Two-Part Analysis Practice Questions — GMAT — Page 7

Question 121Question →

A central logistics facility classifies incoming freight shipments into four priority processing queues—Queue 1 (Highest Priority), Queue 2, Queue 3, and Queue 4 (Lowest Priority)—based on two parameters: Hazard Level (High, Medium, Low) and Perishability Index (1 to 5, where 5 indicates maximum perishability).

The classification rules are as follows:
- Queue 1: Any shipment with a High Hazard Level, OR a Perishability Index of 4 or 5.
- Queue 2: Any shipment NOT in Queue 1 that has a Medium Hazard Level AND a Perishability Index of 3.
- Queue 3: Any shipment NOT in Queue 1 or 2 that has a Low Hazard Level with a Perishability Index of 3, OR a Medium Hazard Level with a Perishability Index less than 3.
- Queue 4: Any shipment NOT in Queue 1, 2, or 3 that has a Low Hazard Level with a Perishability Index of 1 or 2.

Four shipments are evaluated:
- Shipment Alpha: Low Hazard Level, Perishability Index of 5
- Shipment Beta: Medium Hazard Level, Perishability Index of 3
- Shipment Gamma: Medium Hazard Level, Perishability Index of 2
- Shipment Delta: Low Hazard Level, Perishability Index of 1

Based on the classification rules, in what sequence should these four shipments be arranged from highest priority queue (Queue 1) to lowest priority queue (Queue 4)?

Drag items to arrange them in the correct order

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Answer

The correct sequence from highest to lowest priority is Shipment Alpha (Queue 1), Shipment Beta (Queue 2), Shipment Gamma (Queue 3), and Shipment Delta (Queue 4).

Evaluating each shipment according to the rules places Shipment Alpha in Queue 1 (highest priority due to Perishability Index 5), Shipment Beta in Queue 2 (Medium Hazard and Perishability Index 3), Shipment Gamma in Queue 3 (Medium Hazard and Perishability Index 2), and Shipment Delta in Queue 4 (Low Hazard and Perishability Index 1). Thus, the correct priority order is Shipment Alpha, Shipment Beta, Shipment Gamma, Shipment Delta.

Step-by-Step Solution

1

Evaluate Shipment Alpha against the priority queue rules.

Shipment Alpha has Low Hazard Level and Perishability Index 5. Since Perishability Index is 5, it satisfies Queue 1.

Queue 1 includes any shipment with a High Hazard Level or a Perishability Index of 4 or 5.

2

Evaluate Shipment Beta against the priority queue rules.

Shipment Beta has Medium Hazard Level and Perishability Index 3. It does not meet Queue 1 criteria, but meets Queue 2 criteria.

Queue 2 specifies Medium Hazard Level and Perishability Index of 3.

3

Evaluate Shipment Gamma against the priority queue rules.

Shipment Gamma has Medium Hazard Level and Perishability Index 2. It falls under Queue 3.

Queue 3 includes Medium Hazard Level shipments with a Perishability Index less than 3.

4

Evaluate Shipment Delta against the priority queue rules.

Shipment Delta has Low Hazard Level and Perishability Index 1. It falls under Queue 4.

Queue 4 specifies Low Hazard Level with a Perishability Index of 1 or 2.

Key Concept

Categorical Classification and Multi-Attribute Priority Sorting

Question 122Question →

A regional logistics operator manages cargo transit on a river corridor using two types of vessels: Express Barges ( $E$ ) and Heavy-Haul Barges ( $H$ ). Weekly operational constraints are defined as follows:

- Fuel Allowance: Each Express Barge consumes $8$ metric tons of fuel per trip, and each Heavy-Haul Barge consumes $12$ metric tons. Total weekly fuel consumption cannot exceed $144$ metric tons.
- Crew Availability: Each Express Barge requires $3$ crew shifts, and each Heavy-Haul Barge requires $2$ crew shifts. Total available crew shifts per week cannot exceed $30$ .
- Service Minimum: The operator must deploy at least $2$ Heavy-Haul Barges ( $H \ge 2$ ) per week to maintain baseline commercial obligations.
- Profit Structure: Each Express Barge generates a net profit of $\$5,000$ , and each Heavy-Haul Barge generates a net profit of $\$6,000$ .

Match each operational metric on the left with its correct optimal value under the profit-maximizing schedule of integer barge deployments.

Click a left item, then click its matching right item

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Answer

The profit-maximizing deployment requires

3

Express Barges and

10

Heavy-Haul Barges, yielding a total maximum profit of

75

thousand dollars (

\$75,000

).

