Transcript 1 Einleitung - univie.ac.at

Flow Shop Production
http://business.mrwood.com.au/unit3/opstrat/opstrat1.asp
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
Flow shop production

Object-oriented

Assignment is derived from the item´s work plans.
Uniform material flow:



Linear assignment (in most cases)
Useful if (and only if) only one kind of product or a limited
amount of different kinds of products is manufactured (i.e. low
variety – high volume)
(c) Prof. Richard F. Hartl
Flow shop production
According to time-dependencies we distinguish between

Flow shop production without fixed time restriction for
each workstation („Reihenfertigung“)

Flow shop production with fixed time restriction for each
workstation (Assembly line balancing,
„Fließbandabgleich“)
(c) Prof. Richard F. Hartl
Flow shop production

No fixed time restriction for the workload of each workstation:

Intermediate inventories are needed
 Material flow should be similiar for all products
 Some workstations may be skipped, but going back to a previous department is
not possible
 Processing times may differ between products
Inventory
Station 1
(c) Prof. Richard F. Hartl
Int. inventory
Station 2
...
Station m
Inventory
Flow shop production

Fixed time restricition (for each workstation):

Balancing problems
 Cycle time („Taktzeit“): upper bound for the workload of each workstation.
 Idle time: if the workload of a station is smaller than the cycle time.

Production lines, assembly lines

automated system (simultaneous shifting)
Station 1
(c) Prof. Richard F. Hartl
Station 2
Station 3
...
Assembly line balancing

Production rate = Reciprocal of cycle time
The line proceeds continuously.
Workers proceed within their station parallel with their workpiece
until it reaches the end of the station; afterwards they return to the
beginning of the station.

Further possibilites:



Line stops during processing time
 Intermittent transport: workpieces are transported between the stations.
(c) Prof. Richard F. Hartl
Assembly line balancing




„Fließbandabstimmung“, „Fließbandaustaktung“,
„Leistungsabstimmung“, „Bandabgleich“
The mulit-level production process is decomomposed into n
operations/tasks for each product.
Processing time tj for each operation j
Restrictions due to production sequence of precedences may occur
and are displayed using a precedence graph:

Directed graph witout cyles G = (V, E, t)
 No parallel arcs or loops
 Relation i < j is true for all (i, j)
(c) Prof. Richard F. Hartl
Example
Operation j
Predecessor
tj
1
-
6
2
-
9
3
1
4
4
1
5
5
2
4
6
3
2
7
3, 4
3
8
6
7
9
7
3
10
5, 9
1
11
8,1
10
12
11
1
(c) Prof. Richard F. Hartl
Precedence graph
t1=6
1
t2=9
2
4
3
5
4
4
5
2
6
3
7
7
8
10
11
3
9
..1
10
1
12
Flow shop production



Machines (workstations) are assigned in a row, each station
contains 1 or more operations/tasks.
Each operation is assigned to exactly 1 station
i before j , (i, j)  E:

i and j in same station or
 i in an earlier station than j

Assignment of operations to stations:

Time- or cost oriented objective function
 Precedence conditions
 Optimize cycle time
 Simultaneous determination of number of stations and cycle time
(c) Prof. Richard F. Hartl
Single product problems


Simple assembly line balancing problem
Basic model with alternative objectives
(c) Prof. Richard F. Hartl
Single product problems
Assumptions:









1 homogenuous product is produced by performing n operations
given processing times ti for operations j = 1,...,n
Precedence graph
Same cycle time for all stations
fixed starting rate („Anstoßrate“)
all stations are equally equipped (workers and utilities)
no parallel stations
closed stations
workpieces are attached to the line
(c) Prof. Richard F. Hartl
Alternative1
Minimization of number of stations m (cycle time is
given):
Cycle time c:
 lower bound for number of stations
 n

mmin :   t j c
 j 1


upper bound for number of stations
 n

mmax :   t j c  1  t max   1
 j 1

(c) Prof. Richard F. Hartl
Alternative 1
Derivation of upper bound:
t(Sk) … workload of station k
Sk, k = 1, ..., m
Integer property
 tmax + t(Sk) > c
i.e. t(Sk)  c + 1 - tmax  k =1,...,m-1
Sum of inequalities

