Unit 4 - 4.1 - 4.36

B-Senior Unit 11, σελίδα 36 1,4-14

A-Senior Unit 13 σελίδα 36 22-29 Unit 14 σελίδα 38 1-4

B-Junior Unit 13, σελίδα 36 9,10 Unit 14, σελίδα 38 1-4

Unit 4: Vocabulary: Phrasal Verbs (1), Page: 36-37

In this video we take a look at the 0:02 fetch to code 0:03 execute cycle including its effect on 0:06 the various registers we've previously 0:12 [Music] 0:14 discussed a computer is defined Definition 0:17 as an electronic device that takes an 0:20 input 0:22 processes data 0:25 and delivers output 0:29 in this simple example you can see we're 0:31 taking the input 5 0:35 we're multiplying it by 2 that's our 0:37 process 0:39 and we're outputting 10. 0:44 but this could be way more complex for 0:46 example of a game console 0:48 the input could be the buttons you press 0:50 on a controller 0:53 the processes would then be carried out 0:55 by the console itself 0:59 and the output would be some form of 1:01 update to a monitor 1:02 and sound out for a speaker possibly 1:04 vibration feedback through the 1:06 controller 1:10 to process data a computer follows a set 1:13 of instructions 1:14 known as a computer program 1:18 if we take the lid off a typical desktop 1:20 computer we can identify 1:22 two critical components the memory 1:26 that stores the program and the central 1:29 processing unit or processor 1:31 which is under this large fan and 1:33 carries out the instructions 1:37 a computer carries out its function by 1:40 fetching 1:41 instructions decoding them and then 1:43 executing them 1:44 in a continuous repetitive cycle 1:46 billions of times a second 1:48 let's look at each of these stages in a 1:50 little more detail Fetch 1:53 so let's start with the fetch stage the 1:55 very first thing that happens 1:57 is the program counter is checked as it 2:00 holds the address 2:01 of the next instruction to be executed 2:07 the address stored is then copied into 2:09 the memory address register 2:14 the address is then sent along the 2:16 address bus to main memory 2:18 where it waits to receive a signal from 2:21 the control 2:22 bus so it knows what to do 2:27 as we want to read the data that's 2:29 stored in memory address 2:30 0 0 0 0 the control unit sends 2:34 a read signal along the control bus to 2:36 main memory 2:41 now main memory knows the data needs to 2:44 be read 2:45 the content stored in memory address 000 2:49 can be sent along the data bus to the 2:51 memory data register 2:56 now as we're currently in the process of 2:58 fetching an instruction 3:00 the data received by the memory data 3:03 register gets copied 3:04 into the current instruction register 3:11 the instruction effectively has now been 3:14 fetched from memory 3:16 just before we proceed to the decode 3:18 phase we now 3:19 increment the program counter so that 3:22 the address it contains 3:24 points to the address of the next 3:26 instruction which will need to be 3:30 executed 3:32 the instruction now being held in the 3:33 current instruction register 3:35 is ready to be decoded 3:39 now as we mentioned in the previous 3:41 video the instruction is made up of two 3:43 parts 3:44 we have the op code that's what it is we 3:47 need to do 3:50 and we have the operand what are we 3:53 going to do it to 3:55 now the operand could contain the actual 3:57 data 3:58 or indeed it could contain an address of 4:01 where the data is to be found 4:06 by decoding this instruction we can see 4:08 the operation we need 4:10 is a load operation so we need to load 4:14 the contents of memory location0101 4:18 into the cpus accumulator 4:25 in the exam a simple model will be used 4:27 to describe the 4:29 structure of any given instruction 4:32 you're not going to be expected to 4:34 define how an opcode is made up 4:36 but simply to interpret opcodes in the 4:39 given context of an exam 4:40 question in the example here 4:44 you can see there's a total of 16 4:46 different opcodes available 4:48 and this is because we're using four 4:50 bits for our representation 4:56 so now we've fetched the instruction and 4:59 we've decoded it so we know what we need 5:00 to do 5:01 we're finally ready to execute it 5:05 so we now send address 0101 5:08 to the memory dress register 5:13 now we're in the memory address register 5:15 we can finally send the address 5:18 down the address bus to main memory 5:24 this time we want to read the data 5:26 that's stored in memory 5:28 and so the control unit again sends a 5:30 read signal along the control bus 5:36 so main memories now receive an address 5:38 and a read signal 5:40 so the content stored at memory location 5:43 0101 5:44 can now be sent along the data bus back 5:46 to the cpu 5:47 and into the memory data register 5:54 finally the contents of the memory data 5:56 register are copied to the accumulator 5:59 and this is one of a number of general 6:00 purpose registers found in the cpu 6:04 this first instruction is now complete Branching 6:11 so what does this program actually do 6:14 you should be able to work it through 6:16 carefully and figure it out 6:19 we're now pointing instructions zero 6:21 zero zero one in the program counter 6:23 and we're ready to fetch the second 6:25 instruction 6:27 at the end