Question

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

• A. S1
• B. S2
• C. S3
• D. S4

Hint:

Set up an M/M/1 queue for each shovel and add its waiting cost, from the expected number of dumpers in the system, to its operating cost.

Answer 1

The Correct Option is B

Step 1: Recognise the queueing model.
Dumpers arrive at random (exponential inter-arrival time) and each shovel loads one dumper at a time, also at a random (exponential) rate. That makes this a single server queue, model M/M/1, one queue for each shovel choice.
The mean arrival rate is \( \lambda = 5 \) dumpers per hour, and each shovel model has its own mean service (loading) rate \( \mu \).

Step 2: Write the total cost formula.
For an M/M/1 queue, the expected number of dumpers in the system, waiting plus being loaded, is
\[ L_s = \frac{\lambda}{\mu - \lambda} \]
The total cost per hour for a shovel model is its fixed operating cost plus the waiting cost charged on every dumper present in the system:
\[ TC = C_{operating} + C_{wait} \times L_s \]
Here \( C_{wait} = 200 \) rupees per dumper per hour.

Step 3: Compute \( L_s \) and the waiting cost for each model.
For S1: \( \mu = 8 \), so \( \mu - \lambda = 3 \), \( L_s = 5/3 = 1.667 \), waiting cost \( = 200 \times 1.667 = 333.33 \).
For S2: \( \mu = 9 \), so \( \mu - \lambda = 4 \), \( L_s = 5/4 = 1.25 \), waiting cost \( = 200 \times 1.25 = 250 \).
For S3: \( \mu = 10 \), so \( \mu - \lambda = 5 \), \( L_s = 5/5 = 1 \), waiting cost \( = 200 \times 1 = 200 \).
For S4: \( \mu = 11 \), so \( \mu - \lambda = 6 \), \( L_s = 5/6 = 0.833 \), waiting cost \( = 200 \times 0.833 = 166.67 \).

Step 4: Add operating cost to get total cost, and compare.
S1: \( 800 + 333.33 = 1133.33 \)
S2: \( 850 + 250 = 1100.00 \)
S3: \( 920 + 200 = 1120.00 \)
S4: \( 1000 + 166.67 = 1166.67 \)

Final Answer:
S2 gives the lowest total cost, ₹1100 per hour, even though its operating cost is not the cheapest of the four. A slower, cheaper shovel like S1 looks good on paper, but it makes dumpers queue up for longer, and that waiting cost outweighs the saving on the shovel. A faster, pricier shovel like S3 or S4 clears the queue fast, but its high operating cost is not worth paying once the queue is already short. \[ \boxed{S2} \]