m 1
 t Sk   m  1  c  1  tmax 
k 1
n
m 1
 t j   t S k 
j 1
k 1
 upper bound
(c) Prof. Richard F. Hartl
and integer property of m
Alternative 2
Minimization of cycle time
(i.e. maximization of prodcution rate)
lower bound for cycle time c:

tmax = max {tj  j = 1, ... , n} … processing time of longest operation 
c  tmax


Maximum production amount qmax in time horizon T is given

c  T qmax 
Given number of stations m
 n

 c    t j m
 j 1

(c) Prof. Richard F. Hartl
Alternative 2

lower bound for cycle time:

 n



c  cmin : maxt max , T qmax ,   t j m 

 j 1




upper bound for cycle time
c  T /q min 
(c) Prof. Richard F. Hartl
Alternative 3
Maximization of efficiency („Bandwirkungsgrad“)

Determination of:


Cycle time c
Number of stations m
 Efficiency („BG“)
n
1
BG 
t j
m  c j 1

BG = 1  100% efficiency (no idle time)
(c) Prof. Richard F. Hartl
Alternative 3

Lower bound for cycle time: see Alternative 2
Upper bound for cycle time cmax is given

Lower bound for number of stations

 n

mmin :   t j cmax 
 j 1


Upper bound for number of stations
 n
mmax :   t j
 j 1
(c) Prof. Richard F. Hartl

cmin  1  t max   1

ExampIe



T = 7,5 hours
Minimum production amount qmin = 600 units
cmax : T qmin   7,5 * 3600/ 600  45 seconds/unit
t1=6
1
t2=9
2
(c) Prof. Richard F. Hartl
4
3
5
4
4
5
2
6
3
7
7
8
10
11
3
9
..1
10
1
12
ExampIe
Arbeitsgang j
Vorgänger
tj
1
-
6
2
-
9
3
1
4
4
1
5
5
2
4
6
3
2
7
3, 4
3
8
6
7
9
7
3
10
5, 9
1
11
8,1
10
12
11
1
Summe
(c) Prof. Richard F. Hartl
55
tj = 55
 n

 mmin :   t j cmax   55 45  2
 j 1

No maximum production
amount
 Minimum cycle time
cmin = tmax = 10 seconds/unit
ExampIe
m
BG = 1
7
BG = 0.982
6
5
4
3
2
1
0
10
20
30
40
50
Combinations of m and c leading to feasible solutions.
(c) Prof. Richard F. Hartl
60 c
ExampIe

maximum BG = 1
(is reached only with invalid values m = 1 and c = 55)

Optimal BG = 0,982
(feasible values for m and c: 10  c 45 und m  2)
 m = 2 stations
 c = 28 seconds/unit
(c) Prof. Richard F. Hartl
Example
Possible cycle times c for varying number of stations m
# Stationen
m
theoretisch min
Taktzeit 55 m
minimale realisierbare Taktzeit
c
Bandwirkungsgrad
55/cm
1
55
nicht möglich da c  45
-
2
28
28
0,982
3
19
19
0.965
4
14
15
0,917
5
11
12
0.917
6
10
10
0,917
Increasing cycle time  Reduction of BG (increasing idle time) until 1 station
can be omitted.
BG has a local maximum for each number of stations m with the minimum
cycle time c where a feasible solution for m exists.
(c) Prof. Richard F. Hartl
Further objectives
Maximization of BG is equivalent to



Minimization of total processing time („Durchlaufzeit“):
D=mc
Minimization of sum of idle times:
Minimization of ratio of idle time:
n
 tj
L  mc 
LA =
j 1
L
= 1 – BG
mc
n

Minimization of total waiting time:
W  D  t j  L
j 1
(c) Prof. Richard F. Hartl
LP formulation
We distinguish between:

LP-Formulation for given cycle time

LP-Formulation for given number of stations

Mathematical formulation for maximization of efficiency
(c) Prof. Richard F. Hartl
LP formulation for given cycle time