of this video we're gonna 6:29 provide you with the answer 6:34 so let's talk a second about programs 6:37 that branch 6:40 on the left here we have a very simple 6:42 piece of pseudo code 6:44 line zero says first execute this line 6:46 of code 6:47 line 1 now execute this line and then 6:50 line 2 says 6:52 if the age is greater than 18 then 6:56 we're going to execute lines 3 and 4 6:58 otherwise 6:59 we're going to execute lines six and 7:02 seven 7:03 so this program doesn't necessarily 7:05 follow strictly in sequence from line 7:07 zero through to seven there's a chance 7:10 here the program may branch and jump 7:14 around 7:16 so we're going to pretend that this 7:17 program has been loaded into memory 7:20 each line of code on the left here has 7:23 ended up 7:24 as a location in memory now this is not 7:27 strictly how this would happen in this 7:28 one-to-one way 7:29 but for the purpose of example it's 7:31 absolutely fine 7:35 so the program counter starts by 7:37 pointing to memory address zero 7:39 and we fetch the first instruction 7:41 decode it and execute it 7:44 it then updates and tells us the next 7:47 instruction 7:48 is zero zero zero one because remember 7:50 the program counter is being incremented 7:52 so we fetch it decode it and we execute 7:55 line one of our program 7:59 we then fetch line two which in binary 8:01 is one 8:02 zero 8:06 now at this point depending on what 8:10 happens during the execution 8:11 of line two the program may be required 8:15 to fetch line three from memory or 8:18 line five from memory 8:25 so let's look at how this actually works 8:27 because we've said the program counter 8:28 simply gets incremented 8:31 well in the current instruction register 8:33 we have an instruction with the op code 8:36 0 1 1 0. 8:41 now when we look this up in the decode 8:43 unit we discover that this 8:45 code means branch always 8:51 this replaces the value held in the 8:54 program counter 8:56 with the contents of the operand that's 8:58 the second part of the instruction 9:01 from the current instruction register so 9:03 this case 9:04 one zero zero one 9:09 now when the next fetch cycle begins the 9:12 program counter is obviously checked 9:14 and as its contents have been previously 9:16 updated to a new memory location 9:19 and not simply incremented the program 9:22 effectively is able to jump 9:24 around memory 9:28 so having watched this video you should 9:30 be able to answer the following key 9:32 question 9:33 how does a cpu work 9:39 okay so let's um answer the question we 9:41 posed 9:42 earlier what did that program actually 9:48 do 9:50 so this is the first fetch to code 9:53 execute cycle 9:55 and this is the one that we ran through 9:57 in detail earlier 9:58 it effectively loaded the contents of 10:01 the memory 10:02 stored at location location0101 10:05 into the accumulator in other words 10:08 the dna number 3 is moved 10:11 from memory into the cpu 10:18 we then proceed onto the second fetch 10:20 decode execute cycle 10:23 now this one adds the contents of memory 10:27 located at 0 1 1 0 10:30 to the current contents of the 10:32 accumulator 10:34 so in other words the dna number one 10:38 because that's what's stored at address 10:40 zero one one zero 10:43 is added to the number three that was in 10:45 the accumulator 10:46 the results are stored back over the 10:48 accumulator 10:49 so effectively we've done three plus one 10:53 equals four 10:58 the third fetch to code execute cycle 11:00 stores the contents which are in the 11:02 accumulator 11:03 into memory location zero one one one 11:07 and that's because the op code the first 11:09 part of this current instruction 11:10 zero zero one one is the command to 11:13 store when we look it up in the decoder 11:15 unit 11:16 so in other words the result of the 11:17 previous calculation three plus one 11:19 equals four 11:20 is now written back into main memory 11:28 the fourth fetch decode execute cycle 11:30 outputs the contents of the accumulator 11:33 remember they were copied into main 11:34 memory but they're still held in the 11:35 accumulator 11:37 so in this simple abstraction the number 11:40 four is now 11:41 output to the user so they can see the 11:43 result of the calculation 11:49 the fifth and final fetch code execute 11:51 cycle 11:52 brings a halt to the current program 11:58 so this very simple program which has 12:01 five 12:02 fetch decode execute cycles has 12:04 performed the calculation 12:06 three plus one is then stored the result 12:09 in main memory 12:10 and displayed the result four to the 12:12 user 12:13 and in a high-level language this may 12:15 look something very similar to the 12:17 following two lines of code 12:20 sum variable equals num1 plus num2 12:24 print sum to the user 12:27 so you can start to get an appreciation 12:29 here of how the high level code you 12:32 write actually ends up being fetched 12:34 decoded 12:35 and executed inside a processor 12:38 of course your processor is doing 12:40 billions and billions of these 12:42 operations a second 12:43 which when you think about it is really 12:45 very impressive 12:52 [Music] 13:03 you. make 10 questions for a standerd of a level

