Number of locksmiths on duty. Calculation of annual output before and after the introduction of automation. Repair Methods
It has, but only if the plumber on duty does not have to perform work that, in accordance with the requirements of labor protection, he does not have the right to perform without a partner, or without issuing a work permit and as part of a team of at least three people.
Personnel on duty - persons on duty in a shift, admitted to the management and switching of equipment.
Duty personnel are subject to different requirements depending on the scope of their activities by the relevant normative documents. The scope of duties, rights and responsibilities of duty personnel should be defined in job descriptions, as well as in:
a) operating instructions;
b) instruction on labor protection and fire safety;
c) instructions for liquidation of accidents and other local regulatory documents.
The number of required duty personnel is determined by technology, official (production) duties and is established by the head of the organization.
See, for example, the order of the Ministry of Energy of Russia dated March 24, 2003 No. 115 "On approval of the Rules technical operation thermal power plants", paragraph 9.1.54.: "The need for personnel on duty at the thermal point and its duration are established by the management of the organization, depending on local conditions."
So, Rules and norms of technical operation housing stock(Decree of the Gosstroy of Russia dated September 27, 2003 No. 170) provides for the availability of a duty schedule for maintenance personnel (plumbers) - clause 5.2.4., clause 5.2.6.
The number of plumbers on duty is not regulated.
If the duties of on-duty plumbers working in the evening and night shifts include only duties such as:
systematically conduct a tour and inspection of facilities and equipment;
report to the higher duty officer about all deviations from the specified modes of operation of structures and equipment;
strictly observe and require compliance with other rules and instructions established in this area;
do not allow persons to enter your site without special permits or permission from the administration.
in the event of an accident, immediately report the accident to the foreman (chief) of the shift or the dispatcher;
take measures to eliminate the accident in accordance with the production instructions;
v next steps guided by production instruction or instructions of the foreman (chief) of the shift or administration,
In this case, it may be one on-duty plumber.
See also TI-130-2002. Typical instruction on labor protection for a plumber" (approved by Gosstroy of Russia on November 21, 2002), p. 1.23. inform the foreman (chief) of the shift and take preventive measures according to the circumstances, ensuring their own safety.
If the duties of the plumbers on duty include the performance of the necessary repair and maintenance work during duty, then there must be at least 2 such workers.
See, for example, Order of the Ministry of Communications of Russia No. 271 dated 05.12.1994 "On Approval and Enactment of the Rules for Occupational Safety at Radio Enterprises of the Ministry of Communications of Russia" (together with "POT RO-45-002-94").
Extracts:
Item 8.14. Repair and maintenance work at radio enterprises should be carried out by teams (groups) of at least 2 people.
The foreman must have an electrical safety group for work performed on equipment with voltages above 1000 V - not lower than IV, and on equipment with voltages up to 1000 V - not lower than III.
Members of the operational repair team (group) performing work on the equipment must have an electrical safety group of at least III, and those performing mechanical work (locksmith, plumbing) - with electrical safety group I.
In addition, for example, such work as the production of work on the repair, inspection and prevention of sewer and water wells, pits, tanks and structures are high-risk works, and therefore they must be carried out by a team of workers of at least 3 people after receiving security briefing labor and work permit for the production of work.
See the Decree of the Ministry of Labor of Russia dated August 16, 2002 No. 61 "On Approval of the Intersectoral Rules for Occupational Safety in the Operation of Water Supply and Sewerage", clause 5.2.1., clause 5.2.4.
1. Purpose and structure of the RMS plant
The purpose of the RMS of the plant is the organization of a complex of works for the operation, maintenance and repair of the main technological and mechanical equipment, as well as the management of this complex of works.
The structure of the RMC of enterprises consists of three forms of organization of the repair service:
decentralized;
Centralized;
Mixed.
The decentralized form is characterized by the fact that repairmen and technical means dispersed throughout technological workshops in which they independently organize and carry out repairs of their equipment.
Advantages: high mobility, efficiency of the beginning of repair.
Disadvantages: - the need for a large machine park and production facilities.
The centralized form is characterized by the fact that all repair personnel and all material and technical means are concentrated in other centralized work shops, divisions, and repair organizations.
Advantages: - high quality of repairs.
Disadvantages: - high cost repair work.
A mixed form is a combined form that combines both of the above listed forms carrying out repair work. With such
system of work, in addition to the centralization of the mechanical service outside the workshop, repairmen work in the workshop itself and have a repair and mechanical workshop, with a small machine park, and a technical archive.
Dignity: - repair is carried out by repairmen of shop.
Maintenance and repair is organized by the mechanical repair service of the plant, which is headed by the OGM.
For this workshop, we accept a mixed form.
2. Appointment and structure of the OGM of the enterprise.
Appointment of OGM
1) Organization of work on the operation, maintenance and repair of equipment in the shops of the enterprise.
2) Coordination of the activities of the shops and services of the plant for the preparation of annual and monthly PPR schedules.
3) Development of measures for the implementation of repairs and control of their implementation.
OGM activities
The activities of the OGM include the processing of requests for repairs, applications for spare parts, as well as the execution and production of orders and work orders for the manufacture, and the purchase of parts and spare parts. The OGM is headed by the chief mechanic, who directly reports to the chief engineer of the enterprise.
The structure of the enterprise's OGM
3. Form of repairs at the enterprise
There are five options for repairing the main and auxiliary equipment:
1) Factory repair;
2) Repair at a repair plant;
3) Repair by special repair organizations;
4) Repair in the RMC of the plant;
5) Repair by the RMC of the shop.
4. Purpose and structure of the RMS shop
The workshop mechanic is at the head of the RMS of the workshop, who manages the repairs carried out in the workshop, and also monitors the condition of the workshop equipment. Subordinate to the shop mechanic are: locksmiths, welders who carry out minor repairs.
The purpose of the workshop is to carry out all types of planned equipment repairs, as well as unplanned ones. The repair service must always work efficiently and smoothly, avoiding equipment downtime, which can lead to material losses.
The structure of the mechanical repair service of the shop.
5. Calculation of the number of workshop repairmen
The calculation is made according to the following formula
where: K cn - coefficient of the payroll,
K cn \u003d 1.4 ÷ 1.8; accept K cn = 1.4;
F n - the nominal annual fund of working hours of one repairman, F n = 2002 h;
K m - coefficient of mechanization of work, K m =1.3;
K n - coefficient of fulfillment of the norms, K n =1.2;
Total labor costs for repairs and inspections of all workshop equipment, including auxiliary equipment and their quantity,
N 1 -τ | +n 2 ∙τ 2 +... + n m -τ n, (62) where:
n m - the number of machines of the same type;
τ n - laboriousness of equipment repairs, average annual (person ∙ hour);
for small businesses
τ = τ then ∙τ t ∙τ k; (63)
13262 hours - according to practice.
According to the calculation, Ch cn \u003d 5.8. We accept H cn \u003d 6.
6. Number of locksmiths on duty
The duties of the locksmith on duty include carrying out Maintenance equipment. If a mechanical equipment failure occurs on a shift, the locksmith on duty is the first to repair the failure.
The number of locksmiths on duty is determined by the formula:
where K d is the coefficient for accounting for the number of locksmiths on duty, K d \u003d 0.2.
The remaining indicators are similar to those given when calculating the number of workshop repairmen.
We take the number of locksmiths on duty equal to 1, i.e. 1 locksmith per shift.
5.6. Types, methods and systems of repairs in the workshop
5.6.1. Types of repairs
In non-ferrous metallurgy, there are two types of repairs, current (T) and capital (K).
During the current repair, the machine is partially disassembled into components and parts, individual parts and components are replaced or restored, the mechanisms for their regulation are revised, the lining is replaced or restored, and the machine is tested.
If the machine has parts with different service life, then two routine repairs are carried out.
