Modeling with Linear Systems - Linear Algebra

1. A metabolic pathway involves three enzymes (E1, E2, E3) that process substrates at rates $v_1, v_2, v_3$ respectively. At steady state, the concentration of each intermediate remains constant. The pathway can be represented as: $S \xrightarrow{v_1} A \xrightarrow{v_2} B \xrightarrow{v_3} P$, where enzyme E2 also has a side reaction that removes intermediate A at rate $0.2v_2$. What is the correct steady-state balance equation for intermediate A?
   Answer: $v_1 = 1.2v_2$ At steady state, the rate of production of intermediate A must equal its rate of consumption. Intermediate A is produced at rate $v_1$ and consumed at rate $v_2$ (main pathway) plus $0.2v_2$ (side reaction). Therefore: $v_1 = v_2 + 0.2v_2 = 1.2v_2$. Choice A ignores the side reaction. Choice B shows the correct relationship but isn't simplified. Choice D incorrectly places the side reaction term on the production side rather than consumption side.
2. A logistics network ships products between 4 cities using a hub-and-spoke model with city 1 as the hub. Direct shipments between non-hub cities are not allowed. The network is represented by a flow matrix where positive entries indicate outbound flows and negative entries indicate inbound flows. At city 2 (a non-hub city), the flow balance equation is $-x_{12} + x_{21} = d_2$, where $d_2 = -50$ represents net outbound demand. If $x_{21} = 200$ units/day, what does this imply about the logistics operation at city 2?
   Answer: City 2 receives 250 units from hub and ships 200 units to hub From the balance equation $-x_{12} + x_{21} = d_2$ with $x_{21} = 200$ and $d_2 = -50$: $-x_{12} + 200 = -50$, so $x_{12} = 250$. Here, $x_{12}$ represents flow from hub (city 1) to city 2 (inbound to city 2), and $x_{21}$ represents flow from city 2 to hub (outbound from city 2). Therefore, city 2 receives 250 units from the hub and ships 200 units back to the hub. The negative demand indicates net outbound flow from city 2. Choice A uses incorrect calculation. Choice C reverses the flow interpretation. Choice D uses the wrong value for inbound flow.
3. A water treatment facility has three treatment stages connected in series with recycling loops. Fresh water enters at 1000 L/min, and treated water exits at 800 L/min. Stage 2 recycles 150 L/min back to stage 1, and stage 3 recycles 100 L/min back to stage 2. If $f_1, f_2, f_3$ represent the total flow rates through stages 1, 2, and 3 respectively, what is $f_1$?
   Answer: 1150 L/min Working backwards from stage 3: $f_3 = 800 + 100 = 900$ L/min (output plus recycled amount). For stage 2: $f_2 = f_3 + 150 = 900 + 150 = 1050$ L/min (flow to stage 3 plus recycling to stage 1). For stage 1: $f_1 = 1000 + 150 = 1150$ L/min (fresh water input plus recycled flow from stage 2). Choice A incorrectly subtracts the recycle flow. Choice B gives the flow through stage 2. Choice C results from adding only half the recycle flows.
4. An economic input-output model for a three-sector economy (agriculture, manufacturing, services) uses the equation $(I - A)\mathbf{x} = \mathbf{d}$, where $A$ is the technology matrix, $\mathbf{x}$ is total output, and $\mathbf{d}$ is final demand. If $a_{21} = 0.3$ and the manufacturing sector increases its output by 1000 units, how much additional input does manufacturing require from agriculture?
   Answer: 300 units In the technology matrix $A$, entry $a_{21} = 0.3$ represents the amount of agriculture output needed per unit of manufacturing output. If manufacturing increases output by 1000 units, the additional agricultural input required is $a_{21} \times 1000 = 0.3 \times 1000 = 300$ units. Choice B incorrectly uses $(1 - a_{21}) \times 1000$. Choice C assumes a one-to-one relationship. Choice D incorrectly adds the coefficient to the output change.
5. If the efficiency matrix $E$ has entry $e_{ij}$ representing units of component $i$ produced per hour on machine $j$, and $\mathbf{h} = [h_1, h_2, h_3, h_4]^T$ represents hours each machine operates, which matrix equation correctly models this optimization problem's constraints?
   Answer: $E\mathbf{h} = [500, 300, 200]^T$ The efficiency matrix $E$ is 3×4 (3 components, 4 machines) where $e_{ij}$ is units of component $i$ per hour on machine $j$. When we multiply $E$ by the hours vector $\mathbf{h}$ (4×1), we get total production of each component: $(E\mathbf{h})_i = \sum_{j=1}^4 e_{ij}h_j$ = total units of component $i$ produced. This must equal demand $[500, 300, 200]^T$. Choice B uses $E^T$ which would have wrong dimensions (4×3 matrix times 4×1 vector is invalid). Choice C creates a row vector output instead of column vector. Choice D has dimension mismatch (1×4 times 3×4 is invalid).
6. A supply chain network distributes products from 2 warehouses to 3 retail stores. The cost matrix $C$ has entry $c_{ij}$ representing the cost per unit to ship from warehouse $i$ to store $j$. If warehouse 1 ships quantities $x_{11}, x_{12}, x_{13}$ to stores 1, 2, 3 respectively, and warehouse 2 ships quantities $x_{21}, x_{22}, x_{23}$, which constraint ensures that store 2 receives exactly its demand $d_2$?
   Answer: $x_{12} + x_{22} = d_2$ Store 2 receives shipments from both warehouses: $x_{12}$ units from warehouse 1 and $x_{22}$ units from warehouse 2. The total supply to store 2 must equal its demand, giving $x_{12} + x_{22} = d_2$. Choice B represents the total shipment from warehouse 2 to all stores, not the total received by store 2. Choice C incorrectly includes costs rather than quantities. Choice D represents total shipment from warehouse 1 to all stores.
7. A traffic network has four intersections connected by one-way streets. Traffic flows are measured in vehicles per hour. At intersection 2, the conservation equation is $x_1 - x_3 + x_4 = 50$, where $x_1, x_3, x_4$ represent flows on streets 1, 3, and 4 respectively, and 50 represents net traffic generation at that intersection. If street 1 carries 200 vehicles/hour toward intersection 2, and street 4 carries 120 vehicles/hour away from intersection 2, what is the flow on street 3?
   Answer: 270 vehicles/hour toward intersection 2 Using the conservation equation $x_1 - x_3 + x_4 = 50$ with $x_1 = 200$ (toward intersection 2) and $x_4 = 120$ (away from intersection 2, so positive in the equation): $200 - x_3 + 120 = 50$, which gives $320 - x_3 = 50$, so $x_3 = 270$. Since $x_3$ appears with a negative sign in the equation, positive $x_3$ represents flow toward the intersection. Choice A uses incorrect arithmetic ($200 + 120 - 50 = 270$, not 30). Choice B has the wrong magnitude and direction. Choice D has correct magnitude but wrong direction interpretation.

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