Question 7 of 10: Monte Carlo Simulation — Hourly Purchase Volume

Question 7: Monte Carlo Simulation — Hourly Purchase Volume (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Arrivals/hr: $P(0)=0.5,\ P(1)=0.3,\ P(2)=0.2$. Purchase prob. per arrival $=0.7$. Purchase value: $\$10$ (0.2), $\$100$ (0.6), $\$1000$ (0.2). 12 four-digit random numbers.

Find. A random-number assignment scheme (flow chart), one simulated hour's number of purchases and dollar volume, applying the given random numbers in order.

Approach. Split each 4-digit random number into two 2-digit numbers (00–99); assign cumulative-probability ranges to each of the three random experiments (arrivals, purchase decision, purchase value) and consume the random-number stream left to right, one draw per experiment needed.

1. Assign 2-digit random-number ranges by cumulative probability (00–99):

   Random-number assignment table

   Experiment	Outcome	Prob.	Cumulative	RN range
   Arrivals	0	0.50	0.50	00–49
   	1	0.30	0.80	50–79
   	2	0.20	1.00	80–99
   Purchase?	Yes	0.70	0.70	00–69
   	No	0.30	1.00	70–99
   Value	$10	0.20	0.20	00–19
   	$100	0.60	0.80	20–79
   	$1000	0.20	1.00	80–99

2. Consume the random-number stream left to right, splitting each 4-digit number into two 2-digit halves ($9508\to95,08$; $3702\to37,02$; $9250\to92,50$; …):

   Simulation walk-through

   Step	RN used	Draw for	Result
   1	95	Arrivals	95∈80–99 → 2 arrivals
   2	08	Arrival 1: purchase?	08∈00–69 → Yes
   3	37	Arrival 1: value	37∈20–79 → $100
   4	02	Arrival 2: purchase?	02∈00–69 → Yes
   5	92	Arrival 2: value	92∈80–99 → $1000

   Both arrivals purchase, so this simulated hour has $2$ purchases totalling $\$100+\$1000=\$1{,}100$. $$\boxed{\text{Simulated purchases/hr}=2,\quad \text{Simulated dollar volume}=\$1{,}100}$$
3. Theoretical benchmark (for context, not required by the question). Expected arrivals $=0(0.5)+1(0.3)+2(0.2)=0.7$; expected purchases/hr $=0.7(0.7)=0.49$; expected value per purchase $=10(0.2)+100(0.6)+1000(0.2)=\$262$; expected dollar volume/hr $=0.49(262)=\$128.38$. A single simulated hour (2 purchases, $1,100) is well above this long-run average, which is expected sampling variability, not an error — more replications would be needed to estimate the average reliably.