Current repairs, depending on the capacity of the enterprise, are carried out either by the shop, factory RMS, or by special repair teams.
Current repairs are carried out in accordance with the scheduled preventive maintenance schedules. The cost of carrying out current repairs is attributed to the shop cost.
During the overhaul, the machine is completely disassembled into parts, cleaning, repair of basic parts, replacement or restoration of parts, assembly, adjustment, idling and under load testing. Depending on the capacity of the enterprise and the type of equipment, this repair is carried out by the workshop, the enterprise, or with the involvement of special repair enterprises. Capital repairs are financed from depreciation charges. It is carried out according to the PPR schedules.
5.6.2. repair methods.
There are three types of repair methods:
Individual,
Impersonal,
Mixed.
1.Individual - with this method, worn parts are removed from the repaired machine and sent to the restoration workshops. After the parts are installed in the same machine. This method is used in small enterprises.
2. Impersonal - with this method, worn parts are removed from the repaired machine and sent to the restoration workshops, and either new or previously restored parts taken from the spare parts warehouse are installed in the machine. It is used in enterprises where large interruptions in the operation of technological lines are unacceptable, as well as in enterprises of high power.
This method has disadvantages, there are two of them:
The so-called "mortification of capital";
The need for additional space for storing parts and
nodes. There are two types of "impersonal" repair method:
a) MANPADS - periodic replacement of the repair kit;
b) Aggregate-nodal method.
3. A mixed method is a combination in certain proportions of individual and “impersonal methods.
5.7. Repair systems
Three systems of planned preventive repairs are used:
Post-examination;
Standard PPR;
Periodic system PPR.
1. Post-inspection - the PPR system provides for a periodic (once a month) by the mechanic of the workshop equipment, during which he concludes that repairs are necessary (used at low-capacity enterprises).
2. Standard PPR - consists in the use of the assigned resource, after which the equipment, despite its technical condition, replaced or repaired. Applies in cases where possible failure can lead to human casualties, or large material losses.
3. Periodic system - it is associated with repairs according to the PPR schedule. It is used at enterprises with a full load of equipment (around the clock).
We accept an individual repair method and a periodic PPR system.
5.8. PPR calculation
The calculation of the PPR schedule is carried out according to the methodology proposed in the "Regulations on the PPR ...".
The initial data for the calculation are the standards for the frequency and complexity of types of repairs, from the "Regulations on the PPR ...". Initial data are given in Table 5.1.
Table 5.2
PPR schedule.
| Equipment and brief technical specifications |
Duration, h |
Number in a loop |
|||
| Elevator H=18m |
|||||
| screw conveyor diameter =0.5 |
|||||
Air separator |
|||||
| Bag filter |
|||||
| Ball mill dry grinding |
|||||
5.5.1. Factory repair
It consists in cleaning the equipment, packing it in a special container and after sending it to the manufacturer.
At the factory, the equipment is unpacked, then it is disassembled into components and parts, and the parts are troubleshooted. Unusable parts are sorted out, worn parts are restored. Then the car is assembled, it is started at idle (running in). Then it is packed in the same container and sent to the consumer.
Advantages:
a) replacement of worn parts with parts with the same tolerance fields;
6) repair in more short time;
c) higher quality of repair.
Flaws: high repair cost.
5.5.2. Repair at a repair plant
The repair plant is usually located near enterprises with the same type of equipment, as well as equipment with similar technological features. This plant is equipped with the necessary machine park, technological park, tools and fixtures, access roads, etc.
There are two options for equipment repair with the involvement of repair plants:
1) When the repaired equipment is sent for repair directly to the factory. At the factory, the equipment is unpacked, disassembled into units and parts, equipment troubleshooting is carried out, worn parts are restored or replaced with new ones. The machine is assembled, run-in is performed, and after the equipment is sent to the consumer.
Disadvantage: poor workmanship compared to the manufacturer.
2) The repair plant sends specialists - repairmen to the enterprise where the repair is carried out. At the same time, the enterprise provides these specialists essential tool, working conditions, materials, and in some cases locksmiths and welders.
5.5.3. Special repair teams (contractors)
These are organizations that have high-class engineers and repairmen in their staff. They are provided necessary equipment for repairs of any complexity. The advantages include the high quality of repairs carried out in a short time, and the disadvantages are the high cost of repairs.
5.5.4. By the forces of the RMC plant
The enterprise organizes a mechanical repair shop equipped with the necessary machine park, fixtures and tools, and
as well as repair sites. RMCs are staffed with the necessary staff of repair personnel.
Their work (the work of the RMC) is built on the PPR graphs. According to this schedule, all types of inspections and repairs of the plant are carried out, subordinate to the RMC to the chief mechanic of the plant.
5.5.5. By the forces of the RMS workshop
The workshop has a certain staff of locksmiths performing current daily work. For the period of repair, a repair team is allocated from this state, headed by a foreman. The remuneration of the members of the repair team is either piecework or piecework - bonus. At the end of the repair, the repairmen perform a daily current work. The advantage is the stimulation of the work of repairmen, and low staff turnover.
5.11. Lubrication facilities of the workshop
The main task of the lubrication service is to provide reliable and economical lubrication at all stages of the equipment life.
The organization of the lubrication economy in the workshop should be based on the certification of equipment with the compilation of maps and tables of lubrication.
Certification covers all technical, lifting and transport equipment and mechanisms, consumers of lubricants. On the basis of lubrication maps and tables, calculations are made of the need for lubricants by type and brand, based on consumption per shift, per month and throughout the year. In accordance with consumption standards and lubrication tables, determine required amount lubricants for each piece of equipment. Calculations of the need for lubricants performed by mechanics of workshops and other manufacturing enterprises, verified and summarized in the GMO, serve as the basis for the preparation of relevant applications. The arrival, storage and delivery of lubricants to production sites is carried out by employees of the warehouse of lubricants of the workshop.
5.11.1 Application for fuel
Table 5.3
Application for fuel
5.12. Locksmith's workplace
The layout of the mechanics' workplace and the arrangement of auxiliary equipment is shown in Fig. 16.
Scheme of a locksmith's room
 |
1 - vertical drilling machine;
2- grinding machine;
3- cabinet for storing tools and inventory;
4- lathe;
5- metal table
6.7- workbench
5.13. Storage facilities workshops
There are three types of warehouses:
1) Open warehouses;
2) Closed not heated;
3) Closed heated.
This workshop has an open warehouse, which is protected from the effects of precipitation by means of a canopy. This warehouse stores parts and assemblies of machines, as well as different kinds consumables that are not afraid of precipitation. Also in the workshop there is a closed warehouse where equipment, parts and assemblies are stored, which are afraid of direct exposure to precipitation and sunlight. Such Consumables how bearings are stored in special cabinets with constant ambient temperature.
5.14. Work permit for repairs
Enterprise, shop _________________________________________________________________
Work permit No. _____
To perform high-risk work
1. Foreman of works ______________________________________________________ (enterprise, shop, position, surname, I, O.)
2. Allowed to perform __________________________________________________
(place of work, name of equipment, summary works)
3. D lowering to work _____________________________________________________
(position, surname, I, O.)
4. Measures to ensure the safety of work:
4.1 Stop__________________________________________________________________
(stop location, position)
4.2. Disable _______________________________________________________________
(knife switch, gate valve, line, etc., remove the tag)
4.3. Install _______________________________________________________________
(short circuit, dead end, plugs, signal lamps, etc.)
4.4 Take a sample for air analysis _____________________________________
4.5. Protect _________________________________________________________________
(work area, hang out posters)
4.6. Provide for safety measures when working at heights and in wells _____________
_____________________________________________________________________________
4.7. Warn _____________________________________________________________
(drivers of neighboring cranes and cranes of adjacent spans with a signature in the logbook)
______________________________________
4.8 Warn safety measures at railway tracks ______________________
______________________________________________________________________________
(installation of signs, posters, fences, dead ends, etc.)