The correct matches establish that deploying

3

Express Barges and

10

Heavy-Haul Barges yields the maximum possible weekly profit of

\$75,000

. This combination uses

2(3) + 3(10) = 36

tons of fuel (using

100\%

of fuel capacity) and

3(3) + 2(10) = 29

crew shifts (out of

30

available), satisfying all joint constraints.

Step-by-Step Solution

1

Formulate the linear programming inequalities from the narrative constraints.

Fuel constraint:

8E + 12H \le 144 \implies 2E + 3H \le 36

.
Crew constraint:

3E + 2H \le 30

.
Bounds:

H \ge 2

,

E \ge 0

, with

E, H \in \mathbb{Z}

.
Profit function:

P(E, H) = 5E + 6H

(in thousands of dollars).

Converting real-world bounds into simplified algebraic inequalities is required to determine the feasible integer region.

2

Identify boundary integer candidate points near the intersection of the constraints.

Solving

2E + 3H = 36

and

3E + 2H = 30

gives

5H = 48 \implies H = 9.6, E = 3.6

.
Testing adjacent integer points

(E, H)

within the feasible region:
- Case 1:

H = 10 \implies 2E + 30 \le 36 \implies E \le 3

. Check crew for

(3, 10)

:

3(3) + 2(10) = 29 \le 30

(Feasible). Profit

P = 5(3) + 6(10) = 75

.
- Case 2:

H = 9 \implies 2E + 27 \le 36 \implies E \le 4

. Check crew for

(4, 9)

:

3(4) + 2(9) = 30 \le 30

(Feasible). Profit

P = 5(4) + 6(9) = 74

.
- Case 3:

H = 12 \implies 2E + 36 \le 36 \implies E = 0

. Check crew for

(0, 12)

:

3(0) + 2(12) = 24 \le 30

(Feasible). Profit

P = 6(12) = 72

.
- Case 4:

H = 11 \implies 2E + 33 \le 36 \implies E \le 1

. Check crew for

(1, 11)

:

3(1) + 2(11) = 25 \le 30

(Feasible). Profit

P = 5(1) + 6(11) = 71

.

Since integer constraints apply, evaluating feasible grid points along the constraint boundary identifies the global maximum.

3

Compare profit outcomes and confirm optimal values.

The maximum profit occurs at

(E, H) = (3, 10)

, giving a profit of

\$75,000

.

Comparing all candidate integer points verifies that

E=3

,

H=10

, and

P=75

is the unique optimal solution.

Key Concept

Integer Optimization under Simultaneous Bounded Constraints

Question 123Question →

A venture capital firm classifies its portfolio companies into four strategic management categories based on two quarterly performance metrics: Burn Multiple ( $B$ ) and Customer Retention Rate ( $R$ ).

Classification Criteria:
- Category 1 (Prime Growth): $B \le 1.5$ and $R \ge 85\%$
- Category 2 (Capital Efficient): $B \le 1.5$ and $R < 85\%$
- Category 3 (Expansion Focused): $B > 1.5$ and $R \ge 85\%$
- Category 4 (High Risk): $B > 1.5$ and $R < 85\%$

Match each portfolio company to its corresponding strategic management category.

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Answer

Company P matches Category 1 (Prime Growth); Company Q matches Category 2 (Capital Efficient); Company R matches Category 3 (Expansion Focused); Company S matches Category 4 (High Risk).

Each company is mapped to its category by evaluating its two metrics against the threshold boundary conditions (

B = 1.5

and

R = 85\%

). Company P satisfies low burn (

B \le 1.5

) and high retention (

R \ge 85\%

), placing it in Category 1. Company Q meets low burn but has lower retention (

R < 85\%

), placing it in Category 2. Company R exceeds the burn threshold (

B > 1.5

) but maintains high retention (

R \ge 85\%

), placing it in Category 3. Company S exceeds the burn threshold and has lower retention, placing it in Category 4.

Step-by-Step Solution

1

Evaluate Company P (

B = 1.2, R = 90\%

) against threshold rules.

B = 1.2 \le 1.5

is true, and

R = 90\% \ge 85\%

is true. This satisfies Category 1 (Prime Growth).

Both metric conditions match the exact criteria for Category 1.

2

Evaluate Company Q (

B = 1.4, R = 80\%

) against threshold rules.

B = 1.4 \le 1.5

is true, but

R = 80\% < 85\%

is true. This satisfies Category 2 (Capital Efficient).

Low burn multiple combined with retention rate below

85\%

places the entity in Category 2.

3

Evaluate Company R (

B = 1.8, R = 88\%

) against threshold rules.

