Machine Setup Time - A Limit Determiner (Tool)
GanChet
98 views
24 slides
Sep 07, 2024
Slide 1 of 24
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
About This Presentation
Machine Setup Time - A Limit Determiner, a tool that will estimate a reduction limit before setting up an operation, to determine your means... with reference to Single-Minute Exchange of Die SMED, for Lean Manufacturing System. It is tool that consists of data (input of up to 12 entries) that will ...
Slide Content
Machine Setup Time –
A Limit Determiner (Transposition of An
Exponential Reduction Equation) (Part 2)
By Gan Chun Chet (Ir., P.Eng(M‘sia))
Machine Setup Time –
A Limit Determiner (Transposition of An
Exponential Reduction Equation) (Part 2)
By Gan Chun Chet (Ir., P.Eng(M‘sia))
MSc Operations Management 1997
[Manchester School Management]
University of Manchester Institute of Science and Technology (UMIST),
United Kingdom.
BEng (Hons) Mechanical Engineering 1996
[Simon Building]
University of Manchester, United Kingdom
MSc Operations Management 1997
[Manchester School Management]
University of Manchester Institute of Science and Technology (UMIST),
United Kingdom.
BEng (Hons) Mechanical Engineering 1996
[Simon Building]
University of Manchester, United Kingdom
A Limit Determiner (Transposition of An
Exponential Reduction Equation) (Part 2)
By Gan Chun Chet (Ir., P.Eng(M‘sia))
Machine Setup Time –
A Limit Determiner (Transposition of An
Exponential Reduction Equation) (Part 2)
By Gan Chun Chet (Ir., P.Eng(M‘sia))
MSc Operations Management 1997
[Manchester School Management]
University of Manchester Institute of Science and Technology (UMIST),
United Kingdom.
BEng (Hons) Mechanical Engineering 1996
[Simon Building]
University of Manchester, United Kingdom
MSc Operations Management 1997
[Manchester School Management]
University of Manchester Institute of Science and Technology (UMIST),
United Kingdom.
BEng (Hons) Mechanical Engineering 1996
[Simon Building]
University of Manchester, United Kingdom
The Objectives
To obtain accurately an equation which represents a reduction trend.
In order to determine the minimum limit in real time for instances of time of a particular period, a
reduction equation is transpose to a predetermine/pre-set data. The reduction trend consists of any
observable trend in real situation.
With a base equation, it is to transpose and locate the minimum reduction time limit, to be determine by
iterating predetermine/pre-set data, and this computer formulation (will be discuss) will calculate the
least error for every possible data-set that is key-in into the system.
The data-set (input to the system) in this situation, range from 1 to 12 numbers. Average is taken when
the user enter specific data value(s).
To obtain accurately an equation which represents a reduction trend.
In order to determine the minimum limit in real time for instances of time of a particular period, a
reduction equation is transpose to a predetermine/pre-set data. The reduction trend consists of any
observable trend in real situation.
With a base equation, it is to transpose and locate the minimum reduction time limit, to be determine by
iterating predetermine/pre-set data, and this computer formulation (will be discuss) will calculate the
least error for every possible data-set that is key-in into the system.
The data-set (input to the system) in this situation, range from 1 to 12 numbers. Average is taken when
the user enter specific data value(s).
Methodology
A proper tool to enable data to be accepted by a system that will
maximize profitability. This means that work is performed to a minimum
acceptable level.
The limit of this invisible horizontal line, below the trending line is located
to achieve an inviscid plan ahead of time.
Trendline below another limit is allowed. With regards to the response
time, the point of interception is easily identified.
A proper tool to enable data to be accepted by a system that will
maximize profitability. This means that work is performed to a minimum
acceptable level.
The limit of this invisible horizontal line, below the trending line is located
to achieve an inviscid plan ahead of time.
Trendline below another limit is allowed. With regards to the response
time, the point of interception is easily identified.
A Tool That Will Do The Job (1)
•The range with respect to the
center cell.
•Add or Minus, as shown.
•The range with respect to the
center cell.
•Add or Minus, as shown.
A Tool That Will Do The Job (2)
(A1-0.5)-0.2A1-0.5(A1-0.5)+0.2
A1-0.2A1A1+0.2
(A1+0.5)-0.2A1+0.5(A1+0.5)+0.2
•The Arithmetic Progressions.
