**Task 2.**

**Determine the amount of annual depreciation, if known** :

d | |

The initial cost of fixed assets at the beginning of the period, thousand rubles. | |

The cost of input fixed assets, thousand rubles. | |

The cost of retired fixed assets, thousand rubles. | |

Duration of operation of funds, months incoming outgoing | |

Depreciation rate, % |

Decision:

**1) The** average annual cost of fixed assets is determined on the basis of their initial cost, taking into account the commissioning and liquidation of fixed assets during the year:

Fsr.g u003d Fo + Fvv * World Cup / 12 – Fvyb * (12-M) / 12

Fsr.g u003d 8960 + 1000 * 6 / 12 – 760 * (12-8) / 12 u003d 9206.67 thousand rubles.

where Фср.г is the average annual cost of fixed assets, thousand rubles.

Fo – the initial cost of fixed assets, thousand rubles.

Fvv is the cost of fixed assets introduced during the year, thousand rubles.

FM – the number of months of operation of the introduced OS

Fvyb – the cost of retiring in tech. OS years, thousand rubles

M – number of months of operation of retired OS.

**2)** Annual depreciation charges:

Ag u003d Fsr.g * Na / 100 u003d 9206.67 * 13% / 100 u003d 1196.87 rubles.

**Answer:** the average annual cost of fixed assets is 9206.67 thousand rubles, annual depreciation charges – 1196.87 rubles.

**Task 4(e)**

**Determine the coefficient of integral use of equipment, if known:**

Installed equipment in the workshop, machine tools: | |

lathes | |

drilling machines | |

milling machines | |

Annual production program, machine-hour | |

for turning works | |

for drilling | |

for milling | |

Heavy use factor | 0.8 |

Turning and milling machines work in two shifts, drilling machines – in one shift. Turning and milling machines stand idle for repairs during the year 365 hours each, drilling machines – 276 hours each. There are 240 working days a year, the shift duration is 8 hours.

Solution algorithm:

1. Determine the actual operating time of each type of equipment.

2. Determine the operating time of the equipment.

3. Find the coefficient of extensive use of equipment

4. Find the coefficient of integral use of equipment

**Decision:**

1. Given:

shift, hour | Repair, hour | Number of machines | Work days | |

Turning | ||||

Milling | ||||

drilling |

Based on these data, we can calculate the volume of the nominal (regime) and effective funds of the equipment operation time. Then (depending on the purpose of the calculation) we can calculate two types of coefficients for extensive loading of equipment: the coefficient of use of the regime fund of time and the coefficient of the effective fund of time, respectively. The nominal regime fund is calculated by the formula T _{nom} = (D _{in the year} – D _{days off} ) * t _{shift mode} ), and the effective fund of the equipment operation time: (T _{eff} = T _{nom} – T _{rem} ).

F _{nom} ^{(current)} = 240*16*25 = 96000

F _{nom} ^{(drills)} = 240*8*12 = 23040

F _{nom} ^{(mills)} = 240*16*10 = 38400

F _{eff} ^{(current)} u003d 240 * 16 * 25 – 365 * 25 u003d 96000 – 9125 u003d 86875

F _{eff} ^{(mills)} = 240 * 16 * 10 – 365 * 10 = 38400 – 3650 = 34750

F _{eff} ^{(drills)} u003d 240 * 8 * 12 – 276 * 12 u003d 23040 – 3312 u003d 19728

Since there is no information about the actual time worked by the equipment, if we take the volume of the annual production program of the enterprise for the time actually worked, according to the formula we get:

k _{e e} u003d 68000 / 86875 u003d 0.78 k _{e e} u003d 120000 / 141353 u003d 0.85

k _{e e} u003d 22000 / 19728 u003d 1.12

k _{e e} u003d 30000 / 34750 u003d 0.86

from here:

k _{and} u003d 0.78 * 0.8 u003d 0.62 **k _{and} u003d 0.85 * 0.8 u003d 0.68**

k _{and} u003d 1.12 * 0.8 u003d 0.90

k _{and} u003d 0.86 * 0.8 u003d 0.69

and:

k _{e e} u003d 68000 / 96000 u003d 0.71 k _{e e} u003d 120000 / 157440 u003d 0.76

k _{e e} u003d 22000 / 23040 u003d 0.95

k _{e e} u003d 30000 / 38400 u003d 0.78

from here:

k _{and} = 0.62 * 0.8 = 0.50 **k _{and} = 0.76 * 0.8 = 0.61**

k _{and} u003d 0.90 * 0.8 u003d 0.72

k _{and} u003d 0.69 * 0.8 u003d 0.55

**Answer:** coefficient of integral use of equipment, excluding time spent on repairs, k _{and} = 0.61 (equipment is used by 61%); and the coefficient of integral use of equipment, taking into account the time spent on repairs, k _{and} = 0.68 (equipment is used by 68%).

**Task 6**

**Determine the change in the duration of the turnover of working capital**

Indicators | d |

Products sold, million rubles | |

Ratio of own working capital, thousand rubles | |

Increase in output, % | |

Increase in working capital ratio, % |

**Decision:**

**1.** Determine the duration of one turnover of the reporting year.

From the formula we get that T _{about} u003d (500 thousand rubles * 360 days) / 15 rubles u003d 12 days.