B = 1.8 > 1.5

is true, and

R = 88\% \ge 85\%

is true. This satisfies Category 3 (Expansion Focused).

Burn multiple exceeding

1.5

combined with high retention places the entity in Category 3.

4

Evaluate Company S (

B = 2.1, R = 75\%

) against threshold rules.

B = 2.1 > 1.5

is true, and

R = 75\% < 85\%

is true. This satisfies Category 4 (High Risk).

Both metrics fail the favorable thresholds, placing the entity in Category 4.

Key Concept

Categorical Classification and Sorting

Question 124Question →

A cloud analytics company configures daily data processing operations using two types of virtual server instances: Compute-Optimized ( $C$ ) and Memory-Optimized ( $M$ ). The operational parameters and constraints are as follows:

- Each $C$ instance processes 50 batch jobs per hour and incurs an operating cost of $15 per hour. - Each$ M $instance processes 30 batch jobs per hour and incurs an operating cost of$ 10 per hour.
- The system must process at least 360 batch jobs per hour in total.
- The total hourly operating budget for server instances cannot exceed $115. - Due to active database caching policies, at least 3$ M $instances must be active at all times ($ M \ge 3 $). -$ C $and$ M$ must both be non-negative integers.

Which of the following combinations of Compute ( $C$ ) and Memory ( $M$ ) server instances maximizes the total hourly job throughput while satisfying all operational constraints?

5 Compute instances and 4 Memory instances

3 Compute instances and 7 Memory instances

6 Compute instances and 2 Memory instances

4 Compute instances and 6 Memory instances

2 Compute instances and 9 Memory instances

Show answer & explanation

Answer: 5 Compute instances and 4 Memory instances

Answer

5 Compute instances and 4 Memory instances

The combination of 5 Compute instances and 4 Memory instances satisfies all stated system bounds (

15 \times 5 + 10 \times 4 = 115 \le 115

,

50 \times 5 + 30 \times 4 = 370 \ge 360

, and

4 \ge 3

) and delivers the maximum feasible throughput of 370 jobs per hour.

Step-by-Step Solution

1

Formulate the linear system of inequalities from the given constraints.

Throughput:

50C + 30M \ge 360 \Rightarrow 5C + 3M \ge 36

.
Budget:

15C + 10M \le 115 \Rightarrow 3C + 2M \le 23

.
Minimum

M

:

M \ge 3

, where

C, M \in \mathbb{Z}_{\ge 0}

.

Converting narrative conditions into simplified algebraic inequalities makes constraint checking straightforward.

2

Determine the set of integer pairs

(C, M)

that satisfy all three inequalities.

Combining

3C + 2M \le 23

and

5C + 3M \ge 36

with

M \ge 3

yields only two valid integer pairs:
- Pair 1:

(C=3, M=7)

with cost

15(3)+10(7)=115

and throughput

50(3)+30(7)=360

- Pair 2:

(C=5, M=4)

with cost

15(5)+10(4)=115

and throughput

50(5)+30(4)=370

Systematic testing of integer values bounded by the linear inequalities identifies all feasible solutions.

3

Compare total throughput across feasible solutions to find the maximum.

Pair 2

(C=5, M=4)

produces 370 jobs per hour, which exceeds Pair 1's 360 jobs per hour.

The question specifically asks to maximize throughput among all valid configurations.

Key Concept

Optimization under linear integer constraints and bounded systems

Question 125Question →

A metropolitan transit authority categorizes electric buses into three maintenance tiers based on two performance metrics: Cumulative Mileage (in thousands of kilometers) and Battery State of Health (SoH) (as a percentage).

The classification policy dictates:
- Tier I (Standard Operation): Cumulative Mileage $\le 200$ thousand km AND Battery SoH $\ge 85\%$ .
- Tier II (Battery Refurbishment): Cumulative Mileage $\le 300$ thousand km with Battery SoH between $70\%$ and $84\%$ (inclusive); OR Cumulative Mileage between $201$ and $300$ thousand km (inclusive) with Battery SoH $\ge 85\%$ .
- Tier III (Fleet Retirement): Cumulative Mileage $> 300$ thousand km OR Battery SoH $< 70\%$ regardless of mileage.

Two buses in the fleet present the following data:
- Bus X: Cumulative Mileage = $160$ thousand km; Battery SoH = $90\%$
- Bus Y: Cumulative Mileage = $180$ thousand km; Battery SoH = $82\%$

Based on the classification policy, which of the following correctly pairs Bus X and Bus Y with their respective maintenance tiers?

Bus X is assigned to Tier II, and Bus Y is assigned to Tier I.

