Chapter 5
1) Consider the following network. With the indicated link costs, use Dijkstra’s
shortest-path algorithm to compute the shortest path from x to all network nodes.
Show how the algorithm works by computing a table similar to Table 4.3.
Answer:
Step
N’
D(t),p(t)
D(u),p(u)
D(v),p(v)
D(w),p(w)
D(y),p(y)
D(z),p(z)
0
x
∞
∞
3,x
6,x
6,x
8,x
1
xv
7,v
6,v
3,x
6,x
6,x
8,x
2
xvu
7,v
6,v
3,x
6,x
6,x
8,x
3
xvuw
7,v
6,v
3,x
6,x
6,x
8,x
4
xvuwy
7,v
6,v
3,x
6,x
6,x
8,x
5
xvuwyt
7,v
6,v
3,x
6,x
6,x
8,x
6
xvuwytz
7,v
6,v
3,x
6,x
6,x
8,x
2) Consider the network shown in Problem 26. Using Dijkstra’s algorithm, and
showing your work using a table similar to Table 4.3, do the following:
a. Compute the shortest path from t to all network nodes.
b. Compute the shortest path from u to all network nodes.
c. Compute the shortest path from v to all network nodes.
d. Compute the shortest path from w to all network nodes.
e. Compute the shortest path from y to all network nodes.
f. Compute the shortest path from z to all network nodes.
Answer:
a)
Step
N’
D(x), p(x)
D(u),p(u)
D(v),p(v)
D(w),p(w)
D(y),p(y)
D(z),p(z)
0
t
∞
2,t
4,t
∞
∞
1
tu
∞
2,t
4,t
2
tuv
7,v
2,t
4,t
3
tuvw
7,v
2,t
4,t
4
tuvwx
7,v
2,t
4,t
5,u
5,u
5,u
5,u
7,t
7,t
7,t
7,t
7,t
5
tuvwxy
7,v
2,t
4,t
tuvwxyz
7,v
2,t
4,t
7,t
7,t
15,x
6
5,u
5,u
N’
D(x), p(x)
D(t),p(t)
D(v),p(v)
D(w),p(w)
D(y),p(y)
D(z),p(z)
u
∞
2,u
3,u
∞
∞
ut
∞
2,u
3,u
6,v
2,u
3,u
9,t
9,t
∞
utv
3,u
3,u
3,u
utvw
6,v
2,u
3,u
6,v
2,u
3,u
utvwxy
6,v
2,u
3,u
utvwxyz
6,v
2,u
3,u
9,t
9,t
9,t
9,t
∞
utvwx
3,u
3,u
3,u
3,u
N’
D(x), p(x)
D(u),p(u)
D(t),pt)
D(w),p(w)
D(y),p(y)
D(z),p(z)
v
3,v
3,v
4,v
3,v
3,v
4,v
vxu
3,v
3,v
4,v
vxut
3,v
3,v
4,v
vxutw
3,v
3,v
4,v
vxutwy
3,v
3,v
4,v
vxutwyz
3,v
3,v
4,v
8,v
8,v
8,v
8,v
8,v
8,v
8,v
∞
vx
4,v
4,v
4,v
4,v
4,v
4,v
4,v
∞
∞
∞
15,x
15,x
b)
Step
∞
14,x
14,x
14,x
c)
Step
11,x
11,x
11,x
11,x
11,x
11,x
d)
Step
N’
D(x), p(x)
D(u),p(u)
D(v),p(v)
D(t),p(t)
D(y),p(y)
D(z),p(z)
w
6,w
3,w
4,w
∞
∞
∞
wu
6,w
3,w
4,w
∞
∞
wuv
6,w
3,w
4,w
6,w
3,w
4,w
wuvtx
6,w
3,w
4,w
wuvtxy
6,w
3,w
4,w
12,v
12,v
12,v
12,v
∞
wuvt
5,u
5,u
5,u
5,u
5,u
wuvtxyz
6,w
3,w
4,w
5,u
12,v
14,x
N’
D(x), p(x)
D(u),p(u)
D(v),p(v)
D(w),p(w)
D(t),p(t)
D(z),p(z)
y
6,y
∞
8,y
∞
12,y
yx
6,y
∞
8,y
yxt
6,y
9,t
8,y
yxtv
6,y
9,t
8,y
12,x
12,x
12,x
7,y
7,y
7,y
7,y
yxtvu
6,y
9,t
8,y
6,y
9,t
8,y
yxtvuwz
6,y
9,t
8,y
7,y
7,y
7,y
12,y
yxtvuw
12,x
12,x
12,x
N’
D(x), p(x)
D(u),p(u)
D(v),p(v)
D(w),p(w)
D(y),p(y)
D(t),p(t)
z
8,z
∞
∞
∞
12,z
∞
zx
8,z
∞
11,x
14,x