Binary variables:
1 if operation j is assigned tostationk
x jk  
otherwise
0

mmax
 k  x jk
k 1

 j = 1, ..., n
 k = 1, ..., mmax
= number of station, where operation j is
assigned to
Assumption: Graph G has only 1 sink, which is node n
(c) Prof. Richard F. Hartl
LP formulation for given cycle time
Minim ize Z x  
Objective function:
mmax
 kx
k 1
nk
Constraints:
mmax
 x jk  1
 j = 1, ... , n
... j on exactly 1 station
 x jk  t j  c
k = 1, ... , mmax
... Cycle time
k 1
n
j=1
mmax
mmax
k 1
k 1
h,j  E
 k  xhk   k  x jk

x jk 0,1
 j and k
(c) Prof. Richard F. Hartl
… Precedence cond.
... Binary variables
Notes
Possible extensions:
 Assignment restrictions (for utilities or positions)


elimination of variables or fix them to 0
Restrictions according to operations

Operations h and j with (h, j)   are not allowed to be assigned
to the same station.
m
k  x
k 1
(c) Prof. Richard F. Hartl
hk
m
 1   k  x jk with (h,j) E
k 1
LP formulation for given number of stations

Replace mmax by the given number of stations m

c becomes an additional variable
(c) Prof. Richard F. Hartl
LP formulation for given number of stations
Objective function: Minimize Z(x, c) = c
c  0 and integer
Constraints:
m
 x jk
1
k 1
n
 x jk  t j  c
… cycle time
 j = 1, ... , n
... j on exactly 1 station
 k = 1, ... , m
... cycle time
j=1
m
k  x
k 1
hk
m
  k  x jk
k 1
x jk 0,1
(c) Prof. Richard F. Hartl

h,j  E
 j und k
... precedence cond.
... binary variables
LP formulation for maximization of BG

If neither cycle time c nor number of stations m is given
 take the formulation for given cycle time.
Objective function (nonlinear):
Minimize Z x, c   c
mmax
 kx
k 1
Additional constraints:
c  cmax
c  cmin
(c) Prof. Richard F. Hartl
nk
LP formulation for maximization of BG

Derive a LP again
 Weight cycle time and number of stations with factors w1
and w2
Objective function (linear):
Minimize Z(x,c) = w1(kxnk) + w2c
 Large Lp-models!
 Many binary variables!
(c) Prof. Richard F. Hartl
Heuristic methods in case of given cycle
time

Many heuristic methods
(mostly priorityrule methods)

Shortened exact methods

Enumerative methods
(c) Prof. Richard F. Hartl
Priorityrule methods

Determine a priortity value PVj for each operation j

Prioritiy list

A non-assigned operation j can be assigned to station k
if


all his precedessors are already assigned to a station 1,..k and
the remaining idle time in station k is equal or larger than the
processing time of operation j
(c) Prof. Richard F. Hartl
Priorityrule methods

Requirements:




Cycle time c
Operations j=1,...,n with processing times tj  c
Precedence graph, defined by a set of precedessors
Variables

k

c

Lp
Ls

(c) Prof. Richard F. Hartl
number of current station
idle time of current station
set of already assigned operations
sorted list of n operations in respect to priority value
Priorityrule methods

Operation j  Lp can be assigned, if tj 
and h  Lp is true for all h  V(j)

Start with station 1 and fill one station after the other

From the list of operations ready to be assigned to the current
station the highest prioritized is taken

Open a new station if the current station is filled to the maximum
(c) Prof. Richard F. Hartl
c
Priorityrule methods
Start: determine list Ls by applying a prioritiy rule; k := 0; LP := <]; ... No operations
assigned so far
Iteration:
repeat
k := k+1; c := c;
while there is an operation in list Ls that can be assigned to station k do
begin
select and delete the first operation j (that can be assigned to) from list Ls;
Lp:= < Lp,j]; c :=- tj
end;
until Ls = <];
Result: Lp contains a valid sorted list of operations with m = k stations.
Single-pass- vs. multi-pass-heuristics
(procedure is performed once or several times)
(c) Prof. Richard F. Hartl
Priorityrule methods

Rule 1: Random choice of operations

Rule 2: Choose operations due to monotonuously decreasing (or
increasing) processing time: PVj: = tj

Rule 3: Choose operations due to monotonuously decreasing (or
increasing) number of direct followers:
PVj : = (j)

Rule 4: Choose operations due to monotonuously increasing depths of
operations in G:
PVj : = number of arcs in the longest way from a source of the graph to j
(c) Prof. Richard F. Hartl
Priorityrule methods

