Faculty of Computers and Information

Information Technology Department

Networks IISecond Semester

Third Year IT

Sheet No.: 1Subject: Topologies and Transmission Media

  1. A tree-topology local network is to be provided that spans two buildings. If permission can be obtained to string cables between the two buildings, then one continuous tree layout will be used. Otherwise, each building will have an independent tree topology network and a point-to-point link will connect a special communication station on one network with a communications station on the other network. What functions must the communications station perform? Repeat for ring and star.

Solution:

Part I

1)Tree:

a) We choose a device from 1st building and anther from the 2nd building(act as communication station)

b) The communication station takes a copy from the frame & sends it to the other communication device.

c) If a frame will be sent between the 2 buildings , the 1st communication station take it & send it to the other communication station.

d) To reduce the traffic between the 2 buildings, check destination Mac add. .if it referred to that this add. is in the same building , so no need for sending it to the other building.

In Case of Data Link (LAN):

+ Communication station acts as a switch. HOW??

Fe frame mn 1st building we ana 3ayz 23rf howa fen, 2shof el “source Mac add.” yro7 ll Table we y7otoh feh(“source Mac add.”) 3ala 2sas 2no fe el building el fade we y3mel forward ll el building el tanee.

Functions of Communication station:

1) Buffering (in case of sending more than one frame at the same time) .

2) Act as a Switch.

3) Forwarding the frames from the source to the destination.

NOTE: STAR THE SAME AS TREE.

2) Ring:

Hena b3d ma el source yb3t el frame ll communication station ht3melo discard 3lashan myr7sh ll source tanee we tb3ato ll communication station el Tanya fa tro7 mnzlah 3ala el ring we lama yrg3lha tane dah m3nah 2no acknowledgement 2no wesal ll destination fa tro7 ba3ta el ack. Dah ll communication station el Tanya 3ashan t2olha 2no wesal.

Functions of Communication station:

1) Forwarding the frames from the source to the destination.

2) Discard the frame that sent two times to avoid duplication.

  1. Consider the transfer of a file containing 1 million 8-bit characters from one station to another. What is the total elapsed time for the following cases?

NOTE: elapsed m3naha mn sa3t ma el frame nzal l7ad ma wesal

  1. A circuit-switched, star topology. Call setup time is negligible, and the data rate on the medium is 64 kbps.

Total elapsed time

= call setup (ignore cause of circuit switching) + transmition time

= (1*8*10^6)/(64*10^3)

  1. A bus topology with two stations a distance D apart, a data rate of B bps, and a frame size P with 80 bits of overhead. Each frame is acknowledged with an 88-bit frame before the next is sent. The propagation speed on the bus is 200 m/µs. Solve for

(1)D = 1 km,B = 1 Mbps,P = 256 bits

Solution

Frame transformation time=2 propagation time

=2* (D/v)

Tot. Time for one frame= transmition time+2 propagation time + ack. Transmition time

=size of frame/R+2D/V + ack. Size/R

# Of frames =8*10^6/ P

Tot. Frame time=(P+80)/B +2 *D/200*10^-6+88/B

Tot. time to send all frames=# Of frames* Tot. Frame time

NOTE: prop. timemultiplied by 2 in case of collision and that reduce the collision since the carrier cable if there’s a coming signal , wait until the other station finish.

(2)D = 1 km,B = 10 Mbps,P = 256 bits

(3)D = 10 km,B = 1 Mbps,P = 256 bits

(4)D = 1 km,B = 50 Mbps,P = 10,000 bits

  1. A ring topology with a total circular length of 2D, with the two stations a distance D apart. Acknowledgment is achieved by allowing a frame to circulate past the destination station, back to the source station. There are N repeaters on the ring, each of which introduces a delay of one bit time. Repeat the calculation for each of b(1) through b(4) for N = 10; 100, 1000.

Solution:

Here no ack.

Tot. Frame time=transmition time+ propagation time+ (N-1) *1/B

And complete the same as previous

NOTE:

+ N-1 : l2n fe repeater 3and kol source we ana 3nd el source el signal lsa generated 3ando y3ne 7asbt el delay m3 el transmition time

+ 1/B: is the delay time(1 bit per B)

  1. At a propagation speed of 200 m/µs. what is the effective length added to a ring by a bit delay at each repeater at;

(1)At 1 Mbps?

Delay time (1 bit) =1/10^6

Delay time By distance= 200*10^-6 *1/10^6

(2)At 40 Mbps?