Homework Solution: What is the total delay associated with sending a 2000 B packet on a 1,000km link with…

    6. What is the total delay associated with sending a 2000 B packet on a 1,000km link with a propagation speed of 2*1048 m/s and a transmission rate of 10Mb/s? (Ignore queueing and processing delays) 7. Now suppose instead of a single 1,000km link there are three 500km links connected by store-and-forward routers. Assuming the propagation speed and transmission rate are the same as before what is the total delay for a 2000 B packet? The picture looks like this (S is the source, D is the destination and R1 and R2 are the two routers.) S→R1 → R2 → D 8.Now suppose a second source S2 sends to R1 over a separate link having the same speed as the S → R1 link. The packets from S2 are also destined for D. How will the delay change relative to question 7? Why? What assumption do you have to make to ensure that no packets are lost?
    What is the total delay associated with sending a 2000 B packet on a 1,000km link with a propagation speed of 2*10^8 m/s and a transmission rate of 10Mb/s? (Ignore queueing and processing delays) Now suppose instead of a single 1,000km link there are three 500km links connected by store-and-forward routers. Assuming the propagation speed and transmission rate are the same as before what is the total delay for a 2000 B packet? The picture looks like this (S is the source, D is the destination and R1 and R2 are the two routers.) S rightarrow R1 rightarrow R2 rightarrow D Now suppose a second source S2 sends to R1 over a separate link having the same speed as the S rightarrow R1 link. The packets from S2 are also destined for D. How will the delay change relative to question 7? Why? What assumption do you have to make to ensure that no packets are lost?

    Expert Answer

     
    6) Per given question,

    6. What is the entirety retreat associated with sending a 2000 B packet on a 1,000km couple with a propagation acceleblame of 2*1048 m/s and a transmission blame of 10Mb/s? (Ignore queueing and processing retreats) 7. Now divine instead of a sole 1,000km couple there are three 500km couples alove by store-and-forward routers. Assuming the propagation acceleblame and transmission blame are the corresponding as precedently what is the entirety retreat restraint a 2000 B packet? The delineate looks love this (S is the commencement, D is the object and R1 and R2 are the couple routers.) S→R1 → R2 → D 8.Now divine a prevent commencement S2 sends to R1 aggravate a sepablame couple having the corresponding acceleblame as the S → R1 couple. The packets from S2 are to-boot destined restraint D. How achieve the retreat shift not-absolute to inquiry 7? Why? What self-confidence do you own to mould to secure that no packets are obsolete?

    What is the entirety retreat associated with sending a 2000 B packet on a 1,000km couple with a propagation acceleblame of 2*10^8 m/s and a transmission blame of 10Mb/s? (Ignore queueing and processing retreats) Now divine instead of a sole 1,000km couple there are three 500km couples alove by store-and-forward routers. Assuming the propagation acceleblame and transmission blame are the corresponding as precedently what is the entirety retreat restraint a 2000 B packet? The delineate looks love this (S is the commencement, D is the object and R1 and R2 are the couple routers.) S rightarrow R1 rightarrow R2 rightarrow D Now divine a prevent commencement S2 sends to R1 aggravate a sepablame couple having the corresponding acceleblame as the S rightarrow R1 couple. The packets from S2 are to-boot destined restraint D. How achieve the retreat shift not-absolute to inquiry 7? Why? What self-confidence do you own to mould to secure that no packets are obsolete?

    Expert Exculpation

     

    6) Per attached inquiry,

    Queuing and Processing Retreat can be ignored.Hence,

    Propagation Retreat = Distance/Speed

    = 1*107/2*108

    = 0.05 sec

    Transmission Blame = 10 Mbps

    Bandwidth Retreat = Transmission Blame * Propagation Retreat

    = 1*106 * .05 = 5*104 bits