Bus X is assigned to Tier I, and Bus Y is assigned to Tier I.

Bus X is assigned to Tier I, and Bus Y is assigned to Tier II.

Bus X is assigned to Tier II, and Bus Y is assigned to Tier II.

Bus X is assigned to Tier III, and Bus Y is assigned to Tier II.

Show answer & explanation

Answer: Bus X is assigned to Tier I, and Bus Y is assigned to Tier II.

Answer

Bus X is assigned to Tier I, and Bus Y is assigned to Tier II.

Evaluating Bus X: Mileage of 160 thousand km is ≤ 200 and Battery SoH of 90% is ≥ 85%, satisfying both criteria for Tier I. Evaluating Bus Y: Mileage of 180 thousand km is ≤ 300 and Battery SoH of 82% is between 70% and 84%, satisfying Tier II. Thus, Bus X is assigned to Tier I and Bus Y to Tier II.

Step-by-Step Solution

1

Evaluate Bus X against the classification rules

Bus X has Cumulative Mileage = 160 thousand km (≤ 200) and Battery SoH = 90% (≥ 85%). Both conditions for Tier I (Standard Operation) are met.

Tier I requires both mileage ≤ 200 thousand km and SoH ≥ 85%.

2

Evaluate Bus Y against the classification rules

Bus Y has Cumulative Mileage = 180 thousand km (≤ 300) and Battery SoH = 82% (between 70% and 84%). This fits the first clause of Tier II (Battery Refurbishment).

Although Bus Y's mileage is under 200 thousand km, its SoH of 82% falls below the Tier I requirement of 85% but fits within the 70%–84% range specified for Tier II.

3

Combine the evaluations to select the correct paired response

Bus X belongs to Tier I and Bus Y belongs to Tier II.

Both individual classifications match this paired statement.

Key Concept

Multi-attribute categorical classification based on compound conditional logic

Estimated Time:1m 30s

Question 126Question →

A municipal transit agency is installing two types of electric vehicle (EV) charging stations at a new central station: Level 2 Chargers ( $L$ ) and Level 3 Fast Chargers ( $D$ ). Each Level 2 charger costs $\$4,000$ to install and draws $10\text{ kW}$ of grid power. Each Level 3 charger costs $\$15,000$ to install and draws $50\text{ kW}$ of grid power. The agency has a total installation budget of $\$100,000$ and a maximum available grid power allocation of $320\text{ kW}$ . To meet minimum service grant requirements, the hub must install at least 5 Level 2 chargers ( $L \ge 5$ ) and at least 3 Level 3 chargers ( $D \ge 3$ ). Each Level 2 charger can serve up to 8 sessions per day, and each Level 3 charger can serve up to 30 sessions per day. Match each operational optimization metric on the left with its corresponding correct value on the right that maximizes the total daily charging sessions served by the hub.

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Answer

The optimal operational configuration consists of 10 Level 2 chargers and 4 Level 3 chargers, achieving a maximum total of 200 daily charging sessions.

To maximize daily sessions

S = 8L + 30D

under budget

4L + 15D \le 100

, power

L + 5D \le 32

, and lower bounds

L \ge 5, D \ge 3

, testing valid integer points shows that setting Level 3 chargers to 4 and Level 2 chargers to 10 satisfies all joint constraints while achieving the peak capacity of 200 total daily sessions.

Step-by-Step Solution

1

Formulate the linear system of constraints and objective function.

Budget constraint:

4,000L + 15,000D \le 100,000 \implies 4L + 15D \le 100

.
Power constraint:

10L + 50D \le 320 \implies L + 5D \le 32

.
Minimum bounds:

L \ge 5

and

D \ge 3

.
Objective function to maximize:

S = 8L + 30D

.

Define the mathematical boundaries and optimization goal based on problem parameters.

2

Evaluate feasible integer values for Level 3 chargers (

D

).

Case 1: If

D = 3

:
- Budget bound:

4L + 15(3) \le 100 \implies 4L \le 55 \implies L \le 13

.
- Power bound:

L + 5(3) \le 32 \implies L \le 17

.
- Max feasible

L = 13

.
- Total sessions

S = 8(13) + 30(3) = 104 + 90 = 194

.

Case 2: If

D = 4

:
- Budget bound:

4L + 15(4) \le 100 \implies 4L \le 40 \implies L \le 10

.
- Power bound:

L + 5(4) \le 32 \implies L \le 12

.
- Max feasible

L = 10

.
- Total sessions

S = 8(10) + 30(4) = 80 + 120 = 200

.

