C-III
• (Reduced) cost 0
• Reduced cost reduction values 0 6 6 (
  Total assignment cost without cost reduction)

I II III
A 0 6 12
B 5 0 7
C 6 7 0
-3
-2
-1
24
Kuhns Algorithm

• Kuhns algorithm
• finds the exact solution
• is based on adding/subtracting values to/from
  given cost factors in order to find the lowest
  opportunity cost (i.e. not-obtained profits)
• 3 steps to be followed
• Cost reduction -gt Generation of 0 elements
• Try to determine the optimal assignment. If this
  is not possible draw the minimum number of lines
  necessary to cover all zeros in the matrix.
• Adapt the cost factors in the matrix and return
  to step 2.

cf. Heizer, J., Render, B., Operations
Management, Prentice Hall, 2006, Chapter 15
25
Kuhns Algorithm

• Step 1 Cost reduction
• Subtract the smallest number in each column from
  every number in that column
• Subtract the smallest number in each row from
  every number in that row.
• We obtain a matrix with a series of zeros,
  meaning zero opportunity costs (at least one zero
  in each column and each row)
• No cost reduction in columns already including
  zero elements

26
Kuhns Algorithm

• Step 2 Optimal solution?
• Start with a column or row having as few as
  possible 0 entries
• Frame one the 0 in this column/row and cross all
  other 0 in the corresponding column and row
• Go on with the next column or row having as few
  as possible not-framed and not-crossed zeros.
• And so on until all zeros are either framed or
  crossed.
• If we are able to make a zero (reduced) cost
  assignment for all machines we found the optimal
  solution!
• Otherwise we have to find the minimum arrangement
  of lines covering all zeros in the matrix.

27
Kuhns Algorithm

• Example
• Step 1

28
Kuhns Algorithm

• Step 2
• Optimal solution!

29
Kuhns Algorithm

• Step 2 If not draw the minimum number of lines
  covering all zero elements
• Mark (for example X) all rows with no framed 0
• Mark all columns having at least 1 crossed 0 in a
  marked row
• Mark all rows having a framed 0 in a marked
  column
• Repeat until there is no column or row left to be
  marked
• Mark each non-marked row and each marked column
  (shaded) with a continuous line.
• -gt this is the minimum arrangement of lines
  needed to cover all 0. If the number of lines is
  equal to the number of rows/columns, an optimal
  assignment is already possible.

30
Kuhns Algorithm

• Step 3 Adapt the cost factors
• The smallest not-covered element is the new
  reduction value (a).
• Subtract a from all not-covered elements in the
  matrix.
• Add a to all elements covered by two lines.
• Elements covered by 1 line remain unchanged.
• Go on with step 2.

31
Kuhns Algorithm
17,5 15 9 5,5 12
16 16,5 10,5 5 10,5
12 15,5 14,5 11 5,5
4,5 8 14 17,5 13
13 9,5 8,5 12 17,5





13
7
0,5
0,5
6,5
-0,5
2
11,5
8,5
0
5
?
7,5
7,5
6
6
0
0
5,5
0
7,5
12,5
8,5
1,5
0
12
7
-4,5
-8
-5
-8,5
-5,5





12,5
6,5
6
0
0
11,5
8,5
2
0
5
?
6
6
0
7,5
7,5
K
4,5 8 8,5 5 5,5 0,5
5,5
12,5
7,5
0
0
32
0
7
12
8,5
1,5
32
Kuhns Algorithm
12,5 6,5 0 0 6
11,5 8,5 2 0 5
7,5 7,5 6 6 0
0 0 5,5 12,5 7,5
8,5 1,5 0 7 12
No zero cost assignment! -gt Find the minimum
arrangement of lines covering all zeros.
33
Kuhns Algorithm
12,5 6,5 0 0 6
11,5 8,5 2 0 5
7,5 7,5 6 6 0
0 0 5,5 12,5 7,5
8,5 1,5 0 7 12
-gt Adapt the cost elements by adding/subtracting
element a (minimum value of all not-covered
elements)
34
Kuhns Algorithm
12,5 6,5 0 0 6
11,5 8,5 2 0 5
7,5 7,5 6 6 0
0 0 5,5 12,5 7,5
8,5 1,5 0 7 12
0 0
2 0
7,5 7,5 0
0 0 7,5
0 7
11
5
4,5
3,5
10
7
?
7,5
7,5
7
14
7
0
10,5
1 additional zero (assignment 5 ? 2) increases
the chance the find an assignment with total
(reduced) costs of 0.
a 1,5
35
Kuhns Algorithm
Start again with step 2 until a zero cost
assignment is possible!
11 5 0 0 4,5
10 7 2 0 3,5
7,5 7,5 7,5 7,5 0
0 0 7 14 7,5
7 0 0 7 10,5
Optimal assignment!
Total cost sum of all reduction values (step 1
and 3)
C
(4,5 8 8,5 5 5,5 0,5) (1,5)
33,5
36
Kuhns Algorithm

• Some assignment problems entail, e.g., maximizing
  profit instead of minimizing cost. To convert a
  maximization problem to an equivalent
  minimization problem, we subtract every number in
  the original matrix from the largest single
  number in that matrix.

  A B C D E F
I1 4 8 16 20 12 0
I2 16 20 8 0 4 12
I3 0 12 4 16 20 8
I4 4 0 16 12 20 8
I5 12 16 0 8 20 4
I6 20 16 12 0 4 8
Maximize the total profit! Cost 20 -
Profit