Datagram Packet Switching

$T=D_1+D_2+D_3+D_4,$ where $4Ds$ number of hops

$D_1=$ Time to Transmit and Deliver all packets through first hop

$D_2=$ Time to Deliver last packet across second hop

$D_3=$ Time to Deliver last packet across third hop

$D_4=$ Time to Deliver last packet across forth hop

here are $P-H=1024-16=1008$ data bits per packet. A message of $3200$ bits

equires four packets $(3200$ bits$/1008$ bits$/$packet $=3\mathrm{.}17$ packets which we round

$p$ to $4$ packets$).$

$D_1=4t+p$ where $4$ number of packets

$t-$ transmission time for one packet

$b_1=$ propagation delay for one hop

$b_1=4(\frac{P}{B})+D$

$4(\frac{1024}{9600})+0\mathrm{.}001$

$=0\mathrm{.}428$

$D_2=D_3=D_4=t+p$

$(\frac{P}{B})+D$

$=(\frac{1024}{9600})+0\mathrm{.}001=0\mathrm{.}108$

$T=0\mathrm{.}428+0\mathrm{.}108+0\mathrm{.}108+0\mathrm{.}108$

For virtual circuit delay, add call setup time to

$=0\mathrm{.}752sec$

datagram delay $)=(0\mathrm{.}952$