4.9. Specify routes to the place of work _____________________________________________
(attach diagram if necessary)
4.10. Additional activities _______________________________________________
______________________________________________________________________________
5. Work permit issued by _______________________________________________________________
(position, full name)
6Measures completed:
7Agreed: the head of the shift (section) _________________________________________
_____________________________________________________________________________
(surname, signature)
7.1. __________________________________________________________________________
(position name, signature)
8 The measures have been taken, the safety of work has been ensured, the foreman has been acquainted and instructed with the working conditions, I permit permission - allowing to work _____________________________________________________________________________
9Familiarized and instructed with the working conditions, checked the preparation, workplace accepted - foreman ______________________________________________
(position, surname, signature, date, time)
________________________________________________________________________________________________________________
1 The work is finished, the work permit from the foreman accepted ______________________
___________________________________________________________________________
(date, time, position, surname, signature of the applicant)
5.15. The project of the organization of work during the repair of the machine
When preparing the repair of a large and complex equipment a work organization project (WOR) is drawn up, which is developed by the executing organization and agreed with the customer.
The project for the organization of repairs should provide for the technical means of mechanization of the work to be performed, the composition and qualifications of the repair personnel. Organization of receipt, delivery and storage of components and parts of equipment. Organization of repair and installation sites and sites at the work site, as well as measures for the safety of their implementation.
POR regulates the procedure for conducting and measures to ensure the operations of the preparatory, dismantling and recovery-assembly periods.
5.16. Calculation of the operational overhaul schedule
1) At the first stage, it is necessary to write out from " standard norms..." a list of all repair operations performed, indicating the number of nodes, the complexity of operations and the category of repairmen.
2) Determine the approximate number of the brigade to perform repair operations according to the formula:
H k \u003d τ k / t K \u003d 25.4 / 8 \u003d 3.175 people. (65)
We accept the nearest value - the composition of the brigade 3 people: locksmiths of 3,4,5 categories,
where τ to - normative complexity of performing one repair, man-hour;
t K - repair time, hour
3) Rules for scheduling:
There must be a strict alternation of operations;
Repairmen must move from one operation to another without interruption;
There should be the same number of repairmen on each repair day;
Observe the logical, from the point of view of safety, operations;
Repairmen should not interfere with each other.
4) The length of the segment on the graph is determined by the formula:
T 0 \u003d (τ 1 + τ 2 + ... + τ n) / n rem (66)
where τ 1 , τ 2 , τ n - the complexity of the operation, indicated on the graph by one segment, man-hour;
Prem - the number of repairmen on this operation, people.
T o1 \u003d 0.6 / 3 \u003d 0.2 hour;
T o2 \u003d (0.1 + 0.5) / 1 \u003d 0.6 hour;
T o3 \u003d (0.8 + 2.0) / 2 \u003d 1.4 hours;
T o4 \u003d 0.8 / 1 \u003d 0.8 hour;
T o5 \u003d (0.5 + 0.5) / 2 \u003d 0.5 hour;
T o6 \u003d 0.5 / 1 \u003d 0.5 hour;
T o7 \u003d 0.6 / 1 \u003d 0.6 hour;
T o8 \u003d 5.0 / 2 \u003d 2.5 hours;
T o9 \u003d (1.5 + 0.4) / 1 \u003d 1.9 hours;
T o10 \u003d 3.0 / 3 \u003d 1 hour;
T o11 \u003d (0.6 + 1.2) / 2 \u003d 0.9 hour;
T o12 \u003d 0.9 / 1 \u003d 0.9 hour;
T o13 \u003d (0.8 + 0.6 + 0.8 + 0.2) / 3 \u003d 0.8 hour;
T o14 \u003d 2.0 / 3 \u003d 0.67 hours;
T o15 \u003d 1.5 / 3 \u003d 1.5 hour
5) When connecting to the performance of repair operations included in the formula (5.11), other repairmen, the length of the segment is adjusted.
6) When repairmen switch from one operation to another, expression (5.11) is corrected.
7) If repairmen who are not related to the activities of the workshop’s mechanical repair service are involved in the performance of certain operations, then a separate line is allocated in the operational schedule for these types of work. The labor intensity of performing the work of these workers is not taken into account in (5.10).
8) If a large volume of welding work, then gas-electric welders are involved in repairs. The labor intensity of the work of gas-electric welders is not taken into account in expression (5.10). If the volume of welding work is not large, then these operations can be performed by locksmiths of 4 and 5 categories. The labor intensity of work in this case is taken into account in expression (5.10).
Table 5.4
List of defects
| No. p / p |
Name and scope of work |
the name of detail |
||||||
| detail |
Quantity, pcs |
Weight, kg |
Name size |
Mark, GOST |
Unit measurements |
Quantity, pcs |
||
| Gearbox repair |
Bearings gear wheels |
Cushioning cardboard Industrial oil I-20 |
GOST 20346-74 GOST 20793-72 |
|||||
| drum body drum cover |
Sheet steel |
GOST 19904-74 |
||||||
| Bearings GOST 5721-75 |
Cushioning cardboard |
GOST 20376-74 |
||||||
Table 5.5
List of wearing parts
| Knots and their details |
Drawing No. or GOST |
Qty per unit |
REPAIR METHODS
Restoring the performance of equipment lost during its operation must be carried out through the widespread introduction of:
1. The method of dispersed overhaul of equipment.
The essence of this method lies in the performance of work on the overhaul of equipment, where it is technically possible and expedient, in parts on the days of its scheduled shutdowns for current repairs. This allows you to either reduce the duration of downtime during major repairs, or completely eliminate the latter.
The length of the period during which overhaul can be carried out by a dispersed method, should not exceed the duration of the equipment repair cycle
2. Method of aggregate replacement.
The use of the aggregate replacement method provides a significant reduction in equipment downtime for repairs. In some cases, it is advisable to replace the entire machine or mechanism, for example, a lifting table, a gearbox, etc.
CALCULATION OF THE NUMBER OF PERSONNEL FOR THE REPAIR OF GPM
The calculation of the number of workers is necessary to determine the need for production in labor force, in the quantitative and professional composition of workers who can ensure the normal functioning of their site, to determine the production standards for each worker (in shifts, in hours). To calculate the number of workers, and then the annual wage fund, it is necessary, first of all, to draw up a balance of working hours for workers employed in continuous production, taking into account the fact that some work in one shift and intermittently (for example, maintenance personnel of the GPM section) table 4.
The annual balance of the contractor's working time Table 4
|
Indicators |
Continuous production |
Discontinuous production |
|
1. Calendar fund of time Tk |
||
|
2. Weekends and holidays |
||
|
3. Nominal fund of time Tn |
||
|
4. Planned respectful absenteeism: |
||
|
4.1 Regular and additional leave |
||
|
4.2 Absence from work due to illness |
||
|
4.3 Student leave |
||
|
4.4 Performing public duties |
||
|
5. actual working time fund (D) |
Calculate the payroll ratio for continuous production using the formula:
Ksp \u003d Tk: D \u003d 365: 225 \u003d 1.62
Calculate the payroll ratio for discontinuous production using the formula:
Ksp \u003d Tn: D \u003d 250: 217 \u003d 1.15
These data are entered in Appendix 8, Col. 6
Calculate payroll employees with the following information:
the number of mechanics-repairmen of the 5th category - 3 people.
the number of mechanics-repairmen of the 4th category - 3 people.
the number of mechanics-repairmen on duty - 6 people.
the number of electric and gas welders of the 5th category - 1 person.
number of electric and gas welders of the 4th category - 1 person.
the number of electric and gas welders on duty -3 people.
1) The payroll (col. 7) is calculated: the number of workers per day (col. 5) multiplied by the coefficient of the payroll (col. 6)
Let's make calculations using the example of on-duty repairmen.
