MARMARA UNIVERSITY INDUSTRIAL ENGINERING

IE 3025 PRODUCTION PLANNING AND CONTROL -I

2020-2021 FALL

TIME STUDY
∑𝒏
𝒊=𝟏 𝒙𝒊
 Avg. Observed Time (or Actual Time (AT)) :
𝒏

 Normal Time (NT) = (Avg. observed time x Frequency of Occurrence) x Performance Rating

 Standard Time (ST):

a. Allowances expressed as a percent of work time ST= NT* (1+ Allowances)

b. Allowances expressed as a percent of shift time ST = NT / (1- Allowances)

𝜎 2
𝑧
 Sample size determination: 𝑛 = [( ) ( ̅ )]
𝑝 𝑡

𝑧 2
 Work Sampling: 𝑛 = ( ) 𝑝 (1 − 𝑝)
𝑒

Course Lecturer: Asst. Prof. Dr. Merve Er Kara

Teaching Assistant: Res. Asst. Zeynep Tuğçe Kalender
Q1. Compute the normal time for drilling a hole in a steel plate if the following observations (in minutes) have
been observed. Use a rating factor of 1.00.

.24 .25 .29 .24 .27 .25 .245 .19 .20 .23

Answer:

𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒 = 𝐴𝑣𝑔. 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑇𝑖𝑚𝑒 𝑥 𝑃𝑒𝑟𝑓𝑜𝑟𝑚𝑎𝑛𝑐𝑒 𝑅𝑎𝑡𝑖𝑛𝑔

∑𝑛
𝑖=0 𝑥𝑖 0.24+0.25+⋯+0.23 2.405
𝐴𝑣𝑔. 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑇𝑖𝑚𝑒 = = = = 0.2405
𝑛 10 10

𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒 = 0.2405 𝑥 1 = 0.2405

Q2. Provide the missing data in the following table (Assume that allowances expressed as a percent of work time);

Avg. Observed Time Normal Time Standard Time Performance Rating Fatigue Allowance

10.6 11.236 13.48 1.06 20%

7.8 7.2 8.28 0.92 15%
6.5 6.83 7.98 1.05 17%
5.47 6.017 6.92 1.10 15%

Answer:

Row 1
o 𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒 = 𝐴𝑣𝑔. 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑇𝑖𝑚𝑒 𝑥 𝑃𝑒𝑟𝑓𝑜𝑟𝑚𝑎𝑛𝑐𝑒 𝑅𝑎𝑡𝑖𝑛𝑔 = 10.6 𝑥 1.06 = 11.236
o 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑇𝑖𝑚𝑒 = 𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒 𝑥 (1 + 𝐴𝑙𝑙𝑜𝑤𝑎𝑛𝑐𝑒𝑠) = 11.236 𝑥 (1 + 0.20) = 13.48
Row 2
o 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑇𝑖𝑚𝑒 = 𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒 𝑥 (1 + 𝐴𝑙𝑙𝑜𝑤𝑎𝑛𝑐𝑒𝑠) = 7.2 𝑥 (1 + 0.15) = 8.28
𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒 7.2
o 𝑃𝑒𝑟𝑓𝑜𝑟𝑚𝑎𝑛𝑒 𝑅𝑎𝑡𝑖𝑛𝑔 = 𝐴𝑣𝑔.𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑇𝑖𝑚𝑒
= 7.8
= 0.92
Row 3
o 𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒 = 𝐴𝑣𝑔. 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑇𝑖𝑚𝑒 𝑥 𝑃𝑒𝑟𝑓𝑜𝑟𝑚𝑎𝑛𝑐𝑒 𝑅𝑎𝑡𝑖𝑛𝑔 = 6.5 𝑥 1.05 = 6.83
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑇𝑖𝑚𝑒
o 𝐴𝑙𝑙𝑜𝑤𝑎𝑛𝑐𝑒𝑠 = − 1 = 0.168 ≅ 0.17 = 17%
𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒
Row 4
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑇𝑖𝑚𝑒 6.92
o 𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒 = = = 6.017
1+ 𝐴𝑙𝑙𝑜𝑤𝑎𝑛𝑐𝑒𝑠 1+ 0.15
𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒 6.017
o 𝐴𝑣𝑔. 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑇𝑖𝑚𝑒 = = = 5.47
𝑃𝑒𝑟𝑓𝑜𝑟𝑚𝑎𝑛𝑐𝑒 𝑅𝑎𝑡𝑖𝑛𝑔 1.10

Course Lecturer: Asst. Prof. Dr. Merve Er Kara

Teaching Assistant: Res. Asst. Zeynep Tuğçe Kalender
Q3. Figure shows a partially completed time-study worksheet. Determine the standard time for this operation.

