Organization and regulation of wages. Labor economics: examples of problem solving Calculation of the planned number of on-duty mechanics
Centralized services have significant organizational and economic advantages. It allows for a more rational use of service workers, concentrating their efforts in certain service areas during the required period, mechanizing labor, and so on. At the same time, the possibilities for organizing in-production planning of maintenance work are improved, which improves its quality, reliability, timeliness, and efficiency.
In a decentralized system, each workshop manager has subordinate auxiliary workers who perform the entire range of necessary work. This ensures timeliness and efficiency of their implementation. However, as practice has shown, with this system it is difficult to ensure normal and stable employment of support personnel and their rational use in accordance with their qualifications.
The most widespread at industrial enterprises is a mixed (combined) service system, in which some service functions are carried out centrally, and the other part - decentralized.
The choice of service system is influenced by the scale and type of production, the production structure of the enterprise, the quality level of available equipment, the complexity of the products, requirements for their quality, and the layout of production areas. However, in all cases the selection criterion optimal system maintenance is a minimum of working time and material resources for service with high quality of the latter.
The planned preventive, preventive nature of maintenance means the coordination of the maintenance system with the operational calendar planning of the progress of the main production, preliminary preparation and delivery of everything necessary for the effective functioning of the workplace, including its completion with materials, workpieces, tools, devices, documentation during the shift and day. The maintenance schedule must be subordinated to the main production schedule. Therefore, work that requires stopping equipment must be performed during intra-shift and inter-shift breaks, or on non-working days.
The flexibility of the system means the possibility of its rapid restructuring in connection with possible deviations from the normal course of the service process.
The complexity of the system presupposes the coordination and linkage of all service functions performed by various services of the enterprise during the same period of time and at the same facilities.
High quality and efficiency are ensured by appropriate qualifications of auxiliary workers, with their optimal number, clear organization of their work, and provision of everything necessary to perform their functions.
Specific maintenance of workplaces can be carried out in one of three main forms: standard, scheduled and preventative and on duty.
Standard (regulated) maintenance allows you to strictly link the work of maintenance personnel with the work schedule of the main production and thereby minimizes downtime of the main workers and equipment. This linkage is ensured by the development of schedules or schedules according to which maintenance functions are carried out in mandatory and volume, in a timely manner. The advantages of this form of service include ensuring full utilization of auxiliary workers, reducing time spent on maintenance, and high quality of work. This system is most appropriate in conditions of mass and large-scale production.
In serial production, where there is a high probability of possible deviations from the normal flow of the production process, it is advisable to use planned preventive maintenance. This form is of a precautionary nature, which is expressed in the preliminary completion of working documentation, tools and devices, workpieces, repairs, adjustments and other work. All kits, in accordance with the calendar plans for the main production of products, are delivered to workplace. This ensures smooth and rhythmic work of maintenance personnel and minimal likelihood of downtime for key workers.
On-duty service is used in single and small-scale production and is characterized by the absence of pre-developed schedules and schedules. It is carried out on calls from key workers as needed. A prerequisite for this system is the availability of operational communication between workplaces and auxiliary services and the control center.
Topic 8. Task 4
Calculate the standard of service and the number of mechanics on duty required to service 250 machines in machine shop, if the shift ratio is 2.4, the real working time is 230, the nominal is 260 days per year. Operating time for servicing one machine is 12 minutes, Totl - 14 minutes, Tpz - 10 minutes per shift.
Solution:
The maintenance rate is calculated using the formula:
Nob = Td / t rev = (230 * 8) / 0.2 = 9200 units,
where Nob is the standard of service, units;
Тд - actual working time fund;
tob - the established time limit for servicing a piece of equipment, h.
MU theme. Problem 9
A worker, whose work is paid according to the piecework-bonus system, performs an operation for which Tsht = 0.12 hours. The hourly tariff rate for work is 90 rubles.
In fact, 1,520 parts were manufactured and passed to the quality control department from the first presentation. For fulfilling the established production norm, a bonus is paid in the amount of 10% of the piecework payment, for each percentage of overfulfillment - 1.5%, but not more than 20%; For the delivery of products to the quality control department, a bonus of 10% is paid from the first presentation.
Solution:
Let's say a worker must work 8 hours a day, the total working days in a month minus weekends and holidays are 22 days. Then the standard number of parts will be: 22 * 8 / 0.12 = 1467 pcs.
Thus, the plan was exceeded by the amount: 1520 / 1467 = 1.0361 or the plan was exceeded by 3.61%.
In fact, 1520 parts were manufactured and handed over to the quality control department from the first presentation, that is, the time spent was:
B = t * Tpc = 0.12 * 1520 = 182.4 hours.
where t is the standard time for the production of one part;
Tsht – actually manufactured.
Monthly wages of a worker under the direct piecework system:
Salary = B * T = 182.4 * 90 = 16,416 rubles.
where B – time consumption;
T – tariff rate.
Award amount:
P = salary * %pr = 16416 * 0.1 = 1641.6 rub.
where ZP is the monthly wage of a worker according to the direct piecework system;
%pr – premium percentage.
For each percentage of overfulfillment, 1.5% of the monthly fee is paid wages, but not more than 20%, that is: 16416 * 0.015 * 3.61 = 888.9 rubles.
For delivery of products to the Quality Control Department from the first presentation, a premium of 10% is paid, that is, 16416 * 0.1 = 1641.6 rubles.
Thus, the monthly wage of a worker under the direct piecework system is 16,416 rubles, the bonus amount is 3,283.2 rubles. (1641.6 + 1641.6 = 3283.2 rubles), the amount of additional earnings is 888.9 rubles, and the total salary is 20588.1 rubles. (16416 + 3283.2 + 888.9 = 20588.1 rub.)
LIST OF REFERENCES USED
- Bychin V.B., Malinin S.V., Shubenkova E.V. Organization and regulation of labor: Textbook. / Ed. SOUTH. Odegova. - M. Exam, 2010;
- Koltsov N.A. Scientific organization of labor. - M.: Higher School, 2010;
- Labor rationing in the transition to a market economy. - M.: Publishing house of the Russian Economic Academy, 2011;
- Organization and regulation of labor: Tutorial for universities / Ed. V.V. Adamchuk. - M.: Finstatinform, 2011;
- Rofe A.I. Scientific organization of labor: Textbook. - M., 2010.