For example, the payroll per day is 6 people, and the payroll ratio is 1.62. 6 x 1.62 = 10, i.e. the payroll of on-duty repairmen will be 10 people.
- 2) To calculate how much the payroll of shifts needs to work (col. 8), you need to: (col. 7) multiply by the actual working time fund in continuous production (because there are 3 shifts in a day) - 225. For example, 10 x 225 = 2250 shifts
- 3) To calculate how many hours the payroll needs to work (col. 9), you need to: (col. 8) multiply by 8 hours. 2250 x 8 \u003d 18000 hours - it is supposed to work out the payroll of on-duty repairmen.
- 4) To calculate how much the payroll of night hours needs to work (col. 10), you need to: (col. 9) divide by 3. 18000: 3 = 6000 hours.
- 5) To calculate how many holiday hours are supposed to be worked (col. 11), you need: the number of workers calculated per day (col. 5) multiplied by 8 hours and multiplied by the number public holidays established by the Government of the Russian Federation (in this work, 11 days).
For example, 6 x 8 x 11 = 528 holidays
By count. 7 we calculate the payroll of the workers of the repair service of the ESPTS-1: 25 people.
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Task 1
Determine the piece, piece-calculation time for manufacturing a car pallet, as well as the time for manufacturing a batch of pallets, if forging time is 15 minutes, auxiliary time is 2 minutes, preparatory and final time is 9 minutes, the ratio of rest time and personal needs is 13%. The batch size is taken equal to 30 pcs. plus no.
Solution:
T o \u003d 15 min. T in \u003d 2 min. T pz = 9 min.
Lot size 43 pcs. (30 pcs +13 option no.)
T op \u003d T o + T in,
T op \u003d 15 + 2 \u003d 17 min.
T exc \u003d 13% of the operational time \u003d (17x13): 100 \u003d 2.2 min.
T orm - time to service the workplace, T op - operational time; T ex - time for rest and personal needs; T Fri - the time of breaks provided by the technology; T pz - time of preparatory and final work. labor wage rate
We determine the piece time.
T pcs. = 17+2.2+ 9= 28.2 min.
T part \u003d T piece H n + T pz \u003d 28.2 x 43 + 9 \u003d 1222,
where T pcs - piece time, n - the number of products in the batch (43 pcs.)
Answer: The norm of piece time - 28.2 min.
The norm of time for a batch of products is 1222 minutes.
Task 2
As a result of the introduction new technology the output of the main workers increased by 15% plus no. Determine how far time limits should be reduced
Solution:
The output rate of the main workers increased by 28% (15% is given + 13 option).
We determine the % decrease in the norm of time according to the formula
% reduction H vr =
where H temp. - the norm of time, N vyr. - production rate
% reduction H vr \u003d (28x100) : (100 + 28) \u003d 21.9%
Answer : The time rate should be reduced by 21.9%.
Task 2
Calculate the norms of piece, piece-calculation time and time per batch of products, if the main work time is 20 minutes, the auxiliary work time is 7 minutes, the standard time for servicing the workplace is 7% of the operational time, and for rest and personal needs - 6% . Preparatory and final time - 25 min. for a batch of parts. The number of parts in the lot is 30 pieces plus no.
Solution:
T o \u003d 20 min. T in = 7 min. T pz = 25 min.
The number of parts in the lot 43 pcs. (30 pcs +13 option no.)
Determine the operating time.
T op \u003d T o + T in,
where T about - the main time; T in - auxiliary time
T op \u003d 20 + 7 \u003d 27 min.
We determine the time for rest and personal needs
T exc \u003d 6% of the operational time \u003d (27x6): 100 \u003d 1.6 min.
Determining the workplace maintenance standard
T orm \u003d 7% of the operational time \u003d (27x7): 100 \u003d 1.9 min.
We determine the time limit.
N vr \u003d T pz + T op + T orm + T exc \u003d 25 + 27 + 1.9 + 1.6 \u003d 55.5 min.
We determine the piece time.
T pcs \u003d T op + T orm + T exc + T pt
T pcs. = 27+1.9+1.6= 30.5 min.
Determines the norm of time for a batch of products T part.
T part \u003d T piece H n + T pz \u003d 30.5 x 43 + 25 \u003d 1336,
where T pcs - piece time, n - the number of products in the batch (31 pcs.)
Answer: The norm of piece time is 30.5 minutes.
The norm of time for a batch of products - 1336 min.
Task 4
Calculate the service rate and the number of mechanics on duty required to service 250 plus number of machines in machine shop, if the shift ratio is 2.4, the real fund of working time is 230, nominal - 260 days per year. Operational time for maintenance of one machine - 12 min, T exc-14 min, T pz- 10 minutes per shift.
Solution
Determine the number of machines 250+13 =263
We determine the time limit.
N vr \u003d T pz + T op + T orm + T ex + T pt,
where; H vr - the norm of time; T orm - time to service the workplace, T op - operational time; T ex - time for rest and personal needs; T Fri - the time of breaks provided by the technology; T pz - time of preparatory and final work.
H vr \u003d 14 + 10 + 12 \u003d 36 min.
where N h - the norm of the number; O - the total number of service units served or the amount of work performed, T cm the length of the working day.
We determine the norm of the number
We determine the rate of service.
Answer : Population rate 36.4, service rate 0.2
Task 5
Calculate the maintenance rate for automatic machine tools, if the operational time for setting up one machine is 7 minutes, the time to move from machine to machine is 1.5 minutes, the preparatory and final time is 10 minutes, the time for rest and personal needs is 18 minutes per shift. Determine the attendance and write-off number of adjusters, if there are 400 machines plus No. in the shop, the shop works in two shifts, T cm- 8 hours. The number of secret days in a year is 230, workers - 260.
Solution:
We determine the number of machines 400 + 13 = 413 pcs.
T o \u003d 7 min. T in cn = 1.5 min. T pz = 10 min.
Determine the operating time.
T op \u003d T o + T in,
where T about - the main time; T in - auxiliary time
T op \u003d 7 + 1.5 \u003d 8.5 min.
Determine the service time
N vr \u003d T pz + T op + T orm + T exc + T pt \u003d 10 + 8.5 + 18 \u003d 36.5
Determining the maintenance time for machines
Determine the number of adjusters
We determine the list number of adjusters
Answer: Turnout number adjusters - 65
List number of adjusters - 58
Machine maintenance rate - 0.4
Task 6
The rate of production of a worker per shift is 60 parts + No. The worker actually made 66 parts. Determine the percentage of performance standards.
Solution:
We determine the % of the fulfillment of the production norm.
where the actual production of the worker is 66 parts, the production rate is 73 parts (60 + 13 option)
P ext \u003d (66: 73) x 100% \u003d 90.4
Answer: Percentage of fulfillment of production norms -90.4
Task 7
Determine the percentage of fulfillment of the production standards by a team of drivers, if the volume of traffic was 22.5 thousand tons, the freight turnover was 250.6 thousand tkm, the norms of time for loading and unloading 1 ton of cargo - 0.067 h, per 1 tkm - 0.013 h, in the brigade 25 drivers, each worked 24 days with a 7-hour working day.
Solution:
We determine the fulfillment of the norms by the actual hours worked:
Pvn = = =0.05
Where Tn - 0.067 - the time spent on the production of suitable products;
Td - additional time spent due to deviations from normal working conditions;
Tsd - 24 x 7 = 168 hours - actual hours worked
The percentage of compliance with the norms for the shift fund is determined by the formula:
where Tpr - intra-shift losses of working time and downtime;
Tp is the time of using piecework workers for temporary work.
Task 7
Determine the percentage of compliance with the norms, if the norm of time for the manufacture of product A is 0.25 standard hours, product B - 0.3, product C - 1 standard hour. According to the plan, it is necessary to manufacture 200 units + product number A, 320 - products B and 30 - products C. In fact, 243 products A, 400 products B, 28 products C were manufactured.
