习 题 解 析

助教：谢 萍

2007.10.24

2.1 Let A ＝ A(1) ， A(2) ，…， (1000) and

B ： B(1) ， B(2) ，…， B(1000) be two vectors(one dimensional arrays) comprising 1000 numbers each that are to be added to form an array C such that C(I) ＝ A(I)+ B(I) for I ＝ 1 ， 2 ，…， 1000 using IAS instruction set, write a program for this problem.

2.1 Reference answer

Memory address contents 1 A(1) 2 A(2) … … 1000 A(1000) 1001 B(1) … … 2000 B(1000) 2001 C(1) … … 3000 C(1000)

Memory address contents

3001 999 3002 1 … … 3100 JUMP M(3100,20:39) LOAD M(1) 3101 JUMP M(3101,20:39) ADD M(1001) 3002 JUMP M(3102,20:39) STOR M(2001) 3103 LOAD M(3100) ADD M(3002) 3104 STOR M(3100) LOAD M(3101) 3105 ADD M (3002) STOR M(3101)

Memory address contents

3106 LOAD M(3102) ADD M(3002) 3107 STOR M(3102) LOAD

M(3001) 3108 SUB M(3002) STOR

M(3001) 3109 JUMP M(3100,0:19)

3.2Consider a hypothetical 32-bit microprocessor having 32-bitinstructions composed of two fields: The first byte contains theopcode and the remainder the immediate operand or an operandaddress.

a. what is the maximum directly addressable memory capacity (in bytes)? Reference answer:

The maximum directly addressable memory capacity is 224 bytes.

opcode Operand (address)

0 7 31

3.2

b. Discuss the impact on the system speed if the microprocessor bus has

1. a 32-bit local address bus and a 16-bit local data bus, or

2. a 16-bit local address bus and a 16-bit local data bus.

Reference answer:1. If system address is 32-bit, an operand address is only 24-bit.

The other mechanism is required to provide 32-bit address, so more time takes to achieve the corresponding address.

2. If system address is 16-bit , the instruction can provide the address directly.

3.2

C. How many bits are needed for the program counter and the instruction register? Reference answer:

The length of PC is usually same to the length of word, so 32 bits are needed for the program counter

The length of IR is usually same to the length of instruction, so 32 bits are needed for the instruction register.

Ex. 3.2 指令中地址字段 24 比特，故可直接寻址 224Byte （ 1 ）系统 32 位地址，而指令中只提供 24 位地

址，因此必须设立至少 8 比特的页面寄存器，或应当像 8086 系统一样设立段寄存器才能一次提供 32 位地址；否则就要输出 2 次才能传输完地址

（ 2 ）系统只 16 位地址，而指令可提供 24 位地址，因此一次就可提供。 [ 此问题答案可以灵活 ]

程序计数器 PC 宽度一般与字长相同，这里为 32比特；指令寄存器宽度应当与指令长度相同，这里为 32 比特。

4.4

Suppose an 8-bit data word stored in memory is 11000010. Using hamming algorithm, determine what check bits would be stored in memory with the data word. Show how you got your answer.

C1=M1⊕M2⊕M4⊕M5⊕M7=0⊕1⊕0⊕0⊕1=0 C2=M1⊕M3⊕M4⊕M6⊕M7=0⊕0⊕0⊕0⊕1=1 C4=M2⊕M3⊕M4⊕M8 =1⊕0⊕0⊕1 =0 C8=M5⊕M6⊕M7⊕M8 =0⊕0⊕1⊕1 =0

Answer:

Result:

C8C8

C4C4

C2C2

C1C1

4.6 How many check bits are needed if the Hamming

error correction code is used to detect single bit errors in a 1024-bit data word? Answer:

According to the formula

kk 102412

∴ 11k

KMk 12

4.7 Develop an SEC code for a 16-bits

data word. Generate the code for the data word 0101000000111001.

Show that the code will correctly identify an error in data bit 4.

4.7 Answer:

(M4)(M4)

4.8 A set associative cache consists of

64 lines divided into four-line sets. Main memory contains 4K blocks of 128 words each. Show the format of main memory addresses.

4.8 Answer:

27=128,

64/4=16=24,4k*128=212*27=219 19-7-4=8

Answer: The total memory space 4k * 128(212 * 27=219)

bytes, 19 bits needed 64/4=16(24) sets, 4 bits needed. 128words (27) each block , 7 bits needed. 19-4-7=8 bits tag field used

TAG (8 bits) SET (4 bits) WORD (7 bits)

4.20 Consider a memory system with the

following parameters:Tc=100ns Cc=0.01cents/bitTm=1,200ns Cm=0.001cents/bit

a. What is the cost of 1Mbyte of main memory?

b. What is the cost of 1Mbyte of main memory using cache memory technology?

c. If the effective access time is 10%greater than the cache access time, what is the hit ratio H?

