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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 Method

2.Divide Task Into Work Elements

3.Time Work Elements

4.Rate Worker’s Performance

5.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 ns=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 = (ns – 1) = 10 – 1 = 9

– α = 0.05, α/2 = 0.025– tα/2 = t0.025 = 2.262

– Number of cycles = nc = (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

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www.stopwatchesusa.com

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Time Study Results(from text)