Rates Need a Shared Clock
An arrival rate and a service rate can be compared only when they use the same time unit. A rate of \(30\) customers per day and a rate of \(40\) customers per hour are not comparable by dividing \(30\) by \(40\): their numerical values refer to different durations. The ratio used in M/M/1 analysis is dimensionless, but it becomes meaningful only after both rates have been expressed using a common clock.
Consider an M/M/1 queue on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with continuous-time number-in-system process \((X_t)_{t\ge0}\) and state space \(S=\{0,1,2,\ldots\}\). Write \(\lambda\) for its arrival rate and \(\mu\) for its service rate. Both have units of “events per time”: \(\lambda\) counts arrivals per unit time, while \(\mu\) is the exponential service rate, the reciprocal of the mean service duration. We assume \(\mu>0\). The usual stationary M/M/1 formulas apply when \(0\le\lambda<\mu\); in particular, when \(\lambda>0\), the mean time in the system is \(1/(\mu-\lambda)\).
A rate is a count divided by a duration. If the time unit changes, the numerical value of the rate changes even though the physical process does not. A mean service duration changes in the opposite way: converting it to a smaller time unit increases its numerical value, so its reciprocal service rate decreases.
Converting a Rate
Suppose the old time unit is \(u\), the new time unit is \(v\), and one old unit contains \(c\) new units, where \(c>0\). A rate \(r_u\) per old unit becomes \(r_v=r_u/c\) per new unit. For example, one hour contains \(60\) minutes, so a rate per hour is divided by \(60\) to obtain a rate per minute. Conversely, converting from per minute to per hour means dividing by \(1/60\), or multiplying by \(60\).
Proof. In a physical duration of \(T_u\) old units, the count represented by rate \(r_u\) is \(r_uT_u\). That same duration is \(T_v=cT_u\) new units. The count per new unit is therefore \(r_uT_u/T_v=r_uT_u/(cT_u)=r_u/c\), as claimed. For a service duration, \(T_u\) old units equal \(cT_u\) new units, so \(s_v=cs_u\). Taking reciprocals gives \(1/s_v=1/(cs_u)=(1/s_u)/c\). These identities hold for any \(c>0\). \(\square\)
The same conversion rule applies to arrivals and service completions. The denominator changes; the underlying count or physical duration does not. In particular, converting both rates using the same pair of time units preserves their ratio.
Proof. By rate conversion, \(\lambda_v=\lambda_u/c\) and \(\mu_v=\mu_u/c\). Hence \(\rho_v=(\lambda_u/c)/(\mu_u/c)=\lambda_u/\mu_u=\rho_u\), and the inequality \(\lambda_u<\mu_u\) is equivalent to \(\lambda_v<\mu_v\). The standard stationary M/M/1 distribution is \(\pi_n=(1-\rho)\rho^n\) for \(n\ge0\), so it is unchanged because \(\rho\) is unchanged; consequently, any quantity determined by that distribution, such as the mean number in the system, is unchanged as well. For \(\lambda>0\), the standard mean time in the system is \(W=1/(\mu-\lambda)\). In the new units, \(W_v=1/(\mu_v-\lambda_v)=1/((\mu_u-\lambda_u)/c)=cW_u\). Thus time quantities convert as durations should, while the stability comparison and dimensionless traffic intensity do not change. \(\square\)
If \(\lambda=0\), the queue has no arrivals and its stationary number in the system is zero. The ratio \(\rho=0\) is still invariant under common rescaling, but a mean time experienced by an arriving customer is not defined when there are no arrivals. This is why time-in-system formulas for customers are stated here for \(\lambda>0\).
Worked Conversions
Worked Example: Arrivals per Hour and Service Time in Seconds
Suppose arrivals occur at \(42\) customers per hour and the mean service duration is \(75\) seconds per customer. Convert both parameters to rates per minute, find \(\rho\), and calculate the mean time in the system.
There are \(60\) minutes in an hour, so the arrival rate per minute is \(42/60=7/10\) customers per minute. There are \(60\) seconds in a minute, so the mean service duration is \(75/60=5/4\) minutes. Its reciprocal service rate is \(\mu=1/(5/4)=4/5\) customers per minute. The common-unit rates satisfy \(7/10<4/5\), so the queue is in the stable range, and
Using the standard M/M/1 mean-time formula, the mean time in the system is \(1/(\mu-\lambda)=1/(4/5-7/10)=1/(8/10-7/10)=10\) minutes. As a check using rates per hour, the service rate is \(3600/75=48\) customers per hour, and the rate difference is \(48-42=6\) per hour. Its reciprocal is \(1/6\) hour, which is \(60/6=10\) minutes. Both unit choices give the same physical duration.