Case 3: If

D = 5

:
- Budget bound:

4L + 15(5) \le 100 \implies 4L \le 25 \implies L \le 6

.
- Power bound:

L + 5(5) \le 32 \implies L \le 7

.
- Max feasible

L = 6

.
- Total sessions

S = 8(6) + 30(5) = 48 + 150 = 198

.

Case 4: If

D = 6

:
- Budget bound:

4L + 15(6) \le 100 \implies 4L \le 10 \implies L \le 2

.
- Violates minimum constraint

L \ge 5

.

Systematically check integer combinations of variables within the feasible region to identify maximum daily sessions.

3

Compare total sessions across valid integer pairs.

The maximum session count occurs when

D = 4

and

L = 10

, producing

S = 200

daily sessions.

Determine the unique optimal combination satisfying all joint constraints.

Key Concept

Linear Optimization under Bounded Integer Constraints

Estimated Time:2m 0s

Question 127Question →

A boutique catering kitchen prepares custom corporate event packages containing two types of dessert selections: Specialty Layer Cakes ( $L$ ) and Miniature Cupcake Sets ( $M$ ). Each Specialty Layer Cake requires $3$ hours of decorating labor and $4$ kg of specialized flour. Each Miniature Cupcake Set requires $1$ hour of decorating labor and $2$ kg of specialized flour. For an upcoming event order, the kitchen has a maximum resource availability of $30$ total hours of decorating labor and $44$ kg of specialized flour. To meet client specifications, the kitchen must produce at least $4$ Specialty Layer Cakes. If each Specialty Layer Cake yields a profit of $\$50$ and each Miniature Cupcake Set yields a profit of $\$20$ , which of the following pairs of $(L, M)$ represents the combination of Specialty Layer Cakes and Miniature Cupcake Sets that maximizes total profit while satisfying all resource constraints?

(8, 6)

(7, 9)

(9, 4)

(6, 12)

(5, 12)

Show answer & explanation

Answer: $(8, 6)$

Answer

The optimal production combination is 8 Specialty Layer Cakes and 6 Miniature Cupcake Sets, yielding a maximum profit of $520.

The pair

(8, 6)

satisfies the minimum requirement of at least 4 Specialty Layer Cakes (

8 \ge 4

), requires

3(8) + 1(6) = 30

labor hours (within the 30-hour limit), and requires

4(8) + 2(6) = 44

kg of flour (within the 44-kg limit). It yields a total profit of

50(8) + 20(6) = \$520

, which is the highest achievable profit among all feasible combinations.

Step-by-Step Solution

1

Formulate the mathematical model for constraints and objective function.

Decorating Labor:

3L + M \le 30

; Specialized Flour:

4L + 2M \le 44

(or

2L + M \le 22

); Minimum Requirement:

L \ge 4

; Profit:

P = 50L + 20M

.

Establishing explicit linear inequality bounds allows systematic evaluation of feasible integer pairs

(L, M)

.

2

Determine upper bound for

M

as a function of integer values of

L \ge 4

.

For

L=4

:

M \le \min(30-12, 22-8) = 14 \implies P = 50(4)+20(14) = \$480

.
For

L=5

:

M \le \min(30-15, 22-10) = 12 \implies P = 50(5)+20(12) = \$490

.
For

L=6

:

M \le \min(30-18, 22-12) = 10 \implies P = 50(6)+20(10) = \$500

.
For

L=7

:

M \le \min(30-21, 22-14) = 8 \implies P = 50(7)+20(8) = \$510

.
For

L=8

:

M \le \min(30-24, 22-16) = 6 \implies P = 50(8)+20(6) = \$520

.
For

L=9

:

M \le \min(30-27, 22-18) = 3 \implies P = 50(9)+20(3) = \$510

.
For

L=10

:

M \le \min(30-30, 22-20) = 0 \implies P = 50(10)+20(0) = \$500

.

Evaluating potential corner points and boundary integer points identifies the maximum profit within the bounded region.

3

Compare maximum calculated profit across all valid integer pairs.

The highest profit occurs at

L = 8, M = 6

with

P = \$520

.

This point satisfies all constraints while maximizing the objective function.

Key Concept

Optimization under multiple joint inequality constraints

Question 128Question →

A printing company produces two types of customized marketing materials: Standard brochures ( $S$ ) and Premium brochures ( $P$ ). Each Standard brochure requires $2$ minutes of printing time and $1$ minute of binding time. Each Premium brochure requires $3$ minutes of printing time and $3$ minutes of binding time. The production facility has a maximum daily capacity of $240$ minutes for printing and $180$ minutes for binding. Due to a recurring client agreement, the facility must produce at least $20$ Standard brochures per day. The profit is $\$4$ per Standard brochure and $\$9$ per Premium brochure. Match each optimization variable or outcome on the left with its correct value on the right that maximizes daily total profit while satisfying all operational constraints.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The optimal production values are

60

Standard brochures and

40

Premium brochures, resulting in a maximum daily profit of

\$600

.