Solution
Determine the rate of production
Product A In fact, 243 products were manufactured, norm 213 ed. (200 + 13 option)
Pvn \u003d (243: 213) x 100 \u003d 114.1
Product B In fact, 400 products were manufactured, the norm is 320.
Pvn \u003d (400: 320) x 100 \u003d 125
Product B In fact, 28 products were manufactured, the norm is 30.
Pvn \u003d (28: 30) x 100 \u003d 93.3
Determine the time limit
Product A Norm of time for manufacturing - 0.25 norm / hour
Pvn \u003d (0.25: 7) x 100 \u003d 3.6
Product B Norm of time for production - 0.3
Pvn \u003d (0.3: 7) x 100 \u003d 4.3
Product B Norm of time for production - 1
Pvn \u003d (1: 7) x 100 \u003d 14.3
Answer: % of the output rate: Product A -114.1ё B-125, C -93.3.
% of time norm: Product A - 3.6, B - 4.3, C - 14.3
Task 8
Shop No. 1 employs 450 workers, of which 20 have 2nd category, 210 -3, 180 plus No.-4 and 20 people - 4th category. The total labor intensity of work in shop No. 2 is 654.0 thousand standard hours, of which 350.5 for category 3, 146.5 for category 4, 112 standard hours for category 5, the rest of the work is carried out for category 4. Determine the average tariff coefficient of the workers of shop No. 1 and the average tariff coefficient of work in shop No. 2. The company has a tariff scale:
|
Tariff category |
|||||||
|
Tariff coefficient |
Solution:
We calculate the average tariff coefficient
K cf \u003d K m + (K b - K m) x (P cf - R m),
where K cf - the average tariff coefficient, K m - the tariff coefficient is less than the average, K b - the tariff coefficient is more than the average, - tariff category average; - the tariff category is less than average; - the tariff category is above average.
We determine the average tariff category
(2x20 + 3x210 + 4x193 + 5x20) : 443=3.5
A digit that is less than the average K 3 = 1.23, a digit that is greater than the average K 4 = 1.36
We calculate the average tariff coefficient for shop No. 1
Kav = 1.23+(1.36-1.23) x (3.47-3)= 1.29
We calculate the average tariff category for shop No. 2
K cf = (3x350.5+4x146.5+5x112+6x45): 654= 3.77
The discharge is less than the average K 3 = 350.5, the discharge is more than the average - K 4 = 146.5
By Wed = 350.5+ (146.5-350.5) x (3.77-3)= 1.13
Answer: Average tariff coefficient for workshop No. 1- 1.29
The average tariff coefficient for shop No. 2 is 1.13
Task 9
The worker processed 185 products A and 900 products B. The norm of time for product A is 55.48 minutes, for product B-5.22 minutes. The hourly wage rate of an employee is 33 rubles, plus No. . Calculate the employee's salary.
Solution:
We determine the hourly tariff rate - T st.day 33 + 13 \u003d 46 rubles.
Determine the piece rate:
Rass. sd. = H time x T st.h. ,
where is the norm of time; T st.h. - hourly rate.
Product A - Rasc. sd. \u003d 0.92 x 46 \u003d 42.3 rubles.
Product B - Rass. sd. \u003d 0.09 x 46 \u003d 4.1 rubles.
We determine wages.
ZP sd. = Ras. sd x O vyp.
Product A - ZP sd. \u003d 42.3 x 185 \u003d 7881 rubles.
Product B - ZP sd. \u003d 4.1 x 900 \u003d 3690 rubles.
Total salary 7881 +3690 = 11571 rubles.
Answer: Salary - 11571 rub.
Task 10
The planned scope of work is 120 products, in fact, the employee has manufactured 130 products, the piece rate is 35 rubles. per piece plus no. For overfulfillment of the plan, a bonus is set in the amount of 2% of piecework earnings for each percentage of overfulfillment. Calculate the employee's salary.
Solution:
We determine the piece rate of 35 rubles. +13 = 48 rubles.
Determining piecework wages
ZP sd. = T h.st. x O vyp. \u003d 130 x 48 \u003d 6240 rubles.
We calculate the percentage of overfulfillment of the plan 130: 120 x 100 = 108%
We calculate the size wages taking into account the premium for overfulfillment of the plan 116 x 6240: 100 \u003d 7238 rubles.
Answer: The salary of an employee is 7238 rubles.
Task 11
The hourly rate of a service worker is 30 rubles. plus number, service rateivaniya - 2 pieces of equipment (respectively 2 main workers), butRma output of the main worker is 6 pcs. at one o'clock. Calculate the amount earnings of a service worker, if the actual volume of work of the main workers is 12304 pieces. per month.
Solution:
We calculate the hourly tariff rate 30 + 13 = 43 rubles.
where is the indirect piece rate; - service rate.
We calculate the amount of work per hour - 12304: (22x 8) = 70 pcs.
Rass. cos. = 43: (70 x 2) = 0.3
Zp.= Rec. cos. x H vys. \u003d 0.3 x 12304 \u003d 3691 rubles.
Answer: The salary of a service worker is 3691 rubles.
Task 12
Calculate the monthly earnings of a worker according to the piece-bonus wage system, if the time rate is 0.8 man-hours, the rate is 42.3 rubles, 250 products are delivered per month plus No. For the implementation of production standards, a bonus of 10% of piecework earnings is provided. For every percentage overfulfillment of norms - 1% of piecework earnings. Worked out 22 working shifts of 8 hours each.
Solution:
We determine the volume of products 250 + 13 \u003d 263 pieces.
where Zpl / sd - piecework earnings; Rast / sd - piece rate; - workload.
Calculate piecework wages
Zp sd. = Ras. sd. x O vyp. \u003d 42.3 x 263 \u003d 11124 rubles.
We calculate the % of the fulfillment of the norms-0.8 man-hours x 263 pcs. = 210.4: (22x8)= 120%
We calculate a bonus in the amount of 10% of piecework earnings - 11124 x 10: 100 \u003d 1112 rubles.
We calculate the premium for overfulfillment of the norms - 11124 x 20: 100 \u003d 2225 rubles.
We calculate the monthly earnings of a worker 11124 + 1112 + 2225 = 14461 rubles.
Answer: Salary -14461 rub.
Task 13
Determine monthly salary a worker whose work is paid according to the piece-progressive system, if the norm of time for the product is 0.2 standard hour. The worker worked 22 days for 8 hours. The hourly tariff rate is 78 rubles plus no. Actually 1520 items were manufactured per month. The bonus is accrued only if the norms are exceeded by more than 100%.
Solution:
We determine the hourly tariff rate of 78 rubles. + 13 \u003d 91 rubles.
Calculates piece rate
Rass. sd. \u003d 0.2 x 91 \u003d 18.2 rubles.
We calculate how many hours the worker worked 22 x 8 \u003d 176 hours.
Zp sd. = Ras. sd x O vyp. \u003d 18.2 x 1520 \u003d 27764 rubles.
We calculate the percentage of overfulfillment of the norms - 1520 x0.2 = 304: 176 = 173
The bonus is not paid because the norms are overfulfilled by 173%, and according to the conditions they must be overfulfilled by more than 100%.
Answer: Salary - 27764 rubles.
Task 14
Determine the price (according to the norm of time and according to the norm of production) if the norm of time is 40 minutes, the corresponding norm of production is 12 pcs. per day (8-hour working day), tariff rate 4 categories 81 rubles plus No.
Solution:
Tariff rate 4 categories 81+13=94 rub.
Hvr = 40: 60 = 0.67
We calculate the rate according to the norm of time -
Rass. = H time xT st. = 0.67h. x 94 rubles. =63 rub.
