Slide 2 Tutorial: Business Academy Topic: Fixed assets -
capacity Prepared by: Ing. Jana ustrov Projekt Anglicky v odbornch
pedmtech, CZ.1.07/1.3.09/04.0002 je spolufinancovn Evropskm socilnm
fondem a sttnm rozpotem esk republiky. Slide 3 Capacity =the
ability of company, business or services to create or provide a
certain amount of products and services for a certain time under
optimal conditions Production capacity = the companys ability to
produce particular products for a certain time under optimal
conditions Optimal conditions = steady flow of quality material,
continual energy supply, low morbidity, low failure rate, etc.
Slide 4 3 data are needed: Capacity standard (for 1 machine, 1 car
etc.) Capacitive standard time (time required for 1 product)
Capacitive performance standard (maximum quantity that can be
produced in a given time) Number of equipment units (number of
machines, cars..) Usable time fund (maximum time that may be in
operation) (number of days in the year public or day holiday,
vacations) x number of hours per day = nominal time fund (NTF) NTF
planned repairs, interruptions in production fund = usable time
fund (in hours) (UTF) Slide 5 Projected year has 365days. There are
114 days of leaves or holidays. The company plans to have 10
working days as company holiday, 110 hours for machine repairing.
The company operates on 2 shifts e.g. 16 hours. Solution: 1. Number
of working days= 365-114-10=241 2. Nominal time fund= 241x16=3856 h
3. Usable time fund= 3856-110=3 746 h Note VTF can be easily
identified from the planning calendar Slide 6 Slide 7 A. If the
capacitive performance standard is given Capacity = UTF x standard
output capacity x number of units of equipment Example:
Manufacturer of soft drinks has 2 filling lines. Each of them is
able to fill 700 bottles per hour. UTF is 3746 hours. Capacity:
3746x70= 2 622 200 per each line x 2 = 5 244 400 bottles a year B.
If the capacitive standard time is given Capacity =(UTF :
capacitive standard time) x number of equipment units Example:
Pottery kiln has UTF 1870 h a year. It is possible to put there 2
400 mugs that have been burnt for 6 hours. We have 2 kilns.
Standard time for 1 mug= 6 : 2 400 = 0,0025 h Capacity: (1870 :
0,0025) x 2 = 1 496 000 mugs a year Slide 8 Capacity is almost
never used at 100 %. Reasons: Unexpected failures Fluctuation in
demand Increased morbidity Actual capacity use: Real production x
100 capacity Planned use of capacity: Planned production x 100
capacity Slide 9 1. Extensive utilization coefficient = real
production time in hours UTF It indicates whether the property has
been in service all the time: Higher shift working Better
organization of work Faster repairs More consistent use of working
time Slide 10 2. Intensive utilization coefficient = real output
standard capacity It specifies how the machine parameters have been
used. The increase can be achieved by following: To reduce the
labour intensity of products To increase qualification 3. Complex
utilization coefficient = intensive utilization c. x extensive
utilization c. Total capacity utilization is influenced by the time
how long the device has been in service and how effectively its
output has been used. Slide 11 By the term capacity we understand
the companys ability to create a particular number of products in a
certain time. We calculate usable time fund (UTF) that from the
number of days in the year we gradually deduct weekends, holidays,
days of planned repairs and the days of estimated downtimes. The
capacity is optimally utilized if it is used at 100%. Slide 12
Search the Internet a planning calendar for the current year and
determine the capacity based on these data: There are 3 production
lines Each of them produces 300 products per hour Company plans 10
days holiday plant shutdown Estimated repair time is 140 hours It
operates on 2 shifts Slide 13 Klnsk P., Mnch O. Ekonomika pro
obchodn akademie a ostatn stedn koly. Praha: EDUKO nakladatelstv,
s. r. o.,2008, ISBN 978-80-87204-03-0