Rule 5 Choose operations due to monotonuously decreasing positional
weight („Positionswert“):
PVj : t j 
t
h
hN mj

Rule 6: Choose operations due to monotonuously increasing upper bound
for the minimum number of stations needed for j and all it´s predecessors:

 

PVj : E j   t j   th  c
 

hV jm 


Rule 7: Choose operations due to monotonuously increasing upper bound
for the latest possible station of j:

 

PVj : L j  1  m   t j   th  c
 

hN mj
 
(c) Prof. Richard F. Hartl
Example – Rule 5
t1=6
1
t2=9
2
4
3
2
6
5
4
7
8
10
11
3
9
3
7
4
5
S1 = {1,3,2,4,6}
1
12
S2 = {7,8,5,9,10,11}
S3 = {12}
..1
10
j
1
2
3
4
5
6
7
8
9
10
11
12
tj
6
9
4
5
4
2
3
7
3
1
10
1
PVj(5)
42
25
31
23
16
20
18
18
15
12
11
1
Cycle time c = 28 -> m = 3 stations
BG = tj / (3*28) = 0,655
(c) Prof. Richard F. Hartl
Example– Regel 7, 6 und 2

m =3
j
1
2
3
4
5
6
7
8
9
10
11
12
PVj(7)
1
2
1
2
2
2
2
2
2
2
2
2
PVj(6)
1
1
1
1
1
1
1
1
1
2
2
2
PVj(2)
6
9
4
5
4
2
3
7
3
1
10
1
Apply rule 7 (latest possible station) at first
If this leads to equally prioritized operatios -> apply rule 6 (minimum number
of stations for j and all predecessors)
If this leads to equally prioritized operatios -> apply rule 2 (decreasing
processing times tj)
Solution: c = 28  m = 2; BG = 0,982
S1 = {1,3,2,4,5} ; S2 = {7,9,6,8,10,11,12}
(c) Prof. Richard F. Hartl
More heuristic methods

Stochastic elements for rules 2 to 7:




Random selection of the next operation (out of the set of
operations ready to be applied)
Selection probabilities: proportional or reciprocally proportional to
the priority value
Randomly chosen priority rule
Enumerative heuristics:



Determination of the set of all feasible assignments for the first
station
Choose the assignment leading to the minimum idle time
Proceed the same way with the next station, and so on (greedy)
(c) Prof. Richard F. Hartl
Further heuristic methods

Heuristics for cutting&packing problems


Precedence conditions have to be considered as well
E.g.: generalization of first-fit-decreasing heuristic for the bin
packing problem.

Shortest-path-problem with exponential number of nodes

Exchange methods:


Exchange of operations between stations
Objective: improvement in terms of the subordinate objective of
equally utilized stations
(c) Prof. Richard F. Hartl
Worst-Case analysis of heuristics
Solution characteristics for integer c and tj
(j = 1,...,n) for Alternative 2:
t Sk   t Sk 1   c  1 for all k=1,...,m-1
t Sk   tmax  c  1
for all k=1,...,m-1
 Total workload of 2 neigboured stations has to exceed the cycle time
Worst-Case bounds for the deviation of a solution with m
Stations from a solution with m* stations:
m/m*  2 - 2/m* for even m and m/m*  2 - 1/m* for odd m
m < cm*/(c - tmax + 1) + 1
(c) Prof. Richard F. Hartl
Determination of cyle time c

Given number of stations

Cycle time unknown


Minimize cycle time (alternative 1) or
Optimize cycle time together with the number of stations trying to
maximize the system´s efficiency (alternative 3).
(c) Prof. Richard F. Hartl
Iterative approach for determination of
minimal cycle time
1. Calculate the theoretical minimal cycle time:
cmin


tj



number
of
stations


(or cmin = tmax if this is larger) and c = cmin
2. Find an optimal solution for c with minimum m(c) by applying
methods presented for alternative 1
3. If m(c) is larger than the given number of stations: increase c by 
(integer value) and repeat step 2.
(c) Prof. Richard F. Hartl
Iterative approach for determination of
minimal cycle time

Repeat until feasible solution with cycle time  c and number of
stations  m is found