**2.** The increase in output amounted to 20%, and the increase in the working capital ratio – 10%. Therefore, the amount of the expected increase in the standard of working capital and finished products can be calculated as follows:

Q _{g pr 2} u003d 15 million rubles + 15 million rubles * 0.2 u003d 18 million rubles

K _{about 2} u003d 500 thousand rubles + 500 thousand rubles * 0.1 u003d 550 thousand rubles

**3.** Determine the required change in the duration of the turnover of working capital.

T _{about 2} u003d (K _{about 2} * F _{pd} ) / Q _{g pr 2} u003d (550,000 * 360) / 18,000,000 u003d 11 days; hence the required change is 12 – 11 = 1 day.

**Answer:** the change in the duration of the turnover of working capital is 1 day.

ALTERNATIVE SOLUTION….

(Q g pr 2 u003d 15 million rubles + 15 million rubles * 0.2 u003d 6 million rubles

K about 2 u003d 500 thousand rubles + 500 thousand rubles * 0.1 u003d 100 thousand rubles

3. Determine the required change in the duration of the turnover of working capital.

T about 2 u003d (K about 2 * F pd) / Q g pr 2 u003d (100000 * 360) / 6,000,000 u003d 6 days); hence required change 12 – 6 = 6 days)

**Task 9**

Determine the reduction in labor intensity, the release of workers and the growth of annual labor productivity due to a number of organizational and technical events in the previous year.

Indicators | d |

Annual production of parts, pcs | |

Labor intensity of the part before the event, min. | |

Labor intensity of the part after the event, min. | |

Effective annual fund of working time, h. | |

Compliance rate | 1.2 |

**Decision:**

**1)** The time spent on the production of the entire volume of products in the current and planned year, (that is, the labor intensity of production in hours), is equal to:

T _{e 1} u003d 56,000 pieces * 29 minutes u003d 1,624,000 minutes u003d 27,067 hours (current year)

T _{e 2} u003d 56,000 pieces * 22 minutes u003d 1,232,000 minutes u003d 20,533 hours (planned year)

**2)** Since in order to calculate the average number of employees, in the condition of the problem there is not enough data (the number of people (by days) who came to work during the year), we determine the attendance number of workers using the formula .

Ryav _{1} u003d 27067 hours / (1750 hours * 1.2) u003d 12.89 (number of workers in the current year)

Ryav _{2} u003d 20533 hours / (1750 hours * 1.2) u003d 9.78 (number of workers in the planned year)

**3)** Release F _{of workers} = 12.89 – 9.78 = 3.11 ≈ 3 people

**4)** Output per worker according to the formula :

B _{1} = 56,000 / 27067 ~ 2 pcs. parts / per hour (current year)

B _{2} = 56,000 / 20533 = 2.7~3 pcs. parts / per hour (planned year)

**5)** Labor intensity according to the formula :

T _{e 1} = 27067 / 56,000 = 0.5 hours / per part (current year)

T _{e 2} u003d 20533 / 56,000 u003d 0.4 hours / per part (planned year)

The reduction in labor intensity is calculated by the formula: I _{tr.} _{u003d} (T current – T _{plan.} ) / T _{plan.} * 100%

I _{tr.} = (27067 – 20533) / 20533 * 100% = 31.9%.

**6)** Growth in annual labor productivity: I _{pr} _{u003d Current 1} / _{Plan 2} * 100%

I _{pr} u003d 2 / 2.7 * 100% u003d 74%.

**Answer:** reduced labor intensity – 31.9%; release of workers ≈ 3 people; growth of annual labor productivity – 74%.

**Task 11**

**Determine the basic monthly earnings of a worker according to the piecework – progressive wage system. According to the regulations in force at the enterprise, an increase in prices for products produced in excess of the original base is provided, if it is overfulfilled up to 5% – by 1.5 times, and if it is overfulfilled by more than 5% – by 2 times. 100 is taken as the initial base – the percentage fulfillment of production standards**

The value of the hourly tariff rate of the first category is 5 rubles.

Indicators | d |

Discharge of the worker | YI |

Production rate, pcs. | |

Norm piece – calculation time, min. | |

Fulfillment of the norm for the month, % |

**Decision:**

**1)** The number of parts made by the worker:

N _{d (pcs) fact.} = H _{ex.} + H _{ex.} * 10% = 240 + 24 = 264

**2)** The time provided by the norm for the manufacture of all parts:

T _{all d (pcs) norms} = H _{vyr.} * H _{piece-calc.} = 240 * 48 min = 11 520 = **192 hours**

**3)** The actual time for the manufacture of parts:

11,520 / 264 = 43.6 min

**4)** The excess of the norm was:

264 – 240 u003d 24 pcs (The percentage excess of the norm was 10%)

**5)** Earnings of a worker at regular rates:

ZP _{sd} u003d T _{1} * k _{III} * H _{vyr. (hour)} = 5 rubles. * 1.8 * 192 hours = 1728 rubles.

**6)** Products produced in excess of the norm: 24 parts. Since the worker’s earnings are formed based on the number of hours worked, let’s first estimate the cost of this product in hours: 24 * 48 min = 1152 min = 19.2 hours. According to the condition of the problem, the bonus part of earnings consists of two parts: for overfulfillment of the norm within 5%, an increase in prices by 1.5 times and for overfulfillment of more than 5% – by 2 times. Our worker overfulfilled the norm by 10%, therefore:

ZP _{premium. 1 part} u003d 9 * 24 u003d 216 rubles.

Basic earnings: ZP _{sdpr} u003d 1728 + 216 u003d 1944 rubles.

**Answer:** the basic monthly salary of a worker under the progressive piecework system of manufacturing products will be 1,944 rubles.

**Task 14**

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