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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