N working hour = 12: 8 = 1.5
We calculate the rate according to the rate of production -
Esc.= H working hours. xT st. \u003d 1.5 x 94 \u003d 141 rubles.
Answer: Rass. present \u003d 63 rubles, Rass. n.vyp. = 141 rubles.
Task 15
Calculate the rate according to the indirect piecework wage system for adjusters of especially complex and unique equipment, if the adjustment work belongs to the 5th category (hourly rate is 100 rubles plus No.), and the adjuster serves four working units with a standard shift output of 20, 26 and 33 PC. for 8 hours.
Solution:
We determine: the hourly tariff rate is 100 rubles. + 13 \u003d 113 rubles, the daily tariff rate is 113x 8 \u003d 904 rubles.
We calculate N vyv. = 20+26+33=79
where is the indirect piece rate; H vyr. - the rate of production.
Rass. to St. = 11.4
Answer: Rass. cos. = 11.4
Task 16
Calculate the amount of the worker's wages for the hours actually worked, if his monthly salary is 21,000 rubles plus No. per month, according to the schedule, 22 shifts of 8 hours were to be worked, 20 shifts of 8 hours were actually worked.
Solution
Determine how many hours actually worked 20 x 8 = 160 hours.
Scheduled hours worked 22 x 8 = 176 hours
(160:176) x 100 = 90.9
Monthly salary 21000 + 13 = 21013 rubles.
We determine the amount of wages for the hours actually worked - 21013 x 90.9 = 19100 rubles.
Answer : The amount of wages - 19100 rubles.
Problem 17
The hourly wage rate of an employee is 75 rubles plus No. 176 hours actually worked per month. A bonus of 20% of wages is set for the quality of work. Calculate the employee's salary.
Solution:
Hourly tariff rate 75 + 13 = 88 rubles.
We determine wages 88 x 176 = 15488 rubles.
We determine the amount of the premium 15448 x 20: 100 = 3098 rubles.
Salary - 15448 + 3098 \u003d 18546 rubles.
Answer: The salary of an employee is 18546 rubles.
Problem 18
The hourly wage rate of a worker is 100 rubles, plus No. per month worked 21 days for 8 hours. Calculate the worker's earnings if a bonus is paid in the amount of 30% of direct time earnings.
Solution :
We determine the tariff rate of the worker - 100 + 13 \u003d 113 rubles.
Hours worked - 21 x 8 = 168 hours
Salary - 113 x 168 = 18984 rubles.
We determine the amount of the premium 18984 x 30: 100 \u003d 5695 rubles.
The salary of a worker is -18984 + 5695 = 24679 rubles.
Answer: The salary of a worker is 24679 rubles.
Problem 19
The employee's salary is 14,000 rubles. plus no. In January, the employee worked 18 days (according to the schedule - 18 days), in February - 20 (21 days), in March - 15 days (22 days). Determine the conditional "cost" of one working day in January, February and March. Calculate the employee's wages for January, February and March, if a bonus of 25% of direct time earnings is provided.
Solution:
We determine the salary of the employee - 14000 + 13 \u003d 14013 rubles.
Actually worked out in January - 18 days, according to the schedule - 18 days.
in February - 20 days, according to the schedule - 21 days
in March - 15 days, according to the schedule 22 days.
The conditional cost of one working day in January 14013: 18 = 779 rubles.
in February 14001: 20 = 701 rubles.
in March 14001: 22 = 637 rubles.
We determine the salary for January 14013x1 \u003d 14013 rubles.x125 \u003d 17516 rubles.
for February 14013x (20:21) x 100 = 13312x 125 = 16640 rubles.
for March 14013 x (15: 22) x100 \u003d 9529 x 125 \u003d 11911 rubles.
Answer: Salary for January - 17516 rubles.
for February - 16640 rubles.
for March - 11911 rubles.
Problem 20
Working 4 category ( hourly tariff rate for a 6-hour working day 107 rubles plus No.). Worked 24 shifts in a month. For the implementation of the plan, the bonus is paid in the amount of 15% of earnings, for each percentage of overfulfillment of the plan - 1.5% of earnings, for saving material - 40% of its cost.
The plan was fulfilled by 108%, materials were saved by 3200 rubles.
Determine your total income.
Solution:
We determine the hourly tariff rate - 107 + 13 = 120 rubles.
120 x 24 x 7 = 20160 rubles
We determine the amount of the bonus for the implementation of the plan -
20160 x 15: 100 \u003d 3024 rubles.
We determine the amount of the premium for each percentage of overfulfillment -
20160 x 8 x 1.5 = 2419 rubles.
We determine the amount of the premium for saving materials
3200 x 40:100 = 1280 rubles
We determine the total earnings -20160 + 3024 + 2419 + 1280 = 26883 rubles.
Answer: Total earnings - 26883 rubles.
Problem 21
An employee with a monthly salary of 14,920 rubles plus No. worked 20 days a month instead of 23 working days according to the schedule. The employee received a bonus in the amount of 35% of monthly earnings. The area is equated to the regions of the Far North, work experience in this area is 7 years. Determine monthly earnings allowances according to the district coefficient (1.4) and for work experience in areas equated to areas of the Far North.
Solution:
We determine the monthly salary 14920 + 13 = 14933 rubles.
We determine the actual wage - 14933: 23 x 20 \u003d 12985 rubles.
We determine the amount of the premium - 12985 x 35: 100 = 4545 rubles.
Determine the wage rate based on district coefficient- (12985 + 4545) x 1.4 = 24542 rubles.
We determine the amount of additional payments for work experience in areas equated to areas of the Far North - 24542 x 50: 100 = 12271 rubles.
We determine the total earnings - 24542 + 12271 \u003d 36813 rubles.
Answer: Earnings per month is 36813 rubles.
Problem 22
The number of working days per month according to the schedule is 20, the duration of the shift is 8 hours.Workerworked 180 hours per month, including overtime for 3days: 2 days - 3 hours and 1 day - 2 hours.The rest of the time is worked out for production needs on weekends.Hourly rate - 100 rubles plus no. The premium (50%) is charged only on directtemporary earnings.Determine the wages of the worker.
Solution:
We determine the hourly tariff rate - 100+ 13 = 113 rubles.
Monthly salary excluding overtime 8x20x113= 18080 rub.
We determine the amount for overtime - 2x113x1.5 = 339 rubles. - 1x113x2 = 226 rubles. - (339 + 226) x 2 \u003d 1130 rubles. for the first two days
2x113x1.5 = 339 rubles.
The total amount for overtime is 1130+ 339 = 1469 rubles.
We determine the amount of the premium - 18080 x 50: 100 = 9040 rubles.
We determine the total earnings - 18080 + 1469 + 9040 = 28589 rubles.
Answer: worker's earnings - 28589 rubles.
Problem 23
The monthly tariff rate of a worker is 10,000 rubles. plus no.The monthly norm of time is 168 hours.The worker worked 176 hours, including 8 hours, a day off.The bonus is charged on all earnings in the amount of 35%. Determine his income.
Solution:
We determine the tariff rate of the worker 10000 + 13 \u003d 10013 rubles.
We determine the hourly tariff rate 10013: 168 = 60 rubles.
We determine the salary for the day off - 8 x 60x 2 \u003d 960 rubles.
We determine the amount of the premium - (10013 + 960) x 35: 100 = 3840 rubles.
We determine the total earnings - 10013 + 960 + 3840 \u003d 14813 rubles.
Answer: Worker's earnings - 14813 rubles.
Problem 24
At night, the worker worked for a month 12 hours. Surcharge is set at 40% of the tariff rate. Hourly tariff rate - 125 rub. plus No. Total worked out 168 hours. All earnings receive a 30% bonus. Determine the wages of the worker, taking into account additional payments for work in night shift and premium payments.
Solution:
We determine the hourly tariff rate - 125 + 13 = 138 rubles.
