Unformatted text preview: (c) The capacity of the cut is Cct + Cet + Cdt = 6 + 2 + 6 = 14. (d) The saturated arcs are se, sb, ae, et, ed, ct and dt. (e) The flow is maximum, by Corollary 14.1.6. (f) (fct ftc) + (fet f te ) + (fdt ftd) = (6  0) + (2  0) + (6  0) = 14. 3. (a) At a, I:v fva = fsa = 11 and I:v fav = fac + fad = 3 + 8 = 11. At c, I: v fvc = fac = 3 and I:v f cv = f cd + f cf = 0 + 3 = 3. At d, I: v f vd = fad + fbd + f cd = 8 + 1 + 0 = 9 and I:v fdv = fdt = 9. (b) The value of the flow is 21. (c) [BB] Cac + Cbe + Cdt = 3 + 9 + 9 = 21. (d) The flow is maximum, by Corollary 14.1.6. (e) (facfca)+(fbefeb)+(fdtftd)+(fdcfcd) = (30)+(90)+(90)+(00)+(30) = 21. 4. (a) At a, I: v f va = f sa = 3 and I: v f av = f ab + f ac + fad = 1 + 1 + 1 = 3. At b, I: v f vb = f ab = 1 and I:v fbv = fbc + fbe + fbt = 1 + 0 + 0 = 1. At e, I:v f ve = fbe + f de = 0 + 3 = 3 and I:v fev = f et = 3....
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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