∞
zxv
8,z
14,v
11,x
14,x
zxvy
8,z
14,v
11,x
zxvyu
8,z
14,v
11,x
zxvyuw
8,z
14,v
11,x
zxvyuwt
8,z
14,v
11,x
14,x
14,x
14,x
14,x
12,z
12,z
12,z
12,z
12,z
12,z
∞
14,x
14,x
e)
Step
12,y
12,y
12,y
12,y
12,y
f)
Step
15,v
15,v
15,v
15,v
15,v
3) P. 450: P-34
Consider Figure 4.31. Suppose there is another router w, connected to router y and
z. The costs of all links are given as follows: c(x, y) = 4, c (x, z) = 50, c (y, w) = 1,
c (z, w) = 1, c (y, z) = 3. Suppose that poisoned reverse is used in the distancevector routing algorithm.
a. When the distance vector routing is stabilized, router w, y, and z inform their
distances to x to each other. What distance values do they tell each other?
b. Now suppose that the link cost between x and y increases to 60. Will there be
a count-to-infinity problem even if poisoned reverse is used? Why or why not?
If there is a count-to infinity problem, then how many iterations are needed for
the distance-vector routing to reach a stable state again? Justify your answer.
c. How do you modify c (y, z) such that there is no count-to-infinity problem at
all if c (y, x) changes from 4 to 60?
Answer:
a)
Router z
Informs w, Dz(x)=
Informs y, Dz(x)=6
Router w
Informs y, Dw(x)=
Informs z, Dw(x)=5
Router y
Informs w, Dy(x)=4
Informs z, Dy(x)=4
b)
Yes, there will be a count-to-infinity problem. The following table shows the routing
converging process. Assume that at time t0, link cost change happens. At time t1, y
updates its distance vector and informs neighbors w and z. In the following table, “”
stands for “informs”.
Time t0
t1
t2
t3
 w, Dz(x)=
 y, Dz(x)=11
Z
 w, Dz(x)=
 y, Dz(x)=6
No change
W
 y, Dw(x)=
 z, Dw(x)=5
 y, Dw(x)=
 z, Dw(x)=10
Y
 w, Dy(x)=4
 z, Dy(x)=4
 w, Dy(x)=9
t4
No change
No change
 z, Dy(x)= 
 w, Dy(x)=14
 z, Dy(x)= 
We see that w, y, z form a loop in their computation of the costs to router x. If we
continue the iterations shown in the above table, then we will see that, at t27, z
detects that its least cost to x is 50, via its direct link with x. At t29, w learns its least
cost to x is 51 via z. At t30, y updates its least cost to x to be 52 (via w). Finally, at time
t31, no updating, and the routing is stabilized.
Time t27
Z
W
Y
t28
t29
t30
t31
via w, 
via y, 55
via z, 50
 w, Dz(x)=50
 y, Dz(x)=50
 y, Dw(x)=
 z, Dw(x)=50
 w, Dy(x)=53
 z, Dy(x)= 
a) cut the link between y and z.
via w, 
via y, 
via z, 51
 y, Dw(x)=51
 z, Dw(x)= 
 w, Dy(x)= 
 z, Dy(x)= 52
via w, 52
via y, 60
via z, 53