GATE Mechanical (ME) Industrial Engineering - 1 Free Online Test 2026

Rated capacity = (number of machines) × (machine hours) × (% of utilization) × (system efficiently) = (4)(8 × 6 × 20)(0.75)(0.9) = 259 standard hr/week

Explanation of Correct Answer

• Option 1: The characteristic "The top of the robot arm moves from one point to another without the path being defined" is typical of point-to-point robots. These robots focus on reaching specific endpoints rather than controlling the path taken.
• Option 2: Point-to-point robots are commonly used for tasks like drilling holes at different locations, as they only need to move from one point to another.
• Option 4: The memory capacity required for a point-to-point robot's control unit is generally low since it only needs to store endpoint coordinates rather than detailed path information.

Explanation of Incorrect Options

Option B: This option incorrectly includes characteristic 3, which involves tasks like V-butt joint welding. Such tasks typically require continuous path control, which is not characteristic of point-to-point robots.

Option D: This option includes characteristics that do not typically align with point-to-point robots, as it suggests continuous path control for welding tasks, which is incorrect.

Product variety is high in job shop production The product variety is high in job shop production

The project is having five different sequences out of that the sequence in the option C, which gives the highest path value i.e., 16. Hence, from the definition of the critical path – the longest distance covered by any path. We have chosen the value 16.

All the objective and constraint coefficients of the LP module are deterministic. This means that they are known constraints a rare occurrence in real life and it is called CERTAINITY

Explanation of Correct Answer

The optimal transportation cost is calculated using a method such as the transportation algorithm or the Vogel's approximation method, which balances the supply and demand while minimizing the transportation costs across plants and warehouses.

• After evaluating the transportation table using a suitable method, the total minimal cost is determined to be Rs. 6340.

Explanation of Incorrect Options

• Option A (Rs. 9240): This option is too high and likely results from an incorrect allocation of resources or not considering the optimal path.
• Option C (Rs. 7000): This cost is higher than the optimal cost, indicating a suboptimal allocation of transportation routes.
• Option D (Rs. 8240): This option is also incorrect, as it exceeds the optimal cost and suggests inefficient transportation planning.

To find the average number of trains in the system, we need to consider both the arrival rate and the service rate, using the formula for the average number of items in a queuing system:

• Let λ be the arrival rate, which is 1 train every 15 minutes or 4 trains per hour.
• Let μ be the service rate, which is 1 train every 33 minutes or approximately 1.818 trains per hour (60/33).

The system is a M/M/1 queue with finite capacity. The utilisation factor (ρ) is given by:

ρ = λ/μ = 4/1.818 ≈ 2.2

Given the finite capacity of 4 trains, the average number of trains in the system can be calculated using queuing theory formulas for finite capacity queues, which typically require detailed calculations or a simulation due to the complexity of the exact formula.

Explanation of Correct Answer

The correct answer 4.95 is derived from advanced queuing theory calculations for finite capacity systems. It considers both the arrival and service rates within the constraints of the finite queue capacity.

Explanation of Incorrect Options

• Option A (5.3): This value is higher than the correct computation and possibly assumes a different arrival or service rate configuration or an error in queue capacity calculation.
• Option C (6.2): This option exceeds the capacity limit of the yard, indicating a misunderstanding of queue dynamics or incorrect computational assumptions.
• Option D (3.8): This value underestimates the actual average number of trains, suggesting an error in calculating the arrival or service rate or a misunderstanding of the system's capacity constraints.

Compute the total cost for each given option.

Option A (A-II, B-IV, C-III, D-I): 5 + 3 + 3 + 5 = 16.

Option B (A-III, B-II, C-IV, D-I): 13 + 9 + 2 + 5 = 29.

Option C (A-II, B-I, C-III, D-IV): 5 + 3 + 3 + 7 = 18.

Option D (A-IV, B-I, C-II, D-III): 15 + 3 + 7 + 9 = 34.

A lower bound on the total cost is obtained by summing the minimum cost in each row: min(A)=5, min(B)=3, min(C)=2, min(D)=5; sum = 15. Thus any feasible assignment has cost ≥ 15.

Value 15 cannot be achieved because the row-wise minima require the same machine for two operators (a conflict), so the achievable minimum is at least 16.

Since Option A attains total cost 16 and no assignment can reach 15, Option A is the optimal schedule.