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Mohamed G. Gouda						CS 356
Spring 2019							Midterm 2
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(1)
Consider a simplified TCP protocol between two sides src and dst. The 
seq# of the first data segment (from src to dst) is 50. Side src sends
two data segments to side dst: the first segment has 100 data bytes and 
the second has 150 bytes. These two segments are neither lost nor
reordered. Side dst replies by sending two ack segments to src: the
first segment is lost but the second segment is not lost. What are the
seq#'s of the two data segments, and what are the ack#'s of the two
ack segments.
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(2)
Given a block of IP addresses whose subnetwork mask is M. Assume that 
this block is partitioned into 2^B smaller blocks where each smaller 
block has the same number K of IP addresses. Compute K as a function
of M and B. Explain your answer.
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(3)
Explain how DHCP can be viewed as an application layer protocol.
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Solutions:
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(1)
	src --> dst: (Data, 50, 100)
	src <-- dst: (Ack, 150)	This segment is lost
	src --> dst: (Data, 150, 150)
	src <-- dst: (Ack, 300)
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(2)
	K = (number of IP addresses in given block)/
	    (number of smaller blocks)
	  
	  = (2^(32-M))/(2^B)

	  = 2^(32-M-B)
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(3)
Both the DHCP-client and the DHCP-server run on top of UDP like other
application layer protocols.
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Mohamed G. Gouda						CS 356
Spring 2019							Midterm 2A
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(1)
Consider a simplified TCP protocol between two sides src and dst. The 
seq# of the first data segment (from src to dst) is 50. Side src sends
two data segments to side dst: the first segment has 100 data bytes and 
the second has 150 bytes. The first data segment is lost but the second
data segment is not lost. After receiving the second data segment, side 
dst sends the first ack segment. Later, side src times-out and resends
the first data segment which is not lost. Finally, side dst sends the 
second ack segment. What are the seq#'s of the two data segments, and 
what are the ack#'s of the two ack segments.
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(2)
Given a block of IP addresses whose subnetwork mask is M. Assume that 
this block is partitioned into B smaller blocks where each smaller 
block has the same number 2^K of IP addresses. Compute B as a function
of M and K. Explain your answer.
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(3)
Explain how DHCP can be viewed as a network layer protocol.
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Solutions:
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(1)
	src --> dst: (Data, 50, 100) This segment is lost
	src --> dst: (Data, 150, 150)
	src <-- dst: (Ack, 50)	
	src --> dst: (Data, 50, 100)
	src <-- dst: (Ack, 300)
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(2)
	B = (number of IP addresses in given block)/
	    (number of IP addresses in each smaller block)
	  
	  = (2^(32-M))/(2^K)

	  = 2^(32-M-K)
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(3)
The objective of DHCP is to allocate temporary IP addresses to 
DHCP-clients. Thus, DHCP deals with IP addresses like other network
layer protocols.
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