If  > 1, an interval reduction can be applied:
if for c a solution with number of stations  m has been found and for
c- not, one can try to find a solution for c-/2 and so on…
(c) Prof. Richard F. Hartl
Example – rule 5
m = 5 stations
Find: maximum production rate, i.e. minimum cycle time
j
1
2
3
4
5
6
7
8
9
10
11
12
tj
6
9
4
5
4
2
3
7
3
1
10
1
PVj(5)
42
25
31
23
16
20
18
18
15
12
11
1
cmin = tj/m = 55/5 = 11 (11 > tmax = 10)
(c) Prof. Richard F. Hartl
Example – rule 5
Solution c = 11:
{1,3}, {2,6}, {4,7,9}, {8,5}, {10,11},
{12}
Needed: 6 > m = 5 stations
 c = 12, assign operation 12 to
station 5
 S5 = {10,11,12}
t1=6
1
t2=9
2
4
3
5
4
4
5
2
6
3
7
7
8
10
11
3
9
1
12
.1
10
For larger problems: usually, c leading to an assignment for the given
number of stations, is much larger than cmin. Thus, stepwise increase of
c by 1 would be too time consuming -> increase by  > 1 is
recommended.
(c) Prof. Richard F. Hartl
Classification of complex line balancing
problems
Parameters:

Number of products

Assignment restrictions

Parallel stations

Equipment of stations

Station boundaries

Starting rate

Connection between items and transportation system

Different technologies

Objectives
(c) Prof. Richard F. Hartl
Number of products

Single-product-models:

1 homogenuous product on 1 assembly line
 Mass production, serial production

Multi-product models:

Combined manufacturing of several products on 1 (or more) lines.


Mixed-model-assembly:
Products are variations (models) of a basic product
 they are processed in mixed sequence
Lot-wise multiple-model-production:
Set-up between production of different products is necessary
 Production lots (the line is balanced for each product separately)
 Lotsizing and scheduling of products  TSP
(c) Prof. Richard F. Hartl
Assignment restrictions

Restricted utilities:

Stations have to be equipped with an adequate quantity of utilities
 Given environmental conditions

Positions:


Given positions of items within a station
 some operation may not be performed then (e.g.: underfloor
operations)
Operations:

Minimum or maximum distances between 2 operations (concerning time
or space)
  2 operations may not be assigned to the same station

Qualifications:

Combination of operations with similiar complexity
(c) Prof. Richard F. Hartl
Parallel stations

Models without parallel stations:


Heterogenuous stations with different operations  serial line
Models with parallel stations:

At least 2 stations performing the same operation
 Alternating processing of 2 subsequent operations in parallel stations

Hybridization: Parallelization of operations:

Assignment of an operation to 2 different stations of a serial line
(c) Prof. Richard F. Hartl
Equipment of stations

1-worker per station

Multiple workers per station:

Different workloads between stations are possible
 Short-term capacity adaptions by using „jumpers“

Fully automated stations:

Workers are used for inspection of processes
 Workers are usually assigned to several stations
(c) Prof. Richard F. Hartl
Station boundaries

Closed stations:

Expansion of station is limited
 Workers are not allowed to leave the station during processing

Open stations:
Workers my leave their station in („rechtsoffen“) or in reversed
(„linksoffen“) flow direction of the line
 Short-term capacity adaption by under- and over-usage of cycle time.
 E.g.: Manufacturing of variations of products

(c) Prof. Richard F. Hartl
Starting rate

Models with fixed starting rate:


Subsequent items enter the line after a fixed time span.
Models with variable starting rate:

An item enters the line once the first station of the line is idle
 Distances between items on the line may vary (in case of multipleproduct-production)
(c) Prof. Richard F. Hartl
Connection between items and
transportation systems

Unmoveable items:

Items are attached to the transportation system and may not be
removed
 Maybe turning moves are possible

Moveable items:

Removing items from the transportation system during processing is
allowed



Post-production
Intermediate inventories
Flow shop production without fixed time constraints for each station
(c) Prof. Richard F. Hartl
Different technologies

Given production technologies


Schedules are given
Different technologies

Production technology is to be chosen
 Different alternative schedules are given (precedence graph)
and/or

different processing times for 1 operation
(c) Prof. Richard F. Hartl
Objectives

Time-oriented objectives

Minimization of total cycle time, total idle time, ratio of idle time, total
waiting time
 Maximization of capacity utilization (system`s efficieny) – most relevant
for (single-product) problems
 Equally utilized stations

