LECTUTRE # 04
COURSE- ENGINEERING MANAGEMENT
CHAPTER - 05
MANPOWER PLANNING AND CONTROL
COURSE INSTRUCTOR:
PROF. DR. SHAHAB KHUSHNOOD
MECHANICAL ENGINEERING DEPARTMENT
UNIVERSITY OF ENGINEERING AND TECHNOLOGY
TAXILA
Manpower Planning And Control
Manpower planning and control is one of the essential
activities of management for the effective functioning of
an organization, whether it is an engineering or nonengineering organization.
The poor management of manpower planning and
control activity may be reflected in the quality of output
of a company, morale of manpower employed,
efficiency of the organization and so on.
Classification Of Published Literature On Manpower
Planning And Control
 A large number of publications have appeared on the topic of manpower
planning and control in many forms, e.g. books, journal publications and
conference papers.
 Most of the published literature may be grouped into nine categories. These are
as follows
 Span Of Control
Concerned with the supervision of manpower by one individual
 Organizational Size and efficiency
Concerned with the size of organization and its efficiency
 Labor Stability
Concerned with the manpower turnover
 Manpower Planning and Forecasting
Concerned with strictly planning and forecasting manpower need
Classification Of Published Literature On
Manpower Planning And Control(cont…..)
 Manpower selection and recruitment
Deals with selection and recruitment of manpower
 Probabilistic and stochastic models
Concerned with manpower planning
 Case studies
These report the result of real life studies conducted in various parts of the
world and within specific industries
 Review articles
These articles review various aspects of manpower planning and control
 Miscellaneous publications
Concerned with various aspects of manpower planning and control
Manpower Planning And Control
 The objective of manpower planning may be defined as “bringing an
organization’s manpower into line with the present requirement of the company
and its need for a period ahead”.
Approaches For Manpower Planning
Briefly two procedures used for manpower planning which are essentially based on similar
lines
Manpower planning Procedure 1
This approach divided into six steps. These steps are as follows