Calculation of the planned number of on-duty mechanics
The workshop has 250 pieces of equipment. In the planned year, their number increases by 15%. An on-duty mechanic services 20 units. On planning period the standard of service increases by 10%. The plant operates in 2 shifts. In the planned balance of working time, absenteeism of workers for all reasons is 12%.Determine the planned number of mechanics on duty
Task 7 Control phases
Learn concepts:
1. life cycle organizations
2. stages of the life cycle, their characteristics
3. methods of organizational diagnostics of an organization
4. levels of enterprise management
5. display a picture of the organization's life cycle
Life cycle of an organization- a set of development stages that a company goes through during its existence.
The life cycle is as follows:
1. formation (birth)
2. growth - when a company actively fills its chosen market segment.
4. decline (old age) - when a company quickly loses its market share and is crowded out by competitors. Subsequently, the organization is either liquidated, merged into a larger one, or split into smaller organizations, which, depending on the situation, may find themselves at the stages of growth or maturity (less often, at other stages).
1. Becoming
The organization is in its infancy and is being formed product life cycle.Goals are still unclear, the creative process flows freely, progress to the next stage requires stable support. This stage includes the following phenomena: inception, search for like-minded people, preparation for the implementation of the idea, legal registration organization, recruitment of operational personnel and release of the first batch of product.
An organization is being created. The founder is an entrepreneur who, alone or with several associates, performs all the work. At this stage, people often come to the company who are attracted by the very personality of the creator and share his ideas and hopes.
Communication between employees is easy and informal. Everyone works long hours, overtime is compensated by modest salaries.
At this stage, the intangible component of motivation is especially important: the opportunity for self-realization, interesting work, recognition of success. Recommended to use simple circuits remuneration, in particular for sales managers.
Control is based on the personal participation of the manager in all work processes. The organization is not formalized or bureaucratic; it is characterized by a simple management structure. The focus is on creating a new product or service and gaining a place in the market. It should be noted that some companies may stop in their development at this stage and exist with this management style for many years.
At this stage, the organization is a sociosystem, since it consists of people belonging to the same or similar paradigms. Each member of the organization has its own cultural beliefs and value system. The joint activities that members of the organization begin to conduct launches processes of knowledge formation at the individual level, when the experience gained by each member of the organization is processed in accordance with personal beliefs and ideas. At the same stage, the knowledge fair begins, when during joint activities each member of the team, voluntarily or involuntarily, demonstrates own system ideas, skills and abilities.
Firms arise voluntarily because they represent more effective method organization of production. At the first stage of its development, the company behaves as Gray mouse- picks up seeds that are overlooked by larger market structures.
At the stage of emergence of a company, it is very important to determine the strategy competition:
The first strategy is force, operating in the field of large-scale production of goods and services.
Second strategy - adaptive:
The tasks of such companies: meet the individual needs of a particular person.
Third strategy: niche deep specialization of production- what an organization can do better than others.
1. Purpose and structure of the RMS plant
The purpose of the plant's RMS is to organize a set of works for the operation, maintenance and repair of main technological and mechanical equipment, as well as to manage this set of works.
The structure of the RMS 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 material technical means are dispersed across technological workshops, in which they independently organize and carry out repairs of their equipment.
Advantages: high mobility, quick start of repairs.
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 carrying out repairs.
Disadvantages: - high cost repair work.
A mixed form is a combined form that combines both of the above forms of repair work. With such
system of work, in addition to the centralization of mechanical services outside the workshop, repairmen work in the workshop itself and have a mechanical repair shop, with a small machine park, and a technical archive.
Advantage: - repairs are carried out by workshop repairmen.
The organization of maintenance and repair is carried out by the plant's mechanical repair service, which is headed by the OGM.
For this workshop we accept a mixed form.
2. Purpose and structure of the OGM of the enterprise.
Purpose of OGM
1) Organization of work on operation, maintenance and repair of equipment in the workshops of the enterprise.
2) Coordination of the activities of workshops and plant services in preparing annual and monthly maintenance schedules.
3) Development of measures to carry out repairs and monitoring their implementation.
Activities of OGM
The activities of OGM include processing requests for repairs, requests for spare parts, as well as placing and preparing orders and work orders for the manufacture and acquisition of parts and spare parts. The OGM is headed by the chief mechanic, who reports directly to the chief engineer of the enterprise.
Structure of the OGM enterprise
3. Form of repairs at the enterprise
There are five options for repairing main and auxiliary equipment:
1) Repair at the manufacturer;
2) Repair at a repair plant;
3) Repair by special repair organizations;
4) Repair at the plant's RMC;
5) Repair by the RMC workshop.
4. Purpose and structure of the RMS workshop
The RMS of the workshop is headed by a workshop mechanic who supervises the repairs carried out in the workshop and also monitors the condition of the workshop equipment. Subordinate to the workshop mechanic are: mechanics, welders who carry out minor repairs.
The purpose of the RMS 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.
Structure of the workshop's repair and mechanical service.
5. Calculation of the number of workshop repair workers
The calculation is made using the following formula
where: K sp - payroll coefficient,
K sp = 1.4 ÷1.8; we take K sp = 1.4;
F n - nominal annual working time of one repairman, F n = 2002 hours;
K m - coefficient of mechanization of work, K m = 1.3;
K n - coefficient of fulfillment of norms, K n = 1.2;
The 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 - number of cars of the same type;
τ n - labor intensity 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 calculations, Hsp = 5.8. We accept Hsp = 6.
6. Number of mechanics on duty
The duties of the duty mechanic include Maintenance equipment. If a mechanical equipment failure occurs during a shift, the mechanic on duty is the first to arrive to repair the failure.
The number of mechanics on duty is determined by the formula:
where K d is the coefficient for taking into account the number of mechanics on duty, K d = 0.2.
The remaining indicators are similar to those given when calculating the number of workshop repair workers.
We take the number of mechanics on duty equal to 1, i.e. 1 mechanic 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 routine repairs, the machine is partially disassembled into components and parts, individual parts and components are replaced or restored, their control mechanisms are inspected, the lining is replaced or restored, and the machine is tested.
If the machine has parts with different service lives, then two routine repairs are carried out.
