Chapter 9, Problem 9.11HW

A.

Program Plan Intro

Virtual address:

Virtual page number (VPN) and virtual page offset (VPO) are the two components of virtual address. Virtual address is “14 bits” format.

Explanation of Solution

Given data:

Virtual address = 0x027C

The given virtual address is in hexadecimal format; convert it into binary format. Convert each Hexadecimal digit to a 4 bit binary equivalent:

0	2	7	C
0000	0010	0111	1100

The above binary values are filled in the virtual address 14 bits format as follows,

Virtual address format:

13	12	11	10	9	8	7	6	5	4	3	2	1	0
0	0	0	0	1	0	0	1	1	1	1	1	0	0

B.

Program Plan Intro

The page size (P) is 64bytes; using the page size finds the number of bits in VPN.

The formula for page size as follows:

$P=2^p$

Substitute “P = 64” in the above formula

$\begin{array}{l}64=2^p\\ 2^6=2^p\\ \boxed{p=6}\end{array}$

Therefore, the “p” value represents virtual page offset (VPO) and physical page offset (PPO).

Number of bits in VPN is calculated as follows:

n = 14

p = 6

$\begin{array}{c}NumberofbitsinVPN=n-p\\ =14-6\\ =8\end{array}$

The VPN has additional two more components. They are TLB tag (TLBT) and TLB index (TLBI). The TLB is “4 ways” associative with “16” entries totally. Using the TLB find the value of TLBI and TLBT.

The formula for TLB as follows:

$T=2^t$

Substitute “T = 4” in the above formula

$\begin{array}{l}T=2^t\\ 4=2^t\\ 2^2=2^t\\ \boxed{t=2}\end{array}$

Therefore, the “t” value represents TLBI.

TLBI and TLBT are calculated as follows:

The “t” value represents TLBI. Therefore, the value of TLBI is “2”.

TLBI = 2

VPN = 8

$\begin{array}{c}TLBT=VPN-TLBI\\ =8-2\\ =6\end{array}$

Therefore, the value of TLBT is “6”.

Explanation of Solution

Virtual address format:

13	12	11	10	9	8	7	6	5	4	3	2	1	0
0	0	0	0	1	0	0	1	1	1	1	1	0	0

The higher-order “8 bits” of the virtual address represents “VPN” and the lower-order “6 bits” of the virtual address represents “VPO”. The higher-order “6 bits” of the VPN represents “TLBT” and the lower-order “2 bits” of the VPN represents “TLBI”

Value of VPN:

13	12	11	10	9	8	7	6
0	0	0	0	1	0	0	1

VPN = 0000 1001 = 0x9

Value of TLB index:

7	6
0	1

TLB index = 01 = 0x1

Value of TLB tag:

13	12	11	10	9	8
0	0	0	0	1	0

TLB tag = 00 0010 = 0x2

TLB hit: No

There is a no valid match in the virtual address.

Page fault: No

There is a valid PTE and returns the cached PPN from the page table (PTE).

Value of PPN:

The cached PPN from the page table is “0x17”.

Value of VPO:

5	4	3	2	1	0
1	1	1	1	0	0

VPO = 11 1100 = 3C

Completed Table:

Parameter	Value
VPN	0x9
TLB index	0x1
TLB tag	0x2
TLB hit? (Y/N)	N
Page fault? (Y/N)	N
PPN	0x17

C.

Program Plan Intro

Physical address:

Physical page number (PPN) and physical page offset (PPO) are the two components of physical address. Physical address is “12 bits” format. The physical page offset (PPO) is identical to the virtual page offset (VPO).

Explanation of Solution

Physical address format:

To form the physical address, concatenate the physical page number (PPN) from the PTE with virtual page offset (VPO) from the virtual address, which forms “0x05FC”.

Physical address = 0x05FC

The physical address is in hexadecimal format; convert it into binary format. Convert each Hexadecimal digit to a 4 bit binary equivalent:

0	5	F	C
0000	1011	1111	1100

The above binary values are filled in the physical address 12 bits format as follows,

11	10	9	8	7	6	5	4	3	2	1	0
0	1	0	1	1	1	1	1	1	1	0	0

D.

Program Plan Intro

The physical address has additional three more components. They are Byte offset within cache block (CO), Cache index (CI) and Cache tag (CT).

Explanation of Solution

Physical address format:

11	10	9	8	7	6	5	4	3	2	1	0
0	1	0	1	1	1	1	1	1	1	0	0

The lower-order “2 bits” of the physical address represents “CO”, the next “4 bits” represents “CI” and the remaining “6 bits” represents “CT”.

Value of Byte offset (CO):

1	0
0	0

CO = 11 = 0x0

Value of Cache index (CI):

5	4	3	2
1	1	1	1

CI = 1111 = 0xF

Value of Cache tag (CT):

11	10	9	8	7	6
0	1	0	1	1	1

CT = 01 0111 = 0x17

Cache hit: No

The cache index is not matches the cache tag.

Value of Cache byte:

The cached byte is not returned because the cache index is not matches the cache tag.

Completed Table:

Parameter	Value
Byte offset	0x0
Cache index	0xF
Cache tag	0x17
Cache hit? (Y/N)	N
Cache byte returned	---