Skip to main content

Engineering Management - short answer questions from AMIE exams (Summer 2019)

Write short notes on

Collective bargaining
Collective bargaining, the ongoing process of negotiation between representatives of workers and employers to establish the conditions of employment.

Quality control
Quality control (QC) is a procedure or set of procedures intended to ensure that a manufactured product or performed service adheres to a defined set of quality criteria or meets the requirements of the client or customer. In order to implement an effective QC program, an enterprise must first decide which specific standards the product or service must meet. Then the extent of QC actions must be determined -- for example, the percentage of units to be tested from each lot.

Basic EOQ model
EOQ model. It is essentially a single formula for determining the optimal order size that minimizes the sum of carrying costs and ordering costs. Its formula is √(2DS/H) where Q = EOQ units; D = Demand in units (typically on an annual basis); S = Order cost (per purchase order); H = Holding costs (per unit, per year).


Marginal productivity
Marginal productivity or marginal product refers to the extra output, return, or profit yielded per unit by advantages from production inputs. Inputs can include things like labour and raw materials.

Fixed and current liability items
Current liabilities, also known as short-term liabilities, are debts or obligations that need to be paid within a year. Current liabilities should be closely watched by management to ensure that the company possesses enough liquidity from current assets to guarantee that the debts or obligations can be met. Examples of current liabilities: Accounts payable; Interest payable; Income taxes payable.
Fixed liabilities are debt, bonds, mortgages or loans that are payable over a term exceeding one year. These debts are better known as non-current liabilities or long-term liabilities. Debts or liabilities due within one year are known as current liabilities.

Fixed and current asset items
Fixed assets, also known as property, plant, and equipment (PP&E) and as capital assets, are tangible things that a company expects to use for more than one accounting period.
Current assets, such as cash and inventory, are items that the company expects to use up or sell within a year.

Bill of material
A bill of materials (BOM) is an extensive list of raw materials, components, and instructions required to construct, manufacture, or repair a product or service.

Master production schedule
Master Production Schedule (MPS) is the part of production planning that outlines which products need to be manufactured, in which quantity, and when. The Master Production Schedule forms the basis of communication between sales and manufacturing. Using the MPS as a contract between sales and production means that sales can make valid order promises.

Floats in project management
Project management float is the amount of time a given task can be delayed without causing a delay in the entire project. But, there’s a little more to it than that. In fact, there are two distinct types of float that project managers use to manage task timelines: total float and free float. 


Planned V/S scheduled receipts in MRP.
  • A planned order receipt is a future-projected receipt based on the generation of a planned order, that has not yet been firmed into a scheduled receipt. It is the amount of inventory finally ordered after considering the remaining stock, and also the ones that are scheduled to arrive later.
  • The scheduled receipt is a receipt for a released order.
---
The study material for AMIE/B Tech/Junior Engineer exams is available at https://amiestudycircle.com


Comments

Popular posts from this blog

Mechanics of Fluids (Solved Numerical Problems)

Numerical The surface Tension of water in contact with air at 20°C is 0.0725 N/m. The pressure inside a droplet of water is to be 0.02 N/cm² greater than the outside pressure. Calculate the diameter of the droplet of water. (7 marks) (AMIE Summer 2023) Solution Surface tension, σ = 0.0725 N/m Pressure intensity, P = 0.02 N/m 2 P = 4σ/d Hence, the Diameter of the dropd = 4 x 0.0725/200 = 1.45 mm Numerical Find the surface tension in a soap bubble of 40 mm diameter when the inside pressure is 2.5 N/m² above atmospheric pressure. (7 marks) (AMIE Summer 2023) Answer: 0.0125 N/m Numerical The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm² (atmospheric pressure). Calculate the pressure within the droplet if surface tension is given as 0.0725 N/m of water. (AMIE Summer 2023, 7 marks) Answer: 0.725 N/cm 2   Numerical An open lank contains water up to a depth of 2 m and above it an oil of specific gravity 0.9 for a depth of 1 m. Find the pressure intensity (i) at t...

Energy Systems (Solved Numerical Problems)

Wind at 1 standard atmospheric pressure and \({15^0}C\) has velocity of 15 m/s, calculate (i) the total power density in the wind stream (ii) the maximum obtainable power density (iii) a reasonably obtainable power density (iv) total power (v) torque and axial thrust Given: turbine diameter = 120 m, and turbine operating speed = 40 rpm at maximum efficiency. Propeller type wind turbine is considered. (AMIE Winter 2023) Solution For air, the value of gas constant is R = 0.287 kJ/kg.K 1 atm = 1.01325 x 105 Pa Air density \(\rho  = \frac{P}{{RT}} = \frac{{1.01325x{{10}^5}}}{{287}}(288) = 1.226\,kg/{m^3}\) Total Power \({P_{total}} = \rho A{V_1}^3/2\) Power density \(\begin{array}{l}\frac{{{P_{total}}}}{A} = \frac{1}{2}\rho {V_1}^3\\ = \frac{1}{2}(1.226){(15)^3}\\ = 2068.87{\mkern 1mu} W/{m^2}\end{array}\) Maximum power density \(\begin{array}{l}\frac{{{P_{\max }}}}{A} = \frac{8}{{27}}\rho A{V^3}_1\\ = \frac{8}{{27}}(1.226){(15)^3}\\ = 1226{\mkern 1mu} W/{m^2}\end{array}\) Assuming eff...

Design of Electrical Systems (Solved Numerical Problems)

Important note There is something wrong with this question paper. It seems that instead of "Design of Electrical Systems" the IEI has given problems from "Electrical Machines". You should raise a complaint to director_eea@ieindia.org in this regard. Numerical A 120 V DC shunt motor draws a current of 200A. The armature resistance is 0.02 ohms and the shunt field resistance is 30 ohms. Find back emf. If the lap wound armature has 90 slots with 4 conductors per slots, at what speed will the motor run when flux per pole is 0.04 Wb?​ (AMIE Summer 2023, 8 marks) Solution The back EMF (E b ) of a DC motor can be calculated using the formula: E b = V - I a R a   Given: V = 120 V I a = 200 A R a = 0.02 ohms Substituting the values into the formula: E b = 120 − 200 × 0.02 = 120 − 4​ = 116 V Now, let's calculate the speed (N) at which the motor will run using the given flux per pole (φ p ). The formula to calculate the speed of a DC motor is: N = 60×E b /(P×φ p ) Wh...