What is the time taken by the server to send the file to all clients

Question

Suppose 1000 clients are trying to download a 10MB file from a server that has 100Mbps access link. The clients have 2Mbps downstream access rate. What is the time taken by the server to send the file to all clients.(ignore TCP connection time and assume ideal conditions exist.)( Marks: -0.33 )

1.   5 sec
2.   40 sec
3.   800 sec
4.   0.8 sec

--------------------------------------------------------------------------------------------------------------

Total number of bits sent by server = 10M * 8 * 1000=80 gigabits
So transmission time = 80 G/(100-2)Mbps

or

if we send to all the links at the same time then transmission time will be 10MB/Bnadwidth.

bandwdith taken will be 2 Mbps or (100-2)Mbps??

Comments

40 sec??

---

i also thought that it is 40 sec..as since 100Mbps will be divided among 1000 client which gives 0.1*10  Mbps.

download rate is 2Mbps.

hence effective rate is 2-0.1=1.9 Mbps

so,time = 10*8*10⁶/1.9 * 10⁶ =4o sec

have u solved like this only??

but answer given is 800 sec and explanantion is total data transmittes is 10MB*1000*8 bits/100Mbps

Answer 1

Total capacity of link is 100Mbps.

Total no of clients=1000.

If we divide entire bandwidth to each client then each one is having capacity=100Mbps/1000=0.1Mbps.

Server downstream is 2Mbps and client capacity is only 0.1Mbps.

If server send at the rate of 2Mbps packets are lost.So send according to minimum rate then only all packets are safely reach to client.

SO finally bandwidth from server to client is 0.1Mbps.

File size we want to download is 10MB.

Total time=10MB/0.1Mbps

               =$\frac{10*10^{6}*8}{0.1*10^{6}}$

              =800sec.

Comments

simple thing find transmission time.

T.t=L/R.

L=10MB and R=0.1Mbps.

each client download 10MB file.

If only one client use entire link capacity i.e)100Mbps.

---

why R = 0.1 here then..because of the same reason you mentioned above..right?

and if client download rate was suppose less then 0.1 Mbps then we would have considered that..we should tak min of two..right??

thanks a lot..:-)

---

minimum of 2 banwidth's.

it is like same congestion policy

for instance  sender window=4KB and cwnd=2KB.

then sender send min(swnd,cwnd).same concept here also.