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RPSC ME Lecturer 2011: Official Paper

Option 1 : The probability the system is full-is 0.1054

__Concept:__

The probability that there are k customers in the system = (ρ)k(1 - ρ) = \(\frac{\lambda }{\mu }\left( {\frac{1}{{\mu - \lambda }}} \right)\)

No. of customers in the system or length of the system \({L_S} = \frac{\lambda }{{\mu - \lambda }}\)

No. of customers in the queue or length of the queue \({L_q} = \frac{{{\lambda ^2}}}{{\mu \left( {\mu - \lambda } \right)}}\)

Average arrival time and the time spent in the queue (before being served) (Waiting time in queue) = \({W_q} = \frac{\lambda }{\mu }.\frac{1}{{\mu - \lambda }}\)

__Calculation:__

Arrival rate λ = 45 units/hour, Service rate, μ = 60 units/hour

**Now, **

\({\rho} = \frac{\lambda}{{\mu }}\)= 0.75

Now,

The probability that the system is full can be calculated as we know the maximum quantity of the system i.e k = 3

∴The probability that there are 3 customers in the system = (ρ)3(1 - ρ)

**∴The probability that there are 3 customers in the system = 0.1054**

**Now,**

The second option asks about the **length of the system **

The third one is **waiting time in the queue**

The fourth option is the **length of the queue**

**All of the above can be calculated easily by the formulae mentioned in the concept part and any of them did not match with the options provided hence the option 1 is the correct answer.**

__Important Points__

- Average arrival time and the time spent in the system (Waiting time in system) = \({W_s} = \frac{1}{{\mu - \lambda }}\)
- Average arrival time and the time spent in the queue (before being served) (Waiting time in queue) = \({W_q} = \frac{\lambda }{\mu }.\frac{1}{{\mu - \lambda }}\)
- Average number of customers in the system = \({L_s} = \lambda {W_s} = \lambda .\frac{1}{{\mu - \lambda }}\)
- Average number of customers in the queue = Average queue length = \({L_q} = \lambda {W_q} = \lambda .\frac{\lambda }{\mu }.\frac{1}{{\mu - \lambda }} = \frac{{{\lambda ^2}}}{\mu }\left( {\frac{1}{{\mu - \lambda }}} \right)\)
- Probability that there are k customers in the system = (ρ)k(1 - ρ) = \(\frac{\lambda }{\mu }\left( {\frac{1}{{\mu - \lambda }}} \right)\)

- Probability that there are more than k customers = \({\left( {\frac{\lambda }{\mu }} \right)^{k + 1}}\)

ST 1: Engineering Materials (Crystal Geometry)

465

16 Questions
8 Marks
20 Mins