H – 1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Measuring Output Rates H For Operations Management, 9e by Krajewski/Ritzman/Malhotra

H – 1Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Measuring Output RatesMeasuring Output RatesH

For For Operations Management, 9eOperations Management, 9e by by Krajewski/Ritzman/Malhotra Krajewski/Ritzman/Malhotra © 2010 Pearson Education© 2010 Pearson Education

PowerPoint Slides PowerPoint Slides by Jeff Heylby Jeff Heyl


Work Standards Work Standards

A work standard is the time required for a trained worker to perform a task following a prescribed method with normal effort and skill

Used in the following ways: Establishing prices and costs Motivating workers Comparing alternative process designs Scheduling Capacity planning Performance appraisal


Methods of Work Measurement Methods of Work Measurement

The time study method

The elemental standard data approach

The predetermined data approach

The work sampling method


The Time Study MethodThe Time Study Method

Time study is the method used most often Step 1: Selecting work elements

Step 2: Timing the elements

Step 3: Determining sample size2

tpz

n

wheren =required sample sizep =precision of the estimate as a proportion of the true valuet =select time for a work element =standard deviation of representative observed times for a work elementz =number of normal standard deviations needed for the desired confidence



Typical values of z for this formula are as follows:

Desired Confidence (%) z

90 1.65

95 1.96

96 2.05

97 2.17

98 2.33

99 2.58


Estimating the Sample Size in a Estimating the Sample Size in a Time StudyTime Study

EXAMPLE H.1

A coffee cup packaging operation has four work elements. A preliminary study provided the following results:

Work Element

Standard Deviation, , (min)

Select Time, t, (min)

Sample Size

1. Get two cartons 0.0305 0.50 5

2. Put liner in carton 0.0171 0.11 10

3. Place cups in carton 0.0226 0.71 10

4. Seal carton and set aside 0.0241 1.10 10

Work element 1 was observed only five times because it occurs once every two work cycles. The study covered the packaging of 10 cartons. Determine the appropriate sample size if the estimate for the select time for any work element is to be within 4 percent of the true mean 95 percent of the time.


Estimating the Sample Size in a Estimating the Sample Size in a Time StudyTime Study

SOLUTION

For this problem,

Work Element

Standard Deviation, , (min)

Select Time, t, (min)

Sample Size

1. Get two cartons 0.0305 0.50 5

2. Put liner in carton 0.0171 0.11 10

3. Place cups in carton 0.0226 0.71 10

4. Seal carton and set aside 0.0241 1.10 10

Work element 1: n =

Work element 2: n =

Work element 3: n =

Work element 4: n =

2

500003050

040961

..

.

.= 9

2

11001710

040961

..

.

.= 58

2

71002260

040961

..

.

.= 3

2

10102410

040961

..

.

.= 2



Step 4: Setting the standard (ST)

Apply subjective performance rating factor (RF), calculate normal times (NT), normal time for the cycle (NTC), and adjust for allowances

NTC = NT

ST = NTC(1 + A)

where F = the frequency of occurrenceA = proportion of the normal time added for allowances

NT = t(F)(RF)


Determining the Normal TimeDetermining the Normal Time

EXAMPLE H.2

Suppose that 48 additional observations of the coffee cup packaging operation were taken and the following data were recorded:

Work Element t F RF

1 0.53 0.50 1.05

2 0.10 1.00 0.95

3 0.75 1.00 1.10

4 1.08 1.00 0.90

Because element 1 occurs only every other cycle, its average time per cycle must be half its average observed time. That is why F1 = 0.50 for that element. All others occur every cycle. What are the normal times for each work element and for the complete cycle?


Determining the Normal TimeDetermining the Normal Time

SOLUTION

The normal times are calculated as follows:

Work element 1: NT1 =




0.53(0.50)(1.50) = 0.28 minute

0.10(1.00)(0.95) = 0.10 minute

0.75(1.00)(1.10) = 0.83 minute

1.08(1.00)(0.90) = 0.97 minute

The normal time for the complete cycle is 2.18 minutes


Determining the Standard Time

EXAMPLE H.3

Management needs a standard time for the coffee cup packaging operation. Suppose that A = 0.15 of the normal time. What is the standard time for the coffee cup packaging operation, and how many cartons can be expected per 8-hour day?

