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Response Time equation calculator
General instructions
Open de Response Time Equation Calculator. Excel will ask you if you want to
enable macros. Press 'Yes'. The file will consist of 3 sheets, each sheet serves for one
of the following cases, as indicated by the sheet name (left below).
The Response Time equation calculator works with 3 types of process variability: the
natural process variability, the variability from breakdowns and the variability from
setups.
Before looking at the questions, check if the numbers in the response time equation
calculator excel sheet correspond with your data. Only the colored cells (yellow in
excel) are input cells and may be changed. Do not change the formulas in the other
cells. The blue cells of the excel sheet contain the output values.
Only for the first workstation (column C) the arrival rate and its C² (squared
coefficient of variation) need to be specified. The arrival rate will be the same for all
workstations (no loss) and the arrival Ca² (row 3) for a workstation is the departure
Cd² from the previous workstation (row 19), according to the linking equation.
Cell definitions
The notation used should be clear from the lecture notes on factory fysics. The
formulas in the sheet do use approximately the same notation. Take for instance the
availability of workstation 1. If you click on the corresponding cell (C16), "
=m_f/(m_f+m_r) " appears in the formula bar above.
In normal excel code there would have been written " = C14/(C14+C15) ", because
these cells correspond with mttf / (mrrf+mttr) of workstation 1. However excel offers
the possibility to define cells (or rows, columns,..) with names. So row 14 is logically
defined as m_f , the MTTF, and row 15 is defined as m_r, the MTTR. You can use
this row name now in a formula. This gives the value of the cell which is determined
by the row of the row name and the column of the current formula (column C in this
case). These cell names can be created via insertnamedefine. An overview of all
the cell names can be obtained by clicking the popup menu just above column A.
Remarks
1) Jobs between setups and MTTF should never equal zero, since otherwise some
formulas in excel will have a zero in the denumerator, resulting in error messages.
So if no breakdowns happen, put the MTTR equal to zero and not the MTTF. If no
setups take place, put the setup time equal to zero, not the jobs between setups.
2) The controlbutton 'factory fysics' in the menu bar allows you to choose the
number of stations you want to work with. It actually only hides the columns that
are not needed. The maximum amount of workstations is five.
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3) The next 3 cases don't make use of parallel machines. So the m in row 13, which
stands for the number of parallel machines, should always equal to 1.
4) The different CV 's that you can find in the first column (A) of the sheet are
actually squared CV's ,as can be seen in the second column.
5) Remark that it is not necessary to work with hours as a time unit. Working with
another time unit (e.g. minutes or seconds) is allowed, as long as it is done in a
consistent way.
6) It depends on your settings if decimals are displayed with dots (american) or
commas (european). If you are not sure if 10.000 stands for ten or ten thousand,
click the cell and look in the formula bar.
Case 1
The production line
A company produces a product at five serial workstations. The characteristics for
workstations 1, 4 and 5 are the same. For workstation 2 the MTTR is only two hours
and the setup structure is different. Workstation 3 has a larger natural process time.
(Select cell E5, in the formula bar you will see " = 1/11" , while in the cell E5 "0.091"
will be displayed, the result of 1/11.)
Sheet "case 1" should look like the table below. In the beginning of this document,
there are some general instructions about the response time equation calculator!
MEASURE:
Arrival Rate (parts/hr)
time between arrivals (hr)
STATION:
ra
1
10.000
2
10.000
3
10.000
4
10.000
5
10.000
ta
0.100
0.100
0.100
0.100
0.100
2
Arrival CV
ca
1.000
1.786
1.102
1.794
1.992
Natural Process Time (hr)
t0
Natural Process CV
c02
0.083
(=1/12)
0.100
0.083
(=1/12)
0.100
0.091
(=1/11)
0.100
0.083
(=1/12)
0.100
0.083
(=1/12)
0.100
Jobs Between Setups
Ns
500.000 1000.000 500.000 500.000 500.000
Setup Time (hr)
ts
2
0.500
1.000
0.500
0.500
0.500
1.000
1.000
1.000
1.000
1.000
Setup Time CV
cs
Eff Process Time (setups only)
te
0.084
0.084
0.092
0.084
0.084
Eff Process Time Var (setups
only)
Eff Process CV (setups only)
e
2
0.002
0.003
0.002
0.002
0.002
Ce2
0.238
0.379
0.216
0.238
0.238
Number of Machines
MTTF (hr)
m
Mf
1
200
1
200
1
200
1
200
1
200
MTTR (hr)
Mr
4
2
4
4
4
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Questions case 1:
1) It takes on average …. hours to finish a product. Explain
………………………………………………………………………………
………………………………………………………………………………
2) The effective process time variance of workstation 1 is ….. . What would have
been the effective process time variance of workstation 1 if no breakdowns occur?
What can you conclude? (Remember: the standarddeviation is the Coefficient of
variation * average. Remark: do not put MTTF = 0, this yields error messages.)