(A1-0.5)-0.2A1-0.5(A1-0.5)+0.2
A1-0.2A1A1+0.2
(A1+0.5)-0.2A1+0.5(A1+0.5)+0.2
•The Arithmetic Progressions.
A Tool That Will Do The Job (3)
(1-0.5)-0.21-0.5(1-0.5)+0.2
1-0.211+0.2
(1+0.5)-0.21+0.5(1+0.5)+0.2
•The Number Progression, in
Sequence if A1 = 1
(1-0.5)-0.21-0.5(1-0.5)+0.2
1-0.211+0.2
(1+0.5)-0.21+0.5(1+0.5)+0.2
•The Number Progression, in
Sequence if A1 = 1
A Tool That Will Do The Job (4)
0.30.50.7
0.811.2
1.31.51.7
•The Number Progression, the
Answer
0.30.50.7
0.811.2
1.31.51.7
•The Number Progression, the
Answer
A Tool To Iterate the Steepness, by defining the
Minimum Limit and Start Point (1)
•Initializing a unique number for:
Minimum Limit = 1
Start Point = 10
•Select the possible Upper/Lower Range.
•Work Pages To Identify all range set, +/-0.5
(deviation) including the center number, as
shown.
Work Pages+/- 0.5+/-0.5
1Work Page 30.5 9.5
2Work Page 11 10
3Work Page 21.5 10.5
4Work Page 41 9.5
5Work Page 51 10.5
6Work Page 60.5 10
7Work Page 70.5 10.5
8Work Page 81.5 9.5
9Work Page 91.5 10
Minimum Limit and Start Point (1)
•Initializing a unique number for:
Minimum Limit = 1
Start Point = 10
•Select the possible Upper/Lower Range.
•Work Pages To Identify all range set, +/-0.5
(deviation) including the center number, as
shown.
Work Pages+/- 0.5+/-0.5
1Work Page 30.5 9.5
2Work Page 11 10
3Work Page 21.5 10.5
4Work Page 41 9.5
5Work Page 51 10.5
6Work Page 60.5 10
7Work Page 70.5 10.5
8Work Page 81.5 9.5
9Work Page 91.5 10
A Tool To Iterate the Steepness, by defining
the Minimum Limit and Start Point (2)
•Assuming Work Page 9 is selected from the list, the possible set is given,
-Minimum Limit = 1.5
-Start Point = 10
•There will be 10 possible, increment of 0.1, for the steepness value, in each
of these selection.
the Minimum Limit and Start Point (2)
•Assuming Work Page 9 is selected from the list, the possible set is given,
-Minimum Limit = 1.5
-Start Point = 10
•There will be 10 possible, increment of 0.1, for the steepness value, in each
of these selection.
A Tool To Iterate the Steepness, by defining
the Minimum Limit and Start Point (3)
•However, the deviation, as shown is +/-0.2 of the “selected” value.
•The value selected is;
-Minimum Limit = 1.5
-Start Point = 10
•Thus, all possibilities based on a given unique number are as shown in
the following slides, on the left section.
•The, by substitution, the numbers are calculated as shown on the
right section.
the Minimum Limit and Start Point (3)
•However, the deviation, as shown is +/-0.2 of the “selected” value.
•The value selected is;
-Minimum Limit = 1.5
-Start Point = 10
•Thus, all possibilities based on a given unique number are as shown in
the following slides, on the left section.
•The, by substitution, the numbers are calculated as shown on the
right section.