SOLUTION

For A = 0.15 of the normal time,

ST = 2.18(1 + 0.15) = 2.51 minutes/carton


Application H.1Application H.1

Lucy and Ethel have repetitive jobs at the candy factory. Management desires to establish a time standard for this work for which they can be 95% confident to be within ± 6% of the true mean. There are three work elements involved:

SOLUTION

Step 1: Selecting work elements

#1: Pick up wrapper paper and wrap one piece of candy

#2: Put candy in a box, one at a time

#3: When the box is full (4 pieces), close it and place on conveyor



Step 2: Timing the elements. Select an average trained worker, Lucy will suffice.

Element

Initial Observation Cycle Number, Minutes

1 2 3 4 5 6 7 8 9

Wrap #1: .10 .08 .08 .12 .10 .10 .12 .09 .11

Pack #2: .10 .08 .08 .11 .06 .98* .17 .11 .09

Close #3: .27 ... ... ... .34 ... ... ... .29

SelectTime, t

Standard Dev,

* Lucy had some rare and unusual difficulties; don't use this observation.



Step 2: Timing the elements. Select an average trained worker, Lucy will suffice.

Element

Initial Observation Cycle Number, Minutes

1 2 3 4 5 6 7 8 9

Wrap #1: .10 .08 .08 .12 .10 .10 .12 .09 .11

Pack #2: .10 .08 .08 .11 .06 .98* .17 .11 .09

Close #3: .27 ... ... ... .34 ... ... ... .29

SelectTime, t

Standard Dev,

0.1 0.015

0.1 0.03295

0.3 0.03606

Step 3: Determining sample size. First calculate t for each element in Step 2. Assume a 95% confidence interval, with z = 1.96. The precision interval of ± 6% of the true mean implies p = 0.06. To determine the sample size, use the largest value of / t .

116100

032950060961

22

..

.

.tp

zn


Element Select Time, t Frequency Rating Factor Normal Time

Wrap #1: 1.00 1.2

Pack #2: 1.00 0.9

Close #3: 0.25 0.8


Step 4: Setting the standard.

a. The analyst subjectively assigns a rating factor.

b. Determine the normal time (NT) for each work element, given the following rating factors.

.10

.10

.30

0.12

0.09

0.06

Note that for the third element, the frequency is 0.25 because closing the box occurs only once every four cycles.

c. Determine the normal time for the cycle.

NTC = NT = 0.12 + 0.09 + 0.06 = 0.27 minutes


Element Select Time, t Frequency Rating Factor Normal Time

Wrap #1: 1.00 1.2

Pack #2: 1.00 0.9

Close #3: 0.25 0.8


d. Subjectively determine the proportion of the normal time to be added for allowance, and then calculate standard time ST. Let the allowance be 18.5% of the normal time (A = .185).

.10

.10

.30

0.12

0.09

0.06

ST = NTC(1 + A) = 0.27(1.185) = 0.32 minutes


Overall AssessmentOverall Assessment

Most frequently used method for setting time standards

Qualified analysts can typically set reasonable standards Not appropriate for “thinking” jobs Not appropriate for non-repetitive jobs Inexperienced persons should not conduct time

studies because errors can result in unreasonable standards

Workers may object to judgment and subjectivity involved


Elemental Standard Data ApproachElemental Standard Data Approach

Useful for processes with high divergence, but when a high degree of similarity exists for basic elements of work for different services and processes Time standards are developed for common work

elements Study results are stored in a database for later use in

establishing standards for jobs requiring those elements Allowances must still be added An equation may be used to account for the effect on

time required by certain variable characteristics of the jobs

This approach reduces the number of time studies needed, but does not eliminate time studies


