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Time Study of Revised ProcessTime Study of Revised Process
EXAMPLE 4.1
A process at a watch assembly plant has been changed. The process is divided into three work elements. A time study has been performed with the following results. The time standard for process previously was 14.5 minutes. Based on the new time study, should the time standard be revised?
SOLUTION
The new time study had an initial sample of four observations, with the results shown in the following table. The performance rating factor (RF) is shown for each element, and the allowance for the whole process is 18 percent of the total normal time.
Time Study of Revised ProcessTime Study of Revised Process
Obs 1 Obs 2 Obs 3 Obs 4 Average (min) RF Normal
Time
Element 1 2.60 2.34 3.12 2.86 2.730 1.0 2.730
Element 2 4.94 4.78 5.10 4.68 4.875 1.1 5.363
Element 3 2.18 1.98 2.13 2.25 2.135 0.9 1.922
Total Normal Time = 10.014
The normal time for an element in the table is its average time, multiplied by the RF. The total normal time for the whole process is the sum of the normal times for the three elements, or 10.01 minutes. To get the standard time (ST) for the process, just add in the allowance, or
• Definition of Leaning CurvesDefinition of Leaning Curves• Importance of LC.Importance of LC.• LC Fog Linear FunctionLC Fog Linear Function• Operational Application of a Operational Application of a Leaning CurveLeaning Curve
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Learning CurveLearning Curve
Learning Curve
Cumulative Production
Ho
urs
Re
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ire
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to
Pro
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th
e M
ost
Re
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nit
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ExampleExample
Consider a product with the following data about the hours of labor required to produce a unit:
Hours required to produce 1-st unit: 100
Hours required to produce 10-th unit: 48
Hours required to produce 25-th unit: 35
Hours required to produce 75-th unit: 25
Hours required to produce 200-th unit: 18
As more and more units are produced, the hours of labor required to produce the most recent unit is lower and lower.
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Graph for ExampleGraph for Example
1 10010 4825 3575 25200 18
Hours Required to Produce
Most Recent UnitCumulative Production
Learning Curve
0102030405060708090
100110
0 25 50 75 100 125 150 175 200 225
Cumulative Production
Ho
urs
Re
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d
to
Pro
du
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M
ost
Re
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nit
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Reasons for Continual Decease in Reasons for Continual Decease in the Number of Hours Required to the Number of Hours Required to
Produce the Most Recent UnitProduce the Most Recent Unit
On the previous slide, we observed that, as more and more units are produced, the hours required to produce the most recent unit is lower and lower.
What are some potential reasons why this occurs?
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What happens whenWhat happens whencumulative production doubles?cumulative production doubles?
The concept of a Learning Curve is motivated by the observation (in many diverse production environments) that, each time the cumulative production doubles, the hours required to produce the most recent unit decreases by approximately the same percentage.
For example, for an 80% learning curve,
If cumulative production doubles from 50 to 100, then the hours required to produce the 100-th unit is 80% of that for the 50-th unit.
If cumulative production doubles from 100 to 200, then the hours required to produce the 200-th unit is 80% of that for the 100-th unit.
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The Functional FormThe Functional Formof a Learning Curveof a Learning Curve
To model the behavior described in the previous slides, we proceed as follows:
Let x = cumulative production
y = hours required to produce the x-th unit
Then, y = ax-b
where a and b are parameters defined as follows:
a = hours required to produce the 1-st unit
b = a value related to the percentage associated with the Learning Curve
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An 80% Learning CurveAn 80% Learning Curve
Assume that production of the first unit required 100 hours and that there is an 80% Learning Curve.
Again, let
x = cumulative production
y = hours required to produce the x-th unit
Then, mathematicians can show that the Learning Curve is
y = 100x-0.322
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An 80% Learning CurveAn 80% Learning Curve(continued)(continued)
Hours RequiredCumulative to ProduceProduction Most Recent Unit
The Relationship BetweenThe Relationship Betweenb and p (continued)b and p (continued)
There is a direct mathematical relationship between the exponent b in the equation y = ax-b and (p/100)%, where p is the percentage associated with the learning curve:
)2ln(*)%100/()2ln(
)100/ln( beppb ly,equivalentor,
For example, if p=75%, then 415.0)2ln(
)75.0ln( b
For example, if b=0.737, then 60.0)2ln(*737.0)%100/( ep
NOTE: e=2.7183… (never ending, like ¶)
ln(x) is the exponent of e that yields x.