Solving the system of simultaneous constraints yields the optimal vertex at

S = 60

Standard brochures and

P = 40

Premium brochures. At this point, both printing (

2(60) + 3(40) = 240

) and binding (

60 + 3(40) = 180

) capacities are fully utilized, achieving the maximum daily profit of

4(60) + 9(40) = \$600

.

Step-by-Step Solution

1

Formulate the linear programming model with constraints and objective function.

Constraints:

2S + 3P \le 240

(printing time),

S + 3P \le 180

(binding time),

S \ge 20

(contractual minimum), and

S, P \ge 0

. Objective: Maximize

Z = 4S + 9P

.

Clear mathematical formulation is required to define the feasible region bounded by resource limits.

2

Calculate the intersection points (vertices) of the boundary lines of the feasible region.

Subtracting

S + 3P = 180

from

2S + 3P = 240

yields

S = 60

. Substituting

S = 60

into

S + 3P = 180

gives

3P = 120 \implies P = 40

. Corner point 1:

(60, 40)

.

The fundamental theorem of linear programming states that maximum profit occurs at a vertex of the feasible region.

3

Find remaining boundary vertices satisfying all constraints.

Corner point 2:

(120, 0)

from printing limit when

P=0

(satisfies binding limit

120 \le 180

). Corner point 3:

(20, 53)

when

S=20

(integer constraint on

P

where

20 + 3P \le 180 \implies P \le 53.33

). Corner point 4:

(20, 0)

.

All feasible vertices must be identified to compare their objective function values.

4

Evaluate the profit function

Z = 4S + 9P

at each vertex point.

At

(60, 40)

:

Z = 4(60) + 9(40) = 240 + 360 = \$600

. At

(120, 0)

:

Z = 4(120) = \$480

. At

(20, 53)

:

Z = 4(20) + 9(53) = 80 + 477 = \$557

. At

(20, 0)

:

Z = 4(20) = \$80

.

Direct substitution reveals which corner point yields the global maximum value inside the bounded constraints.

Key Concept

Linear programming optimization under joint resource and integer bound constraints

Question 129Question →

A high-tech manufacturing firm produces custom drone components: Micro-Sensors ( $S$ ) and Control Units ( $U$ ). Each Micro-Sensor requires 2 hours of precision calibration and 1 GB of firmware memory allocation. Each Control Unit requires 5 hours of precision calibration and 4 GB of firmware memory allocation. For an upcoming production batch, the facility has a maximum of 40 hours of calibration time available and a maximum total memory allocation limit of 30 GB. If the firm earns a profit of $100 per Micro-Sensor and$ 350 per Control Unit, which of the following combinations of Micro-Sensors ( $S$ ) and Control Units ( $U$ ) maximizes total batch profit while satisfying all operational bounds?

2 Micro-Sensors and 7 Control Units

6 Micro-Sensors and 6 Control Units

0 Micro-Sensors and 8 Control Units

5 Micro-Sensors and 6 Control Units

20 Micro-Sensors and 0 Control Units

Show answer & explanation

Answer: 2 Micro-Sensors and 7 Control Units

Answer

The combination of 2 Micro-Sensors and 7 Control Units maximizes total profit ($2,650) under the given calibration and memory constraints.

The pair with 2 Micro-Sensors and 7 Control Units fully satisfies both constraints (

2(2) + 5(7) = 39 \\leq 40

hours and

2 + 4(7) = 30 \\leq 30

GB) and yields the maximum profit of $2,650.

Step-by-Step Solution

1

Set up the linear programming model with objective function and system of inequalities.

Maximize

P = 100S + 350U

subject to:

2S + 5U \\leq 40

(Calibration limit),

S + 4U \\leq 30

(Memory limit), and non-negative integers

S, U \\geq 0

.

Establishing explicit bounds is necessary to evaluate boundary integer points.

2

Solve for the intersection of the boundary equations

2S + 5U = 40

and

S + 4U = 30

.

Substituting

S = 30 - 4U

into the first equation yields

2(30 - 4U) + 5U = 40 \\implies 60 - 3U = 40 \\implies U = 20/3 \\approx 6.67

and

S = 3.33

.

The continuous optimal point occurs at the intersection of the binding constraints.