Predetermined Data ApproachPredetermined Data Approach

Break each work element into micromotions: reach, move, disengage, apply pressure, grasp, position, release, and turn

Step 1: Break each work element into its basic micromotions

Step 2: Find the proper tabular value of time for each micromotion

Step 3: Normal times of micromotions are added for the task

Step 4: Adjust for allowances to arrive at the standard time


TABLE H.1 | MTM PREDETERMINED DATA FOR THE MOVE MICROMOTION

Time TMU Wt. Allowance

Distance Moved

(in.)A B C

Hand in Motion

B

Wt. (lb) Up to

Dynamic Factor

Static Constant (TMU)

3/4 or less 2.0 2.0 2.0 1.7

1 2.5 2.9 3.4 2.3 2.5 1.00 0

2 3.6 4.6 5.2 2.9

3 4.9 5.7 6.7 3.6 7.5 1.06 2.2

4 6.1 6.9 8.0 4.3

5 7.3 8.0 9.2 5.0 12.5 1.11 3.9

6 8.1 8.9 10.3 5.7

7 8.9 9.7 11.1 6.5 17.5 1.17 5.6

8 9.7 10.6 11.8 7.2

9 10.5 11.5 12.7 7.9 22.5 1.22 7.4




Advantages Standards can be set for new jobs Work methods can be compared without a

time study Greater consistency of results Reduces the problem of biased judgment



Disadvantages Impractical for jobs with low repeatability Data may not reflect the actual situation in a

specific plant Performance time variations can result from

many factors Actual time may depend on the specific

sequence of motions Considerable skill is required to achieve good

standards


Work Sampling Method Work Sampling Method

Results in a proportion of time spent doing an activity, rather than a standard time for the work

Requires a large number of random observations spread over the length of the study

Proportion of observations in which the activity occurs is assumed to be the proportion of time spent on the activity in general



Step 1. Define the activities

Step 2. Design the observation form

Step 3. Determine the length of the study

Step 4. Determine the initial sample size

Step 5. Select random observation times using a random number table

Step 6. Determine observer schedule

Step 7. Observe the activities and record the data

Step 8. Decide whether further sampling is required



Select a sample size so that the estimate of the proportion of time spent on a particular activity that does not differ from the true proportion by more than a specified error, so

eppep ˆˆˆ

where=sample proportion (number of occurences divided by the sample size)e =maximum error in the estimate

p̂



As the binomial distribution applies, the maximum error of the estimate is

n

ppze

ˆˆ

1

wheren =sample sizez =number of standard deviations needed to achieve the desired confidence

Solving for n

ppez

n ˆˆ

1

2



n

ppze

ˆˆ

1

Confidence interval

Probability that trueproportion will fall

within confidence interval

ep ˆ p̂ ep ˆ

Figure H.1 – Confidence Interval for a Work Sampling Study


Using Work Sampling DataUsing Work Sampling Data

EXAMPLE H.4

The hospital administrator at a private hospital is considering a proposal for installing an automated medical records storage and retrieval system. To determine the advisability of purchasing such a system, the administrator needs to know the proportion of time that registered nurses (RNs) and licensed vocational nurses (LVNs) spend accessing records. Currently, these nurses must either retrieve the records manually or have them copied and sent to their wards. A typical ward, staffed by eight RNs and four LVNs, is selected for the study.



a. The hospital administrator estimates that accessing records takes about 20 percent of the RNs’ time and about 5 percent of the LVNs’ time. The administrator wants 95 percent confidence that the estimate for each category of nurses falls within 0.03 of the true proportion. What should the sample size be?

b. The hospital administrator estimates that the annual amortization cost and expenses for maintaining the new automated medical records storage and retrieval system will be $150,000. The supplier of the new system estimates that the system will reduce the amount of time the nurses spend accessing records by 25 percent. The total annual salary expense for RNs in the hospital is $3,628,000, and for LVNs it is $2,375,000. The hospital administrator assumes that nurses could productively use any time saved by the new system. The pilot work sampling study resulted in the data shown in Figure H.2. Should the administrator purchase the new system?