That is, eln(x)=x
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Operational ApplicationOperational Applicationof the Leaning Curveof the Leaning Curve
Assume that production of the 1-st unit required 100 hours, and assume that there is an 80% learning curve. Then, y = 100x-0.322.
Also, assume that cumulative production to date is 150 units.
The learning curve can be used to provide estimates of answers to questions about the production of the next 100 units.
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Operational Application of a Leaning Operational Application of a Leaning Curve Curve (continued)(continued)
Question 1: To produce the next 100 units, how many hours of labor will be required?
Question 2: With a labor force of 6 workers each working 40 hours per week, how long will it take to produce then next 100 units?
Question 3: To produce 100 units in 5 weeks with each worker working 40 hours per week, what should be the size of the labor force?
Question 4: To produce 100 units in 5 weeks using a work force of 60 workers, how many hours per week should each worker work?
Hours RequiredCumulative to ProduceProduction Most Recent Unit
Effect of Sales’Effect of Sales’Annual Growth RateAnnual Growth Rate
Assume that:
Three firms have the same 80% learning curve: y=100x-0.322
During Year 1, all three firms sold 5000 units.
The three firms have respective annual growth rates in sales of 5%, 10%, and 20%.
Compare the three firms at the end of Year 4.
Conclusion?
Cummulative Production At End of Year 4Hours Required to Produce
Most Recent Unit
x y =100 x -0.322
A 5% x = [1.00+(1.05)+(1.05)2+(1.05)3](5000) = 15,764 4.453
B 10% x = [1.00+(1.05)+(1.05)2+(1.05)3](5000) = 16,551 4.384
C 20% x = [1.00+(1.05)+(1.05)2+(1.05)3](5000) = 18,202 4.252
Firm
Annual Growth Rate
in Sales
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Effect of Sales’Effect of Sales’Annual Growth Rate Annual Growth Rate (continued)(continued)
Learning Curve
0
2
4
6
8
5,000 10,000 15,000 20,000
Cumulative Production
Hou
rs R
equi
red
to P
rodu
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M
ost R
ecen
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Learning Curve Firm A Firm B Firm C
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Strategic ApplicationsStrategic Applicationsof a Learning Curveof a Learning Curve
Frequent Decreases in Selling Price.
Each decrease in selling price increases your market share, which in turn leads to a “faster ride” down the learning curve, which in turn makes it tougher for your competitors.
Reinvest Increased Profits
As the hours required to produce the most recent unit continually decreases, the cost to produce the unit continually decreases. Therefore, your profits increase. You can reinvest the incremental profit to improve the product or the production process, or you can reinvest the incremental profit in another area of the firm.
As the hours required to produce the most recent unit continually decreases, the cost to produce the unit continually decreases. Therefore, you can frequently decrease the selling price without decreasing total profit.
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How do we determine the How do we determine the parameters of a Learning Curve?parameters of a Learning Curve?
From previous slides, we know that, to model a learning curve, we proceed as follows:
Let x = cumulative production
y = hours required to produce the x-th unit
Then, y = ax-b
where a and b are parameters defined as follows:
a = hours required to produce the 1-st unit
b = a value related to the percentage associated with the learning curve
For a given set of data, how do we determine the specific values of a and b?
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ExampleExample
For the Learning curve y=ax-b, how do we determine the specific values of a and b?
We begin by taking the natural logs of both sides of y=ax-b.
1 4000
7 255025 185065 1500180 1170
Cumulative Production
Hours Required to Produce
Most Recent Unit
Learning Curve Data
0500
10001500
20002500
30003500
40004500
0 50 100 150 200
Cumulative Production
Ho
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to P
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the
Mo
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Un
it
Note the linear relationship between ln(x) and ln(y).
This suggests taking the natural logs of the data.
)*ln()ln( xbaybaxy
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Example Example (continued)(continued)
Natural Log Natural Log
0.000 8.2941.946 7.8443.219 7.5234.174 7.313
5.193 7.065
Cumulative Production
Hours Required to Produce
Most Recent Unit
Note the approximate linear relationship between ln(Cumulative Production) and ln (Hours Required).