Here’Transformation,Ratio,Proportion, Fractions and Algebraic Expressions,Transformation 1. Translation 2. Reflection 3. Rotation 4. Enlargement 5. Transformation 6. Congruence 7. Similarity 8. Scale Factor 9. Image 10. Pre-image 11. Symmetry 12. Isometry 13. Ratio 14. Proportion 15. Equivalent Ratios 16. Simplify 17. Unit Ratio 18. Scale 19. Part-to-Part 20. Part-to-Whole 21. Rate 22. Comparison 23. Proportional Relationship 24. Cross Multiplication 25. Direct Proportion 26. Inverse Proportion 27. Constant of Proportionality 28. Golden Ratio 29. Linear Relationship 30. Equal Proportions 31. Proportional Constant 32. Scale Drawing 33. Word Problems 34. Unitary Method 35. Percentage 36. Double Number Line 37. Fraction 38. Numerator 39. Denominator 40. Improper Fraction 41. Proper Fraction 42. Mixed Number 43. Simplified Fraction 44. Reciprocal 45. Least Common Denominator (LCD) 46. Greatest Common Factor (GCF) 47. Equivalent Fractions 48. Decimal 49. Variable 50. Coefficient 51. Constant 52. Algebraic Term 53. Polynomial 54. Monomial 55. Binomial 56. Expression 57. Equation 58. Like Terms 59. Simplify 60. Substitution -