Current repairs, depending on the capacity of the enterprise, are carried out either by workshop forces, factory RMS, or by special repair teams.
Current repairs are carried out according to scheduled preventive maintenance schedules. The costs of carrying out current repairs are included in the workshop cost.
During a major overhaul, the machine is completely disassembled into parts, cleaned, repaired basic parts, replaced or restored parts, assembled, adjusted, tested at idle and under load. 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. Financing of capital repairs is made from depreciation charges. Executed according to 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 machine being repaired and sent to restoration shops. Afterwards the parts are installed in the same machine. This method is used in low-power enterprises.
2. Impersonal - with this method, worn parts are removed from the machine being repaired and sent to restoration shops, and parts either new or previously restored, taken from the spare parts warehouse, are installed in the machine. It is used in enterprises where long interruptions in the operation of technological lines are unacceptable, as well as in large-capacity enterprises.
This method has two disadvantages:
The so-called “death of capital”;
The need for additional storage space for parts and
nodes There are two types of “impersonal” repair method:
a) MANPADS - periodic replacement of the repair kit;
b) Aggregate-node method.
3. A mixed method is a combination in certain proportions of individual and “impersonal” methods.
5.7. Repair systems
Three systems of scheduled preventive maintenance are used:
Post-examination;
Standard PPR;
Periodic PPR system.
1. Post-inspection - the PPR system provides for a periodic (once a month) inspection by a mechanic of the workshop equipment, during which he makes a conclusion about the need for repairs (used in low-power enterprises).
2. Standard maintenance work - consists of using an assigned resource, upon exhaustion of which the equipment, despite its technical condition, is replaced or repaired. It is used in cases where a possible failure could lead to human casualties or major material losses.
3. Periodic system - it is associated with carrying out repairs according to the maintenance schedule. It is used in 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
We calculate the PPR schedule using the methodology proposed in the “Regulations on PPR...”.
The initial data for the calculation are the standards for the frequency and labor intensity of types of repairs, from the “Regulations on PPR...”. The initial data is 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 |
|||||
| Dry grinding ball mill |
|||||
5.5.1. Factory repair
It consists of cleaning the equipment, placing it in a special container and then sending it to the manufacturer.
At the manufacturing plant, the equipment is unpacked, then disassembled into assemblies and parts, and defect detection of parts is carried out. Unusable parts are sorted out, worn parts are restored. Then the machine is assembled and started at idle speed (run-in). Then it is packaged in the same container and sent to the consumer.
Advantages:
a) replacement of worn parts with parts with the same tolerance ranges;
6) repairs in more short time;
c) higher quality of repairs.
Flaws: high cost of repairs.
5.5.2. Repair at a repair plant
A repair plant is usually located near enterprises that have 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 devices, access roads, etc.
There are two options for equipment repair using repair plants:
1) When the equipment being repaired is sent directly to the factory for repair. At the factory, the equipment is unpacked, disassembled into assemblies and parts, equipment is troubleshooted, worn parts are restored or replaced with new ones. The machine is assembled, run-in is carried out, and then the equipment is sent to the consumer.
Disadvantage: low quality of workmanship compared to the manufacturer.
2) The repair plant sends repair specialists to the enterprise where the repairs are carried out. At the same time, the company provides these specialists necessary tool, working conditions, materials, and in some cases, mechanics and welders.
5.5.3. Special repair teams (contractors)
These are organizations that employ high-class engineers and repairmen on their staff. They are provided necessary equipment for carrying out repairs of any complexity. The advantages include the high quality of repairs carried out in a short time, and the disadvantages high cost repair.
5.5.4. By the forces of the plant's RMC
The enterprise organizes a mechanical repair shop, equipped with the necessary machine tools, fixtures and tools, and
also repair sites. RMCs are staffed with the necessary maintenance personnel.
Their work (the work of the RMC) is based on PPR graphs. According to this schedule, all types of inspections and repairs of the plant are carried out, and the RMC reports to the chief mechanic of the plant.
5.5.5. By the forces of the RMS workshop
The workshop has a certain staff of mechanics performing routine daily work. For the period of repairs, a repair team is allocated from this staff, led by a foreman. Remuneration for the members of the repair team is either piecemeal or piecework - bonus. Upon completion of the repair, repairmen perform daily routine 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 equipment life.
The organization of lubrication facilities in the workshop should be based on certification of equipment with the preparation of maps and lubrication tables.
Certification covers all technical, lifting and transport equipment and consumer mechanisms lubricants. Based on 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, the required amount of lubricants is determined 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 OGM, serve as the basis for drawing up relevant applications. Arrival, storage and delivery of lubricants to production areas is carried out by employees of the workshop's lubricants warehouse.
5.11.1 Application for fuel and lubricants
Table 5.3
Application for fuel and lubricants
5.12. Mechanics' workplace
The diagram of the mechanics' workplace and the arrangement of auxiliary equipment is shown in Fig. 16.
Locksmith room diagram
 |
1 - vertical drilling machine;
2- sharpening machine;
3- cabinet for storing tools and equipment;
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 a canopy. This warehouse stores machine parts and assemblies, as well as different kinds consumables that are not afraid of precipitation. There is also a closed warehouse in the workshop where equipment, parts and components are stored that are protected from direct exposure to precipitation and sunlight. Such Consumables how bearings are stored in special cabinets with a constant ambient temperature.
5.14. Work permit for repairs
Enterprise, workshop_________________________________________________________________
Work permit No._____
To carry out work increased danger
1. Work performer ______________________________________________________________ (enterprise, workshop, position, last name, I, O.)
2. Allowed for execution_______________________________________________________________
(place of work, name of equipment, summary works)
3. D lowering to work _____________________________________________________
(position, surname, I, O.)
4. Measures to ensure work safety:
4.1 Stop_________________________________________________________________
(stop place, position)
4.2. Disable ______________________________________________________________
(switch, valve, line, etc., remove the tag)
4.3. Install _______________________________________________________________
(short circuit, dead end, plugs, warning lights, etc.)
4.4 Take a sample for air analysis _____________________________________
4.5. Protect _________________________________________________________________
(work area, put up posters)
4.6. Provide safety measures when working at heights and in wells _____________
_____________________________________________________________________________
4.7. Warn _____________________________________________________________
(drivers of adjacent cranes and cranes of adjacent spans with a signature in the logbook)
______________________________________
4.8 Warn safety measures near railway tracks ______________________
______________________________________________________________________________
(installation of signs, posters, fences, dead ends, etc.)
