class 09.1 - Time Study - Oregon State...
Transcript of class 09.1 - Time Study - Oregon State...
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IE 366
Chapter 25
Time Study
Supplementary Material from:Groover, M.P. (2007). Work Systems and the Methods, Measurement, and Management of Work, Upper Saddle River, NJ: Pearson Prentice Hall, pp. 319 - 360.
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Time Study● Also known as
– Direct Time Study– Stopwatch Time Study
● Involves– direct, continuous observation of a task– using a time measurement instrument– to record time taken to complete task.
● Allowances made for– personal needs– fatigue– unavoidable delays
● Dates back to 1883● Inextricably connected with origins and early history of IE
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Direct Time Study Procedure
1.Define and Document Standard Method2.Divide Task Into Work Elements3.Time Work Elements4.Rate Worker’s Performance5.Apply Allowances
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Define and Document Standard Method● Goal: “one best method”● Seek worker’s advice, if appropriate● Elements of Document
– Procedure (steps, actions, work elements, hand/body motions)
– Tools, equipment– Machine settings (e.g., feeds, speeds)– Workplace layout– Frequency of irregular elements– Working conditions– Setup
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Reasons For Thorough Documentation● Batch production (likely to be repeated)● Methods improvement by operator● Disputes about method (too tight?)● Data for standard data system
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Divide Task Into Work Elements● Series of motion activities logically grouped because
of unified purpose.● Guidelines
– Each work element should consist of a logical group of motion elements.
● e.g., reach, grasp, move, place– Beginning point of one element should be end of previous.
● No time gap between elements.– Each element should have readily identifiable end point.
● i.e., easily detected, no ambiguity– Work elements should not be too long.
● < “several” min
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Divide Task Into Work Elements● Guidelines (continued)
– Work elements should not be too short.● > 3 sec
– Irregular elements should be identified & distinguished.● i.e., not every cycle
– Manual elements should be separated from machine elements.
● generally constant values– Internal elements should be separated from external
elements.● i.e., performed by operator during machine cycle
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“Irregular” and “Foreign” Elements● Elements that occur routinely, but not every cycle – should be
included: irregular elements (Groover)● Elements that the observer didn’t anticipate –probably should
be included: irregular elements (Konz & Johnson)● Elements that are not normal work – should not be included:
foreign elements (Konz & Johnson)
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Time Work Elements● Collect data on time study form (on clipboard).
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Time Work Elements (2)● Collect data on time study form (on clipboard).
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Time Work Elements (3)● Use stopwatch calibrated in 0.01 minutes:
– Snapback method● Start watch at beginning of every element.● “Snap” watch back to zero at end of element.● Record time.● Advantages
– Element variations easily observable– No subtraction
– Continuous method● Start watch at beginning of observation (or beginning of each
cycle)● Record elapsed time at end of each element.● Let it run …● Advantages
– Not so much manipulation of stopwatch– Elements not so easily omitted– Regular/irregular elements more readily distinguished (?)