A Tool To Iterate the Steepness, by defining
the Minimum Limit and Start Point (4)
To calculate all possibilities based on any given unique
number, with a deviation of +/-0.2
1 1.5 10.0
With 10 Possible Rows Each (Increment of 0.1)
2 1.5+0.110.0
With 10 Possible Rows Each (Increment of 0.1)
3 1.5-0.1 10.0
With 10 Possible Rows Each (Increment of 0.1)
4 1.5+0.210.0
With 10 Possible Rows Each (Increment of 0.1)
5 1.5-0.2 10.0
With 10 Possible Rows Each (Increment of 0.1)
6 1.5+0.110+0.1
With 10 Possible Rows Each (Increment of 0.1)
7 1.5+0.110-0.1
With 10 Possible Rows Each (Increment of 0.1)
8 1.5+0.110+0.2
With 10 Possible Rows Each (Increment of 0.1)
9 1.5+0.110-0.2
With 10 Possible Rows Each (Increment of 0.1)
10 1.5-0.110+0.1
With 10 Possible Rows Each (Increment of 0.1)
11 1.5-0.110-0.1
With 10 Possible Rows Each (Increment of 0.1)
12 1.5-0.110+0.2
With 10 Possible Rows Each (Increment of 0.1)
13 1.5-0.110-0.2
With 10 Possible Rows Each (Increment of 0.1)
14 1.5+0.210+0.1
With 10 Possible Rows Each (Increment of 0.1)
15 1.5+0.210-0.1
With 10 Possible Rows Each (Increment of 0.1)
16 1.5+0.210+0.2
With 10 Possible Rows Each (Increment of 0.1)
17 1.5+0.210-0.2
With 10 Possible Rows Each (Increment of 0.1)
18 1.5-0.210+0.1
With 10 Possible Rows Each (Increment of 0.1)
19 1.5-0.210-0.1
With 10 Possible Rows Each (Increment of 0.1)
20 1.5-0.210+0.2
With 10 Possible Rows Each (Increment of 0.1)
21 1.5-0.210-0.2
With 10 Possible Rows Each (Increment of 0.1)
22 1.5 10+0.1
With 10 Possible Rows Each (Increment of 0.1)
23 1.5 10-0.1
With 10 Possible Rows Each (Increment of 0.1)
24 1.5 10+0.2
With 10 Possible Rows Each (Increment of 0.1)
25 1.5 10-0.2
With 10 Possible Rows Each (Increment of 0.1)
Substituting by data, results as follows:-
1 1.5 10.0
2 1.6 10.0
3 1.4 10.0
4 1.7 10.0
5 1.3 10.0
6 1.6 10.1
7 1.6 9.9
8 1.6 10.2
9 1.6 9.8
10 1.4 10.1
11 1.4 9.9
12 1.4 10.2
13 1.4 9.8
14 1.7 10.1
15 1.7 9.9
16 1.7 10.2
17 1.7 9.8
18 1.3 10.1
19 1.3 9.9
20 1.3 10.2
21 1.3 9.8
22 1.5 10.1
23 1.5 9.9
24 1.5 10.2
25 1.5 9.8
0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1
the Minimum Limit and Start Point (4)
To calculate all possibilities based on any given unique
number, with a deviation of +/-0.2
1 1.5 10.0
With 10 Possible Rows Each (Increment of 0.1)
2 1.5+0.110.0
With 10 Possible Rows Each (Increment of 0.1)
3 1.5-0.1 10.0
With 10 Possible Rows Each (Increment of 0.1)
4 1.5+0.210.0
With 10 Possible Rows Each (Increment of 0.1)
5 1.5-0.2 10.0
With 10 Possible Rows Each (Increment of 0.1)
6 1.5+0.110+0.1
With 10 Possible Rows Each (Increment of 0.1)
7 1.5+0.110-0.1
With 10 Possible Rows Each (Increment of 0.1)
8 1.5+0.110+0.2
With 10 Possible Rows Each (Increment of 0.1)
9 1.5+0.110-0.2
With 10 Possible Rows Each (Increment of 0.1)
10 1.5-0.110+0.1
With 10 Possible Rows Each (Increment of 0.1)
11 1.5-0.110-0.1
With 10 Possible Rows Each (Increment of 0.1)
12 1.5-0.110+0.2
With 10 Possible Rows Each (Increment of 0.1)
13 1.5-0.110-0.2
With 10 Possible Rows Each (Increment of 0.1)
14 1.5+0.210+0.1
With 10 Possible Rows Each (Increment of 0.1)
15 1.5+0.210-0.1
With 10 Possible Rows Each (Increment of 0.1)
16 1.5+0.210+0.2
With 10 Possible Rows Each (Increment of 0.1)
17 1.5+0.210-0.2
With 10 Possible Rows Each (Increment of 0.1)
18 1.5-0.210+0.1
With 10 Possible Rows Each (Increment of 0.1)
19 1.5-0.210-0.1
With 10 Possible Rows Each (Increment of 0.1)
20 1.5-0.210+0.2
With 10 Possible Rows Each (Increment of 0.1)
21 1.5-0.210-0.2
With 10 Possible Rows Each (Increment of 0.1)
22 1.5 10+0.1
With 10 Possible Rows Each (Increment of 0.1)
23 1.5 10-0.1
With 10 Possible Rows Each (Increment of 0.1)
24 1.5 10+0.2
With 10 Possible Rows Each (Increment of 0.1)
25 1.5 10-0.2
With 10 Possible Rows Each (Increment of 0.1)
Substituting by data, results as follows:-
1 1.5 10.0
2 1.6 10.0
3 1.4 10.0
4 1.7 10.0
5 1.3 10.0
6 1.6 10.1
7 1.6 9.9
8 1.6 10.2
9 1.6 9.8
10 1.4 10.1
11 1.4 9.9
12 1.4 10.2
13 1.4 9.8
14 1.7 10.1
15 1.7 9.9
16 1.7 10.2
17 1.7 9.8
18 1.3 10.1
19 1.3 9.9
20 1.3 10.2
21 1.3 9.8
22 1.5 10.1
23 1.5 9.9
24 1.5 10.2
25 1.5 9.8
0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1
A Tool To Iterate the Steepness, by defining
the Minimum Limit and Start Point (5)
•From here, the extended range of the center value is defined herein
as additional or subtraction of 0.7
•This points to the fact that the key-in value should be within the
upper/lower limit range, a positive or negative acclimation, that
should be exemplify to its best possible solution.