4.9. Indicate routes to the place of work _____________________________________________
(attach a diagram if necessary)
4.10. Additional events ________________________________________________
______________________________________________________________________________
5. The permit was issued by __________________________________________________________
(position, full name)
6Activities completed:
7 Agreed: shift (section) supervisor _________________________________________
_____________________________________________________________________________
(surname, signature)
7.1. __________________________________________________________________________
(position surname, signature)
8 The measures have been completed, work safety has been ensured, the work foreman has been familiarized and instructed with the working conditions, I have granted permission - allowing him to work ________________________________________________________________________________
9Familiarized and instructed with the working conditions, checked the training, accepted the workplace - the work foreman_______________________________________________
(position, surname, signature, date, time)
________________________________________________________________________________________________________________
1 The work has been completed, the work permit has been accepted by the work manufacturer ______________________
___________________________________________________________________________
(date, time, position, surname, signature of the person admitting)
5.15. Project for organizing work during car repairs
When preparing to repair large and complex equipment, a work organization project (WOP) is drawn up, which is developed by the performing organization and agreed with the customer.
The project for organizing repairs must provide for technical means of mechanization of the work to be performed, the composition and qualifications of repair personnel. Organization of receipt, delivery and storage of equipment units and parts. Organization of repair and installation areas and sites at the work site, as well as safety measures for their implementation.
The ERP regulates the procedure for carrying out and measures to ensure operations of the preparatory, disassembly and restoration and assembly periods.
5.16. Calculation of the operational schedule for major repairs
1) At the first stage, it is necessary to write out from the “standard norms...” a list of all repair operations performed, indicating the number of components, the labor intensity of the operations and the category of repairmen.
2) Determine the approximate size of the team to perform repair operations using the formula:
H k = τ k / t K =25.4/8 =3.175 people. (65)
We accept the closest value - the composition of the team is 3 people: mechanics of 3, 4, 5 categories,
where τ k is the standard labor intensity of performing one repair, person-hour;
t K - repair time, hour
3) Rules for drawing up a schedule:
There must be a strict rotation of operations;
Repairers must move from one operation to another without interruption;
On each repair day there should be the same number of repairmen;
Observe logical, from a safety point of view, operations;
Repairers should not interfere with each other.
4) The length of the segment on the graph is determined by the formula:
T 0 = (τ 1 + τ 2 + ... +τ n)/n rem (66)
where τ 1, τ 2, τ n is the labor intensity of the operation, indicated on the graph by one segment, person-hour;
Prem - the number of repairmen at this operation, people.
T o1 = 0.6 / 3 = 0.2 hour;
T o2 = (0.1+0.5)/1 =0.6 hour;
T o3 = (0.8 + 2.0)/2 = 1.4 hours;
T o4 = 0.8/1 =0.8 hour;
T o5 = (0.5+ 0.5)/2=0.5 hour;
T o6 = 0.5 / 1 = 0.5 hour;
T o7 = 0.6 /1 = 0.6 hour;
T o8 = 5.0 / 2 = 2.5 hour;
T o9 = (1.5 + 0.4)/1 = 1.9 hours;
T o10 =3.0/3 = 1 hour;
T o11 = (0.6+ 1.2)/2 = 0.9 hour;
T o12 = 0.9/1 =0.9 hour;
T o13 = (0.8 + 0.6 + 0.8 + 0.2) /3 = 0.8 hour;
T o14 = 2.0 / 3 = 0.67 hour;
T o15 = 1.5 /3 = 1.5 hour
5) When other repairmen are involved in the repair operations included in formula (5.11), the length of the segment is adjusted.
6) When repairmen switch from one operation to another, expression (5.11) is adjusted.
7) If repairmen who are not associated with the activities of the repair and mechanical service of the workshop are involved in performing individual operations, then a separate line is allocated in the operational schedule for these types of work. The complexity of performing the work of these workers is not taken into account in (5.10).
8) If repairing equipment requires a large amount of welding work, then gas-electric welders are involved in the 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 mechanics of 4 and 5 categories. The complexity of the work in this case is taken into account in expression (5.10).
Table 5.4
List of defects
| No. |
Name and scope of work |
the name of detail |
||||||
| detail |
Quantity, pcs |
Weight, kg |
Name size |
Brand, GOST |
Unit measurement |
Quantity, pcs |
||
| Gearbox repair |
Bearings Gears |
Cardboard spacer Industrial oil I-20 |
GOST 20346-74 GOST 20793-72 |
|||||
| Drum body Drum cover |
Sheet steel |
GOST 19904-74 |
||||||
| Bearings GOST 5721-75 |
Cardboard spacer |
GOST 20376-74 |
||||||
Table 5.5
List of wearing parts
| Units and their details |
Drawing No. or GOST |
Quantity per unit |
Page
9
Number of instrumentation mechanics
The number of instrumentation mechanics is determined based on labor intensity production program workshop, calculated in paragraphs 2.2 and 2.3 and presented in the form of table 9.
Based on the labor intensity of maintenance and current repairs, the number of mechanics on duty in the operation group is determined (at posts in the instrumentation and automation workshop production workshops). The number of instrumentation mechanics in repair shops and the calibration laboratory of the instrumentation and automation workshop is calculated on the basis of the labor intensity of major repairs, installation and inspections. The estimated number of workers by category is determined by dividing the labor intensity of work of a certain category by the effective time fund of one payroll worker. To determine the accepted number of workers, the estimated number of workers is rounded to a whole number, while it is allowed to transfer part of the work of a lower category to mechanics of a higher category.
Number of metal workers
The number of metalworkers is determined based on the labor intensity of manufacturing spare parts and is presented in the form of a table. 10.
The list of required professions and their categories are established according to the data technological maps for processing parts. For each profession of a certain category, the estimated number of workers () is determined by the formula:
where Ni is the number of parts of the i-th name in the annual production program, pcs. (i= 1,2,3, .,l);
Norm of piece-calculation time for i-th operation, min;
60 - number of minutes in an hour;
Effective time fund of one payroll worker, h;
1.3 - planned coefficient of fulfillment of time standards.
