Essentials of Computer Organization and Architecture
5th Edition
ISBN: 9781284123036
Author: Linda Null
Publisher: Jones & Bartlett Learning
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Chapter 4, Problem 29E
Explanation of Solution
MARIE assembly instruction for given code segment:
ORG 100
If, Load X / Load value of X
Subt One /Subtract 1, and store the data in accumulator
SkipCond 800 /If AC>0 (X>1)
Jump Endif /Jump the loop
Then, Load X /Reload the value of X
Add X /Add the X value by X
Store Y /Y=X+X
Clear /Move 0 into accumulator
Store /Move X to 0
Endif, Load Y /Load Y into accumulator
Add One /Add the value 1 to Y
Store Y /Y=Y+1
Halt /Stop the program
X, Dec ? /Starting value of X
Y, Dec ? /Starting value of Y
One, Dec 1 /Constant value
Explanation:
- Initially, the assembler loads the value “1” to accumulator, then the value stored to X...
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I need help understanding how could I got the IP Address for the 1st Host on this Subnet, the IP Address for the Last Host on this Subnet & the Broadcast Address for this Subnet
My Explanation:8 bits --> 17 & 0
128 64 32 16 | 8 4 2 1 0 0 0 1 0 0 0 1 -->17+ 0 0 0 0 0 0 0 0 --> 0 0 0 0 0 0 0 0 0 --> 0 (The Major (Base) Network Address): 142.20.33.0
To find the broadcast address, we can set all the bits in the host portion of the address to 1, which gives us 142.20.33.255. Based on the subnet mask 255.255.255.0, no bits were borrowed from the host portion, meaning 8 bits remain in the host portion. Using the formula 2^h – 2 (where h represents the number of host bits), we get 2^8 - 2 = 254 hosts.
8 bits --> 17 & 252
128 64 32 16 | 8 4 2 1 0 0 0 1 0 0 0 1 --> 17+ 1 1 1 1 1 1 0 0 --> 252 0 0 0 1…
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Chapter 4 Solutions
Essentials of Computer Organization and Architecture
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