An algorithmic trading firm executes orders using two automated execution venues, Venue X … — GMAT

An algorithmic trading firm executes orders using two automated execution venues, Venue X and Venue Y. Venue X processes orders at a constant rate of $R_X$ orders per second with an execution slippage rate of $2\%$ . Venue Y processes orders at a constant rate of $R_Y$ orders per second with an execution slippage rate of $5\%$ .

During a $10$ -minute trading window when both venues operate simultaneously and continuously, a total of $18,000$ orders are processed, and the overall combined slippage rate across all orders processed by both venues is $3\%$ .

Based on the given information, which of the following correctly identifies the order processing rate for Venue X ( $R_X$ , in orders per second) and Venue Y ( $R_Y$ , in orders per second)?

Venue X rate: 20 orders per second; Venue Y rate: 10 orders per secondAnswer
B
Venue X rate: 10 orders per second; Venue Y rate: 20 orders per second
C
Venue X rate: 15 orders per second; Venue Y rate: 15 orders per second
D
Venue X rate: 25 orders per second; Venue Y rate: 5 orders per second
E
Venue X rate: 12 orders per second; Venue Y rate: 18 orders per second

Answer

Venue X rate: 20 orders per second; Venue Y rate: 10 orders per second

The correct answer specifies Venue X rate as 20 orders per second and Venue Y rate as 10 orders per second. Over 600 seconds, Venue X processes 12,000 orders (yielding 240 slipped orders at 2%) and Venue Y processes 6,000 orders (yielding 300 slipped orders at 5%). The total volume of 18,000 orders produces 540 slipped orders, which exactly equals a 3% overall slippage rate.

Step-by-Step Solution

Convert time window to seconds and establish total processing rate

Time

T = 10 \text{ minutes} = 600 \text{ seconds}

. Combined rate

R_X + R_Y = \frac{18,000}{600} = 30 \text{ orders/second}

Processing rates are requested in orders per second, so total volume must be divided by total seconds.

Set up weighted average slippage equation

Total slipped orders =

3\% \times 18,000 = 540

. Let

N_X

be total orders from X and

N_Y

be total orders from Y.

0.02 N_X + 0.05 N_Y = 540

The sum of slipped orders from both venues equals the overall combined slipped orders.

Solve simultaneous linear system for total order volumes

N_X

and

N_Y

Substitute

N_X = 18,000 - N_Y

0.02(18,000 - N_Y) + 0.05 N_Y = 540 \implies 360 + 0.03 N_Y = 540 \implies 0.03 N_Y = 180 \implies N_Y = 6,000

. Thus

N_X = 12,000

Algebraic elimination isolated

N_Y

and

N_X

order volumes.

Determine individual processing rates

R_X

and

R_Y

R_X = \frac{12,000}{600} = 20 \text{ orders/second}

and

R_Y = \frac{6,000}{600} = 10 \text{ orders/second}

Dividing each venue's total order volume by 600 seconds yields its constant rate.

Key Concept

Weighted Average Ratios and Simultaneous Rate Equations

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