To determine the accepted number of workers, the estimated number of workers is rounded to the nearest whole number. When rounding, the following combination of professions (transfer of work) is allowed:
a) work of a lower category within one profession is transferred to workers of a higher category (within two adjacent categories);
b) the work of the marker is transferred to the mechanic and vice versa; The work of the boring man is transferred to the turner and vice versa.
Calculation of the number of instrumentation mechanics
Table 10
| Device names | Technological service | Current repairs | Installation work, start-up and commissioning |
||||||||||||||
| labor intensity per person | category of work | labor intensity per person | category of work | labor intensity per person | category of work | labor intensity per person | category of work | labor intensity per person | category of work |
||||||||
|
thermometers | |||||||||||||||||
|
thermocouples | |||||||||||||||||
|
pressure gauges | |||||||||||||||||
|
draft and pressure meters | |||||||||||||||||
|
electronic level gauges | |||||||||||||||||
|
differential pressure gauges | |||||||||||||||||
|
"E" type devices | |||||||||||||||||
|
EVP type devices | |||||||||||||||||
|
ring pressure gauges | |||||||||||||||||
|
flow meters | |||||||||||||||||
|
Total labor intensity person/hour | |||||||||||||||||
| by category | |||||||||||||||||
|
III category | |||||||||||||||||
|
Effective time fund of one payroll person-hour | |||||||||||||||||
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Send your good work in the knowledge base is simple. Use the form belowStudents, graduate students, young scientists who use the knowledge base in their studies and work will be very grateful to you. Posted on http://www.allbest.ru/ Posted on http://www.allbest.ru/ Problem 1 Determine the piece, piece-calculation time for manufacturing a car pallet, as well as the time for producing a batch of pallets, if the forging time is 15 minutes, the auxiliary time is 2 minutes, the preparatory and final time is 9 minutes, the coefficient of rest time and personal needs is 13%. The batch size is assumed to be 30 pcs. plus no. Solution: T o = 15 min. T in = 2 min. T pz = 9 min. Lot size 43 pcs. (30 pcs +13 option no.) T op = T o + T v, T op = 15+2 = 17 min. T exc = 13% of operating time = (17x13):100 = 2.2 min. T orm - time for servicing the workplace, T op - operational time; T exc - time for rest and personal needs; T pt - time of breaks provided by the technology; T pz - time of preparatory and final work. standard wages and labor We determine piece time. T pcs. = 17+2.2+ 9= 28.2 min. T part = T piece H n + T pz = 28.2 x 43 + 9 = 1222, where T pcs is the piece time, n is the number of products in the batch (43 pcs.) Answer: The standard piece time is 28.2 minutes. The standard time for a batch of products is 1222 minutes. Problem 2 As a result of implementation new technology the output of the main workers increased by 15% plus no. Determine how much time standards should be reduced Solution: The production rate of the main workers increased by 28% (15% given + option 13). We determine the % reduction in the time norm using the formula % reduction in Нр = where N vr. - time norm, N vyr. - production rate % reduction in H time = (28x100) : (100+28) = 21.9% Answer : The time limit should be reduced by 21.9%. Problem 2 Calculate norms of piece, piece-calculation time and time for a batch of products, if the main work time is 20 minutes, time auxiliary work- 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 minutes. for a batch of parts. The number of parts in the batch is 30 pcs. plus no. Solution: T o = 20 min. T in = 7 min. T pz = 25 min. Number of parts per batch: 43 pcs. (30 pcs +13 option no.) We determine the standard operating time. T op = T o + T v, where T o - main time; T in - auxiliary time T op = 20+7 = 27 min. Determine time for rest and personal needs T exc = 6% of operating time = (27x6) : 100 = 1.6 min. We determine the standard for workplace maintenance T orm = 7% of operational time = (27x7): 100 = 1.9 min. We determine the standard time. N time = T pz + T op + T orm + T exc = 25+27+1.9+1.6 = 55.5 min. We determine piece time. T pcs = T op + T orm + T exc + T pt T pcs. = 27+1.9+1.6= 30.5 min. Determines the time standard for a batch of products T batches. T part = T piece H n + T pz = 30.5 x 43 + 25 = 1336, where T pcs is the piece time, n is the number of products in the batch (31 pcs.) Answer: The standard piece time is 30.5 minutes. The standard time for a batch of products is 1336 minutes. Problem 4 Calculate the standard of service and the number of mechanics on duty required to service 250 plus number of machines in a machine shop, if the shift ratio is 2.4, the real working time is 230, the nominal one is 260 days a year. Operating time for servicing one machine - 12 min, T excellent-14 min, T pz- 10 minutes per shift. Solution Determine the number of machines 250+13 =263 We determine the standard time. N time = T pz + T op + T orm + T ex + T pt, Where; N time - standard time; T orm - time for servicing the workplace, T op - operational time; T exc - time for rest and personal needs; T pt - time of breaks provided by the technology; T pz - time of preparatory and final work. N time = 14+10+12 = 36 min. where N h is the population norm; O - the total number of service units serviced or the volume of work performed, T cm length of the working day. Determining the population norm We determine the standard of service. Answer : Headcount norm 36.4, service norm 0.2 Problem 5 Calculate the rate of maintenance of automatic machines if the operational time for setting up one machine is 7 minutes, the time for moving 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 written-off number of adjusters, if the workshop has 400 machines plus No., the workshop operates in two shifts, T cm- 8 hours. The number of attendance days per year is 230, working days - 260. Solution: We determine the number of machines 400+13 = 413 pcs. T o = 7 min. T in sp = 1.5 min. T pz = 10 min. We determine the standard operating time. T op = T o + T v, where T o - main time; T in - auxiliary time T op = 7+1.5 = 8.5 min. Determining the standard service time N time = T pz + T op + T orm + T exc + T pt = 10+8.5+ 18 = 36.5 We determine the standard maintenance time for machines Determining the availability of service technicians We determine