A regional logistics operator manages cargo transit on a river corridor using two types of… — GMAT

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.

Number of Express Barges ( $E$ ) in the profit-maximizing schedule $3$
Number of Heavy-Haul Barges ( $H$ ) in the profit-maximizing schedule $10$
Maximum total weekly profit achieved (in thousands of dollars) $75$

Cevap

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.

Adım Adım Çözüm

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.

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.

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.

Anahtar Kavram

Integer Optimization under Simultaneous Bounded Constraints

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