Sample solved problems on Productivity Measurement : OM Assignments

Sample solved problems on Productivity Measurement : OM Assignments
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In our previous text, we covered the basic solved problems based on productivity measurement highlighting the importance of single factor productivity and multi factor productivity. As we progress in the text, we will discover how important it is to measure process productivity and how it is proportional to the efficiency of the company. Here we start with some simple and then advance level solved problems that will clear your fundamentals regarding this one of the most touched topic in operations management.

Operations Management

It basically includes the set of all the activities that produce valuable outputs from inputs. From Output, we mean goods and services. Whereas input includes various factors such as material, labor, labor hours, capital and much more.

For solving problems based on Productivity, we will make use of simple formula given below

Productivity Measurement (P. M) = Output / Input

Practically, we need to improve productivity and this can be achieved by two different ways and these are :

  • Reduce the Input and keeping the output constant
  •  Reduce the Output and keeping the Input constant

Book Referred, in this case, is Operations Management by Jay Heizer and Barry Render 

operations_ management
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Productivity Measurement Problems # 1

Productivity can be measured in a  variety of ways, such as by labor, capitol, energy, material usage, and so on. At Modern Lumber.Inc., Art Binley, president and producer of apple crates sold to growers, has been able, with his current equipment to produce 240 crates per 100 logs. He currently purchases 100 logs per day and each log requires 3 labor hours to process .he believes that he can hire a professional buyer who can buy a better quality log at the same cost. If this is the case, he can increase his production to 260 crates per 100 logs. His labor hours will increase by 8 hours per day. What will be the impact on productivity ( measurement in crates per labor hour) if the buyer is hired?

Solution # 1

Productivity   =    Units produced/labor hours used

Case One : Current Labor Productivity ( when no buyer ) =Output / Input

Here Output = 240 crates

Input = Material required + labor hours

One log requires 3 labor hour. Therefore 100 logs require 300 labor hour.

Current Labor Productivity = 240 crates / ( 100 logs * 3 hours/ log ) = 240 / 300 = .8

Answer is : .8 crates per labor hour

Case Two : 

Labor Productivity  with buyer= Output / Input

New Output = 260 crates

Increase in labor hours = 8

Labor Productivity = 260 crates / { ( 100 logs * 3 hours/ log ) + 8 hours } = 240 / 300

= 260 / 308

= .844 crates per labor hour

As a result, there is an increase in productivity from .8 to .844 crates per labor hour.

Percentage increase in productivity = .844 / .8 = 1.055 =5.5 % increase .


SONAR Tricky questions pdf download 

Answers for Unsolved Productivity Measurement Numericals 

Operations Management Solved Problems # 2

Art Binley has decided to look at his productivity from a multifactor (total factor productivity) perspective (refer to previous Solved Problem 1.1). To do so, he has determined his labor, capital, energy, and material usage and decided to use dollars as the common denominator. His total labor-hours are now 300 per day and will increase to 308 per day. His capital and energy costs will remain constant at $350 and $150 per day, respectively. Material costs for the 100 logs per day are $1,000 and will remain the same. Because he pays an average of $10 per hour (with fringes), Binley determines his productivity increase as follows:


We will find multifactor productivity for the current and proposed system :


Productivity with Current System = Output / Input

Output with Current System = 240 crates

Input = labor cost + material cost + capital + energy

Given: Labor hours are 300 per day

Cost of labor = 300 * 10 dollars ( as he pays 10 dollars per hour )

Material cost for 100 logs per day = $ 1000

Capital = 350

Energy =  150

Therefore total input =  $4,500

Hence , Multifactor productivity = 240 crates/4,500 = .0533 crates/dollar


Productivity with Professional Buyer = Output / Input

Output = 260 crates

Given: Labor hours are 308 per day

Cost of labor = 308 * 10 dollars ( as he pays 10 dollars per hour )

Material cost, Capital and energy cost remain the same

Therefore total input is  = 3080 + 1000+ 350 + 150 = 4580

Hence , Multifactor productivity  with this new buyer = 260 crates/4,580 = .0568 crates/dollar

Result :

With the new system, we observe an increase in productivity from  .0533 crates/dollar to  .0568 crates/dollar.

The increase is ( .0568 – .0533  )/ .0533  = .066

Therefore , the percentage increase in productivity is =  6.6%


These are the two basic examples showing the changes in multifactor productivity when we make changes in labor and other factors with the introduction of new systems.

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