3) Explain why Ca² for workstation 3 is low compared to Ca² for workstation 2.
4) Compute the cumulative cycle time upto (and) including workstation 2 by means
of the formula's developed in the class notes and check with the results given by
the calculator.
5) Set MTTR = 4 hours for all workstations, the cumulative cycle time is …… . Set
MTTR = 2 hours for all workstations, the cumulative cycle time is ……. . Set
MTTR = 0, the cumulative cycle time is ……. . One can conclude that if MTTR
decreases, the cumulative cycle time …… ………(decreases / increases). This is
mainly due to a decrease in …….(utilization / process variability). Compare the 3
different cycle times, what can you conclude
6) Suppose we were able to "balance" the line (all workstations have a processing
time of 0.083 hours and a MTTR of 4). The cumulative cycle time …… ……
(decreases / increases) because the …… …… ….(process time variability /
utilization / both) …. ………..(increases / decreases).
7) Apply Little's law to the original numerical example. Compute the cumulative
WIP, given the cumulative cycle time.
8) Compare the 3 following series: 1,1,1,1 and 0,0,2,2 and 0,0,0,4. It's clear they
have the same mean but the last serie the hihghest variance. Following the same
logic, the setup structure of workstation ….( 1 / 2) is to be preferred. This means
that if there are more jobs between setups and larger setup times, process
variability …………(increases / decreases), and so cumulative cycle times
…….(increases / decreases).
9) If we increase the processing time of workstation 3 to 0.099 hours / unit, there will
be overload. The reason is …………..( a, b, c, a & b, a & c, b & c, a & b & c )
a) The natural process time (0.099) is smaller than the time between arrivals (0.1)
b) The utilization is too large
c) The process variability is too large.
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Case 2
Description:
Consider the two machine problem as illustrated in the figure below:
T0 = 25 min
C0² = 0.8
buffer1
Machine1
T0 = 28 min
buffer2
Machine2
Arrivals are exponentially distributed with an expected interarrival time of 30
minutes.
Processing time data:
Machine1:
The natural time equals t0 = 25 min. and the squared coefficient of
variation equals C0² = 0.8. This machine is characterized by a lot of outages. The
expected time between failures equals 4 hours, and the time to repair equals 10
minutes. Moreover, there are quality problems: 1 product out of 10 requires repair; the
repair time equals on average 25 min. and has a squared coefficient of variation of
0.5.
Machine2:
The natural processing time equals 28 min. (deterministic). There are
however quality problems: one out of ten needs rework. The rework time is
exponentially distributed with an average of 28 minutes.
Open the excel-file "response time equation calculator" and go to sheet "case 2". In
the beginning of this document, there are some general instructions about the response
time equation calculator!
Questions case 2:
a) How is the rework problem analysed in the Response Time equation calculator ?
b) Do all the following calculations yourself and compare them with the response
time equation calculator.
1 Compute E(w), expected time in the system for machine 1 in minutes.
2 Compute the average number of units in buffer1 (buffer of machine1)
3 Compute the squared coefficient of variation of the arrivals at the second machine.
4 Compute E(w) for machine 2 in minutes
Are their any differences between your results and the excel sheet ?
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Case 3
1 Description:
In a company the production line consists of 5 machines in a row. Between these
machines there are buffers with almost infinite capacity . The products arrive on
average every 50 seconds. The arrival times are exponentially distributed. The process
time for each machine is on average 40 seconds, also exponentially distributed. The
first and the last machine are exchangeable since they both perform the same job. One
of these machines, the one which is placed first in the production line, is quite
obsolete. This results in a lot of failures. The mean time to failure is 900 minutes on
average and the mean time to repair is 100 minutes. Both are also exponentially
distributed.
In this case we will illustrate the use of the Response Time equation calculator (based
on the mathematical approximation of factory fysics). Furthermore, the influence of
the place of the defective machine and the influence of process times on waiting times
is evaluated.
Finally the results are compared with the simulation results of the F&H Taylor II
software (only available in the PC-class of the faculty of economics). In the beginning
of this document, there are some general instructions about the response time equation
calculator!
2 Response time equation calculator (ResponseTimeEC)
Check if the excel-sheet "case 3" looks as the table below.
