Lesson 8.5 — Queues: Arrivals versus Service Capacity
排队问题：到达速度与服务能力
A queue is not fixed while service is happening: new people may keep arriving. Learn to compare the rate entering the line with the rate leaving through ticket windows or entrance gates.
Meet the three parts of a queue
Use a steady-rate model: arrivals and each open window’s service rate stay constant. Windows work without breaks while anyone is waiting, and nobody leaves except through service. Alternative trials start with the same queue and use identical windows.
A queue problem has a starting line, people arriving, and people being served. The line changes only because of those three quantities.
The queue itself is a stock. Arrivals add to it; service removes from it.
Three quantities
Starting queue, Q: people already waiting when service begins.
Arrival rate, a: new people joining each minute.
Service capacity: people checked or admitted each minute.
Identify each quantity
Build the balance equation
If one window serves r people per minute and m windows are open, their combined service rate is mr.
At the instant the queue disappears
Q + at = mrtThe left side counts everyone who must be served:
- Q people who were already waiting;
- at people who arrive during t minutes.
The right side is what the open windows can serve in that time.
t = Q ÷ (mr − a)This works only when:
mr > aLive queue laboratory
Check the model
Compare two complete service scenarios
Sometimes the number of people one window serves per minute is not given. We can define one window’s one-minute work as one service unit.
Scenario A
Four ticket windows clear the line in 30 minutes.
4 × 30 = 120 service unitsScenario B
Five ticket windows clear the line in 20 minutes.
5 × 20 = 100 service unitsThese are separate trials. Both contain the same starting queue and arrival rate, and the stated time is when the line first clears. The 30-minute scenario also includes ten more minutes of arrivals.
The bar lengths are scaled to their service-unit totals.
Build the comparison
Reveal the arrival rate and starting queue
Subtracting the two scenarios cancels the same starting queue. What remains is the work caused by ten extra minutes of arrivals.
Step 1 — arrivals
(120 − 100) ÷ (30 − 20) = 2The arrival rate is 2 service units per minute.
Step 2 — starting queue
120 − 2 × 30 = 60So the line already contained 60 service units of work when ticket checking began.
The shared blue part cancels; the leftover orange part is ten minutes of arrivals.
Check the hidden quantities
How long with seven windows?
Two window-units of capacity are needed just to keep up with new arrivals. Only the capacity beyond that reduces the original queue.
The seven-window calculation
Net reduction = 7 − 2 = 5 Time = 60 ÷ 5 = 12 minutesImagine that two of the seven windows spend all their capacity serving people who keep arriving. The remaining five window-units reduce the original line.
Boundary cases:
- 1 window: the line grows.
- 2 windows: the line stays the same size.
- 3 or more: the line eventually clears.
Finish the example
Use actual passengers per minute
In two separate trials, a station starts with 945 waiting passengers. Four windows first clear the line after 15 minutes; eight identical windows first clear it after 7 minutes. New passengers arrive at a constant rate.
Let r be one window’s rate and a be arrivals per minute
Divide each equation by its time:
Subtract:
4r = 72, so r = 18Then:
a = 4×18 − 63 = 9In five minutes, 45 more passengers arrive, so 990 passengers must be served.
How many windows for a five-minute target?
945 + 9×5 = 990 peopleOne window serves 18×5 = 90 people in 5 minutes990 ÷ 90 = 11 windowsCheck the station model
Decide whether a queue can clear
A train station receives 15 new passengers each minute. One window serves 30 passengers each minute. With one window, the queue disappears in 6 minutes.
Recover the starting queue
Net reduction = 30 − 15 = 15 people/min Starting queue = 15 × 6 = 90 peopleOpen two windows
Net reduction = 2×30 − 15 = 45 people/min Time = 90 ÷ 45 = 2 minutesCapacity test
Check the 90-person queue scenario
Work backward to when the queue began
An exhibition starts admitting visitors at 9:00. Before that, visitors join an initially empty line at the same constant rate as after opening, and nobody is admitted early. In separate trials with the same starting queue, three identical gates first clear it in 9 minutes or five first clear it in 5 minutes.
Find the arrival rate in gate-service units
3×9 = 275×5 = 25The 9-minute scenario contains four more minutes of arrivals but only two more service units:
a = (27−25) ÷ (9−5) = 0.5Find how long people arrived before 9:00
Let x be the minutes before opening.
27 = 0.5(x+9) x = 45The first visitor arrived at 8:15.
Schematic timeline: intervals are not drawn to a common time scale. The queue at 9:00 was created by 45 minutes of arrivals before the gates opened.
Check the backward reasoning
Plan a different ticket queue — independent investigation
Try the eight steps yourself, using a small hint if needed. Check an attempt before opening worked review. Correct all eight to complete this investigation.
In separate trials starting with 240 people, three identical windows first clear the queue in 8 minutes or five first clear it in 4 minutes. Arrivals are constant and each window serves equally fast. Let r be people served by one window per minute, and a the arrivals per minute.
Write quantities in one consistent unit. Enter decimals or fractions such as 30/7; a decimal within 0.001 is accepted.
Hint
Divide the starting queue by the time needed to remove it.
Worked solution
240 ÷ 8 = 30, so 3r − a = 30.
Hint
Divide the same starting queue by the four-minute clearing time.
Worked solution
240 ÷ 4 = 60, so 5r − a = 60.
Hint
Subtract the net-rate equations so that the same arrival rate cancels.
Worked solution
Subtract the equations: 2r = 30, so r = 15.
Hint
Total service minus net queue reduction equals arrivals.
Worked solution
3 × 15 − 30 = 15.
Hint
Multiply one window’s service rate by four windows.
Worked solution
4 × 15 = 60.
Hint
Subtract arrivals per minute from all four windows’ service per minute.
Worked solution
60 − 15 = 45 people/min.
Hint
Subtract four minutes of net reduction from the initial 240 people.
Worked solution
240 − 45 × 4 = 60.
Hint
Include three minutes of new arrivals, divide by one window’s three-minute capacity, then round up to a whole number of windows.
Worked solution
Need (240 + 15 × 3) ÷ (15 × 3) = 285 ÷ 45 = 6⅓ windows. Round UP to 7; 6 cannot meet the deadline.
Exit ticket
Answer all five questions. A perfect score completes the exit ticket. The certificate also requires the other core missions and the independent investigation.
Certificate of Mastery
Queue-Flow Systems Engineer
This certifies that the learner can separate a starting queue from continuing arrivals, calculate service capacity, test whether a queue can clear, and work backward from multiple service scenarios.
Q + at = mrtLesson 8.5 • Grade 5 Math Enrichment