•Thus, to incline the result(s) towards the real situation, decision-
making at your finger tips, achieving it in the following slides.
the Minimum Limit and Start Point (5)
•From here, the extended range of the center value is defined herein
as additional or subtraction of 0.7
•This points to the fact that the key-in value should be within the
upper/lower limit range, a positive or negative acclimation, that
should be exemplify to its best possible solution.
•Thus, to incline the result(s) towards the real situation, decision-
making at your finger tips, achieving it in the following slides.
How the Program Calculates The Input(s) and Returns an
Accurate Data Trend to the User(s)
•1. To Calculate the "Limit Factored" based on define formula.
-Calculate all possibilities base on the following formula:
y(t) = A + B * [exp ( -C * t )]
•2-1. The "Prediction" is based on the difference between independent result of the calculated
"Limit Factored" and the user entry at respective interval(s)
-The "Limit Factored" is based on an observe trend.
-The User(s) enter any value in each of the given cells (in this case, any number within the 12
intervals. The request can be randomly entered.
•2-2. The errors are calculated independently, the program subsequently select the least error,
either based on e or e' and advise the trend.
e' : Then if the value is negative, merely taking the modulus of the negative number (i.e. ignore
the negative (-) sign), and then summing the difference, herein, known as the "error",
e : Then summing the difference, herein, known as the "error", and then if the value is negative,
merely taking the modulus of the negative number (i.e. ignore the negative (-) sign)
•3. By having a tool, to iterate even more accurately –generally values close to these range are
key-in into Excel to obtain a stable result, with a possible minimum“error” value.
Accurate Data Trend to the User(s)
•1. To Calculate the "Limit Factored" based on define formula.
-Calculate all possibilities base on the following formula:
y(t) = A + B * [exp ( -C * t )]
•2-1. The "Prediction" is based on the difference between independent result of the calculated
"Limit Factored" and the user entry at respective interval(s)
-The "Limit Factored" is based on an observe trend.
-The User(s) enter any value in each of the given cells (in this case, any number within the 12
intervals. The request can be randomly entered.
•2-2. The errors are calculated independently, the program subsequently select the least error,
either based on e or e' and advise the trend.
e' : Then if the value is negative, merely taking the modulus of the negative number (i.e. ignore
the negative (-) sign), and then summing the difference, herein, known as the "error",
e : Then summing the difference, herein, known as the "error", and then if the value is negative,
merely taking the modulus of the negative number (i.e. ignore the negative (-) sign)
•3. By having a tool, to iterate even more accurately –generally values close to these range are
key-in into Excel to obtain a stable result, with a possible minimum“error” value.
The Definition of Method 1 and 2 -
Calculation of Errors by Summation (1)
•Method 1
Modulus of “averaging the sum
of errors (e)”, identifying the
least value of the key-in value
(base on basis equation y1)
•Method 2
summing the “modulus of the
errors (e’)”, without averaging,
exemplifying the results.
Calculation of Errors by Summation (1)
•Method 1
Modulus of “averaging the sum
of errors (e)”, identifying the
least value of the key-in value
(base on basis equation y1)
•Method 2
summing the “modulus of the
errors (e’)”, without averaging,
exemplifying the results.