THE BATTLE OF THE PHILIPPINE SEA Occured in the Philippine Sea and Marinas The battle engaged the bulk of thepagan American forces, and prevented the Japanese from reinforcing, their fleet in the •Marianas. •A month after, the Japanese LOST THEIR 4 best aircraft carriers in the Battle of Midway, an island northwest of Pearl "Harbor. 1. The Batle of the Philipine Sea lasted just one day June 19- June 20 1944 3. is also called the "Marianas Turkey Shoot” The Battle of the Coral Sea The Coral Sea is Between New Guinea and Australia The Japanese Were Trying to Attack Australia! The U.S. Navy and the Japanese Navy Fought! Nobody Won! U.S. Was Able to STOP the Attack on Australia! (Victory!) The American fleet defeated the Japanese. American planes bombarded Japanese installation in Manila from the air. Air strikes were also carried out in the Visayas. Battle of Okinawa: Battle Details The attack on Okinawa took heavy toll on both sides of the fighting... The Americans lost 7,373 men killed and 32,056 wounded on land. At sea, the Americans lost 5,000 killed and 4,600 wounded. The Japanese lost 107,000 killed and 7,400 men taken prisoner. The Japanese may have lost another 20,000 dead as a result of American tactics whereby Japanese troops were incinerated where they fought. The Americans also lost 36 ships. 368 ships were also damaged. 763 aircraft were destroyed. The Japanese lost 16 ships sunk and over 4,000 aircraft were lost. battle facts -the japanese launched fierce kamikaze attacks l-arge amouunts of civillian deaths -japanese soldiers using civilians as human shields Americans ended with more triumphs in the battles like in: General MacArthur and the Allies next turned to the Island of Iwo Jima The island was critical to the Allies as a base for an attack on Japan It was called the most heavily defended spot on earth Allied and Japanese forces suffered heavy casualties IWO JIMA American soldiers plant the flag ol the Island of Iwo Jima after their victory Battle of Leyte Gulf "Second Battle of the Philippine Sea". Time: October 23 - 26, 1944 during WW. Location: Leyte Gulf in Philippines (East coast), Philippine islands of Leyte, Samar, and Luzon. Largest naval battle in WWII. Leyte was secured and was liberated from the hands of the Japanese Americans decided to launch their attack in Leyte since the weak side of the Japanese fleet was in Visayas. Heavy bombing at the Leyte beaches cleared the way for the landing in Palo, Leyte. Leading the American troops were General Douglas MacArthur and President Osmeña, who took over after the death of President Quezon in Saranak Lake in New York. the battle for the liberation of manila The commonwealth government capital was transferred from tacloban to Manila. Manila once again became the seat of the national leadership. ON july 4, 1945, general macarthur announced the total liberation of the Philippines • The Commonwealth government capital was transferred from Tacloban to Manila. Manila once again became the seat of the national leadership. On July 4, 1945, General MacArthur announced the total liberation of the Philippines. Americans surprised the Japanese with the landing of troops in Lingayen Gulf in Pangasinan. • The Filipino guerillas had already cleared the area and neutralized many of the Japanese forces. The first target was the UST, which was used by the Japanese as a camp for civilian prisoners of war, and they were able to free them. • More than 1000 POWs from Bataan and Corregidor were also freed from the Bilibid Prisons. • The battle of Manila was recorded as the fiercest urban fighting in the entire Pacific War. WATERLOO DAILY COURIER-NEWSPAPER “PEACE! WAR ENDS; JAPANESE ACCEPT ALLIED TERMS. ON EMPEROR" On August 6, due to persistent refusal of Japan to yield, another atomic bomb was dropped in the shipbuilding city of Nagasaki. On August 15, V-J Day (Victory in Japan), Emperor Hirohito finally admitted defeat and on September 3, 1945 the document of surrender was signed on board of the U.S. battleship Missouri in Tokyo Bay. With the liberation of the Philippines, the Americar troops moved on to finally end the war in Asia. The Japanese cities of Tokyo, Osaka, Nagoya and Kure were bombed. In July 26, the allies demanded surrender but the Japanese continued to hold on to their belief that surrender is a dishonor. Atomic bomb was dropped on the populous city of Hiroshima, 60% of the city was destroyed. It was estimated that there were more than 80 000 people died on the spot and more than 37 000 suffered severe injuries. Today, the HIROSHIMA SHRINE serves as a reminder of th horrors of war and the need to preserve PEACE in the world. THE POSTWAR PHILIPPINE COMMONWEALTH-• Philippine Commonwealth resumed. Osmeña was confronted with a war - ravaged country with no financial resources for PROBLEMS • Poverty • Destruction of Properties • Unemployment • Price Increase • Hoarding • Graft and Corruption HINDRANCES -Rehabilitation of INDUSTRIES COULD NOT BE DONE BECAUSE OF LACK OF FUND -RAILWAYS WERE DESTROYED THAT LED TO SLOW PRODUCTION AND TRANSPORTATION SOLUTION -PCAU (Philippine Civil Affairs Unit) was established by MacArthur to provide emergency relief in areas liberated by the Americans. - It organized food distribution centers. CHALLENGES TO INDEPENDENCE • On April 30, 1946, the Philippine the US President. BIASED AGREEMENTS: Rehabilitation or the Tydings Act of 1946, passed by the US Congress, was approved by Commission • This Act created the US Philippine War Damage The Act also provided for the transfer of $100,000,000 surplus property of the United States to the Philippines. The Philippine Armed Forces received large quantities of valuable military equipment and supplies. BIASED AGREEMENTS: • The United States Congress offered $800 million for post World War Il rebuilding funds if the Bell Trade Act was ratified by the Philippine Congress Parity rights granting U.S. citizens and corporations rights to Philippine natural resources equal to (in parity with) those of Philippine citizens The Philippines used to celebrate its Independence Day on July 4, and not June 12, by virtue of the Truman Proclamation in 1946. In the early 1960s, however, the Philippine Historical Association lobbied to bring back June 12 as our Independence Day. In 1962, President Diosdado Macapagal issued a proclamation to make the change official. DECLARATION OF PHILIPPINE INDEPENDENCE • On July 4, 1946, the Americans granted independence to the Philippines.

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