Further objectives

Minimization of number of stations in case of given cycle time
 Minimization of cycle time in case of given number of stations
 Minimization of sum of weighted cycle time and weighted number of
stations
(c) Prof. Richard F. Hartl
Objectives

Profit-oriented approaches:

Maximization of total marginal return
 Minimization of total costs




Machines- and utility costs (hourly wage rate of machines depends on the
number of stations)
Labour costs: often identical rates of labour costs for all workers in all
stations
Material costs: defined by output quantity and cycle time
Idle time costs: Opportunity costs – depend on cycle time and number of
stations
(c) Prof. Richard F. Hartl
Multiple-product-problems
Mixed model assembly:
Several variants of a basic product are processed in mixed
sequence on a production line.
 Processing times of operations may vary between the models
 Some operations may not be necessary for all of the variants
 Determination of an optimal line balancing and of an optimal
sequence of models.

(c) Prof. Richard F. Hartl



multi-model
Lot-wise
mixed-model
production
With machine set-up
(c) Prof. Richard F. Hartl
Set-up from type „X“
to type „Y“ after 2
weeks



mixed-model
Without set-up
Balancing for a
„theoretical
average model“
(c) Prof. Richard F. Hartl
Balancing mixed-model assembly lines

Similiar models:

Avoid set-ups and lot sizing
 Consider all models simultaneously

Generalization of the basic model






Production of p models of 1 basic model with up to n operations;
production method is given
Given precedence conditions for operations in each model j = 1,...,n 
aggregated precendence graph for all models
Each operation is assigned to exactly 1 station
Given processing times tjv for each operation j in each model v
Given demand bv for each model v
Given total time T of the working shifts in the planning horizon
(c) Prof. Richard F. Hartl
Balancing mixed-model assembly lines

Total demand for all models in planning horizon
b
p
 bv
v 1

Cumulated processing time of operation j over all
models in planning horizon:
tj 
p
 bv t jv
v 1
(c) Prof. Richard F. Hartl
LP-Model

Aggregated model:

Line is balanced according to total time T of working shifts in the
planning horizon.
1 if operationj  Sk
x jk  
otherwise
0

for all j=1,...,n and k=1,...,m
Same LP as for the 1-product problem, but cycle time c
is replaced by total time T
(c) Prof. Richard F. Hartl
LP-Model
Objective function:
m
Minim izeZ x    k  xnk
k 1
… number of the last station (job n)
Constraints:
m
x
k 1
jk
1
for all j = 1, ... , n
... Each job in 1 station
for all k = 1, ... , n
... Total workload in station k
n
 x jk  t j  T
j=1
m
k  x
k 1
m
hk
  k  x jk
for all
h,j  E
k 1
x jk 0,1
(c) Prof. Richard F. Hartl
for all j and k
... Precedence conditions
Example
v = 1, b1 = 4
t11=6
1
7
2
4
3
5
4
v = 2, b2 = 2
2
6
4
7
7
8
10
11
3
9
5
5
1
12
11
2
1
10
v = 3, b3 = 1
t13=8
1
13
2
0
3
5
4
t12=5
1
6
3
1
6
4
8
5
4
1
7
4
9
3
5
11
11
0
12
70
11
7
12
1
10
aggregated model
4
6
13
8
3
7
1
9
2
5
(c) Prof. Richard F. Hartl
8
11
3
12
28
t1=42 3
1
35
4
1
10
63
2
28
5
14
6
49
8
21
7
21
9
7
10
Example
Applying exact method:

given: T = 70

Assignment of jobs to stations with m = 7 stations:
S1 = {1,3}
S2 = {2}
S3 = {4,6,7}
S4 = {8,9}
S5 = {5,10}
S6 = {11}
S7 = {12}
(c) Prof. Richard F. Hartl
Parameters
n
 kv  bv  t jv x jk
...
Workload of station k for model v in T
 v  bv  t jv / m
...
Average workload of m stations for model v in T
j 1
n
j 1
Per unit:
n
 kv
 