The Definition of Method 1 and 2 -
Calculation of Errors by Summation (2)
e (n) = | ( e1 + e2 )| / ( n )
Lowest error considered, just by taking the modulus of
“averaging the sum of errors”
Also possible to take the sum of positive errors
If <=2 or > 2 points, e.g. in this situation, 2 or 3 data
points as shown below the errors
When it is
negative,
modulus of
the error is
taken in
account
Calculation of Errors by Summation (2)
e (n) = | ( e1 + e2 )| / ( n )
Lowest error considered, just by taking the modulus of
“averaging the sum of errors”
Also possible to take the sum of positive errors
If <=2 or > 2 points, e.g. in this situation, 2 or 3 data
points as shown below the errors
When it is
negative,
modulus of
the error is
taken in
account
The Basis of the Exponential Reduction
Formula
•As determine, it is as shown below with the variables obtain from time
motion study in a foreign base company while reducing machine setup
time.
•y(t) = A + B * [exp ( -C * t )]
•A = 6.3
B = 3.2
C = 0.2
•y1 = 6.3 + 3.2 exp(-0.2 x)
•In any case for example, simulating 3 data points within a define set -
instances of time (in this situation; 12 interval), the trendline can be
transpose.
Formula
•As determine, it is as shown below with the variables obtain from time
motion study in a foreign base company while reducing machine setup
time.
•y(t) = A + B * [exp ( -C * t )]
•A = 6.3
B = 3.2
C = 0.2
•y1 = 6.3 + 3.2 exp(-0.2 x)
•In any case for example, simulating 3 data points within a define set -
instances of time (in this situation; 12 interval), the trendline can be
transpose.
Transposing The Equation (1-1)
y1 = 6.3 + 3.2 exp (-0.2 x)
>>y2 = 1.2 + 11.8 exp (-0.3 x)
3<
x y1y2 ( e')
y3', y3" &
y3'''y2-y3
18.9199389.941655
28.4450247.675977
38.0561975.9975226-0.00248
47.7378534.754092
57.4772143.832936
67.2638213.1505274-0.84947
7 7.089112.644986
86.9460692.270472
96.8289561.9930252-0.00697
106.7330731.787487
116.654571.635221
126.5902971.52242
No. of Data Points (Normally : 2, 3 or 4)
y1 = 6.3 + 3.2 exp (-0.2 x)
>>y2 = 1.2 + 11.8 exp (-0.3 x)
3<
x y1y2 ( e')
y3', y3" &
y3'''y2-y3
18.9199389.941655
28.4450247.675977
38.0561975.9975226-0.00248
47.7378534.754092
57.4772143.832936
67.2638213.1505274-0.84947
7 7.089112.644986
86.9460692.270472
96.8289561.9930252-0.00697
106.7330731.787487
116.654571.635221
126.5902971.52242
No. of Data Points (Normally : 2, 3 or 4)
Transposing The Equation (1-2)
y1 = 6.3 + 3.2 exp (-0.2 x)
>>y2 = 0.2 + 8 exp (-1 x)
3<
x y1y2 ( e')
y3', y3" &
y3'''y2-y3
18.9199383.143036
28.4450241.282682
38.0561970.5982970.6-0.0017
47.7378530.346525
57.4772140.253904
67.2638210.219830.4-0.18017
7 7.089110.207295
86.9460690.202684
96.8289560.2009870.20.000987
106.7330730.200363
116.654570.200134
126.5902970.200049
No. of Data Points (Normally : 2, 3 or 4)
y1 = 6.3 + 3.2 exp (-0.2 x)
>>y2 = 0.2 + 8 exp (-1 x)
3<
x y1y2 ( e')
y3', y3" &
y3'''y2-y3
18.9199383.143036
28.4450241.282682
38.0561970.5982970.6-0.0017
47.7378530.346525
57.4772140.253904
67.2638210.219830.4-0.18017
7 7.089110.207295
86.9460690.202684
96.8289560.2009870.20.000987
106.7330730.200363
116.654570.200134
126.5902970.200049
No. of Data Points (Normally : 2, 3 or 4)
Transposing The Equation (2-1)
•By entering 3 data points, the average error (e) is calculated
(summing the modulus of the errors);
•Similarly, summing the modulus of the errors (e’) will be sufficient
without averaging the errors, is also acceptable (error value
exemplified).
•In this example,
Item IntervalValue (Hours)
13 6
26 4
39 2
•Each interval can be one month, making this a complete year.
