一直不清楚EIGRP 的 cost 计算:5 o! [- V1 g1 H) Y& ]6 c. J/ X0 O
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在这做个总结, 这个是EIGRP的基础知识, 属于实验考试范围.
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7 _" g% o" i0 I0 p& U4 N; X2 lThe formulae for calculating EIGRP metric is:
. f" G0 b& l5 w" O; qmetric = [K1 * bandwidth + (K2 * bandwidth) / (256 - load) + K3 * delay] * [K5 / (reliability + K4)]
0 `" W5 ~0 s! [$ X }' Fk1=bandwidth k2=load k3=delay k4=reliability k5=MTU 1 Q9 ~, d% L) \. Q
usuallly only Bandwidth and Delay are considered for Bandwidth calculation (by default) . so K1 = 1 and K3 = 1 and rest all are Zero. (although you can change these metric weights (k- values))
* m, V' r& K+ m( r, D |9 Papplying that on the formula. / B7 A, y7 e2 r1 [9 T! Q2 S, _
the Metric would be % {3 O# U6 B* m; B- V
256 *( (k1 * BW) + (K3 * Delay)) - }; }4 p) Y7 m0 z. X3 O
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----------------------------------- For example:
$ I. F# Y1 K5 f1 j6 r& uk1=bandwidth 1000 k k2=load 10 k3=delay 2000 k4=reliability 255 k5=MTU 1500 # ?; T a3 H9 F5 Z3 _) {* d' }
so by default the cost shoud be : 256 *( (1 * 1000) + (1 * 2000)) == 768000
/ q4 t. p9 P$ h- W" cif change the default value for k1-k5 to all 1 (metric weights 0 1 1 1 1 1) + j5 _/ {4 h! o, f( m& ?
then the cost shoud be : Metric = 256*[(1 * 1000) + (1 * 1000)/(256-10) + (1*2000)]*(1/(255 + 1)) == 3004
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