Worked Example: Rates Per Eight-Hour Shift
A service desk receives \(96\) customers during an eight-hour shift. Its mean service duration is \(4\) minutes. Assuming the shift duration is the relevant operating time, convert to rates per hour and per minute and calculate the traffic intensity.
The arrival rate per hour is \(96/8=12\) customers per hour. The mean service duration is \(4/60=1/15\) hour, so the service rate per hour is \(1/(1/15)=15\) customers per hour. Therefore \(\rho=12/15=4/5\), and the queue satisfies \(12<15\).
In minutes, the arrival rate is \(12/60=1/5\) customer per minute. The mean service duration is already \(4\) minutes, so the service rate is \(1/4\) customer per minute. The ratio is \((1/5)/(1/4)=(1/5)\cdot4=4/5\), confirming that the change of clock leaves \(\rho\) unchanged. As a further check, the mean time in the system is \(1/(1/4-1/5)=1/(5/20-4/20)=20\) minutes. In hourly units, the same calculation is \(1/(15-12)=1/3\) hour, and \((1/3)\cdot60=20\) minutes.
A shift-based rate must not be silently treated as a calendar-time rate if the desk is closed outside its shifts. The conversion above uses eight operating hours as the specified exposure time; a different operating schedule requires a rate based on that schedule.
Worked Example: A Unit Mismatch Changes the Prediction
Suppose a queue receives \(30\) customers per 24-hour day, and the mean service duration is \(90\) seconds. Compare the correct calculation with the result of dividing the raw numerical values \(30\) and \(40\) without converting their units.
Convert arrivals to an hourly rate: \(30/24=5/4\) customers per hour. There are \(3600\) seconds in an hour, so the service rate is \(3600/90=40\) customers per hour. Both rates now use hours, and
The stable-range condition is satisfied. The difference between service and arrival rates is \(40-5/4=160/4-5/4=155/4\) per hour. Thus the mean time in the system is \(1/(155/4)=4/155\) hour, or \((4/155)\cdot60=240/155=48/31\) minutes. The mean number in the system is \(\lambda/(\mu-\lambda)=(5/4)/(155/4)=5/155=1/31\). As a check, the stationary mean \(\rho/(1-\rho)\) is \((1/32)/(31/32)=1/31\) as well.
The tempting but incorrect calculation \(30/40=3/4\) treats both inputs as if they were per hour. It would predict a different traffic intensity and, using the same-unit formula, a mean time in the system of \(1/(40-30)=1/10\) hour, or \(6\) minutes. The inconsistent calculation also gives the mean number \(30\cdot(1/10)=3\). Those predictions do not match the correctly converted values \(48/31\) minutes and \(1/31\) customer. The error is not in the M/M/1 formulas: it is the use of mismatched units in their inputs.
A Reliable Unit Check
Before calculating an M/M/1 performance measure, write the time unit beside each rate. For example, \(\lambda=12\) customers per hour and \(\mu=15\) customers per hour can be compared directly; \(\lambda=12\) customers per hour and \(\mu=15\) customers per minute cannot. If service is reported as a mean duration, convert that duration first and then take its reciprocal. Do not use the duration itself as the service rate.
A practical check is to compute \(\rho\) twice after converting both rates to two different common units. The two answers should agree exactly. If they do not, check the conversion direction: hourly rates become smaller when expressed per minute, while per-minute rates become larger when expressed per hour. Finally, check whether the calculated units make sense: \(\rho\) has no time unit, a mean time has a time unit, and a mean number of customers has no time unit.
Common mistake. Dividing the numerical arrival rate by the numerical service rate before comparing their denominators can give a plausible-looking ratio, such as \(30/40=3/4\). That arithmetic is correct but does not calculate \(\lambda/\mu\) when the first number is per day and the second is per hour. Convert one rate so both denominators match, then divide. A dimensionless answer is not automatically a valid answer: the inputs must first describe rates on the same clock.
Check Your Understanding
For each question, keep the time unit attached to every rate until the final calculation.
- An M/M/1 queue receives \(18\) customers per hour, and the mean service duration is \(3\) minutes. Find \(\mu\) per hour and calculate \(\rho\).
- A process receives \(48\) customers in a \(6\)-hour operating period. Express its arrival rate per minute. What assumption about the operating period does this conversion use?
- Explain why converting both rates from per hour to per minute does not change \(\rho\).
- An arrival rate is reported per day and a service rate per hour. What conversion must be made before using the stability condition \(\lambda<\mu\)?
- A mean time in the system is calculated as \(1/(\mu-\lambda)\) using rates per minute. What unit does the resulting duration have, and how would it be converted to hours?