•By entering 3 data points, the average error (e) is calculated
(summing the modulus of the errors);
•Similarly, summing the modulus of the errors (e’) will be sufficient
without averaging the errors, is also acceptable (error value
exemplified).
•In this example,
Item IntervalValue (Hours)
13 6
26 4
39 2
•Each interval can be one month, making this a complete year.
Transposing The Equation (2-2)
•By entering 3 data points, the average error (e) is calculated
(summing the modulus of the errors);
•Similarly, summing the modulus of the errors (e’) will be sufficient
without averaging the errors, is also acceptable (error value
exemplified).
•In this example,
Item IntervalValue (Hours)
13 0.6
26 0.4
39 0.2
•Each interval can be one month, making this a complete year.
•By entering 3 data points, the average error (e) is calculated
(summing the modulus of the errors);
•Similarly, summing the modulus of the errors (e’) will be sufficient
without averaging the errors, is also acceptable (error value
exemplified).
•In this example,
Item IntervalValue (Hours)
13 0.6
26 0.4
39 0.2
•Each interval can be one month, making this a complete year.
Transposing The Equation (3-1)
ty1y2 ( e' )y4( e )IterationActual Time (Data Entry)
18.9199389.94165511.9797511.43413- -
28.4450247.67597710.08019.579001- -
38.0561975.9975228.5247898.060145-6-
47.7378534.7540927.2514116.816612- -
57.4772143.8329366.2088575.798493- -
67.2638213.1505275.3552864.964928-4-
77.089112.6449864.6564414.282462- -
86.9460692.2704724.0842753.723706 - -
96.8289561.9930253.6158263.266236-2-
106.7330731.7874873.2322922.891691- -
116.654571.6352212.918282.585039- -
126.5902971.522422.661192.333974- -
(Notice: Iterating y2 ( e' ) with real data)
GRAPHICAL
REPRESENTATION
oABC
y16.33.20.2<< Model Equation
y2 ( e' )1.211.80.3
y4 ( e )1.512.80.2
Iteration1.212.50.2< Enter A Value Near the Given (or Simulation) Number, C set to default (equals to y1)
ty1y2 ( e' )y4( e )IterationActual Time (Data Entry)
18.9199389.94165511.9797511.43413- -
28.4450247.67597710.08019.579001- -
38.0561975.9975228.5247898.060145-6-
47.7378534.7540927.2514116.816612- -
57.4772143.8329366.2088575.798493- -
67.2638213.1505275.3552864.964928-4-
77.089112.6449864.6564414.282462- -
86.9460692.2704724.0842753.723706 - -
96.8289561.9930253.6158263.266236-2-
106.7330731.7874873.2322922.891691- -
116.654571.6352212.918282.585039- -
126.5902971.522422.661192.333974- -
(Notice: Iterating y2 ( e' ) with real data)
GRAPHICAL
REPRESENTATION
oABC
y16.33.20.2<< Model Equation
y2 ( e' )1.211.80.3
y4 ( e )1.512.80.2
Iteration1.212.50.2< Enter A Value Near the Given (or Simulation) Number, C set to default (equals to y1)
Transposing The Equation (3-2)
GRAPHICAL
REPRESENTATION
ty1y2 ( e' )y4( e )IterationActual Time (Data Entry)
13.6199383.1430365.8587767.222958- -
23.1450241.2826823.7109155.931784- -
32.7561970.5982972.4081714.874661-0.6-
42.4378530.3465251.6180184.009162- -
52.1772140.2539041.1387653.300551- -
61.9638210.219830.8480842.72039-0.4-
71.789110.2072950.6717762.245394- -
81.6460690.2026840.5648411.8565- -
91.5289560.2009870.4999811.5381-0.2-
101.4330730.2003630.4606421.277417- -
111.354570.2001340.4367811.063987- -
121.2902970.2000490.4223090.889246- -
(Notice: Iterating y2 ( e' ) with real data)
oABC
y113.20.2<< Model Equation
y2 ( e' )0.281
y4 ( e )0.490.5