3

Evaluate feasible integer coordinates near the intersection boundary.

For

U = 7

:

S \\leq \\min(\\frac{40 - 35}{2}, 30 - 28) = \\min(2.5, 2) = 2 \\implies (2, 7)

. Profit =

100(2) + 350(7) = \\

2,650

. For

U = 6

:

S \\leq \\min(\\frac{40 - 30}{2}, 30 - 24) = \\min(5, 6) = 5 \\implies (5, 6)

. Profit =

100(5) + 350(6) = \\

2,600

.

Since production quantities must be integers, testing feasible boundary pairs determines the global optimum.

Key Concept

Optimization under Bounded Linear Constraints

Question 130Question →

A regional emergency health network deploys two types of mobile medical units for event coverage: Rapid Response Units ( $R$ ) and Heavy Support Units ( $H$ ).

The deployment is governed by the following staffing and operational constraints:
- Each Rapid Response Unit ( $R$ ) requires $1$ paramedic and $2$ EMTs.
- Each Heavy Support Unit ( $H$ ) requires $3$ paramedics and $1$ EMT.
- On any given shift, a maximum of $21$ paramedics and $22$ EMTs are available.
- Operational policy mandates deploying at least $3$ Rapid Response Units ( $R \ge 3$ ) and at least $2$ Heavy Support Units ( $H \ge 2$ ).

Each Rapid Response Unit can treat $10$ patients per hour, and each Heavy Support Unit can treat $15$ patients per hour.

Match each operational metric on the left to its corresponding optimal value on the right that maximizes total patient treatment capacity per hour.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The optimal operational configuration pairs 9 Rapid Response Units with 4 Heavy Support Units, achieving a maximum treatment capacity of 150 patients per hour.

The optimal pairing matches 9 Rapid Response Units, 4 Heavy Support Units, and a total capacity of 150 patients per hour. The optimal point

(9, 4)

fully utilizes both available resource pools (

1(9) + 3(4) = 21

paramedics and

2(9) + 1(4) = 22

EMTs), maximizing total throughput

10R + 15H

.

Step-by-Step Solution

1

Set up the linear inequality constraints and objective function.

Paramedic resource constraint:

R + 3H \le 21

; EMT resource constraint:

2R + H \le 22

; Operational bounds:

R \ge 3, H \ge 2

; Objective function to maximize:

C = 10R + 15H

.

Translating the narrative conditions into algebraic forms defines the feasible region of operation.

2

Calculate the corner vertices of the bounded feasible polygon.

Solving

R + 3H = 21

and

2R + H = 22

simultaneously yields

R = 9

and

H = 4

. Other extreme vertices defined by lower bounds are

(3, 6)

,

(3, 2)

, and

(10, 2)

.

Linear programming optimization principles guarantee that the maximum objective value occurs at one of the corner vertices of the feasible region.

3

Evaluate the capacity objective function

C(R, H) = 10R + 15H

at each vertex.

At

(3, 2): 10(3) + 15(2) = 60

; At

(3, 6): 10(3) + 15(6) = 120

; At

(10, 2): 10(10) + 15(2) = 130

; At

(9, 4): 10(9) + 15(4) = 150

.

Comparing capacities across all vertices establishes

(9, 4)

as the global maximum.

Key Concept

Linear Programming Optimization under Bounded System Constraints

Question 131Question →

A renewable energy company designs micro-grid installations containing two types of modular components: Battery Storage Units ( $B$ ) and Solar Inverter Modules ( $I$ ). The installation configuration is subject to the following system constraints:

1. Budget Bound: Each Battery Storage Unit costs \ $4,000 and each Inverter Module costs \$2,000. The total installation budget cannot exceed \$32,000, giving the constraint$ 2B + I \le 16 $. 2. **Peak Power Requirement**: To meet grid stability requirements, the combined system rating must satisfy$ 3B + 4I \ge 30 $. 3. **Physical Space Limit**: Due to floor load capacity, a maximum of 6 Battery Storage Units can be installed ($ B \le 6$).

The total daily credit rating generated by the installation is given by the objective function $E = 12B + 5I$ .

Which combination of Battery Storage Units ( $B$ ) and Solar Inverter Modules ( $I$ ) satisfies all system constraints while maximizing the total daily credit rating $E$ ?

