1. OST: Chapter 3 (Part 1) Tactical planning: Layout
Fixed-position layout
Job shop production 1 (LAP)

2. 1. Introduction Strategic decisions (e.g. location problems)
Concretization 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, bridges, buildings, …)
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
Example: Production and assembly
of 4 parts (A, B, C, D)
A: saw -> turn -> mill -> drill
B: saw -> mill -> drill -> paint
C: grind -> mill -> drill -> paint
D: weld -> grind -> turn -> drill
Minimum equipment:
1 weld
1 grind
1 saw
1 turn
2 mills
2 drills
1 paint
Equipment requirements
Part
Weld
Grind
Saw
Turn
Mill
Drill
Paint A -
0.5
0.3
0.2 B
0.4 C D
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 1.1 Fixed-postion layout Drill Mill Drill Mill Turn Grind
Workpiece
Stores
Warehouse
Saw
Assembly
Weld
Paint
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. Fixed-position layout

7. Introduction 1.2 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

8. Job shop production

9. Introduction 1.3 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

10. Cellular manufacturing system
(c) Prof. Richard F. Hartl

11. Manufacturing cells
(c) Prof. Richard F. Hartl

12. Manufacturing cells
(c) Prof. Richard F. Hartl

13. Introduction 1.4 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

14. Flow Shop Production
(c) Prof. Richard F. Hartl

15. Introduction Selection of layout: Combinations:
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

16. 2. Fixed-Position Layout
Workpiece too large or cumbersome to be moved trough its processing steps -> 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

17. 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

18. 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

19. 3. 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

20. Job shop production Advantages:
Better utilization of machines can result -> 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

21. Job shop production Limitations:
Longer flow lines are needed -> 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

22. Job shop production Eliminiation of negative effects
Production planning (operational level)
Optimized machine allocation (tactical level)
-> Assignment problems:
LAP (linear assignment problem)
QAP (quadratic assignment problem)

23. 3.1 Linear Assignment Problem
Simplest optimization problem in intra-company location planning
Given:
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.

24. Linear Assignment Problem
3 machines, 4 locations and the following costs cij
location
j =
i \ j 1 2 3 4
machine 13 10 12 11
i = 15  20 5 7 6
Machine 2 cannot be assigned to location 2 -> cost 

25. 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 13 10 12 11
i = 15  20 5 7 6
dummy
Location with dummy machine -> location stays empty

26. 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: m=n basis variables; TP: m+n-1 basis variables)
TP:

27. 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

28. 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: 1+2+3=6 I II
III A 1 8 15 B 6 2 10 C 7 9 3

29. Linear Assignment Problem
Cost reduction (Columns -> 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: = 6 (= Total assignment cost without cost reduction) I II
III A 6 12 B 5 7 C -1 -2 -3

30. Kuhn´s Algorithm Kuhn´s algorithm 3 steps to be followed:
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 -> 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

31. Kuhn´s 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 or rows already including zero elements

32. Kuhn´s Algorithm Step 2: Optimal solution?
Start with a column or row having as few as possible 0 entries
Frame one of 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.

33. Kuhn´s Algorithm
Example:
Step 1:

34. Kuhn´s Algorithm Step 2: Optimal solution! Reduced cost = 0\ j 1 2 3 4 s 7 5 d
Reduced cost = 0
Original cost = 0 + reduction constant = 28

35. Kuhn´s 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.
-> 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.

36. Kuhn´s 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.

37. Kuhn´s Algorithm
17,5 15 9
5,5 12 16
16,5
10,5 5
15,5
14,5 11
4,5 8 14 13
9,5
8,5 13 7
0,5
0,5
6,5
-0,5
11,5
8,5 2 5 
7,5
7,5 6 6
5,5
12,5
7,5
8,5
1,5 7 12
-4,5 -8
-8,5 -5
-5,5
12,5
6,5 6
11,5
8,5 2 5 
7,5
7,5 6 6
K =
4, , ,5 + 0,5
5,5
12,5
7,5
= 32
8,5
1,5 7 12

38. Kuhn´s Algorithm
12,5
6,5 6
11,5
8,5 2 5
7,5
5,5
1,5 7 12
No zero cost assignment! -> Find the minimum arrangement of lines covering all zeros.

39. Kuhn´s Algorithm
12,5
6,5 6
11,5
8,5 2 5
7,5
5,5
1,5 7 12
-> Adapt the cost elements by adding/subtracting element a (minimum value of all not-covered elements)

40. Kuhn´s Algorithm -a +a
12,5
6,5 6
11,5
8,5 2 5
7,5
5,5
1,5 7 12 2
7,5 7 11 5
4,5 10 7
3,5 
7,5
7,5 7 14 7
10,5
1 additional zero (assignment 5  2) increases the chance to find an assignment with total (reduced) costs of 0.
a = 1,5

41. Kuhn´s Algorithm 11 5 4,5 10 7 2 3,5 7,5 14 10,5 Optimal assignment!
Start again with step 2 until a zero cost assignment is possible! 11 5
4,5 10 7 2
3,5
7,5 14
10,5
Optimal assignment!
Total cost: sum of all reduction values (step 1 and 3): C=
(4, , ,5 + 0,5) + (1,5) =
33,5

42. LAP with Maximization
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 I2 I3 I4 I5 I6
Maximize the total profit! [„Cost“ = 20 - Profit]

43. Transformed Matrix Minimize cost: - 4 - 8 A B C D E F I1 16 12 4 8 20A B C D E F I1 16 12 4 8 20 I2 I3 I4 I5 I6 A B C D E F I1 16 12 8 I2 4 20 I3 I4 I5 I6
- 4
- 8

44. Total Profit = Max Profit - total cost = 5*20 – 16 = 184A B C D E F I1 16 12 8 I2 4 20 I3 I4 I5 I6 A B C D E F I1 16 12 I2 4 8 20 I3 I4 I5 I6
a = 4 A B C D E F I1 16 12 I2 4 8 20 I3 I4 I5 I6
Total Cost = (4 + 8) + (4) = 16
Total Profit = Max Profit - total cost = 5*20 – 16 = 184