124 258 223 83 688

28 251 46 19 344


Activity

Accessing records

Attending to patients

Other support

activitiesIdle or break

Total observations

RN

LVN

Figure H.2 – Results of the Initial Study



SOLUTION

a. Using estimates for the proportion of time spent accessing records of 0.20 for RNs and 0.05 for LVNs, an error of ± 0.03 for each, and a 95 percent confidence interval (z = 1.96), we recommend the following sample sizes:

683800200030961 2

..

.

.n

203950050030961 2

..

.

.n

RN:

LVN:



Eight RNs and four LVNs can be observed on each trip. Therefore, 683/8 = 86 (rounded up) trips are needed for the observations of RNs, and only 203/4 = 51 (rounded up) trips are needed for the LVNs. Thus, 86 trips through the ward will be sufficient for observing both nurse groups. This number of trips will generate 688 observations of RNs and 344 observations of LVNs. It will provide many more observations than are needed for the LVNs, but the added observations may as well be recorded as the observer will be going through the ward anyway.



b. Before using the estimates from the work sampling study, we must be sure that additional sampling is not required. Figure H.2 shows that RNs accessed records 124 times and LVNs only 28 times. The computer output shows that the proportion of working time spent on accessing records is 0.1802 for the RNs and 0.0814 for the LVNs. Thus, the original estimates were off the mark. The computer uses the new estimates for the proportions in the same formulas we used in part (a) to revise the sample sizes. However, the new sample sizes are smaller than those already used, so no additional sampling is required. If the sample sizes were too small for the proportions found, additional sampling would have to be performed. In addition, the confidence interval shows the range possible in the “true” proportions, based on the results of the pilot study. For example, the actual proportion of time spent by the RNs on accessing records could be as low as 0.15 and as high as 0.21.



Confidence Interval

Workgroup Total Obs.

Activity Obs.

Proportion of Total Lower Upper

Required Sample

Size

RN 688 124 0.1802 0.15151 0.2090 631

LVN 344 28 0.0814 0.05250 0.1113 320

Because the nurses will not be using the system all the time, we accept the supplier’s estimate of 25 percent to determine the value of the time spent accessing records. Estimated annual net savings from the purchase of the automatic medical records storage and retrieval system are

Net savings = 0.25[($3,628,000)(0.18) + ($2,375,000)(0.08)] - $150,000

= $60,760



Major League Baseball (MLB) is concerned about excessive game duration. Batters now spend a lot of time between pitches when they leave the box to check signals with coaches, and then go through a lengthy routine including stretching and a variety of other actions. Pitching routines are similarly elaborate. In order to speed up the game, it has been proposed to prohibit batters from leaving the box and to prohibit pitchers from leaving the mound after called balls and strikes. MLB estimates the proportion of time spent in these delays to be 20% of the total game time. Before they institute a rules change, MLB would like to be 95% confident that the result of a study will show a proportion of time wasted that is accurate within ± 4% of the true proportion.



SOLUTION

Steps 1 and 2. Define the activities and design the observation form.

Step 3. Determine the length of the study. Suppose that ten games (or 32 hours) are appropriate.

Step 4. Determine the initial sample size.

nsobservatio 3852001200040961 2

..

.

.n

Steps 5 and 6. Determine the observer schedule.

hour per nsobservatio 12 or 12.0332

385



Step 7. Observe the activities and record the data. You find 96 unacceptable delays for pitchers and 46 unacceptable delays for batters.

Step 8. Check to see whether additional sampling is required.

nsobservatio 4502501250040961 2

..

.

.

nsobservatio 2541201120040961 2

..

.

.