Natural Log of Learning Curve Dataln(Cumulative Production) versus ln(Hours Required)
7.0
7.5
8.0
8.5
0 1 2 3 4 5 6
ln(Cumulative Production)
ln(H
ou
rs R
eq
uir
ed
)
We can use the statistical technique of Regression to determine the straight line that “best fits” the data.
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Example Example (continued)(continued)
Best Linear Fit (via Regression)ln(Cumulative Production) versus ln(Hours Required)
7.0
7.5
8.0
8.5
0 1 2 3 4 5 6
ln(Cumulative Production)
ln(H
ours
Req
uire
d)
ln(Data) Best Linear Fit
Using Excel’s Regression Tool, we obtain
ln(y) = 8.29642 – 0.23694 ln(x)
Intercept=8.29642
Negative of Slope = 0.23694
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Example Example (continued)(continued)
From the previous slide, we know
ln(y) = 8.29642 – 0.23694 ln(x)
So,
eln(y) = e[8.29642 – 0.23694 ln(x)]
or, equivalently, the equation for the Learning Curve is
So, in our example, we have a Learning Curve that is close to but just below an 85% learning curve.
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Excel TemplateExcel Templatefor a Learning Curvefor a Learning Curve
A B C D E F G H I2
3 Hours Required LN of Regression4 Cumulative to Produce Cumulative LN of Estimate of5 Production Most Recent Unit Production Hours Required Hours Required6 1 4000 0.000 8.294 4009.57 7 2550 1.946 7.844 2528.48 25 1850 3.219 7.523 1870.19 65 1500 4.174 7.313 1491.2
10 180 1170 5.193 7.065 1171.4
1112131415 SUMMARY OUTPUT1617 Regression Statistics18 Multiple R 0.99987412719 R Square 0.9997482720 Adjusted R Square 0.9996643621 Standard Error 0.00876481622 Observations 52324 ANOVA25 df SS MS F Significance F26 Regression 1 0.915298765 0.915298765 11914.53938 1.69521E-0627 Residual 3 0.000230466 7.6822E-0528 Total 4 0.9155292312930 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%31 Intercept 8.296416964 0.007427526 1116.982525 1.58245E-09 8.272779238 8.320054689 8.272779238 8.32005468932 X Variable 1 -0.236941604 0.002170714 -109.1537419 1.69521E-06 -0.243849792 -0.230033415 -0.243849792 -0.23003341533 4009.48134
Rows 15-32 generated using the menu selection "Tools, Data Analysis, Regression" with Input X-Range of E6:E10 Input Y-Range of F6:F10 Output Range of A15
=LN(B8) =LN(C8) =$B$33*(B8^$B$32)
=EXP(B31)
NOTE: Regression output in cells B31 and B32 shows that LN(Hours Required) = 8.296 - 0.237*LN(Cumulative Production) or, equivalently, (Hours Required) = 4009.5*[(Cumulative Production)^(-0.237)]
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A “Not So Nice” ExampleA “Not So Nice” Example
In our example, there was a very close linear relationship between
ln(Cumulative Production) and ln(Hours Required)
This is NOT the typical situation.
A more typical situation is shown on the next slide.
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A “Not So Nice” ExampleA “Not So Nice” Example(continued)(continued)
Pareto Chart for a RestaurantPareto Chart for a Restaurant
EXAMPLE 4.2
The manager of a neighborhood restaurant is concerned about the smaller numbers of customers patronizing his eatery. Complaints have been rising, and he would like to find out what issues to address and present the findings in a way his employees can understand.
SOLUTION
The manager surveyed his customers over several weeks and collected the following data:
Pareto Chart for a RestaurantPareto Chart for a Restaurant
50 –45 –40 –35 –30 –25 –20 –10 –5 –0 –
Fa
ilu
res
Discourteous server
Slow service
Cold dinner
Cramped tables
Atmosphere
Failure Name
Figure 4.9 – Bar Chart
Figure 4.9 is a bar chart and Figure 4.10 is a Pareto chart, both created with OM Explorer’s Bar, Pareto, and Line Charts solver. They present the data in a way that shows which complaints are more prevalent (the vital few).