We determine the amount of wages without taking into account work at night - 168-12 \u003d 156 x 138 \u003d 21528 rubles.
We determine the amount of wages for work at night - 12x138x2 \u003d 3312 rubles.
We determine the amount of the premium - (21528 + 3312) x30: 100 = 7452 rubles.
We determine the amount of wages - 21528 + 3312 + 7452 = 32292 rubles.
Answer: Salary - 32292 rubles.
Problem 25
The tariff rate of a worker per hour of work is 126 rubles, 22 shifts of 7.4 hours each were worked out in conditionswith a high levelnoise and increased dustiness of the working shift zone. For This workplace provides for an additional payment for working conditions in the amount of 8%. The employee received a 40% bonus on all earnings. Determine the worker's monthly salarywork.
Solution:
We determine the size 126x22x7.4 \u003d 20513 rubles.
We determine the amount of the surcharge - 20513x8: 100 = 1641 rubles.
We determine the amount of the premium - (20513 + 1641) x 40: 100 = 8862 rubles.
We determine the amount of wages - 20513 + 1641 + 8862 \u003d 31016 rubles.
Answer: Salary - 31016 rubles.
Problem 26
Employee worked 22 daysaccording to the schedule for 8 hours and 4 hours in a row overtime according to production needs.His daily tariff rate is 1000 rubles plus no. For all earnings, including additional payments for workWeekends are subject to a 30% bonus. The employee is given a personal allowance of 2,500 rubles.The company where he is employed the employee is located in the area where the district coefficient is 1.5., the area belongs to the southern regions Far East. The work experience of the worker in this area is 5 years. Determine earnings.
Solution:
We determine the daily tariff rate 1000 + 13 \u003d 1013 rubles.
We determine wages excluding overtime - 22x1013 \u003d 22286 rubles.
We determine the hourly tariff rate 1013:8 = 127 rubles.
We determine overtime wages - 2x127x1.5 \u003d 381 rubles, 82x127x2 \u003d 508 rubles, 381 + 508 \u003d 889 rubles.
We determine earnings - 22286 + 889 + 2500 = 25675 rubles.
We determine earnings, taking into account the regional coefficient 25675x 1.5 = 38512 rubles.
We determine wages for the southern regions of the Far East 38512 x 30: 100 = 11554 rubles.
We determine the total earnings 38512 + 11554 \u003d 50066 rubles.
Answer: Total earnings - 50066 rubles.
Problem 27
The firm sets the salaries of its managersdepending on the turnover achieved in their sector in the previous month. With a turnover of 100.0 thousand rubles. salary is 8000 rubles. plus no. Each percentage increase in trade yields an increase in wages by 0.7%. Determine the salary of the manager (for each month) according to the following data on the turnover in his sector:
Solution:
We determine the salary 8000 + 1 \u003d 8013 rubles.
where % TO - % increase in turnover, % increase. Zp. - % increase in wages from % turnover = 0.7%
We calculate wages taking into account the % increase in turnover.
September
Problem 28
The audit firm employs 5 people: a director, an accountant and three specialists. They are assigned constant coefficients dependingon the significance of the work performed: director - 0.3; accountant - 0.15; 1st specialist - 0.2; 2nd - 0.15; 3rd - 0.2. For a month, funds for the remuneration of employees of the company amounted to 300 thousand rubles. RUB plus no. Determine the salaries of the firm's employees.
Solution:
We determine the amount of funds for wages 300,000 + 13 = 300,013 rubles.
We determine wages.
Director - 300013 x 0.3 = 90003 rubles.
Accountant - 300013x 0.15 \u003d 45002 rubles.
1 specialist - 300013 x 0.2 = 60003 rubles
2 specialist - 300013 x 0.15 = 45002 rubles.
3 specialist - 300013 x 02 = 60003 rubles.
Answer: Salary: Director - 90003 rubles, accountant 45002 rubles, 1st specialist 60003 rubles, 2nd specialist 45002 rubles, 3rd specialist 60003 rubles.
Problem 29
Employees are paid monthly wagesfees of 150,000 rub. plus no. There are 4 people working in the group, the ratio of their wages is as follows:1.95; 2.1;2.25;4.0. Determine the salaries of the firm's employees.
Solution:
We determine the monthly wage fund 150,000 + 13 = 150,013 rubles.
Determine the average wage
150013: (1.95 + 2.1 + 2.25 + 4.0) \u003d 14564 rubles.
We determine wages
14564x1.95 = 28400 rubles
14564x2.1= 30584 rubles
14564х 2.25=32769 rub.
14564x4 = 58260 rubles.
Answer: Salary: 28400 rubles, 30584 rubles, 32769 rubles, 58260 rubles.
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Practical lesson 4. Substantiation of the number of workers.
Solve the proposed tasks. Analyze the obtained results. Draw conclusions. In accordance with the version of the course work, the task on this topic is also performed (for example, option number 10, and the task number is solved accordingly 10, after 25 the task number, for option number 26 will correspond to task number 1, etc.)
Execution technique control task 4
Justification of the number of workers.
Justification of the number of workers is the starting point for the formation of savings in labor costs and the growth of labor efficiency. The calculation of the estimated number of workers is carried out according to the formula:
where ∑T epp- total labor intensity of the production program (n/h);
O R- the planned scope of work (in rubles, pieces, pieces of equipment);
F 1 - fund of time of one worker corresponding to the planned period (hour, min);
V 1 - production of one worker for the corresponding planning period (rubles, units);
TO vnpl- the planned coefficient of fulfillment of the norms is equal to 1.1.
Estimated number (H R) is adjusted depending on the changing organizational and technical conditions of production. With an increase in the volume of production, the estimated number increases by the growth rate of the volume of production. An increase in the coefficient of compliance with the norms leads to a decrease in the number, since the denominator in the calculation formula (1) increases. The presence of loss of working time (K P) leads to a decrease in the employment rate of the worker (K zan), coefficient of use of working time (K and= 1-K P) and leads to an increase in the required (attendance) number of workers, since it reduces the denominator in the calculation formula (1). Therefore, the ratio between the calculated (H R) and secret (H I'm in) the number of workers is determined by the formula:
(2)
Example 1
Determine the number of technologists at the enterprise if the planned annual production of equipment and tooling for which it is necessary to develop technological processes is equal to 5000 thousand USD; the conditional annual output of one technologist, calculated on the basis of the actual annual production volume in monetary terms, is 25 thousand USD; the coefficient taking into account additional work performed during the development of the technological process is 1.15.
Chr \u003d ∑Or / B \u003d 5000/25 \u003d 200 people.
Crv \u003d Cr * Kd \u003d 200 * 1.15 \u003d 230 people, where Kd is a coefficient that takes into account additional work.
Answer: the number of technologists at the enterprise is 230 people.
Example 2
In the base year, the volume of marketable output at the plant with a workforce of 730 people. was 2650 c.u. In the planned year, with an increase in production by 8%, a relative reduction in the number of employees should be made due to the following factors: modernization of equipment - 26 people, organization of integrated teams and combination of professions - 12, reorganization of the departments of the chief mechanic and power engineer - 3 people. At the same time, due to production needs, the tool area is expanding with an increase in the number of workers by 11. Calculate the change in labor productivity from the change in output per worker.
Determine labor productivity in the base year:
Pr \u003d TP / Chr \u003d 2650/730 \u003d 3.63 c.u. / person
Determine the number of workers in the planned year:
Chr \u003d 730-26-12-3 + 11 \u003d 700 people.
Determine the volume of marketable products in the planned year:
TP \u003d 2650 * 1.08 \u003d 2862 c.u.
We determine labor productivity in the planned year:
Pr \u003d TP / Chr \u003d 2862/700 \u003d 4.088 USD / person
Change in labor productivity in comparison with the planned with the base in real terms:
∆Pr=4.088-3.63=+0.458 c.u./person
Change in labor productivity compared to the planned with the base in%:
∆Pr=4.088/3.63*100%=+112.62% - 100%=+12.62%
Answer: Change in labor productivity in comparison with the planned one in %: 12.62% and in physical terms: +0.458 c.u./person.