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Rate Worker’s Performance● Standard performance = 100%● Rate
– Individual elements– Or entire work cycle
● Most difficult & controversial step in time study● Requires analyst’s judgment
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Apply AllowancePure Manual Work
Work Element a b c d*
Obs. Time 0.56 min 0.25 min 0.50 min 1.10 min
Perf. Rating 100% 80% 110% 100%
PFD Allowance = 15%
* irregular element performed every 5 cycles* irregular element performed every 5 cycles
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Apply AllowancePure Manual Work
Work Element a b c d*
Obs. Time 0.56 min 0.25 min 0.50 min 1.10 min
Perf. Rating 100% 80% 110% 100%
PFD Allowance = 15%
Normal Time:NT = 0.56(1.00) + 0.25(0.80) + 0.50(1.10) + 1.10(1.00)/5 = 0.56 + 0.20 + 0.55 + 0.22 = 1.53 min
Standard Time:ST = 1.53(1 +0.15) = 1.76 min
* irregular element performed every 5 cycles* irregular element performed every 5 cycles
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Apply AllowanceTask Including Machine Cycle
Work Element a b c d*
Obs. Time 0.22 min 0.65 min 0.47 min 0.75 min
Perf. Rating 100% 80% 100% 100%
Mach. Time (idle)m
1.56 min (idle) (idle)
PFD Allowance = 15%Machine Allowance = 20%
* irregular element performed every 15 cycles* irregular element performed every 15 cycles
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Apply AllowanceTask Including Machine Cycle
Work Element a b c d*
Obs. Time 0.22 min 0.65 min 0.47 min 0.75 min
Perf. Rating 100% 80% 100% 100%
Mach. Time (idle)m
1.56 min (idle) (idle)
PFD Allowance = 15%Machine Allowance = 20%
Normal Time:NT = 0.22(1.00) + Max{0.65(0.80), 1.56} + 0.47(1.00) + 0.75(1.00)/15 = 0.22 + 1.56 + 0.47 + 0.05 = 2.30 min
Standard Time:ST = (0.22 +0.47 + 0.05)(1 + 0.15) + Max{0.52(1 + 0.15), 1.56(1 + 0.20)} = 0.85 + 1.87 = 2.72 min
* irregular element performed every 15 cycles* irregular element performed every 15 cycles
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Number of Cycles To Be Timed
α/2 α/2
1-α
x
μX
Overall, P(μx lies within x + zα/2 [σ/√n
c ]) = 1 – α
where nc = number of cycles timed
But σ unknown, so take preliminary sample of ns observations and use
∑(x-x)2
s = n
s-1
Let X be a random variable, time of one work element in a task.Time several cycles to estimate true mean:
α/2 α/2
1-α
x
α/2 α/2
1-α
x
low estimate high estimate
close estimate
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Number of Cycles To Be Timed (2)
P(μ lies within x + tα/2 [s/√nc ]) = 1 – α
Interval size = x + kx
where k = proportion of sample mean (e.g., if k = 10%, interval size = x + 0.10 x)
kx = tα/2 [s/√nc ] (remember, s is an estimate of σ based on preliminary sample of n
s)
So, rearranging,
nc = (tα/2s / kx)2
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Number of Cycles To Be Timed: Example● From preliminary study, engineer has collected n
s=10 samples on one work
element– x = 0.40 min – s = 0.07 (an estimate of σ based on preliminary sample of n
s=10)
● How many cycles should be timed to ensure actual element time is + 10% of the sample mean, with 95% confidence?– df = (n
s – 1) = 10 – 1 = 9
– α = 0.05, α/2 = 0.025– tα/2 = t0.025 = 2.262– Number of cycles = n
c = (tα/2s / kx)2 = [2.262(0.07) / 0.10(0.40)]2
= 15.7 ≈ 16 cycles● If 16 observed cycles yields
– x = 0.45 min
P(μx lies within x + kx) = P(μ
x lies within 0.45 + 0.10(0.45))
= P(μx lies within [0.405, 0.495]) = 95%
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Performance Rating● Also called performance leveling.● Performance relative to engineer’s concept of “standard” performance.● Most common method based on speed or pace: speed rating.
– > 100% means faster than standard pace– engineer must use judgment– must consider
● degree of difficulty of work element● worker’s pace relative to standard
● Standards– Walk 3 mi/hr on flat, level ground, no load, 27-in steps.
● Problem: few work situations lend themselves to such precise measurement.● However, many situations in which experts judge (e.g., gymnastics, dog shows)● Solutions
– Experience (including feedback)– Training (e.g., using training films)
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Performance Rating (2)
● Pace depends on worker’s– skill– experience– exertion level– attitude toward time study
● So, select skilled worker– familiar with job– Accepts time study as necessary management tool
● Characteristics of good performance rating system– consistency among tasks (one task to another)– consistency among engineers– easily understood– related to standard performance (well-defined concept)– machine-paced elements rated at 100% (no worker control of machine)– rating recorded during observation, not after– worker notification
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Time Study Issue
● Why is time study important to the organization?
● What are some worker concerns?● How can they be resolved?
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Time Study Equipment: Mechanical Stopwatch 1
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Time Study Equipment: Mechanical Stopwatch 2