timeout timeout Sender Consider the figure below (simplex communication scenario). 23 Seq1, 10 bytes of data Seq2, 10 bytes Seq3, 10 bytes ACK1 ACK2 XA ACK4 Seq4 ACK3 The initial SEQ Number (Seq1) is 567. 1. 123 2. 3. What is Seq2 and Seq3 What is ACK1, ACK2, Ack3 What is Seq4 and ACK4 Receiver 3
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- Consider a "little-endian" byte-addressable computer which has the a 0x01 at address 0, 0x23 at address 1, 0x45 at address 2, and a 0x67 at address 3. If I tell the computer to print the integer at address O in hex format, what will the computer print out? 67452301 O 01234567 O 23016745 O 45670123 O 76543210 O 45670123Part 1 A small computer on a smart card has four page frames. At the first clock tick, the R bits are 0111 (page 0 is 0, the rest are 1). At subsequent clock ticks, the values are 1011, 1010, 1101, 0010, 1010, 1100, and 0001. If the aging algorithm is used with an 8-bit counter, give the values of the four page counters after the last tick. Part 2 A computer has four page frames. The time of loading, time of last access, and the R and M bits for each page are as shown below (the times are in clock ticks): Page Loaded Last Referenced R M 0 176 330 1 0 1 280 315 0 1 2 190 320 0 0 3 160 335 1 1 (a) Which page will NRU replace? (b) Which page will FIFO replace? (c) Which page will LRU replace? (d) Which page will second chance replace?The Hi-Fi modem company has designed a new frequency modulation that can transmit 38400 bits/sec. The baud rate is 4800. How many frequency levels can each interval have? 3. 4. Given a location with 7 bits what is the minimum and maximum value that it can hold A) As an unsigned location B) As a signed location
- Next time you snap a picture, you realize you are capturing millions of pixels into a buffer. The buffer data is read and converted into JPEG in real time. Each pixel in that buffer is an unsigned int (four bytes ): Alpha, Blue, Green, and Red. Let us ignore alpha for now. As you know, a unsigned byte can have a value 0 to 255. In remote sensing jargons, it is called blue channel, green channel, and red channel. Each channel provides valuable information such as, say farm lands, forest fire, drought, landscape, diseases , If a pixel has a value (say in hex) = 0x00a1b1c1 , then 0xc1 is the red pixel, 0xb1 is the green pixel, 0xa1 is the blue and 00 is the alpha. #define RED 1 #define GREEN 2 #define BLUE 3 then, develop a function void calculateSum ( unsigned int *ptr , int count , unsigned char channel , unsigned int *sum, float *average ) { *sum = 0; if (channel == RED ) calculate sum and average for red channel else if (channel == GREEN )…Used C PLUS PLUS LANGUAGE. all instructions are given in question. Kindly follow it. When a message is transmitted in secret code over a transmission channel, it is usually sent as a sequence of bits, that is, 0s and 1s. Due to noise in the transmission channel, the transmitted message may become corrupted. That is, the message received at the destination is not the same as the message transmitted; some of the bits may have been changed. There are several techniques to check the validity of the transmitted message at the destination. One technique is to transmit the same message twice. At the destination, both copies of the message are compared bit by bit. If the corresponding bits are the same, the message received is error-free. Response Required: Write a program to check whether the message received at the destination is error-free. For simplicity, assume that the secret code representing the message is a sequence of digits (0 to 9) and the maximum length of the message is 250…Assume a fixed priority scheme where Processor3> Processor2 > Processor1. Processor3 makes a request and is granted access. Then, Processor2 and Processor1 make requests. Which processor is granted access first? O Processor 1. Processor2 O Processor3 O An error will occur because two requests are pending
- This is what my professor said: Items are 4 bytes, pages are 2k, so ~512 items fit on a page. Even if we start in the beginning of a page, the first access will be a miss followed by ~200 hits, then a TLB miss followed by ~500 hits, so the hit rate is ~100%5. Page size 512 means 9 bits of offset and 7 bits of VPN. The offset is just the decimal version of 1 0111 0111 [unsigned] which is 375The offset is 0011 100 which is 28. Is he correct?Representation in the Hamming coding system (7.3). It is expressed as 3 control (test) bits and 7 total bits (data+control blts). In the literature, control ends are shown as (Cl, C2,C3,) or (p! p2, p3,). The bit string sent by Hamming coding (15,4) is encoded on the receiving side, eito It goes down as 0011001011100 (plp2..,al). According to this series, re Which option is given according to the control bits (plp2p3.p4) sequence calculated on the receiving side? 1100 A B oon C 011 D 1110 0001a. Given the logical address 0x 69656, what will be the logical page number issued by a process P? b. What is the corresponding physical frame number? c. What is the hexadecimal physical address? Note: Your responses should be given in hexadecimal format!
- For a computer system using a 64 bit width data path, how many number of a bits will this computer process if it will undergo a combination of 14 Tydbits, 26 Playtes, 13 Bytes and 3 Words,Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. How many bits are used for the page number? How many bits are used for the offset? 8 bits each. With this system, what’s the maximum number of pages that a process can have? 256 Suppose that each entry in the page table comprises 4 bytes (including the frame number, the valid bit, and miscellaneous “bookkeeping bits”). An OS uses an array to store the page table. What is the size of the page table? 1024 Bytes Furthermore, suppose the first 6 pages of a process map to frames 222 to 227 (as decimal numbers), and the last 5 pages of the process map to frames 1 to 5 (also decimal numbers). All other pages are invalid. Draw the page table, including the valid bit and the frame number. DONE Translate the following virtual addresses to physical addresses, and show how you obtain the answers. (Hint: You do not need to convert…Please solve it correctly and please provide explanation of your answers. Please answer parts g, h and i. A PC and a Web Server are communicating over a TCP connection. The PC had started the three way handshake with the initial sequence number of 3069 . The Web Server's initial sequence number is 4830 . The window size of the PC is 815 bytes and the window size of the Web Server is 463 bytes. Using the third TCP handshake ack segment the PC sends the http request of the size 396 bytes to the Web Server. Then the Web Server answers with 3 segments containing the requested data. The first segment size is 42 bytes and the second segment size is 276 bytes and the third segment size is 146 bytes respectively. The PC receives all three segments within the timer. But unfortunately the second segment was corrupted. So the PC immediately sends an acknowledgement segment. Assume that the PC uses Selective Repeat/Reject ARQ. g) The second segment from webserver was corrupted. So the PC…