the number of service technicians Answer: Available number of adjusters - 65 The number of service technicians is 58 Machine maintenance rate - 0.4 Problem 6 The production rate for a worker per shift is 60 parts + no. The worker actually produced 66 parts. Determine the percentage of fulfillment of production standards. Solution: We determine the % of fulfillment of the production norm. where the actual output of the worker is 66 parts, the production rate is 73 parts (60 + 13 option) P in = (66: 73) x 100% = 90.4 Answer: Percentage of fulfillment of production standards -90.4 Problem 7 Determine the percentage of fulfillment of production standards by a brigade of drivers, if the volume of transportation was 22.5 thousand tons, the cargo turnover was 250.6 thousand tkm, the time standards for loading and unloading 1 ton of cargo was 0.067 hours, for 1 tkm - 0.013 hours, in a team 25 drivers, each worked 24 days with a 7-hour working day. Solution: We determine compliance with standards based on actual time worked: Pvn = = =0.05 Where Tn - 0.067 is the time required to produce suitable products; Тд - additional time costs due to deviations from normal working conditions; Tsd - 24 x 7 = 168 hours - actual time worked The percentage of compliance with the standards for the replacement stock is determined by the formula: where Tpr is intra-shift loss of working time and downtime; Tp is the time of use of piece workers for time-based work. Problem 7 Determine the percentage of standards fulfilled if the standard time for manufacturing product A is 0.25 standard hours, product B is 0.3, product C is 1 standard hour. According to the plan, it is necessary to produce 200 units + number of product A, 320 - product B and 30 - product C. In fact, 243 products A, 400 products B, 28 products C are manufactured. Solution Determining the production rate Product A In fact, 243 products were manufactured, standard 213 ed. (200 + 13 option) Pvn = (243: 213) x 100 = 114.1 Product B In fact, 400 products were manufactured, the norm is 320. Pvn = (400: 320) x 100 = 125 Product B In fact, 28 products were manufactured, the norm is 30. Pvn = (28: 30) x 100 = 93.3 Determining the standard time Product A Production time standard - 0.25 standard/hours Pvn = (0.25: 7) x 100 = 3.6 Product B Production time standard - 0.3 Pvn = (0.3: 7) x 100 = 4.3 Product B Standard time for production - 1 Pvn = (1: 7) x 100 = 14.3 Answer: % production rate: Product A -114.1е B- 125, B -93.3. % time norm: Product A - 3.6, B - 4.3, C - 14.3 Problem 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 workshop No. 2 is 654.0 thousand standard hours, of which for category 3 - 350.5, for category 4 - 146.5, for category 5 - 112 standard hours, the rest of the work is carried out according to category 4. Determine the average tariff coefficient for workers in workshop No. 1 and the average tariff coefficient for work in workshop No. 2. The company has a tariff schedule:
Solution: We calculate the average tariff coefficient K av = K m + (K b -K m) x (P av -R m), where K av - the average tariff coefficient, K m - the tariff coefficient is less than the average, K b - the tariff coefficient is greater than the average, - the tariff category is average; - tariff category is less than average; - the tariff category is higher than average. Determining the average tariff category (2x20 + 3x210 +4x193 +5x20) : 443=3.5 A rank that is less than the average K 3 = 1.23, a rank that is greater than the average K 4 = 1.36 We calculate the average tariff coefficient for workshop No. 1 K av = 1.23+(1.36-1.23) x (3.47-3)= 1.29 We calculate the average tariff category for workshop No. 2 K av = (3x350.5+4x146.5+5x112+6x45) : 654= 3.77 Rank below average K 3 = 350.5, rank above average - K 4 = 146.5 By Wed. = 350.5+ (146.5-350.5) x (3.77-3) = 1.13 Answer: The average tariff coefficient for workshop No. 1 is 1.29 The average tariff coefficient for workshop No. 2 is 1.13 Problem 9 The worker processed 185 products A and 900 products B. The standard time for product A is 55.48 minutes, for product B is 5.22 minutes. The hourly wage rate for an employee is 33 rubles plus No. . Calculate the employee's earnings. Solution: We determine the hourly tariff rate - T standard daily 33+13 = 46 rubles. We determine the piece rate: Dist. sd. = N time x T st.h. , where is the time standard; T st.h. - hourly tariff rate. Product A - Dist. sd. = 0.92 x 46 = 42.3 rub. Product B - Dist. sd. = 0.09 x 46 = 4.1 rub. We determine wages. Salary sd. = Dist. sd x O vyr. Product A - ZP sd. = 42.3 x 185 = 7881 rub. Product B - ZP sd. = 4.1 x 900 = 3690 rub. Total salary 7881 +3690 = 11571 rubles. Answer: Salary - 11571 rub. Problem 10 The planned volume of work is 120 products, in fact the employee produced 130 products, the piece rate is 35 rubles. per piece plus no. For exceeding the plan, a bonus is established in the amount of 2% of piecework earnings for each percentage of overfulfillment. Calculate the employee's earnings. Solution: We determine the piece rate of 35 rubles. +13 = 48 rub. Determining piecework wages Salary sd. = T h.st. x About vyr. = 130 x 48 = 6240 rub. We calculate the percentage of exceeding the plan 130: 120 x 100 = 108% We calculate the amount of wages taking into account the bonus for exceeding the plan 116 x 6240: 100 = 7238 rubles. Answer: The employee's salary is 7238 rubles. Problem 11 The hourly wage rate for a service worker is 30 rubles. plus No., service normequipment - 2 units of equipment (respectively 2 main workers), butRThe production output of the main worker is 6 pcs. at one o'clock. Calculate amount earnings of a service worker, if the actual volume of work of the main workers is 12,304 pcs. per month. Solution: We calculate the hourly tariff rate 30+13 = 43 rubles. where is the indirect piece rate; - standard of service. We calculate the amount of work per hour - 12304: (22x 8) = 70 pcs. Dist. kosv. = 43: (70 x 2) = 0.3 Salary = Calculation kosv. x N vyr. = 0.3 x 12304 = 3691 rub. Answer: The salary of a service worker is 3,691 rubles. Problem 12 Calculate the worker’s monthly earnings according to the piecework-bonus wage system, if the standard time is 0.8 person-hour, the price is 42.3 rubles, 250 products were delivered per month plus no. For compliance with production standards, a bonus of 10% of piecework earnings is provided. For every percent exceeding standards - 1% of piecework earnings. 