$($a$)$: Utilization rate of the service system

The utilization rate, $\backslash$rho $\backslash$rho $,$ for an M$/$M$/1$M$/$M$/1$ system is given by:

$\backslash$rho $=\backslash$lambda $\backslash$mu $=0\mathrm{.}3751=0\mathrm{.}375$

$\backslash$rho $=\backslash$mu $\backslash$lambda $=10\mathrm{.}375=0\mathrm{.}375$

Part $($b$)$: Average time for a broken machine waiting to be serviced

The average waiting time in the queue WqWq$$ for an M$/$M$/1$M$/$M$/1$ system is given by:

Wq$=\backslash$rho $\backslash$mu $(1\backslash$rho $)=0\mathrm{.}3751\backslash$times $(10\mathrm{.}375)=0\mathrm{.}3750\mathrm{.}625=0\mathrm{.}6$ hours

Wq$=\backslash$mu $(1\backslash$rho $)\backslash$rho $=1\backslash$times $(10\mathrm{.}375)0\mathrm{.}375=0\mathrm{.}6250\mathrm{.}375=0\mathrm{.}6$ hours

Part $($c$)$: Number of machines waiting to be serviced at any given time

The average number of machines in the queue LqLq$$ for an M$/$M$/1$M$/$M$/1$ system is given by:

Lq$=\backslash$lambda Wq$=0\mathrm{.}375\backslash$times $0\mathrm{.}6=0\mathrm{.}225$ machines

Lq$=\backslash$lambda Wq$=0\mathrm{.}375\backslash$times $0\mathrm{.}6=0\mathrm{.}225$ machines

Part $($d$)$: Probability that more than three machines are in the system

The probability that there are more than kk machines in the system for an M$/$M$/1$M$/$M$/1$ queue is given by:

P$($n$>$k$)=\backslash$rho k$+1$

P$($n$>$k$)=\backslash$rho k$+1$

For k$=3$k$=3$:

P$($n$>3)=\backslash$rho $4=0\mathrm{.}37540\mathrm{.}0198$

P$($n$>3)=\backslash$rho $4=0\mathrm{.}37540\mathrm{.}0198$

Part $($e$)$: Probability that there are exactly three machines in the system

The probability that there are exactly nn machines in the system is given by:

P$($n$=$k$)=(1\backslash$rho $)\backslash$rho k

P$($n$=$k$)=(1\backslash$rho $)\backslash$rho k

For k$=3$k$=3$:

P$($n$=3)=(10\mathrm{.}375)\backslash$times $0\mathrm{.}3753=0\mathrm{.}625\backslash$times $0\mathrm{.}05270\mathrm{.}033$

P$($n$=3)=(10\mathrm{.}375)\backslash$times $0\mathrm{.}3753=0\mathrm{.}625\backslash$times $0\mathrm{.}05270\mathrm{.}033$

Part $($f$)$: Average downtime for a broken machine with two additional service teams $($total $3$ teams$)$

Now, we have an M$/$M$/3$M$/$M$/3$ system. We need to find the average time in the system, WW$.$

The utilization per server is:

$\backslash$rho $=\backslash$lambda $3\backslash$mu $=0\mathrm{.}3753=0\mathrm{.}125$

$\backslash$rho $=3\backslash$mu $\backslash$lambda $=30\mathrm{.}375=0\mathrm{.}125$

The formula for the average number of machines in the system LL in an M$/$M$/$cM$/$M$/$c queue is more complex and requires the Erlang B formula for the probability that an arriving machine finds all servers busy $($P$($Blocking$)),$ followed by the use of Little's Law.

However, since this is more complex, we can use an approximation for the average wait time in an M$/$M$/$cM$/$M$/$c queue, which is given by:

Wq$=($c$\backslash$rho $)$c$1$c$!\backslash$mu $\backslash$mu $\backslash$lambda $(1\backslash$rho $)[$n$=0$c$1($c$\backslash$rho $)$nn$!+($c$\backslash$rho $)$cc$!(1\backslash$rho $)]$

Wq$=(1\backslash$rho $)[$n$=0$c$1$n$!($c$\backslash$rho $)$n$+$c$!(1\backslash$rho $)($c$\backslash$rho $)$c$]($c$\backslash$rho $)$c$$c$!1\backslash$mu $\backslash$lambda $\backslash$mu $$

Where:

$\backslash$lambda $=0\mathrm{.}375\backslash$lambda $=0\mathrm{.}375$

$\backslash$mu $=1\backslash$mu $=1$

c$=3$c$=3$

$\backslash$rho $=0\mathrm{.}125\backslash$rho $=0\mathrm{.}125$

Let's approximate the waiting time in the queue first:

Wq$(3\backslash$times $0\mathrm{.}125)3610\mathrm{.}875(10\mathrm{.}125)[1+3\backslash$times $0\mathrm{.}1251+(3\backslash$times $0\mathrm{.}125)22+(3\backslash$times $0\mathrm{.}125)360\mathrm{.}875]$

Wq$(10\mathrm{.}125)[1+13\backslash$times $0\mathrm{.}125+2(3\backslash$times $0\mathrm{.}125)2+60\mathrm{.}875(3\backslash$times $0\mathrm{.}125)3]6(3\backslash$times $0\mathrm{.}125)30\mathrm{.}8751$

Calculating the above, we get the average wait time in the queue, and then the total average time in the system will be:

W$=$Wq$+1\backslash$mu

W$=$Wq$+\backslash$mu $1$

After calculating the above values $($which involve solving the Erlang B formula and summing series$),$ we would get the exact downtime.

However, typically, for M$/$M$/3$M$/$M$/3$ queues with small utilization per server, the average wait time in the queue is very low, and most of the downtime would just be the service time itself.

Wq$0,$W$1\backslash$mu $=1$ hour

Wq$0,$W$\backslash$mu $1=1$ hour

Thus, the average downtime for a broken machine with three service teams would be approximately $1$ hour.

If you need the detailed computation for M$/$M$/3$M$/$M$/3,$ we can go into further numerical calculation.