Time Study Chart

Work Normal
1 2 3 4 5 6 7 8 9 Sum Avg Rating
Element Time
Cumulative Time 0,09 0,12 0,08 0,11 0,10 0,09 0,13 0,12 0,13
A 0.97 0.11 1,05 0.1155
Element Time 0.09 0,12 0,08 0,11 0,10 0,09 0,13 0,12 0,13
Cumulative Time 0,23 0,28 0,21 0,2 0,24 0,22 0,26 0,25 0,25
B 1.17 0.13 1 0.23
Element Time 0.14 0.16 0.13 0.09 0.14 0.13 0.13 0.13 0.12
Cumulative Time 0,46 0,49 0,46 0,44 0,47 0,47 0,49 0,46 0,48
C 2.08 0.23 0,9 0.21
Element Time 0.23 0.21 0.25 0.24 0.23 0.25 0.23 0.21 0.23
Cumulative Time 0,61 0,66 0,62 0,59 0,69 0,67 0,67 0,66 0,7
D 1.65 0.183 0,85 0.16
Element Time 0.15 0.17 0.16 0.15 0.22 0.20 0.18 0.20 0.22
Cumulative Time 0,7 0,74 0,72 0,68 0,79 0,8 0,76 0,78 0,81
E 0.91 0.10 1 0.10
Element Time 0.09 0.08 0.1 0.09 0.1 0.13 0.09 0.12 0.11
Cumulative Time 1 1,02 0,98 0,99 1,07 1,09 1,02 1,06 1,09
F 2.54 0.28 1,1 0.31
Element Time 0.3 0.28 0.26 0.31 0.28 0.29 0.26 0.28 0.28
Sum 1.0255
Allowance: Personal %5, Fatigue %5, Delay %5 (percent of worktime)

Total Allowance = 0.05 +0.05 +0.05 =0.15

Normal Time = 1.0255

Standard Time = 1.0255 x ( 1+ 0.15) = 1.179 ≅ 1.18

Q4. XYZ chocolate company wants to measure their casting and packing processes. Casting employee is working
with a %10 faster performance and packing employee is working with a %20 slower performance. Allowances
determined as a percentage of shift time and calculated as %15 by the company. According to the observations
given below, calculate the standard time for each process.

Observations (min.)
Casting 26 30 29 31
Packing 4 5 4 3

Casting
∑𝑛
𝑖=0 𝑥𝑖 26+30+29+31
 𝐴𝑣𝑔. 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑇𝑖𝑚𝑒 = = = 29 𝑚𝑖𝑛.
𝑛 4
 𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒 = 𝐴𝑣𝑔. 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑇𝑖𝑚𝑒 𝑥 𝑃𝑒𝑟𝑓𝑜𝑟𝑚𝑎𝑛𝑐𝑒 𝑅𝑎𝑡𝑖𝑛𝑔 = 29 𝑥 1.10 = 31.9 𝑚𝑖𝑛.
𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒 31.9
 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑇𝑖𝑚𝑒 = (1−𝑎𝑙𝑙𝑜𝑤𝑎𝑛𝑐𝑒𝑠) = = 37.5 𝑚𝑖𝑛.
1−0.15

Course Lecturer: Asst. Prof. Dr. Merve Er Kara

Teaching Assistant: Res. Asst. Zeynep Tuğçe Kalender
 Packing
∑𝑛
𝑖=0 𝑥𝑖 4+5+4+3
 𝐴𝑣𝑔. 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑇𝑖𝑚𝑒 = 𝑛
= 4
= 4 𝑚𝑖𝑛.
 𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒 = 𝐴𝑣𝑔. 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑇𝑖𝑚𝑒 𝑥 𝑃𝑒𝑟𝑓𝑜𝑟𝑚𝑎𝑛𝑐𝑒 𝑅𝑎𝑡𝑖𝑛𝑔 = 4 𝑥 0.80 = 3.2 𝑚𝑖𝑛.
𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒 3.2
 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑇𝑖𝑚𝑒 =
(1−𝑎𝑙𝑙𝑜𝑤𝑎𝑛𝑐𝑒𝑠)
=
1−0.15
= 3.76 𝑚𝑖𝑛.

Q5. A manufacturing company has conducted a time study for 10 cycles of a job. The job has five elements, and
the total elemental times (minutes) for each element and performance rating factors are as follows;

Element ∑ 𝒕 (𝒎𝒊𝒏) RF
1 3.61 1.05
2 4.84 0.90
3 2.93 1.00
4 4.91 1.10
5 1.78 0.95

a. Compute the standard time using an allowance factor of 18%.

