Circuit switching vs packet switching

Question

Suppose that x bit  of message is transmitted over a k -hop path in a  network.The circuit set up time is s seconds,the propagation delay is d seconds per hop.The packet size is p bits and data rate is b bps.At what time user will receive the message  if

a.) Circuit switching is used

b:) Packet switching is used

Comments

Packet : x/p[dk+p/b]

Circuit : x/p[dk+p/b]+s?

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I dint have answer.Please tell you logic.I will verify it with my answer

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x/p is the no of packets dk is the total propogation delay and p/b is the transmission time. Addition of s is for circuit switching which is obvious.

Answer 1

Circuit switching--> total time = s+(x/b)+kd

Packet switching-- no. of packets,n =x/p

total time = (kp/b)+(n-1)p/b + kd

Comments

Please explain the total time for packet switched network.

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In packet switching when the router will forward the packet they will allow them in the router queue and then they will forward it on the same path as it is mentioned that the network is lightly loaded. 

So the expression contains.  x/b+ (k-1)p/b +kd

x/b is the transmission time by sender for the complete message.

(k-1)p/b is the transmission time for a single packed by the k hops excluding sender?(I don't understand this one).

kd is the propagation delay between k hops. 

Somebody please explain how come (k-1)p/b but not (k-1)x/b because every router/hop will transmit every packet of the message. 

 

Answer 2

Circuit Switching:-

setup time = s , transmission time = x/b , propagation time= kd

so , total time = s + x /b + kd

Packet Switching:-

no setup time , the last bit is sent at t = x /b , the last packet must be retransmitted k − 1 times by intermediate routers, with each retransmission taking p /b sec. , propagation time= kd

so , total time =  x /b + (k − 1)p /b + kd

Packet switching is faster if s > (k − 1)p /b. 

Hence,

In Circuit switching total time t = s+(x/b)+kd

In Packet switching total time t = x /b + (k − 1)p /b + kd