An earth-observation satellite payload operator allocates integer numbers of channels to t… — GMAT

An earth-observation satellite payload operator allocates integer numbers of channels to two operational payload modes: High-Resolution Imaging ( $I$ ) and Atmospheric Sounding ( $S$ ). Operational parameters and resource bounds are specified as follows:

- Bandwidth: Each imaging channel requires $15\text{ MHz}$ and each sounding channel requires $10\text{ MHz}$ . Total available payload bandwidth is at most $210\text{ MHz}$ .
- Power: Each imaging channel consumes $40\text{ W}$ and each sounding channel consumes $15\text{ W}$ . Total available power budget is at most $440\text{ W}$ .
- Mission Minimums: The satellite must operate at least $2$ imaging channels ( $I \ge 2$ ) and at least $4$ sounding channels ( $S \ge 4$ ).
- Buffer Constraint: The number of sounding channels cannot exceed twice the number of imaging channels ( $S \le 2I$ ).

Match each payload optimization target on the left with its corresponding integer channel count on the right.

Maximum possible number of Imaging channels ( $I$ )9
Maximum possible number of Sounding channels ( $S$ )12
Number of Sounding channels ( $S$ ) that maximizes daily data return ( $R = 70I + 20S$ Gbit/day)5

Answer

The maximum possible number of Imaging channels (

I

) is 9. The maximum possible number of Sounding channels (

S

) is 12. The number of Sounding channels (

S

) that maximizes daily data return is 5.

Matching the maximum imaging channels (

I

) to 9 is correct because

I = 10

forces

S \le 2

, violating

S \ge 4

, whereas

I = 9

is fully feasible with

S = 5

. Matching the maximum sounding channels (

S

) to 12 is correct because at

(6, 12)

, bandwidth, power, and buffer constraints are simultaneously satisfied. Matching the sounding channels for maximum data return to 5 is correct because the function

R = 70I + 20S

reaches its peak of 730 Gbit/day at

(9, 5)

Step-by-Step Solution

Formulate the algebraic system of inequality constraints.

Bandwidth:

15I + 10S \le 210 \implies 3I + 2S \le 42

. Power:

40I + 15S \le 440 \implies 8I + 3S \le 88

. Bounds:

I \ge 2

S \ge 4

, and

S \le 2I

Converting narrative bounds into simplified algebraic inequalities facilitates evaluating integer extreme points.

Find the maximum feasible value for Imaging channels (

I

I = 10

, power requires

8(10) + 3S \le 88 \implies 3S \le 8 \implies S \le 2

, which violates the minimum requirement

S \ge 4

. If

I = 9

, power requires

8(9) + 3S \le 88 \implies 3S \le 16 \implies S \le 5

. With

S = 4

S = 5

, bandwidth gives

3(9) + 2(5) = 37 \le 42

and

S \le 2(9)

holds. Thus, max

I = 9

Determining the upper bound of

I

requires testing larger integer values of

I

while ensuring

S \ge 4

is not violated.

Find the maximum feasible value for Sounding channels (

S

Since

S \le 2I

S

is constrained by

I

. Testing

I = 6

: bandwidth gives

3(6) + 2S \le 42 \implies 2S \le 24 \implies S \le 12

. Power gives

8(6) + 3S \le 88 \implies 3S \le 40 \implies S \le 13.33

. Buffer gives

S \le 2(6) = 12

. At

(6, 12)

, all constraints hold (

18+24=42 \le 42

48+36=84 \le 88

12 \le 12

). Thus, max

S = 12

Maximizing

S

requires balancing the ratio constraint

S \le 2I

against bandwidth and power limits.

Maximize the objective function

R(I, S) = 70I + 20S

over all feasible boundary points.

Evaluating candidate integer vertex points:

R(9, 5) = 70(9) + 20(5) = 730

;

R(9, 4) = 710

;

R(8, 8) = 720

;

R(7, 10) = 690

;

R(6, 12) = 660

. The maximum data return occurs at

(9, 5)

, where

S = 5

Comparing total return at all boundary points identifies the global integer maximum.

Key Concept

Linear Programming and Integer Bounded Optimization

Question

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