22 work shifts of 8 hours each were worked. Solution: We determine the volume of products 250 + 13 = 263 pieces. where salary/sd is piecework earnings; Rast/sd - piece rate; - workload. We calculate piecework wages Salary SD = Dist. sd. x About vyr. = 42.3 x 263 = 11124 rub. We calculate the % of fulfillment of standards - 0.8 person hour x 263 pieces. = 210.4: (22x8)= 120% We calculate a bonus in the amount of 10% of piecework earnings - 11124 x 10: 100 = 1112 rubles. We calculate the bonus for exceeding the standards - 11124 x 20: 100 = 2225 rubles. We calculate the worker’s monthly earnings as 11124+1112+2225=14461 rubles. Answer: Salary -14461 rub. Problem 13 Determine monthly salary a worker whose work is paid according to a progressive piece-rate system, if the standard time for a product is 0.2 standard hours. The worker worked 22 days for 8 hours. The hourly tariff rate is 78 rubles plus no. Actually 1520 products were manufactured per month. The bonus is awarded only if the standards are exceeded by more than 100%. Solution: We determine the hourly tariff rate of 78 rubles. + 13= 91 rub. Calculates the piece rate Dist. sd. = 0.2 x 91 = 18.2 rub. We calculate how many hours a worker worked 22 x 8 = 176 hours. Salary SD = Dist. sd x O vyr. = 18.2 x 1520 = 27764 rub. We calculate the % of exceeding the standards - 1520 x0.2 =304: 176 =173 The bonus is not paid since the standards were exceeded by 173%, and according to the conditions they should be exceeded by more than 100%. Answer: Salary - 27,764 rubles. Problem 14 Determine the price (based on the time standard and the production rate) if the time standard is 40 minutes, the corresponding production rate 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. N time = 40: 60 = 0.67 We calculate the price according to the time standard - Dist. = N time xT st. = 0.67h. x 94 rub. =63 rub. N calculated hour = 12: 8 = 1.5 We calculate the price based on the production rate - Calc. = N calculated hour. xT st. = 1.5 x 94 = 141 rub. Answer: Dist. present =63 rub., Calc. n.vy. = 141 rub. Problem 15 Calculate the price according to the indirect piecework wage system for adjusters of particularly complex and unique equipment, if the adjustment work is classified as category 5 (hourly rate - 100 rubles plus No.), and the adjuster services four operating units with a standard shift output of 20, 26 and 33 PC. in 8 hours. Solution: We determine: the hourly tariff rate is 100 rubles. + 13 = 113 rubles, daily tariff rate 113x 8 = 904 rubles. We calculate N vyr. = 20+26+33=79 where is the indirect piece rate; N output - production rate. Dist. to osv. = 11.4 Answer: Dist. kosv. = 11.4 Problem 16 Calculate the amount of the worker’s wages for the time actually worked, if his monthly salary is 21,000 rubles plus No. per month, according to the schedule, he should have worked 22 shifts of 8 hours each, but in fact 20 shifts of 8 hours were actually worked. Solution We determine how many hours actually worked: 20 x 8 = 160 hours. According to the schedule, 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 time actually worked - 21013 x 90.9 = 19100 rubles. Answer : The salary amount is 19,100 rubles. Problem 17 The hourly wage rate of an employee is 75 rubles plus No. per month, 176 hours were actually worked. high-quality execution work, a bonus is established in the amount of 20% of tariff earnings. Calculate the amount of the employee's salary. Solution: Hourly tariff rate 75 + 13 = 88 rub. We determine the salary 88 x 176 = 15,488 rubles. We determine the size of the premium 15448 x 20: 100 = 3098 rubles. Salary - 15448+3098 = 18546 rubles. Answer: The employee's salary is 18,546 rubles. Problem 18 The worker's hourly tariff rate is 100 rubles plus No. per month worked 21 days for 8 hours. Calculate the worker's earnings if a bonus payment of 30% of direct time-based earnings is provided. Solution : We determine the worker’s tariff rate - 100 + 13 = 113 rubles. Hours worked - 21 x 8 = 168 hours Salary - 113 x 168 = 18,984 rubles. We determine the size of the premium 18984 x 30: 100 = 5695 rubles. Worker's salary -18984+5695=24679 rub. Answer: The worker's salary is 24,679 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 is provided in the amount of 25% of direct time earnings. Solution: We determine the employee’s salary - 14,000 + 13 = 14,013 rubles. Actually worked in January - 18 days, according to schedule - 18 days. in February - 20 days, according to schedule - 21 days in March - 15 days, according to the schedule 22 days. Conditional cost of one working day in January 14013: 18 = 779 rubles. in February 14001: 20 = 701 rub. in March 14001: 22 = 637 rubles. We determine the salary for January 14013x1 = 14013 rubles x 125 = 17516 rubles. for February 14013x (20:21) x 100 = 13312x 125 = 16640 rubles. for March 14013 x (15: 22) x100 = 9529 x 125 = 11911 rubles. Answer: Salary for January - 17,516 rubles. for February - 16,640 rubles. for March - 11911 rubles. Problem 20 Worker 4th category ( hourly tariff rate for a 6-hour working day is 107 rubles plus no.). worked 24 shifts in a month. For fulfilling the plan, a bonus is paid in the amount of 15% of earnings, for each percentage of exceeding the plan - 1.5% of earnings, for saving material - 40% of its cost. The plan was completed by 108%, materials were saved for 3200 rubles. Determine your total earnings. Solution: We determine the hourly tariff rate - 107 + 13 = 120 rubles. 120 x 24 x 7 = 20,160 rubles. We determine the amount of the bonus for fulfilling the plan - 20160 x 15: 100 =3024 rub. We determine the size of the bonus for each percentage of overfulfillment - 20160 x 8 x 1.5 = 2419 rub. Determining the size of the bonus for saving materials 3200 x 40:100 = 1280 rub. We determine the total earnings -20160+3024+2419+1280=26883 rubles. Answer: Total earnings - 26,883 rubles. Problem 21 An employee with a monthly salary of 14,920 rubles plus No. worked 20 days per month instead of 23 working days according to 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, taking into account bonuses according to the regional coefficient (1.4) and for work experience in areas equated to the regions of the Far North. Solution: We determine the monthly salary 14920 + 13 = 14933 rubles. We determine the actual salary - 14933: 23 x 20 = 12985 rubles. We determine the size of the premium - 12985 x 35: 100 = 4545 rubles. We determine the amount of wages taking into account regional coefficient- (12985+4545) x 1.4 = 24542 rub. We determine the amount of additional payments for work experience in areas equated to the regions of the Far North - 24542 x 50: 100 = 12271 rubles. We determine the total earnings - 24542 + 12271 = 36813 rubles. Answer: Monthly earnings are 36,813 rubles. Problem 22 The number of working days per month according to the schedule is 20, the shift duration is 8 hours.Workerworked 180 hours in a month, including overtime for 3days: 2 days - 3 hours and 1 day - 2 hours.The rest of the time was worked according to production needs on weekends.The hourly tariff rate is 100 rubles plus no. The bonus (50%) is awarded only on directtemporary income.Determine the worker's earnings. 