A shovel needs to be selected from four models of different capacities, S1 to S4, for loading of dumpers. The operational data for the shovel-dumper combination are provided. The mean arrival rate of the dumpers at the loading point is 5 per hour and the waiting cost of dumper is ₹200 per hour. The inter-arrival time of dumpers and loading time of shovel follows exponential distribution. Ignoring the shovel hiring cost, the shovel type that will have minimum total cost (waiting and operating cost) is

Model Operating cost of shovel (₹ per hour) Mean loading rate of shovel (number of dumpers per hour)

S1 800 8

S2 850 9

S3 920 10

S4 1000 11

Question

The primary purpose of mine ventilation is to:

Answer 1

This is a queueing cost problem. Dumpers turn up at random and wait their turn to be loaded by one shovel, so each shovel choice behaves as its own single server queue, an M/M/1 system. Instead of jumping straight to the crowd size formula, it helps to build up through the utilisation and the time a dumper spends in the system.

Utilisation, \( \rho = \lambda/\mu \): with \( \lambda = 5 \), S1 gives \( \rho = 5/8 = 0.625 \), S2 gives \( 5/9 = 0.556 \), S3 gives \( 5/10 = 0.5 \), and S4 gives \( 5/11 = 0.455 \). A higher \( \rho \) means the shovel is busier and dumpers wait longer.
Average time a dumper spends in the system, \( W_s = \dfrac{1}{\mu - \lambda} \): S1 gives \( 1/3 = 0.333 \) hour, S2 gives \( 1/4 = 0.25 \) hour, S3 gives \( 1/5 = 0.2 \) hour, S4 gives \( 1/6 = 0.167 \) hour.
Average number of dumpers in the system, by Little's Law, \( L_s = \lambda W_s \): S1 gives \( 5 \times 0.333 = 1.667 \), S2 gives \( 5 \times 0.25 = 1.25 \), S3 gives \( 5 \times 0.2 = 1 \), S4 gives \( 5 \times 0.167 = 0.833 \). These match the values from the direct formula \( \lambda/(\mu - \lambda) \), as they must, since Little's Law is just another route to the same quantity.
Waiting cost per hour, at ₹200 per dumper present: S1 costs \( 200 \times 1.667 = 333.3 \), S2 costs \( 200 \times 1.25 = 250 \), S3 costs \( 200 \times 1 = 200 \), S4 costs \( 200 \times 0.833 = 166.7 \).
Total cost, operating plus waiting: S1 is \( 800 + 333.3 = 1133.3 \), S2 is \( 850 + 250 = 1100 \), S3 is \( 920 + 200 = 1120 \), S4 is \( 1000 + 166.7 = 1166.7 \).

Let's summarize:

A cheap, slow shovel saves on operating cost but piles up waiting cost.
A fast, costly shovel cuts waiting cost but its own price tag eats the saving.
S2 sits at the sweet spot between the two, giving the lowest total of ₹1100 per hour.

So the shovel to pick is S2.

Answer 2

This is a queueing cost problem. Dumpers turn up at random and wait their turn to be loaded by one shovel, so each shovel choice behaves as its own single server queue, an M/M/1 system. Instead of jumping straight to the crowd size formula, it helps to build up through the utilisation and the time a dumper spends in the system.

Utilisation, \( \rho = \lambda/\mu \): with \( \lambda = 5 \), S1 gives \( \rho = 5/8 = 0.625 \), S2 gives \( 5/9 = 0.556 \), S3 gives \( 5/10 = 0.5 \), and S4 gives \( 5/11 = 0.455 \). A higher \( \rho \) means the shovel is busier and dumpers wait longer.
Average time a dumper spends in the system, \( W_s = \dfrac{1}{\mu - \lambda} \): S1 gives \( 1/3 = 0.333 \) hour, S2 gives \( 1/4 = 0.25 \) hour, S3 gives \( 1/5 = 0.2 \) hour, S4 gives \( 1/6 = 0.167 \) hour.
Average number of dumpers in the system, by Little's Law, \( L_s = \lambda W_s \): S1 gives \( 5 \times 0.333 = 1.667 \), S2 gives \( 5 \times 0.25 = 1.25 \), S3 gives \( 5 \times 0.2 = 1 \), S4 gives \( 5 \times 0.167 = 0.833 \). These match the values from the direct formula \( \lambda/(\mu - \lambda) \), as they must, since Little's Law is just another route to the same quantity.
Waiting cost per hour, at ₹200 per dumper present: S1 costs \( 200 \times 1.667 = 333.3 \), S2 costs \( 200 \times 1.25 = 250 \), S3 costs \( 200 \times 1 = 200 \), S4 costs \( 200 \times 0.833 = 166.7 \).
Total cost, operating plus waiting: S1 is \( 800 + 333.3 = 1133.3 \), S2 is \( 850 + 250 = 1100 \), S3 is \( 920 + 200 = 1120 \), S4 is \( 1000 + 166.7 = 1166.7 \).

Let's summarize:

A cheap, slow shovel saves on operating cost but piles up waiting cost.
A fast, costly shovel cuts waiting cost but its own price tag eats the saving.
S2 sits at the sweet spot between the two, giving the lowest total of ₹1100 per hour.

So the shovel to pick is S2.

Show Hint

The Correct Option is B

Solution and Explanation

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