Review manpower operations
Identify corporate strategy
Forecast the manpower demand
Forecast the manpower supply
Reconcile demand and supply forecasts
Exercise control
Manpower planning procedure 2
This approach has only three phases
• Review of present manpower and study of external factors
• Manpower resources in future
• Monitoring progress results
Selective Mathematical Models
 The mathematical models associated with span of control, labour stability,
organization size and efficiency, and learning process are presented here
Model-1 Span of control
This model can be used to calculate
• The number of persons, P to be supervised by a leader
• The total number of leaders, L in an organization
• Hierarchy Levels, K in an organization
• Work force, Fw of an organization
The total number of leaders in a company is given by
Lt =Fw (Pk-1)/Pk (P-1)
Model-1 Span Of Control
 If the company has one president, then the above equation
reduces to
Lt = (Pk-1)/ (P-1)
Because
Fw/Pk=0
This gives
K= (log Fw) / (log P)
And
P= 𝒌 𝑭𝒘
Model-1 Span Of Control
The following points are to be noted for this method
 Span of control is the same at all levels
 The organizational levels decrease as the span of control increases
 It can be concluded that K is inversely proportional to logP because the Fw is constant
Model-1 Span Of Control
 Example
An engineering organization employs 4000 workers at the shop floor level. The
number of organizational levels above the working level is equal to 20.Assume that
the company has one president and the same number of people are supervised by
each persons at all levels. Calculate the number of persons to be supervised by the
leader.
Here
Fw=4000
K=8
Using formula
P=(Fw)1/k
=(4000)1/8
=3 persons
 Each leader should supervise three persons
Selective mathematical Models (CONT…)
 Model 2-Labour Stability
This is simply an index which is used to assess the stability of the work force in an organization.
This stability index, S, is defined as follows
S=Ts/0.5 m t
Where
m denotes the number of persons employed by the organization at present
Ts denotes the presently employed persons ‘total length of service in years
t denotes the time in years between the mean retirement age of employees and the mean
recruitment age
Model 2-labour Stability
 Example
An electronic components manufacturer employs 400 persons. The total length of service of
all persons employed by the company is 6000 years. The time between the mean retirement
age and the mean recruitment age is 25 years. Calculate the value of the stability index.
Here
Ts=6000 years
T=25 years
m=400 employs
Putting values in equation
S=Ts/0.5 m t
S=6000/ (0.5) (400) (25)
=1.2
 The value of the stability index, S, is 1.2.Therefore it indicates that the company manpower
is excessibly stable.
Selective Mathematical Models (CONT…)
 Model 3-Organization size and efficiency
This model is directed towards organizations which basically deal with paper study oriented tasks such as
research.
The total number of publications or reports, K, produced by a department or an organization annually is given
by
K=(240 β)/Tw + Tr β µ)
Model 3-organization Size And Efficiency(cont…)
 β represents the total number of professional persons employed in the organization or the
department
 µ denotes the fraction of all publications received by the average professional person; in
other words, those reports which the person in question is expected to read
 Trdenotes the mean time to read one report by a professional person
 Tw denotes the mean time to accomplish one report by a professional employee. This time
includes the time spend on investigation, analysis, writing and so on.
 It is assumed that in one year there are 240 workdays. As the number of professional
persons, β,becomes very large the value of K approaches the upper limit; i.e.
Kˊ=240/ µ Tr
 The efficiency, E, of the organization is defined by
E=K/Ko= 𝟏 + (µ𝜷𝑻𝒓 )/𝑻𝒘
 Where Kodenotes the number of reports which can be produced if no time was spent for
reading any report.
Model 3-organization Size And Efficiency(cont…)
 Example
A consulting organization’s basic task is to conduct various types of engineering
research. The company employs 500 professional persons to carry out such tasks.
Each employee works eight hours per day and 240 days per year. In addition, each
professional employee spends on average 2 days to read a report written by others
and 25 days to write his report. The writing time incudes time spent on investigation,
analysis, writing and so on. Each professional worker reads only 1/5 of the total
reports received per year. Calculate the total number of reports to be produced
annually by the company.
 β =500 professional employee
 µ=1/5
 Tr=2 days
 Tw=25 days
Sol:
Using equation
K=(240 β)/Tw + Tr β µ)
=240(500)/25+2(500)(0.2)
=533.33 repots per year
 Therefore the company will produce approximately 533 reports annually.
Selective Mathematical Models (CONT…)
 Model 4-The learning Curve
 This model is based on the fact that the more frequently a person or a worker repeats a
specified task, the more efficient that person will become .In this case the time reduction
results from the learning phenomena.
 The following equation is used to represent the time reduction curve
Z=tf y-α
Where
 Z denotes the cumulative man-hours per item
 Y denotes the quantity of items produced
 tf denotes the time taken to produce the first item
 α represents the curve exponent
Model 4-the Learning Curve (Cont….)
 Example
In the aircraft industry it was found that if an aircraft’s first unit took 2000 man-hours to manufacture,
the second unit took only 1600 man-hours (in other words , to produce two units, the time will be
3200 hours instead of 4000 hours), the forth unit absorbed 1280 man –hours, the eight unit required
1024 man-hours , the 16th unit took 819.2 man-hours and so on. It means that to double production
of the airplane units it required only 80% of the previous time. For example, to produce the forth
unit, it took only 80% of man hours(i.e., 1280 man-hours) of second unit (i.e.,1600 hours).Therefore in
the aircraft industry the 80% learning factor is widely practiced.
 To estimate the value of α for the 80% learning factor we assume that the time to produce the
first unit and the second unit is given by the following equations, respectively:
Z1=tf y1-α
Where
Z2=tf y2-α
 Z1 is the time to produce first unit, y1
 Z2 is the time to produce second unit, y2
Now
Z1/ Z2= y2-α/ y2-α
Model 4-the Learning Curve (Cont….)

To double production at 80% learning factor we let y2=2y1=2,000 man-hours, and z2=1600 man-hours in equation as follows
2000/1600=(2y1)α/y1
5/4=2α

So
α=0.3219
Thus at 80% learning factor
Z=tf y-0.3219

Similarly the calculated values of α at 95%, 90%, ,85%,and 75% learning factors are 0.07,0.15,0.23 and 0.42, respectively. Thus

For 95% learning factor
Z=tf y-0.07

For 90% learning factor
Z=tf y-0.15

For 85% learning factor
Z=tf y-0.23

For 75% learning factor
Z=tf y-0.42

The plot of equation
Z=tf y-α

For the various given values of α are shown as
Model 4-the Learning Curve (Cont….)
Model 4-the Learning Curve (Cont….)
 Example
A company is to produce 16 identical parts of an engineering system. To manufacture the first
part requires 2000 man-hours. When the production of the unit is doubled, the per unit
manufacturing time is reduced by 80% (i.e. the 80% of the time before doubling the
production)
Calculate the total number of man-hours needed to manufacture 16 parts.
 Using equation
Z=tf y-0.3219
= (2000) (16)-0.3219
=819.2 man hours
 This means to produce 16 units, the per unit average manufacturing time will be 819.2
man-hours. Thus to manufacture 16 components, the total time, is equal to
T= (819.2) (16)
=13,107.2 man-hours