Task 4
Solve the proposed tasks
Task 1. In the base year, the volume of marketable output at the plant with 680 workers. was 3250 c.u. In the planned year, with an increase in production by 9%, a relative reduction in the number of employees should be made due to the following factors: modernization of equipment - 28 people, organization of integrated teams and combination of professions - 15, reorganization of the departments of the chief mechanic and power engineer - 5 people. At the same time, due to production needs, the tool area is expanding with an increase in the number of workers by 14. Calculate the change in labor productivity from the change in output per worker.
Task 2. The production labor intensity of the planned volume of production according to the basic norms and conditions of production is 2792 thousand norm-hours. According to the plan for the introduction of new technology, labor intensity will decrease by 4.2% over the year. By improving the organization of labor, the loss of working time will be reduced by 2%. Determine the number of employees if the working time fund per worker per year is 1880 hours.
Task 3. The number of serviced machines in the workshop is 120, the service rate is 8 machines, the number of shifts is 2, the nominal working time fund per worker per year is 262 days, the real one is 232 days. Determine the planned payroll and attendance number of service workers.
Task 4. The workshop has 250 pieces of equipment. In the planned year, their number increases by 15%. The service rate for a locksmith on duty is 20 pieces of equipment, in the planned year it will increase by 10%. The plant operates in two shifts. In the planned balance of working time, absenteeism of workers for all reasons is 12%. Determine the planned payroll number of locksmiths on duty.
Task 5. The number standards for servicing workers are as follows: machine operators - 0.2 people, assemblers - 0.12 people, traffic controllers - 0.08 people. Determine the rate of attendance and payroll number of service workers, if there are 50 machine operators, 40 assemblers, 18 traffic controllers in the workshop. The shift ratio is 1.78, the nominal working time fund per year is 260 working days, the real one is 230 working days.
Problem 6. T the ore intensity of the annual production program of the factory is 2,500 thousand norm-hours. In accordance with the plan for improving production efficiency, it is planned to reduce labor intensity by 15% from July 1. The planned percentage of fulfillment of production norms on average for the factory is 130%; the effective working time fund according to the planned balance of time for one worker is 238 days, and the length of the working day is 7.6 hours. Determine the number of main workers in the planned year.
Task 7. The actual number of industrial and production personnel in the base period amounted to 2800 people. It is planned to increase the volume of production by 105%, and labor productivity - by 106%. Determine the planned number of industrial and production personnel.
Task 8. The total repair complexity of the equipment in the workshop is 5,000 units. The maintenance rate is planned at the level of 500 units. The maintenance rate is planned at the level of 500 units of repair complexity per locksmith per shift; the planned effective and nominal working time funds are equal to 240 and 270 days, respectively. Determine the payroll number of repairmen for two-shift work.
Task 9. The production labor intensity of the planned volume of work according to the basic standards is 2630 thousand norm-hours. It is planned to reduce labor intensity by 120 thousand norm-hours. Determine the planned number of production workers if the working time fund per worker is planned to be 1800 people per hour. per year, and the implementation of production standards by 116%.
Task 10. Determine the number of technologists at the enterprise, if the planned annual production of equipment and tooling for which it is necessary to develop technological processes is 7500 thousand USD; the conditional annual output of one technologist, calculated on the basis of the actual annual production volume in monetary terms, is 5 thousand USD; the coefficient taking into account additional work performed during the development of the technological process is 1.18.
Task 11. In the base year, the volume of marketable output at the plant with 980 workers. was 2890 c.u. In the planned year, with an increase in production by 6%, a relative reduction in the number of employees should be made due to the following factors: modernization of equipment - 28 people, organization of integrated teams and combining professions - 14, reorganization of the departments of the chief mechanic and power engineering - 5 people. At the same time, due to production needs, the tool area is expanding with an increase in the number of workers by 13. Calculate the change in labor productivity from the change in output per worker.
Task 12. The production labor intensity of the planned volume of production according to the basic norms and conditions of production is 3272 thousand norm-hours. According to the plan for the introduction of new technology, labor intensity will decrease by 4.8% over the year. By improving the organization of labor, the loss of working time will be reduced by 3%. Determine the number of employees if the working time fund per worker per year is 1880 hours.
Task 13. The number of serviced machines in the workshop is 100, the service rate is 5 machines, the number of shifts is 2, the nominal working time fund per worker per year is 266 days, the real one is 234 days. Determine the planned payroll and attendance number of service workers.
Task 14. There are 280 pieces of equipment in the shop. In the planned year, their number increases by 12%. The service rate for a locksmith on duty is 20 pieces of equipment; in the planned year it will increase by 12%. The plant operates in two shifts. In the planned balance of working time, absenteeism of workers for all reasons is 10%. Determine the planned payroll number of locksmiths on duty.
Task 15. The number standards for servicing workers are as follows: machine operators - 0.4 people, assemblers - 0.16 people, traffic controllers - 0.08 people. Determine the rate of attendance and payroll number of service workers, if there are 40 machine operators, 30 assemblers, 14 traffic controllers in the shop. The shift ratio is 1.8, the nominal working time fund per year is 262 working days, the real one is 232 working days.
Problem 16. T the ore intensity of the annual production program of the factory is 2,450 thousand norm-hours. In accordance with the plan for improving production efficiency, it is planned to reduce labor intensity by 13% from July 1. The planned percentage of fulfillment of production norms on average for the factory is 120%; the effective working time fund according to the planned balance of time for one worker is 236 days, and the length of the working day is 7.8 hours. Determine the number of main workers in the planned year.
Task 17. The actual number of industrial and production personnel in the base period amounted to 3,400 people. It is planned to increase the volume of production by 104%, and labor productivity - by 103%. Determine the planned number of industrial and production personnel.
Task 18. The total repair complexity of the equipment in the workshop is 6000 units. The maintenance rate is planned at the level of 600 units of repair complexity per locksmith per shift; the planned effective and nominal working time funds are 246 and 278 days, respectively. Determine the payroll number of repairmen for two-shift work.
Task 19. The production labor intensity of the planned volume of work according to the basic standards is 3300 thousand norm-hours. It is planned to reduce labor intensity by 110 thousand norm-hours. Determine the planned number of production workers, if the fund of working time per worker is planned to be 1820 people per hour. per year, and the implementation of production standards by 112%.
Task 20. Determine the number of technologists at the enterprise if the planned annual production of equipment and tooling for which it is necessary to develop technological processes is equal to 6750 thousand USD; the conditional annual output of one technologist, calculated on the basis of the actual annual production volume in monetary terms, is 5.5 thousand USD; the coefficient taking into account additional work performed during the development of the technological process is 1.16.
Table 1
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Indicator |
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Gross output, thousand rubles |
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Table 1
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Indicator |
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Gross output, thousand rubles |
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|
Output per employee PPP, rub. |
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|
Percentage increase in output compared to base period |
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|
Percentage of output growth compared to the base period |
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Table 1
|
Indicator |
||||||
|
Gross output, thousand rubles |
||||||
|
Output per employee PPP, rub. |
||||||
|
Percentage increase in output compared to base period |
||||||
|
Percentage of output growth compared to the base period |
||||||
Table 1
|
Indicator |
||||||
|
Gross output, thousand rubles |
||||||
|
Output per employee PPP, rub. |
||||||
|
Percentage increase in output compared to base period |
||||||
|
Percentage of output growth compared to the base period |
||||||
Table 1
|
Indicator |
||||||
|
Gross output, thousand rubles |
||||||
|
Output per employee PPP, rub. |
||||||
|
Percentage increase in output compared to base period |
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|
Percentage of output growth compared to the base period |
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