Element ∑ 𝒕 (𝒎𝒊𝒏) t RF N*t

1 3.61 (3.61 /10)= 0.361 1.05 0.379
2 4.84 0.484 0.90 0.436
3 2.93 0.293 1.00 0.293
4 4.91 0.491 1.10 0.542
5 1.78 0.178 0.95 0.169

𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝑖𝑚𝑒 = ∑(𝑁 ∗ 𝑡 ) = 1.819

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑇𝑖𝑚𝑒 = 1.819 𝑥 ( 1 + 0.18) = 2.146 𝑚𝑖𝑛.

b. Determine the sample size, n, for a time study so there is 98% confidence that the average time computed
from the time study is within 4% of the actual average cycle time. The sample standard deviation is 0.23.

𝑧 𝜎 2 2.33 0.23 2
𝑛 = [( ) ( )] = [( )( )] = 54.97 ≅ 55 𝑐𝑦𝑐𝑙𝑒𝑠
𝑝 𝑡̅ 0.04 ̅̅̅̅̅̅̅
1.807

𝑡̅ = 𝑠𝑢𝑚𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑡𝑖𝑚𝑒𝑠 ( 𝑐𝑜𝑙𝑢𝑚𝑛 𝑡)

Course Lecturer: Asst. Prof. Dr. Merve Er Kara

Teaching Assistant: Res. Asst. Zeynep Tuğçe Kalender
Q6. Lucy and Ethel have repetitive jobs at the candy factory. There are three work elements involved:

Initial Observation Cycle Number, Minutes

Element 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

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. (Let the allowance be 18.5% of the normal time (A = .185).)

Initial Observation Cycle Number, Minutes

Element 1 2 3 4 5 6 7 8 9

Wrap #1: .10 .08 .08 .12 .10 .10 .12 .09 .11 0.1 0.015
Pack #2: .10 .08 .08 .11 .06 .98* .17 .11 .09 0.1 0.03295
Close #3: .27 ... ... ... .34 ... ... ... .29 0.3 0.03606

*Lucy had some rare and unusual difficulties during this observation so the related time to complete the work is too high. Do
not include 0.98 to your calculations. Sum up your observed times and divide to number of observations which is 8 for this
work. ( HINT: Do not include too low or too high observations –if any-)
Packing;
2
 z    
2
 1.96  0.03295  
n           116
 p  t    0.06  0.10  

Wrap 𝒏 = 𝟐𝟒. 𝟎𝟏 ≅ 𝟐𝟓 𝒄𝒚𝒄𝒍𝒆𝒔

Close 𝒏 = 𝟏𝟓. 𝟒𝟐 ≅ 𝟏𝟔 𝒄𝒚𝒄𝒍𝒆𝒔
**Select the largest number of observations to be sure all estimates will meet or exceed desired precision.

Element Select Time, t Frequency Rating Factor Normal Time

Wrap #1: 0.1 1.00 1.2 0.12

Pack #2: 0.1 1.00 0.9 0.09

Close #3: 0.3 0.25*** 0.8 0.06

Normal Time = 0.27 min.

Standard Time = 0.27 x (1 + 0.185) = 0.32 min.
***Frequency is 0.25 because this observed in every 4 cycles. (Initially in 1st observation then in 5th then in 9th).

Course Lecturer: Asst. Prof. Dr. Merve Er Kara

Teaching Assistant: Res. Asst. Zeynep Tuğçe Kalender
Q7. A technician is conducting a work sampling study of a machine maintenance worker to determine the
proportion of the time the worker spends in one particular department. Management has indicated that they believe
that repairs in this department consume 50% of the maintenance worker’s time, and they want the estimate to be
within ±5 percent of the true proportion, with 95% confidence. Determine the number of observations in the
sample.

𝑧 2 1.96 2
𝑛 = ( ) 𝑝 (1 − 𝑝 ) = ( ) 0.5 (1 − 0.5) = 384.16 ≅ 385 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠
𝑒 0.05

Q8.What sample size should be used for the following work sampling?
a) There should be a 0.95 probability that the value of the sample mean be within 2 minutes if the standard
deviation is 4 minutes.

Standard Deviation 𝜎 = √𝑝 𝑥 (1 − 𝑝)
Variance 𝜎 2 = 𝑝 𝑥 (1 − 𝑝)

𝑧 2 1.96 2 2
𝑛 = ( ) 𝑝 (1 − 𝑝) = ( ) 4 = 15.36 ≅ 16 𝑠𝑎𝑚𝑝𝑙𝑒𝑠
𝑒 2
b) There should be a 90 percent chance that the sample mean provides an error of .10 minutes or less when
the variance is estimated to be 0.50 minutes.

