CMPTR
3rd Edition
ISBN: 9781337681872
Author: PINARD
Publisher: Cengage
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Implementation of an Integrated Inventory Management System at Green Fields Manufacturing” Green Fields Manufacturing is a mid-sized company specialising in eco-friendly home and garden products. In recent years, growing demand has exposed the limitations of their fragmented processes and outdated systems. Different departments manage production schedules, raw material requirements, and finished goods inventory using a patchwork of spreadsheets and older software tools. These silos create inconsistent data, errors in stock levels, delivery delays, and customer dissatisfaction. Green Fields plans to implement an Integrated Inventory Management System to centralise production, procurement, inventory, and sales data to address these challenges. This technology aims to provide real-time visibility into stock levels, automate reorder points, and generate analytical dashboards for managers at both operational and strategic levels. Ultimately, the new system will streamline workflows, reduce…
. Differentiate between continuous and discrete systems. How does their nature affect the selection of simulation techniques?
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- The following table shows the timestamp and actions by two users. Choose the best option that describes the outcome of the actions. Time JohnSara 10:14 select* from hr.employees; 10:15 Update hr.employees set salary= 100 where employee_id= 206; 10:16 Commit: Select* from hr.employees; 10:18 Commit: 10:20 Select* from hr.employees; Commit: John's query willreturn the same results all three times it is executed as they are run in the same session. John's queries run at10:16 and10:20 produce the same result, which is different from the one at 10:14 John's query run at 10:16 waits until 10:18 to produce results, waiting for the commit to happen. John's queries run at 10:14 and 10:16 produce the same result, which is different from the one at 10:20arrow_forwardwhat's the process used to obtain IP configuration using DHCP in Windows Server.arrow_forwardConsider the following sequential circuit: CLOCK a. Define the diagram circuit variables (5 pts) b. Derive the Flip-Flop input equations) (5 pts) c. Derive the circuit output equation (5 pts) d. Derive the state table of the circuit (5 pts) e. Derive the state diagram for this circuit (5 pts) Clk A D B B' CIK Question 2 (25 pts) A sequential circuit with two D flip-flops A and B, two inputs x and y, and one output z is specified by the following next-state and output equations: A(t + 1) = xy' + xB B(t + 1) = xA + xB' z = A a. Draw the logic diagram of the circuit. (5 pts) b. List the state table for the sequential circuit. (10 pts) c. Draw the corresponding state diagram. (10 pts)arrow_forward
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- The next problem concerns the following C code: /copy input string x to buf */ void foo (char *x) { char buf [8]; strcpy((char *) buf, x); } void callfoo() { } foo("ZYXWVUTSRQPONMLKJIHGFEDCBA"); Here is the corresponding machine code on a Linux/x86 machine: 0000000000400530 : 400530: 48 83 ec 18 sub $0x18,%rsp 400534: 48 89 fe mov %rdi, %rsi 400537: 48 89 e7 mov %rsp,%rdi 40053a: e8 d1 fe ff ff 40053f: 48 83 c4 18 add callq 400410 $0x18,%rsp 400543: c3 retq 0000000000400544 : 400544: 48 83 ec 08 sub $0x8,%rsp 400548: bf 00 06 40 00 mov $0x400600,%edi 40054d: e8 de ff ff ff callq 400530 400552: 48 83 c4 08 add $0x8,%rsp 400556: c3 This problem tests your understanding of the program stack. Here are some notes to help you work the problem: • strcpy(char *dst, char *src) copies the string at address src (including the terminating '\0' character) to address dst. It does not check the size of the destination buffer. You will need to know the hex values of the following characters:arrow_forwardA ROP (Return-Oriented Programming) attack can be used to execute arbitrary instructions by chaining together small pieces of code called "gadgets." Your goal is to create a stack layout for a ROP attack that calls a function located at '0x4018bd3'. Below is the assembly code for the function 'getbuf', which allocates 8 bytes of stack space for a 'char' array. This array is then passed to the 'gets' function. Additionally, you are provided with five useful gadgets and their addresses. Use these gadgets to construct the stack layout. Assembly for getbuf 1 getbuf: 2 sub $8, %rsp 3 mov %rsp, %rdi 4 call gets 56 add $8, %rsp ret #Allocate 8 bytes for buffer #Load buffer address into %rdi #Call gets with buffer #Restore the stack pointer #Return to caller. Stack Layout (fill in Gadgets each 8-byte section) Address Gadget Address Value (8 bytes) 0x4006a7 pop %rdi; ret 0x7fffffffdfc0 Ox4006a9 pop %rsi; ret 0x7fffffffdfb8 0x4006ab pop %rax; ret 0x7fffffffdfb0 0x7fffffffdfa8 Ox4006ad mov %rax,…arrow_forwardIn each of the following C code snippets, there are issues that can prevent the compilerfrom applying certain optimizations. For each snippet:• Circle the line number that contains compiler optimization blocker.• Select the best modification to improve optimization.1. Which line prevents compiler optimization? Circle one: 2 3 4 5 6Suggested solution:• Remove printf or move it outside the loop.• Remove the loop.• Replace arr[i] with a constant value.1 int sum( int ∗ ar r , int n) {2 int s = 0 ;3 for ( int i = 0 ; i < n ; i++) {4 s += a r r [ i ] ;5 p r i n t f ( ”%d\n” , s ) ;6 }7 return s ;8 }2. Which line prevents compiler optimization? Circle one: 2 3 4 5 6Suggested solution:• Move or eliminate do extra work() if it’s not necessary inside the loop.• Remove the loop (but what about scaling?).• Replace arr[i] *= factor; with arr[i] = 0; (why would that help?).1 void s c a l e ( int ∗ ar r , int n , int f a c t o r ) {2 for ( int i = 0 ; i < n ; i++) {3 a r r [ i ] ∗= f a c t o r…arrow_forward
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