Arrival Rate (parts/hr)
ra
Time between arrivals (hr)
ta
2
0.020
0.020
0.020
0.020
0.020
50.000 50.000 50.000 50.000 50.000
Arrival CV
ca
Natural Process Time (hr)
t0
Natural Process CV
c02
Jobs Between Setups
Ns
Setup Time (hr)
ts
0.000
0.000
0.000
0.000
0.000
Setup Time CV
cs 2
0.000
0.000
0.000
0.000
0.000
Eff Process Time (setups only)
te
40.000 40.000 40.000 40.000 40.000
Eff Process Time Var (setups only)
e2
1600.0 1600.0 1600.0 1600.0 1600.0
Eff Process CV (setups only)
ce 2
1.000
1.000
1.000
1.000
1.000
Number of Machines
MTTF (hr)
m
mf
1
900
1
200
1
200
1
200
1
900
MTTR (hr)
mr
100
0
0
0
0
1.000
1.356
1.128
1.046
1.017
40.000 40.000 40.000 40.000 40.000
1.000
1.000
1.000
1.000
1.000
10.000 10.000 10.000 10.000 10.000
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3 What should we expect (without the use of the response time equation calc.) ?
a)
Workstation …. will be the bottle neck. Its availability ( = MTTF/
MTTF+MTTR) is …….., its utilization rate is ……, while the utilization rate of all
other machines is …….
b)
The expected waiting time will be …………(the same - different) for each of
the last 4 workstations. (we will come back to this later)
c)
If the products never have to wait before being processed, the average lead
time would have been……. However, the variability of the arrival and production
process causes waiting time and influences the total average lead time. The
……………..(ResponseTimeEC - Simulation - both methods) takes into account this
variability.
d)
If one wants to know the variability of the total lead time, one should use the
………..(ResponseTimeEC - Simulation - both methods).
4 The results of the ResponseTimeEC
Fill in the table below with the result you obtained using the ResponseTimeEC.
Where Ca² is the CV² of the arrival rate, U the utilization rate, CT the waiting time
and WIP the work in progress.
Buffer
1
Ca²
U (%)
Mach
1
Buffer
2
///////
////////
Mach
2
Buffer
3
///////
///////
Mach
3
Buffer
4
///////
///////
Mach
4
Buffer
5
///////
////////
Mach
5
///////
////////
CT
WIP
The average total lead time is……..seconds. ……% of the amount of lead time is
waiting time.
There is clearly a relation between Ca² and CT. This relation can be explained by the
following formula:……………………
5 Alternatives
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From the previous results, it is clear that the main part of the lead time consists of
waiting. Two ways to reduce this waiting time are examined, a reduction of process
times and the replacement of the defective machine to the last workstation.
5a. shorter process times
Suppose that the engineers found a technique to reduce the process times of all
machines to 35 seconds, a decrease of 12.5%. One should expect the lead time to
……..(decrease - increase) with ……..(more -less) than 12.5%.
If we change the process times in the ResponseTimeEC one can calculate the changes.
The total waiting time is ………(decreased - increased) with……%, the total lead
time with……%
5b. changing the bottleneck
Since the first and the last machine perform the same job, we exchange both to see the
impact on the production process, additional to a reduction of the process times. So
the bottleneck will now be at the end. Will this influence the lead times? Motivate
your answer with the linking equation.
To make the changes in the ResponseTimeEC we exchange cells C14:C15 and
G14:G15. One can directly see the changes. Fill in the table and compare with the
previous one.
Buffer
1
Ca²
U (%)
Mach
1
Buffer
2
///////
////////
Mach
2
Buffer
3
///////
///////
Mach
3
Buffer
4
///////
///////
Mach
4
Buffer
5
///////
////////
Mach
5
///////
////////
CT
WIP
In sum: placing the bottle neck at the end of the production line has a ………(positive
- negative) impact on the total lead time.
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6. Simulation results of F&H Taylor II1
This part of the case does not have the intention to learn about F&H Taylor software.
Therefore it is as briefly as possible explained how to get the results of the simulation.
-
-
Go to Programs / Nmenu and open taylor. It may be possible that you than need
to press the spacebar some times.
From the main menu select File / Retrieve and type "case2 " as input name, and
press enter.
A lay-out will appear with 12 elements, 1 input element, 5 buffers and 5
machines, and 1 output element.
Select Go / Simulate. A message box will appear asking to overwrite the trace
file. Press OK, the simulation starts. This can take a little while. When the lay-out
displays some results, it means that the simulation is done.
Press Esc untill you are back in the main menu. Select Results / Reports.
A window with the options View and Edit appears. Click Edit, than Enter and
select Job Report with the arrows -normally it is already selected -. Press Enter,
Esc and than press the letter V on the keyboard, the report will display. Write
down the data you need for the table below and press Esc . Do the same for the
Element report. (Report / Edit / Enter, select element report, Esc, press " V " )
The table below can now be filled in completely. Press Esc, File / Quit to end the
Taylor program.
Buf
fer 1
U (%)
///////
Mach
1
Buf
fer2
///////
Mach
2
Buf
fer3
///////
Mach
3
Buf
fer4
Mach
4
///////
Buf
fer5
Mach
5
///////
Total
produced
Cycle Time
WIP
If you know that 1 day has 8 working hours, can you determine how many days were
simulated?
Compare this table with the results found with the Response Time Equation Editor.
What can you conclude?
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The DOS-version, installed in the PC-classes of FETEW under programs / nmenu