6 Battery Storage Units and 4 Inverter Modules

6 Battery Storage Units and 5 Inverter Modules

5 Battery Storage Units and 6 Inverter Modules

4 Battery Storage Units and 8 Inverter Modules

7 Battery Storage Units and 2 Inverter Modules

Show answer & explanation

Answer: 6 Battery Storage Units and 4 Inverter Modules

Answer

6 Battery Storage Units and 4 Inverter Modules

The combination of 6 Battery Storage Units and 4 Inverter Modules strictly satisfies all three constraints: the space constraint (

6 \le 6

), the budget constraint (

2(6) + 4 = 16 \le 16

), and the peak power requirement (

3(6) + 4(4) = 34 \ge 30

). Substituting these values into the credit rating function yields

E = 12(6) + 5(4) = 92

, which is higher than any other feasible combination.

Step-by-Step Solution

1

Identify the feasible values for

B

based on bounded constraints.

The physical space constraint limits

B

to non-negative integers where

B \le 6

.

Establishing individual bounds reduces the search space for integer solutions.

2

Evaluate upper bounds for

I

using the budget constraint

I \le 16 - 2B

for candidate values of

B

.

For

B = 6

,

I \le 16 - 12 = 4

. For

B = 5

,

I \le 16 - 10 = 6

. For

B = 4

,

I \le 16 - 8 = 8

.

Determines maximum allowable Inverter Modules for each valid count of Battery Storage Units.

3

Check peak power constraint

3B + 4I \ge 30

and compute objective function

E = 12B + 5I

for feasible candidate pairs.

Pair

(6, 4)

:

3(6) + 4(4) = 34 \ge 30

,

E = 12(6) + 5(4) = 92

.
Pair

(5, 6)

:

3(5) + 4(6) = 39 \ge 30

,

E = 12(5) + 5(6) = 90

.
Pair

(4, 8)

:

3(4) + 4(8) = 44 \ge 30

,

E = 12(4) + 5(8) = 88

.

Testing valid candidates against all joint constraints identifies the configuration yielding maximum credits.

4

Compare total daily credit ratings across all valid choices.

The maximum credit rating is 92, achieved at

B = 6

and

I = 4

.

Confirms the optimal bounded solution.

Key Concept

Optimization under Bounded Linear Constraints

Estimated Time:2m 0s

Question 132Question →

A fintech corporation tracks server performance metrics across five regional data centers: Region Alpha, Region Beta, Region Gamma, Region Delta, and Region Epsilon.

The table below presents Q3 transaction processing latency and cluster allocations:

Data Center	Average Transaction Latency (ms)	Active Server Clusters
Region Alpha	120	15
Region Beta	85	10
Region Gamma	150	20
Region Delta	95	12
Region Epsilon	110	14

Additionally, graphical analysis indicates that monthly operational expenses (

y

, in thousands of dollars) correlate linearly with active server clusters (

x

) according to the line of best fit

y = 4.5x + 10

.

Which of the following correctly pairs the data center with the lowest latency per active server cluster and its predicted monthly operational expense (in thousands of dollars)?

Region Gamma; $100 thousand

Region Beta; $55 thousand

Region Gamma; $90 thousand

Region Alpha; $77.5 thousand

Region Delta; $64 thousand

Show answer & explanation

Answer: Region Gamma; $100 thousand

Answer

Region Gamma; $100 thousand

Evaluating the latency per active server cluster for each data center yields

8.0

ms for Region Alpha,

8.5

ms for Region Beta,

7.5

ms for Region Gamma,

7.92

ms for Region Delta, and

7.86

ms for Region Epsilon. Region Gamma exhibits the lowest rate (

7.5

ms per cluster). Applying its

20

active server clusters to the graphical trend line

y = 4.5x + 10

gives

y = 4.5(20) + 10 = 100

thousand dollars.

Step-by-Step Solution

1

Calculate the latency per active server cluster for each data center using table data.

Region Alpha:

120 / 15 = 8.0

ms/cluster; Region Beta:

85 / 10 = 8.5

ms/cluster; Region Gamma:

150 / 20 = 7.5

ms/cluster; Region Delta:

95 / 12 \approx 7.92

ms/cluster; Region Epsilon:

110 / 14 \approx 7.86

ms/cluster.

Dividing latency by active server clusters isolates efficiency per unit cluster.

2

Identify the data center with the minimum ratio.

Region Gamma has the lowest latency per active server cluster at

7.5

ms/cluster.

Comparing calculated rates reveals the minimum value.

3

Use the linear model equation

y = 4.5x + 10

to find the operational expense for Region Gamma.

Substitute

x = 20

active server clusters into the trend equation:

y = 4.5(20) + 10 = 90 + 10 = 100

.

Region Gamma operates

20

active server clusters.

Key Concept

Integrated Table and Graphical Analysis

Estimated Time:2m 0s

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