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 rub. - (339+226)x 2 = 1130 rub. in the first two days 2x113x1.5 = 339 rub. The total amount for overtime is 1130+ 339 = 1469 rubles. We determine the size of the bonus - 18080 x 50: 100 = 9040 rubles. We determine the total earnings - 18080+1469+9040=28589 rubles. Answer: worker's earnings - 28,589 rubles. Problem 23 The monthly tariff rate for a worker is 10,000 rubles. plus no.The monthly time norm is 168 hours.The worker worked 176 hours, including 8 hours on a day off.The bonus is calculated on all earnings in the amount of 35%. Determine his earnings. Solution: We determine the worker’s tariff rate 10000+13= 10013 rubles. We determine the hourly tariff rate 10013: 168 = 60 rubles. We determine the salary for a day off - 8 x 60x 2 = 960 rubles. We determine the size of the bonus - (10013+960) x 35: 100 = 3840 rubles. We determine the total earnings - 10013+960+3840= 14813 rubles. Answer: The worker's earnings are 14,813 rubles. Problem 24 The employee worked 12 hours at night during the month. The surcharge is set at 40% of the tariff rate. Hourly tariff rate - 125 rubles plus No. Total worked 168 hours. A 30% bonus is awarded on all earnings. Determine the worker’s earnings taking into account additional payments for work in night shift and bonus payments. Solution: We determine the hourly tariff rate - 125 + 13 = 138 rubles. We determine the amount of wages without taking into account night work - 168-12 = 156 x 138 = 21,528 rubles. We determine the amount of wages for work at night - 12x138x2 = 3312 rubles. We determine the size of the premium - (21528+3312)x30:100= 7452 rubles. We determine the salary amount - 21528+3312+7452=32292 rubles. Answer: Salary - 32292 rub. Problem 25 The worker's tariff rate per hour of work is 126 rubles, 22 shifts of 7.4 hours each were worked in conditionswith an increased levelnoise and increased dust content of the work shift area. For This workplace provides an additional payment for working conditions in the amount of 8%. The employee received a bonus on all earnings in the amount of 40%. Determine the worker's monthly salarywork. Solution: We determine the size 126x22x7.4 = 20513 rubles. We determine the amount of the additional payment - 20513x8:100 = 1641 rubles. We determine the size of the bonus - (20513 + 1641) x 40:100 = 8862 rubles. We determine the amount of wages - 20513+1641+8862 = 31016 rubles. Answer: Salary - 31,016 rubles. Problem 26 The 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, taking into account additional payments for workon a weekend, a bonus of 30% is charged. The employee is given a personal allowance of 2,500 rubles.The company where you work the employee is located in an area where the regional coefficient is 1.5, the area refers to the southern regions Far East. The employee's work experience in this area is 5 years. Determine earnings. Solution: We determine the daily tariff rate 1000+13=1013 rubles. We determine wages excluding overtime - 22x1013 = 22,286 rubles. We determine the hourly tariff rate 1013:8 = 127 rubles. We determine overtime wages - 2x127x1.5 = 381 rubles, 82x127x2 = 508 rubles, 381+508 = 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 = 50066 rubles. Answer: Total earnings - 50,066 rubles. Problem 27 The company sets the salary of its managersdepending on trade turnover, achieved in their sector in the previous month. With a turnover of 100.0 thousand rubles. salary is 8000 rubles. plus no. Each percent increase in trade turnover results in a 0.7% increase in wages. Determine the manager’s salary (for each month) based on the following data on turnover in his sector: Solution: We determine the salary 8000+1=8013 rubles. where % TO - % increase in trade turnover, % increase in salary. - % 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. During the month, funds for paying the company’s employees amounted to 300 thousand. rub.plus no. Determine the salaries of the company's specialists. Solution: We determine the amount of funds for wages: 300,000 + 13 = 300,013 rubles. We determine wages. Director - 300013 x 0.3 = 90003 rub. Accountant - 300013x 0.15 = 45002 rub. 1 specialist - 300013 x 0.2=60003 rub. 2 specialist - 300013 x 0.15 = 45002 rub. 3 specialist - 300013 x 02 = 60003 rub. Answer: Salary: Director - 90,003 rubles, accountant - 45,002 rubles, 1st specialist - 60,003 rubles, 2nd specialist - 45,002 rubles, 3rd specialist - 60,003 rubles. Problem 29 Employees are paid a monthly wage fundfees in the amount of 150 thousand. 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 company's specialists. Solution: We determine the monthly wage fund 150000+13=150013 rubles. Determining the average salary 150013:(1.95+2.1+2.25+4.0)=14564 rub. Determining wages 14564x1.95 = 28400 rub. 14564x2.1= 30584 rub. 14564x 2.25=32769 rub. 14564x4 = 58260 rub. Answer: Salary: 28,400 rubles, 30,584 rubles, 32,769 rubles, 58,260 rubles. Posted on Allbest.ru ...Similar documentsThe importance of wage level factors for the national economy. Calculation of a worker’s monthly wage under a piece-rate progressive and piece-rate wage system. Analysis of the use of the planned monthly wage fund for workshop workers. course work, added 02/25/2015 Essence, characteristics and types of piecework forms of remuneration. An example of a piecework-bonus and lump sum wage system. Characteristics of the piece-rate progressive and indirect wage system. Efficiency of application of piecework wage systems. abstract, added 02/18/2013 Factors and trends in the development of wage forms. Time-based wages. Features of a tariff-free payment system. Bonus and its economic justification. 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