4.20 Reference answer

a. The cost of 1M byte of main memory is:1M byte * 8 bits /byte * 0.001 cents/bit = 8388.608 cents

b. The cost of 1M byte of cache memory is: 1M byte * 8 bits/byte * 0.01 cents/bit = 83886.08 cents

c. HTc+(1-H)(Tm+Tc)=(1+10%) Tc

∴ H=99.17% Note:

1M * 8 * 0.001 = 223 * 0.001 = 220 (0.001=10-3)

5.4

What is the transfer rate for a nine-track magnetic tape unit whose tape speed is 120 inches per second and whose tape density is 1,600 linear bits per inch?

5.4 Conditions available:

nine-tracked the tape speed 120 inches/second the tape density 1600 linear bits/inch

What is the transfer rate? Answer:

120 inches/s * 1600 linear bits/inch * 9

=1728000 bits/s

5.5 Assume a 2400-foot tape reel; nine-tracked; an Assume a 2400-foot tape reel; nine-tracked; an

interrecord gap of 0.6 inch where the tape stopinterrecord gap of 0.6 inch where the tape stops midway, between reads; that the rate of tape s midway, between reads; that the rate of tape speed increase/decrease during gaps is linear; tspeed increase/decrease during gaps is linear; that the tape speed is 120 inches per second; anhat the tape speed is 120 inches per second; and that the tape density is 1,600 linear bits per ind that the tape density is 1,600 linear bits per inch. Data on the tape are organized in physical rch. Data on the tape are organized in physical records, where each physical record contains a fecords, where each physical record contains a fix number of user-defined units, called logical rix number of user-defined units, called logical records. ecords.

5.5

conditions available: the tape is 2400-foot long

• 1 food=12 inch have nine-tracked length of an interrecord gap is 0.6 inch the tape speed is 120 inches/second the tape density is 1600 linear bits/inch. data on the tape are organized in physical records,

which are formed by N logical records.

5.5 a. How long will it take to read a full tape of 120-byte a. How long will it take to read a full tape of 120-byte

logical records blocked 10 per physical record?logical records blocked 10 per physical record? Answer:Answer: 1food=12inches1food=12inches

The tape speed during gap is 60 inch /second The tape speed during gap is 60 inch /second The length of a physical record excluding its The length of a physical record excluding its

following gap is:following gap is: 120 * 10 * 8 ÷1600 = 6 inch120 * 10 * 8 ÷1600 = 6 inch The physical records and gaps individually are The physical records and gaps individually are 2400 * 12/(62400 * 12/(6 ＋＋ 0.6)≈43630.6)≈4363 So the total time to read the full tape is :So the total time to read the full tape is : 4363 * (6÷1204363 * (6÷120 ＋＋ 0.6÷60)≈261.78 seconds0.6÷60)≈261.78 seconds

5.5 b. How long will it take to read a full tape of 120-byte b. How long will it take to read a full tape of 120-byte

logical records blocked 30 per physical record?logical records blocked 30 per physical record?

Answer:Answer: The length of a physical record excluding its following The length of a physical record excluding its following

gap gap is:is:

120 * 30 * 8 ÷1600 = 18 inch120 * 30 * 8 ÷1600 = 18 inch

The physical records and gaps individually are:The physical records and gaps individually are:

2400 * 12/(182400 * 12/(18 ＋＋ 0.6)≈15480.6)≈1548

So the total time to read the full tape is :So the total time to read the full tape is :

1548 * (18÷1201548 * (18÷120 ＋＋ 0.6÷60)≈247.68 seconds0.6÷60)≈247.68 seconds

5.5 c. How many logical records will the tape hold c. How many logical records will the tape hold

with each of the preceding blocking factors?with each of the preceding blocking factors? Answer:Answer: (1) (2400*12)/(6 +0.6)*10=4363(1) (2400*12)/(6 +0.6)*10=4363 (2) (2400*12)/(18+0.6)*30=1548*30=46440(2) (2400*12)/(18+0.6)*30=1548*30=46440

d. What is the effective overall transfer rate for d. What is the effective overall transfer rate for each of the two preceding blocking factors?each of the two preceding blocking factors? Answer:Answer:

(1) (120*10)/(6/120+0.6/60) = 20000(byte/s)(1) (120*10)/(6/120+0.6/60) = 20000(byte/s) (2) (120*30)/(18/120+0.6/60) = 22500(byte/s)(2) (120*30)/(18/120+0.6/60) = 22500(byte/s)

e. e. What is the capacity of the tape? What is the capacity of the tape? Answer:Answer:

a.(2400*12)/(6+0.6)*10*120=25600000(byte)a.(2400*12)/(6+0.6)*10*120=25600000(byte)

b.(2400*12)/(18+0.6)*30*120=36378947(byte)b.(2400*12)/(18+0.6)*30*120=36378947(byte)

Question & Answer