 t jv x jk
...
Workload of station k for 1 unit of model v
 v 
 t jv / m
...
Avg. workload of m stations for 1 unit of model v
j 1
n
j 1
Aggregated over all models:
p
t(Sk ) 
 t kv
v 1
(c) Prof. Richard F. Hartl
...
Total workload of station k in T
Example – parameters per unit
Station
k
’kv
Avg.
Model v
1
2
3
4
5
6
7
 `v
1
10
7
11
10
6
10
1
7,86
x4
2
11
11
7
8
4
11
0
7,43
x2
3
8
13
12
14
3
8
3
8,71
x1
(c) Prof. Richard F. Hartl
Example - Parameters
Station
k
kv
Avg.
Model v
1
2
3
4
5
6
7
v
1
40
28
44
40
24
40
4
31,43
2
22
22
14
16
8
22
0
14,86
3
8
13
12
14
3
8
3
8,71
t(Sk)
70
63
70
70
35
70
7
55
(c) Prof. Richard F. Hartl
Conclusion



Station 5 and 7 are not efficiently utilized
Variation of workload kv of stations k is higher for the models v as for
the aggregated model t(Sk)
Parameters per unit show a high degree of variation for the models.
Model 3, for example, leads to an high utilization of stations 2, 3, and
4.

If we want to produce several units of model 3 subsequently, the average
cycle time will be exceeded -> the line has to be stopped
(c) Prof. Richard F. Hartl
Avoiding unequally utilized stations

Consider the following objectives

Out of a set of solutions leading to the same (minimal) number of
stations m (1st objective), choose the one minimizing the
following 2nd objective:
m

k 1

p

v 1
kv
  v ...Sum of absolute deviation in utilization
Minimization by, e.g., applying the following greedy heuristic
(c) Prof. Richard F. Hartl
Thomopoulos heuristic
Start: Deviation  = 0, k = 0
Iteration: until non-assigned jobs are available:
increase k by 1
determine all feasible assignments Sk for the next station k
choose Sk with the minimum sum of deviation
 =  + (Sk)
(c) Prof. Richard F. Hartl
 ( Sk ) 
p
  kv   v
v 1
Thomopoulos example
T = 70
m=7
Solution:
9 stations (min. number of stations = 7):
S1 = {1}, S2 = {3,6}, S3 = {4,7}, S4 = {8}, S5 = {2},
S6 = {5,9}, S7 = {10}, S8 = {11}, S9 = {12}
Sum of deviation:  = 183,14
(c) Prof. Richard F. Hartl
Thomopoulos heuristic


Consider only assignments Sk where workload t(Sk)
exceeds a value  (i.e. avoid high idle times).
Choose a value for  :

 small:



well balanced workloads concerning the models
Maybe too much stations
 large:


Stations are not so well balanced
Rather minimum number of stations [very large   maybe no
feasible assignment with t(Sk)  ]
(c) Prof. Richard F. Hartl
Thomopoulos heuristic – Example
 = 49
Solution:
7 stations:
S1 = {2}, S2 = {1,5}, S3 = {3,4},
S4 = {7,9,10}, S5 = {6,8}, S6 = {11}, S7 = {12}
Sum of deviation:  = 134,57
(c) Prof. Richard F. Hartl
Exact solution
7 stations:
S1 = {1,3}, S2 = {2}, S3 = {4,5}, S4 = {6,7,9 }, S5 = {8,10},
S6 = {11}, S7 = {12}
Sum of deviation:  = 126
kv
Station k
Avg.
Modelv
1
2
3
4
5
6
7
v
1
40
28
40
36
32
40
4
31,43
2
22
22
16
12
10
22
0
14,86
3
8
13
7
8
14
8
3
8,71
t(Sk)
70
63
63
56
56
70
7
55
(c) Prof. Richard F. Hartl
Further objectives

Line balancing depends on demand values bj
Changes in demand  Balancing has to be reivsed and
further machine set-ups have to be considered

Workaround:


 
Objectives not depending on demand
m
p
   kv   v
k  1v  1
(c) Prof. Richard F. Hartl
… sum of absolute deviations in utilization per unit
Further objectives

Disadvantages of this objective:

Large deviations for a station (may lead to interruptions in
production). They may be compensated by lower deviations in
other stations
  max  max  kv
   v
k,v
(c) Prof. Richard F. Hartl
... Maximum deviation in utilization per unit