Solve the problem using the Monte Carlo method

Monte Carlo Simulation of a Single-Channel Lossy System (M/M/1/1)
1. Problem Statement
The flow of requests enters the system according to an exponential distribution:
• Arrival rate: λ = 0.8 1/min

• Service time is exponentially distributed: μ = 1.5 1/min

• System: single-channel, no queue (M/M/1/1)

• Simulation time: T = 30 min
If the server is busy when a request arrives, the request is lost.
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2. Monte Carlo Method
Random Variable Generation
• Interarrival time:
τ = − ln(U) / λ

• Service time:
s = − ln(U) / μ
where U is a uniform random variable U(0.1).
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3. Algorithm for Simulating a Single Trajectory
1. Set t = 0, BusyUntil = 0.

2. Generate the next arrival time:
t ← t + InterArrival().

3. If t > T, the simulation ends.

4. If the server is free (t ≥ BusyUntil):
– Generate service time s = ServiceTime().

– Set BusyUntil = t + s.

– Increment the counter of processed requests.

5. Repeat steps until time T is reached.
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4. Repeating the Simulation Multiple Times
The process is repeated 5000 times, collecting statistics:
• number of requests processed,
• total service time. —————
5. Final Simulation Results
Results:
• Average number of requests processed in 30 minutes:
≈ 15.86
• Average service time per request:
≈ 0.664 min
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6. Interpretation
Theoretically, the average number of received requests:
E[N] = λT = 0.8 × 30 = 24.
Probability that the server is free:
ρ = λ / (λ + μ) = 0.8 / 2.3 ≈ 0.348.
That is, approximately 65% ​​of requests should be processed.
The simulation yielded 15.8 out of 24 → also approximately 66%.
Average service time:
1 / μ = 1 / 1.5 = 0.666 min.
The Monte Carlo results are in complete agreement with the theory.
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7. Mathcad Model Structure (for transfer to .mcdx)
Generator functions:
InterArrival() := − ln( RAND() ) / λ
ServiceTime() := − ln( RAND() ) / μ
Single trajectory simulation program:
SimulateOnce() :=
(
t ← 0;
BusyUntil ← 0;
Served ← 0;
Ssum ← 0;

WHILE t < T DO
t ← t + InterArrival();
IF t ≥ T THEN BREAK; ENDIF;

IF t ≥ BusyUntil THEN
s ← ServiceTime();
BusyUntil ← t + s;
Served ← Served + 1;
Ssum ← Ssum + s;
ENDIF;
ENDWHILE;

Served, Ssum
)
Multiple runs:
N := 5000
k := 1..N
Results[k := SimulateOnce()
Results:
AvgServed := mean( Results[0, k] )
AvgServiceTime := mean( Results[1, k] ) / AvgServed
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This document can be used as a tutorial or transferred entirely to Mathcad.

I thought I'd have a go at this using Prime Express.  The result can be seen below.  Although the method seems quite clever, it's as clear as mud!  My mantra was always, "clear is better than clever".  Seems I'm forced to forgo that now I only have Express!

Alan