Analysis of Flight Departure DelaysAnalysis of Flight Departure Delays
EXAMPLE 4.3
The operations manager for Checker Board Airlines at Port Columbus International Airport noticed an increase in the number of delayed flight departures.
SOLUTION
To analyze all the possible causes of that problem, the manager constructed a cause-and-effect diagram, shown in Figure 4.11. The main problem, delayed flight departures, is the “head” of the diagram. He brainstormed all possible causes with his staff, and together they identified several major categories: equipment, personnel, materials, procedures, and “other factors” that are beyond managerial control. Several suspected causes were identified for each major category.
Causes of Headliner Process FailuresCauses of Headliner Process Failures
EXAMPLE 4.4
The Wellington Fiber Board Company produces headliners, the fiberglass components that form the inner roof of passenger cars. Management wanted to identify which process failures were most prevalent and to find the cause.
SOLUTION
Step 1: A checklist of different types of process failures is constructed from last month’s production records.
Step 2: A Pareto chart is prepared from the checklist data.
Step 3: A cause-and-effect diagram for identifies several potential causes for the problem.
Step 4: The manager reorganizes the production reports into a bar chart according to shift because the personnel on the three shifts had varied amounts of experience.
Total cost of “enter, process, and track orders” per $1,000 revenue System costs of processes per $100,000 revenue Value of sales order line item not fulfilled due to stockout, as percentage of
revenue Average time from sales order receipt until manufacturing logistics is
notified Average time in direct contact with customer per sales order line item
Order Fulfillment Process
Value of plant shipments per employee Finished goods inventory turnover Reject rate as percentage of total orders processed Percentage of orders returned by customers due to quality problems Standard customer lead time from order entry to shipment Percentage of orders shipped on time
Figure 4.13 – Illustrative Benchmarking Metrics by Type of Process
Percentage of sales due to services/products launched last year Cost of “generate new services/products” process per $1,000 revenue Ratio of projects entering the process to projects completing the process Time to market for existing service/product improvement project Time to market for new service/product project Time to profitability for existing service/product improvement project
Supplier Relationship Process
Cost of “select suppliers and develop/maintain contracts” process per $1,000 revenue
Number of employees per $1,000 of purchases Percentage of purchase orders approved electronically Average time to place a purchase order Total number of active vendors per $1,000 of purchases Percentage of value of purchased material that is supplier certified
Figure 4.13 – Illustrative Benchmarking Metrics by Type of Process
Systems cost of finance function per $1,000 revenue Percentage of finance staff devoted to internal audit Total cost of payroll processes per $1,000 revenue Number of accepted jobs as percentage of job offers Total cost of “source, recruit, and select” process per $1,000 revenue Average employee turnover rate
Figure 4.13 – Illustrative Benchmarking Metrics by Type of Process
Create a flowchart for the following telephone-ordering process at a retail chain that specializes in selling books and music CDs. It provides an ordering system via the telephone to its time-sensitive customers besides its regular store sales.
The automated system greets customers, asks them to choose a tone or pulse phone, and routes them accordingly.
The system checks to see whether customers have an existing account. They can wait for the service representative to open a new account.
Customers choose between order options and are routed accordingly.
Customers can cancel the order. Finally, the system asks whether the customer has additional requests; if not, the process terminates.
An automobile service is having difficulty providing oil changes in the 29 minutes or less mentioned in its advertising. You are to analyze the process of changing automobile engine oil. The subject of the study is the service mechanic. The process begins when the mechanic directs the customer’s arrival and ends when the customer pays for the services.
SOLUTION
Figure 4.15 shows the completed process chart. The process is broken into 21 steps. A summary of the times and distances traveled is shown in the upper right-hand corner of the process chart.
The times add up to 28 minutes, which does not allow much room for error if the 29-minute guarantee is to be met and the mechanic travels a total of 420 feet.
Vera Johnson and Merris Williams manufacture vanishing cream. Their packaging process has four steps: (1) mix, (2) fill, (3) cap, and (4) label. They have had the reported defects analyzed, which shows the following:
Draw a Pareto chart to identify the vital defects.