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1)
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**a) Choose a number between 51 and 70. This will be number A. Choose another number between 80 and 120. This will be number B. Using 2’s complement 8-bit binary arithmetic, calculate number A – number B giving your answer in binary and decimal. (5 marks)**
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Number A - 64 - 0100 0000
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@@ -9,7 +8,7 @@ Number B - 96 - 0110 0000
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| Num A | 0100 0000 |
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| Num B | 0110 0000 |
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| Flip B | 1001 1111 |
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| Add 1 | 1010 0000 |
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| Add 1 | 1010 0000 |
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| Num A | 0100 0000 |
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| Num B | 1010 0000 |
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@@ -99,4 +98,4 @@ NumberF - 223
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| OR (Hexadecimal) | 0xFF |
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| OR (Octal) | 0377 |
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Here, I have picked two numbers, the first of which is converted to hexadecimal, and the second converted to octal using a division conversion method on the decimal value. Then I have done a bitwise operation for OR on the two binary values to get a binary value, and then converted to decimal, hexadecimal and octal - in this case it is the maximum value for each using 8 bit binary.
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Here, I have picked two numbers, the first of which is converted to hexadecimal, and the second converted to octal using a division conversion method on the decimal value. Then I have done a bitwise operation for OR on the two binary values to get a binary value, and then converted to decimal, hexadecimal and octal - in this case it is the maximum value for each using 8 bit binary.
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1) **Choose a short phrase of between 50-60 characters. You could choose a website headline, song title, etc. Convert this phrase using Caesar shift encryption and a key of 5 to create your ciphertext. Carry out a cryptanalytic attack to try to work out the decryption key and plaintext. (10 marks)**
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1. **Choose a short phrase of between 50-60 characters. You could choose a website headline, song title, etc. Convert this phrase using Caesar shift encryption and a key of 5 to create your ciphertext. Carry out a cryptanalytic attack to try to work out the decryption key and plaintext. (10 marks)**
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F1 veteran to retire and take up new role in 2024.
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@@ -26,7 +26,7 @@ This results in a key of 5, so translating the 2nd word: ajyjwfs -> veteran. Thi
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`K1 ajyjwfs yt wjynwj fsi yfpj zu sjb wtqj ns 2024.` -> `F1 veteran to retire and take up new role in 2024.`
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2) Computer A sends 5 packets of data to computer B using Sliding Windows Flow Control.
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1. Computer A sends 5 packets of data to computer B using Sliding Windows Flow Control.
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- The transmission time (time to put on the network) for a packet of data is 1 'time units'
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- Transmission time for an acknowledgement is 0 ‘time units’ (they are very small)
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- The propagation time (time to travel through network) for any transmission is random (between 3 and 5 ‘time units’, you choose a random time for each packet and acknowledgement sent).
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@@ -40,7 +40,6 @@ Propagation Time = 3-5tu
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Processing Time (B) = 2tu
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Window Size = 2
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For propagation time, I am going to use the following values:
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For propagation time, I am going to use the following values:
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( 0 = 3, 1 = 3, 2 = 4, 3 = 4, 4 = 5), and for simplicity's sake, I will mirror the propagation time there with the propagation time back for the acknowledgement.
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-
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@@ -4,4 +4,4 @@
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- Task 1.3 (3 marks): Using absolute filenames (ones that start with a /), move the grep.txt file into the “Workshop 1” directory. cd into the “PDP” directory. Using relative filenames (.. notation ones that don't start with a /), copy the cv.txt file into the “CV” directory. cd into the “CSIL” directory. Using relative filenames ones that don't start with a /), rename the “Workshop 1” directory so it is called “Tutorial 1” Run the tree command and include images of the commands and the tree/find output. Your final output should also include your command history.
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@@ -5,4 +5,4 @@
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3.3: Change all consoles years from the 20th century (19xx) to “antique”.
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3.4: With lines that contain "Hybrid" append the line "Runs better when plugged into a TV."
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@@ -1,13 +1,17 @@
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## Lecture 1 (12:00)
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IEEE-754
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- 32 and 64 bit numbers
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- Float and double in java
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- Promotes interoperability among programming languages
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- 32 bit
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- Leftmost bit is a sign bit (0=+ve 1=-ve)
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- 8 bit exponent (127 bit unsigned binary number)
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- 23 bit mantissa _without_ the 1 infront of the binary point
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- 23 bit mantissa *without* the 1 infront of the binary point
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### Scientific Notation
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13.5 = 1101.1 = 1.1011x2 ^3
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2.75 = 10110.11 = 1.011011x2 ^4
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0.125 = 0.001 = 1x2 ^-3
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@@ -90,4 +90,4 @@ Flow Control is the mechanism that ensures the sender stops transmitting before
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- Red number = sent but no acknowledge
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- Green number = sent and acknowledged
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@@ -1,2 +1,3 @@
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## Lecture 1 (12:00)
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### sed
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### Sed
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