For pitchers:

For batters:

n

n

Thus 65 more observations are needed (450 – 385)


Overall AssessmentOverall Assessment

Advantages No special training required of observers Several studies can be conducted simultaneously More economical for jobs having long cycle times Workers prefer this method to time studies

Disadvantages A large number of observations are required Usually not used for repetitive, well-defined jobs Workers may increase quantity at the expense of quality


Managerial Considerations Managerial Considerations

Managers should carefully evaluate work measurement techniques to ensure that they are used in ways that are consistent with the firm’s competitive priorities

Technological changes Increased automation There is less need to observe and rate worker

performance, because work is machine paced Work sampling may be electronically

monitored


Solved Problem 1Solved Problem 1

For a time study of a health insurance claims-adjusting process, the analyst uses the continuous method of recording times. The job is divided into four work elements. Shown in Figure H.3 are the performance rating factors, RF, and the continuous method recorded times, r, for each work element.

a. Calculate the normal time for this job.

b. Calculate the standard time for this job, assuming that the allowance is 20 percent of the normal time.

c. What is the appropriate sample size for estimating the time for element 2 within ± 10 percent of the true mean with 95 percent confidence?


Operation: Insurance claim processing Date: 10/07 Observer: Jennifer Johnson

Work Element

Observations

t RF 1 2 3 4 5

1. Check form completion and signatures

t

r

2. Enter claim amounts, check math

t

r

3. Determine proportion of claim to be disallowed

t

r

4. Generate form letter, enter data for check

t

r


0.50

0.50

0.52 1.1 0.0570

0.24 1.2 0.0742

0.65 1.2 0.0791

1.20 0.9 0.1000

0.20

0.70

0.75

1.45

1.30

2.75

0.55

3.30

0.15

3.45

0.60

4.05

1.20

5.25

0.45

5.70

0.25

5.95

0.55

6.50

1.10

7.60

0.60

8.20

0.35

8.55

0.70

9.25

1.10

10.35

0.50

10.85

0.25

11.10

0.65

11.75

1.30

13.05



SOLUTION

a. To get the normal time for this job, we must first determine the observed time, t, for each work element for each cycle. We calculate the time for each observation by finding the difference between successive recorded times, r. For example, the time for the fifth observation of the first work element is the difference between the recorded time when that element was completed and the time when the fourth observation of the fourth work element was completed. With no extreme variation in the observed times for the work elements, they are representative of the process. All the data can be used for calculating the average observed time, called the select time, t, and the standard deviation of the observed times, . The results of those calculations are given in Figure H.3. Every work element occurs during every cycle, so the frequency, F, equals 1.



The normal times are calculated as

RFNT1 Ft

Work element 1: NT1 = (0.52)(1)(1.1) = 0.572 minute



Work element 4: NT4 = (1.20)(1)(0.9) = 1.080 minutes

Total = 2.720 minutes



b. Standard time = (Normal time per cycle)(1.0 + Allowances), or

ST = NTC(1.0 + A) = 2.72(1.0 + 0.2)

= 3.264 minutes

c. The appropriate sample size for 95 percent confidence that the select time for work element 2 is within 10 percent of the true mean is

2

tpz

n 2

24007420

100961

..

.

.

= 36.72, or 37 observations



A library administrator wants to determine the proportion of time the circulation clerk is idle. The following information was gathered randomly by using work sampling:

DayNumber of Time

Clerk BusyNumber of Time

Clerk IdleTotal Number of

Observations

Monday 8 2 10

Tuesday 7 1 8

Wednesday 9 3 12

Thursday 7 3 10

Friday 8 2 10

Saturday 6 4 10

If the administrator wants a 95 percent confidence level and a degree of precision of ± 4 percent, how many more observations are needed?



2506015 ./ˆ p

SOLUTION

The total number of observations made was 60. The clerk was observed to be idle 15 times. The initial estimate of the sample proportion is . The required sample size for a precision of 4 percent is

2

2 1e

ppzn

ˆˆ 2

2

040

750250961

.

...

nsobservatio 451 or 450.19,

As 60 observations have already been made, an additional 391 are needed.


Documents

H – 1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Measuring Output Rates H For Operations Management, 9e by Krajewski/Ritzman/Malhotra