𝑧 2 1.65 2
𝑛 = ( ) 𝑝 (1 − 𝑝) = ( ) 0.50 = 136.125 ≅ 137 𝑠𝑎𝑚𝑝𝑙𝑒𝑠
𝑒 0.10
Q9. Compute the number of observations required in a work sampling study (The standard deviation is 0.2
minutes.) if there should be a 90 percent chance that the sample mean has an error of;

a) 0.15 minutes
𝑧 2 1.65 2
𝑛 = ( ) 𝑝 (1 − 𝑝) = ( ) 0.04 = 4.84 ≅ 5 𝑠𝑎𝑚𝑝𝑙𝑒𝑠
𝑒 0.15
b) 0.10 minutes
𝑧 2 1.65 2
𝑛 = ( ) 𝑝 (1 − 𝑝) = ( ) 0.04 = 10.89 ≅ 11 𝑠𝑎𝑚𝑝𝑙𝑒𝑠
𝑒 0.10

c) 0.005 minutes
𝑧 2 1.65 2
𝑛 = ( ) 𝑝 (1 − 𝑝) = ( ) 0.04 = 4356 𝑠𝑎𝑚𝑝𝑙𝑒𝑠
𝑒 0.005

Course Lecturer: Asst. Prof. Dr. Merve Er Kara

Teaching Assistant: Res. Asst. Zeynep Tuğçe Kalender
Q10. In a work sampling study conducted on a clerical worker in an office, the employee was found to be working
2700 times out of a total of 3000 observations made over a time span of 240 working hours. The output of the
employee was 1800 forms. If a performance rating of 1.05 and an allowance of 15 percent is given, what is the
standard output for this task?

Performance Ratio : 1.05 Normal Time = 0.12 x 1.05 = 0.126 hr.

Allowance = 0.15 0.126
Standard Time = (1−0.15) = 0.1449 ≅ 0.15 ℎ𝑟
2700
Total Actual Time = 2400 𝑥 = 216 ℎ𝑟.
3000
240
216 Standard Output= = 1600 𝑜𝑢𝑡𝑝𝑢𝑡𝑠
Avg. observed time = = 0.12 ℎ𝑟. 0.15
1800

Q11. The Metro Food Services Company delivers fresh sandwiches each morning to vending machines throughout
the city. Workers work through the night to prepare the sandwiches for morning delivery. A worker normally
makes several kinds of sandwiches. A time study for a worker making ham and cheese sandwiches is shown in
table below.

(Allowances = %15)

a. The company wants to know how many ham and cheese sandwiches can be produced in a two-hour period.

Standard Time = Normal Time x (1+ Allw.) = 0.387x ( 1+ 0.15) = 0.445 min.
120 𝑚𝑖𝑛
In 2- hour period = 269.7 ≅ 270 𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ𝑒𝑠
0.445 𝑚𝑖𝑛/𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ

Course Lecturer: Asst. Prof. Dr. Merve Er Kara

Teaching Assistant: Res. Asst. Zeynep Tuğçe Kalender
 b. If the Metro Company pays workers rate of $0.04 per sandwich what would an average worker be paid per
hour?
For the average worker;
60 𝑚𝑖𝑛
= 134.8 ≅ 135 𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ𝑒𝑠
0.445 𝑚𝑖𝑛/𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ

135 𝑥 0.04 = $5.40

c. What would happen if a worker would produce at the average cycle time not adjusted by the rating factor?

For Operator Smith (From Table);

Standard Time = Normal Time x (1+ Allw.) = 0.361 x ( 1+ 0.15) = 0.415 min.

60 𝑚𝑖𝑛
= 144.6 ≅ 145 𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ𝑒𝑠
0.415 𝑚𝑖𝑛/𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ

145 𝑥 0.04 = $5.80

d. Let’s consider that company conducted a time study for 10 cycles of a job assembling ham and cheese
sandwiches, which can be consider to be a sample. The standard deviation of the sample was 0.03 minutes.
The company wants to determine the number of cycles for a time study such that it can be 95% confident
that the average time compared from the time study is within 5% of the true average cycle time. Is 10 cycle
enough?

𝑧 𝜎 2 1.96 0.03 2
𝑛 = [( ) ( )] = [( )( )] = 10.61 ≅ 11 𝑐𝑦𝑐𝑙𝑒𝑠
𝑝 𝑡̅ 0.05 ̅̅̅̅̅̅̅
0.361

The time study should include 11 cycles. The 10 cycles (given in the question) that were used in our time
study were just about right.

Course Lecturer: Asst. Prof. Dr. Merve Er Kara

Teaching Assistant: Res. Asst. Zeynep Tuğçe Kalender