Question text Consider the following circulation with demand flow network.The demand is indicated in each vertex. The lower bound is indicated in blue and the capacity is indicated in black for each edge. The first step is to resolve the lower-bound of every edge, updating the demand of each vertex and the capacity of each edge.Demand for vertex u = Answer 1 Question 1[input] Demand for vertex v = Answer 2 Question 1[input] Demand for vertex w = Answer 3 Question 1[input] Demand for vertex x = Answer 4 Question 1[input] Capacity for edge = Answer 5 Question 1[input] Capacity for edge = Answer 6 Question 1[input] Capacity for edge = Answer 7 Question 1[input] Capacity for edge = Answer 8 Question 1[input] Capacity for edge = Answer 9 Question 1[input] Then the demand can be resolved by creating a source node and a sink node. The source node is connected to [table] u | v | w | x [/table]Mark 1.00 out of 1.00 and the remaining vertices are connected to the sink node.Then we run Ford-Fulkerson and the flow network is Answer 10 Question 1[select: , Feasible, Not Feasible].

Applying the lower-bound reduction rule "-incoming + outgoing" (equivalently: for each lower bound l on edge a->b, decrease a's demand by l and increase b's demand by l, and set the new capacity to c' = c - l): 1) Relevant concepts - For an edge (a->b) with lower bound l and capacity c, replace capacity by c' = c - l and adjust demands: d(a) := d(a) - l, d(b) := d(b) + l. - After adjustments, create a super-source and super-sink: connect supply vertices (negative demand) to the source with capacity equal to -d, and connect demand vertices (positive demand) to the sink with capacity equal to d. The circulation is feasible iff a max flow saturates all these added edges. 2) Step-by-step calculations (using the picture: edges with lower/capacity shown as blue/black: x->u 1/3, x->v 1/3, u->v 1/1, u->w 1/4, v->w 1/2; initial demands du=-3, dv=1, dw=5, dx=-3): - Vertex u: incoming lower bounds = 1 (from x); outgoing lower bounds = 1 (to v) + 1 (to w) = 2. New du = old du - incoming + outgoing = -3 -1 +2 = -2. - Vertex v: incoming = 1 (from u) + 1 (from x) = 2; outgoing = 1 (to w). New dv = 1 -2 +1 = 0. - Vertex w: incoming = 1 (from u) + 1 (from v) = 2; outgoing = 0. New dw = 5 -2 +0 = 3. - Vertex x: incoming = 0; outgoing = 1 (to u) + 1 (to v) = 2. New dx = -3 -0 +2 = -1. 3) Updated capacities (c' = c - l): - : 1 - 1 = 0 - : 4 - 1 = 3 - : 2 - 1 = 1 - : 3 - 1 = 2 - : 3 - 1 = 2 These values match the provided answers: du = -2, dv = 0, dw = 3, dx = -1; capacities 0, 3, 1, 2, 2 respectively. 4) Why the source is connected to u and x After adjustment u and x have negative demand (supplies): du = -2, dx = -1. The construction connects supply nodes to the super-source (with capacities 2 and 1). The remaining vertices (v and w) with nonnegative demand connect to the sink. 5) Feasibility check and why "Feasible" is correct Total supply = 2 + 1 = 3 equals total demand = 3 (w). A feasible flow exists: send 2 units from source->u then u->w (capacity 3) and send 1 unit from source->x then x->v (capacity 2) then v->w (capacity 1). w receives total 3 and the added source/sink edges are saturated, so the circulation is feasible. Therefore "Feasible" is correct; "Not Feasible" is incorrect because it ignores the achievable saturation demonstrated above. Conclusion: The computed demands and capacities and the final verdict "Feasible" match the provided answers.

Question text Consider the following circulation with demand flow network.The demand is indicated in each vertex. The lower bound is indicated in blue and the capacity is indicated in black for each edge. The first step is to resolve the lower-bound of every edge, updating the demand of each vertex and the capacity of each edge.Demand for vertex u = Answer 1 Question 1[input] Demand for vertex v = Answer 2 Question 1[input] Demand for vertex w = Answer 3 Question 1[input] Demand for vertex x = Answer 4 Question 1[input] Capacity for edge <u,v> = Answer 5 Question 1[input] Capacity for edge <u,w> = Answer 6 Question 1[input] Capacity for edge <v,w> = Answer 7 Question 1[input] Capacity for edge <x,u> = Answer 8 Question 1[input] Capacity for edge <x,v> = Answer 9 Question 1[input] Then the demand can be resolved by creating a source node and a sink node. The source node is connected to [table] u | v | w | x [/table]Mark 1.00 out of 1.00 and the remaining vertices are connected to the sink node.Then we run Ford-Fulkerson and the flow network is Answer 10 Question 1[select: , Feasible, Not Feasible].多项填空题

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