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Q # 1:
a) Local Access:
Load instruction = 10 %
Store instruction = 8 %
% for Load and store instructions for local = (10+8) % = 18% No. of Load and Store Instructions = (18/100) * 200, = 36000 Instructions Time to execute Load/Store Inst. = 36000clock cyclesTime of one Inst. Time for One Instruction Execution = 1/ (Processor’s Clock Rate) = 1/ (200 MHz) = 5 ns Time to execute Load/Store Inst. = 36000 * 4 * 5 (ns)
= 720 μs
b) Remote Access:
i) Load Instructions:
Total load instructions = 10% Total Load Instructions in the given instructions = (10/100) * 200, = 20,000 Instructions No. of Load Inst. (for local access) = (85/100) * 20, = 17,000 Instructions No. of Load Inst. (for remote access) = (15/100) * 20, = 3000 Instructions Time for Load inst. (local) = 17,000 * 4 * (1/200 MHz) = 340 μs
Time for Load inst. (remote) =? There are many factors which are required for remote access. Delay for all these factors will be added. These are as given below:
A) Fix overhead to initiate the request = 12 * (1/200 MHz) = 60 ns B) Time for Sending Request Packet =? As Request block is
Source address (1 byte)
Destination address(1 byte)
Address to memory (3bytes)
Time for transmit Request Packet = (1+1+3) / (Bandwidth of IN) = (5 Bytes) / (75 MBytes/s) = 66.67 ns
C) Time for data retrieved at remote CE (processor)=? = (4 cycles) * (1/200 MHz) = 20 ns D) Fix overhead to initiate the reply by responder = 12 * (1/200 MHz)
= 60 ns E) Time for Sending Reply (Data) Packet =? As Response block is
Source address (1 byte)
Destination address (1 byte)
Data from memory (5 bytes)
So Time for Sending Reply (Data) Packet = (5+1+1) / (Bandwidth of IN) = (7 Bytes) / (75 MBytes/s) = 93.33 ns Total Time for Remote Access = Time of (A+B+C+D+E) = (60+66.67+20+60+93.33) ns = 300.00 ns
C) Time for to store data by remote Processor =? = (4 cycles) * (1/200 MHz) = 20 ns D) Fix overhead to initiate the reply by remote processor = 12 * (1/200 MHz) = 60 ns // for sending Acknowledgement back to sender’s Processor
E) Acknowledgment Packet will be like that
Source Address (1 byte)
Destination Address (1 byte)
Acknowledgment (1 byte)
Time for Sending Ack Packet = (1+1+1) / (Bandwidth of IN) = (3 Bytes) / (75 MBytes/s) = 40 ns Total time for one remote store = Time of (A+B+C+D+E) = (60+133.33+20+60+40) ns = 313.33 ns Total Time for Remote Access (All Store inst) = 313.33 ns * No. of Store Inst. (for remote access) = 313.33 ns * 2400 = 751.992 μs
So total time for Store instructions = (local + remote) = (272 μs + 751.992 μs) = 1023.992 μs = 1024 μs
Q # 2:
a)
c) If we go from Node # 6 to Node # 1 then there will be blocking in the path. As shown in the figure
.Source = S = Node # 6 = 0110 Destination = D = Node # 3 = 0001 Source = S = 0 1 1 0 Destination= 0 0 0 1
R=S(xor)D = 0 1 1 1 R3 R2 R1 R Switch Stages 3 2 1 0
Blocking will occure at stage 3 as shown in figure
d) By using Shuffle and Exchange functions. As
Source = S = Node # 13 = 1101 Destination = D = Node # 7 = 0111 R= S (XOR) D So 1 1 0 1 0 1 1 1
1 0 1 0 R3 R2 R1 R
If value of R is 1 then there will be exchange according to formula and if it is 0 then there will be no exchange. However shuffle operation will be implemented at every stage. Now R Shuffle Exchange ---------------------------------------------------------------- R3 1 S(1101) 1011 E(1011) 1010
R2 0 S(1010) 0101 (No E) 0101
R1 1 S(0101) 1010 E(1010) 1011
R0 0 S(1011) 0111 (No. E) 0111 0111
So 0111 is the destination = 7 Path is shown in figure
e) NxN MIN: Switch Cost = (N/2) x Log 2 N Link Cost = N x Log 2 N Delay = Log 2 N NxN Cross Bar: Switch Cost = N^2 Link Cost = N^2 Delay = 1
For 1024 x 1024 MIN: Switch Cost = 5120 Link Cost = 10240 Delay = 10
Cross Bar: Switch Cost = 1048576 Link Cost = 2097152
Delay = 1
4- For i= n-1 to 0
If ri= 1
move to adjacent node in dim di to Ci( )
Else
don’t move to adjacent node in dim di
End if
End for
Source = S = Node # 14 = 1110 Destination = D = Node # 3 = 0011 Distance = d = S (XOR) D
Source S 1 1 1 0
Destination D 0 0 1 1
Routing vector R = S (xor)E 1 1 0 1 r 3 r 2 r 1 r 0 dimension d 3 d 2 d 1 d 0 Now Source R 3 1 C 3 =(1110) 0110
R 2 1 C 2 (0110) 0010
R 1 0 C 1 (0010) 0010 (don’t move)
R 0 1 C 0 (0010) 0111 Destination
So path will be like this
Q # 4:
a) K-d mesh (no wrap round)
For 3-d mesh Where three dimensions are x, y, z
Routing Algorithm (S, D, k, d)
Source = S = (Sx,Sy, Sz)
Destination =D= (Dx, Dy, Dz)
R = D-S = ( (Dx-Sx),(Dy – Sy ), (Dz-Sz) )
( rx , ry , rz )
For i= x, y, z do
Move ri steps in dim di
End for
Using Algorithms for 2x3x5 Mesh
As given in the question
S = (1, 2, 0)
D = (2, 3, 4)
R = (2-1, 3-2, 4-0)
This means that we have to move 1 step in x-axis, 1 step in y-axis and 4 steps in z-axis direction from the source node, and we will reach at destination node.
b)