Iteration0.18.70.2< Enter A Value Near the Given (or Simulation) Number, C set to default (equals to y1)
GRAPHICAL
REPRESENTATION
ty1y2 ( e' )y4( e )IterationActual Time (Data Entry)
13.6199383.1430365.8587767.222958- -
23.1450241.2826823.7109155.931784- -
32.7561970.5982972.4081714.874661-0.6-
42.4378530.3465251.6180184.009162- -
52.1772140.2539041.1387653.300551- -
61.9638210.219830.8480842.72039-0.4-
71.789110.2072950.6717762.245394- -
81.6460690.2026840.5648411.8565- -
91.5289560.2009870.4999811.5381-0.2-
101.4330730.2003630.4606421.277417- -
111.354570.2001340.4367811.063987- -
121.2902970.2000490.4223090.889246- -
(Notice: Iterating y2 ( e' ) with real data)
oABC
y113.20.2<< Model Equation
y2 ( e' )0.281
y4 ( e )0.490.5
Iteration0.18.70.2< Enter A Value Near the Given (or Simulation) Number, C set to default (equals to y1)
Reducing the Setup Time by
Transposition of An Equation :
Manual Calculations (1
st
Set)
•Initial Setup Time : 32 Hours (4 working days)
•1
st
50% Setup Reduction : 0.5 * 32 = 16 Hours
2
nd
50% Setup Reduction : 0.5 * 16 = 8 Hours
3
rd
50% Setup Reduction : 0.5 * 8 = 4 Hours
4
th
50% Setup Reduction : 0.5 * 4 = 2 Hours
5
th
50% Setup Reduction : 0.5 * 2 = 1 Hour
•5 times setup reduction(s) will deliver the setup time
from 32 hours to an hour (1 hour) of supply.
Transposition of An Equation :
Manual Calculations (1
st
Set)
•Initial Setup Time : 32 Hours (4 working days)
•1
st
50% Setup Reduction : 0.5 * 32 = 16 Hours
2
nd
50% Setup Reduction : 0.5 * 16 = 8 Hours
3
rd
50% Setup Reduction : 0.5 * 8 = 4 Hours
4
th
50% Setup Reduction : 0.5 * 4 = 2 Hours
5
th
50% Setup Reduction : 0.5 * 2 = 1 Hour
•5 times setup reduction(s) will deliver the setup time
from 32 hours to an hour (1 hour) of supply.
Reducing the Setup Time by
Transposition of An Equation :
Manual Calculations (2
nd
Set)
•Initial Setup Time : 1 Hour
•1
st
50% Setup Reduction : 0.5 * 1 = 0.5 Hours
2
nd
50% Setup Reduction : 0.5 * 0.5 = 0.25 Hours
3
rd
50% Setup Reduction : 0.5 * 0.25 = 0.125 Hours
4
th
50% Setup Reduction : 0.5 * 0.125 = 0.0625 Hours
5
th
50% Setup Reduction : 0.5 * 0.0625 = 0.03125 Hour
•5 times setup reduction(s) will deliver the setup time
from 1 hour to 0.03125 hour (1.875 minutes or
approximately 1 minute 53 seconds) of supply.
Transposition of An Equation :
Manual Calculations (2
nd
Set)
•Initial Setup Time : 1 Hour
•1
st
50% Setup Reduction : 0.5 * 1 = 0.5 Hours
2
nd
50% Setup Reduction : 0.5 * 0.5 = 0.25 Hours
3
rd
50% Setup Reduction : 0.5 * 0.25 = 0.125 Hours
4
th
50% Setup Reduction : 0.5 * 0.125 = 0.0625 Hours
5
th
50% Setup Reduction : 0.5 * 0.0625 = 0.03125 Hour
•5 times setup reduction(s) will deliver the setup time
from 1 hour to 0.03125 hour (1.875 minutes or
approximately 1 minute 53 seconds) of supply.
Categories
TechnologyDownload
Download Slideshow Get the original presentation fileQuick Actions
Statistics
- Views 98
- Slides 24
- Age 451 days
Related Slideshows
11
8-top-ai-courses-for-customer-support-representatives-in-2025.pptx
JeroenErne2
46 views
10
7-essential-ai-courses-for-call-center-supervisors-in-2025.pptx
JeroenErne2
46 views
13
25-essential-ai-courses-for-user-support-specialists-in-2025.pptx
JeroenErne2
37 views
11
8-essential-ai-courses-for-insurance-customer-service-representatives-in-2025.pptx
JeroenErne2
33 views
21
Know for Certain
DaveSinNM
20 views
17
PPT OPD LES 3ertt4t4tqqqe23e3e3rq2qq232.pptx
novasedanayoga46
24 views