Official Solution for Final Exam
CS 15-347
Spring 1997

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PROBLEM 1. 

A. phi | tf - ta | <= phi Emax ta clocks
B. phi | tf - ta | <= phi Emax (tm + delta) clocks
C. Setting lhs of B to 1 yields Emax = 1 / [phi ( tm + delta)]
D. No units

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PROBLEM 2. 

A. 14 clocks (AS runs in parallel)
B. L1, L2, A1, MS, A2, A3, SLT, BNE, AS
C. 9 clocks 

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PROBLEM 3.  

A. 512 meters 
B. 128 bytes

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PROBLEM 4. 

A. 24 bytes
B. doubleword alignment
C. CNT = 9
C. a_struct = double, int[0], int[1], int[2], int[3]

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PROBLEM 5.

A. VPN:  [13-6] VPO:  [5-0]
   TLBT: [13-8] TLBI: [7-6]

B. PPN:  [11-6] PPO:  [5-0]
   CT:   [11-6] CI :  [5-2] CO: [1-0] 

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PROBLEM 6.

A. 00 0010 0111 1100

B. VPN:      0x9
   TLBI:     0x1
   TLBT:     0x2
   TLB hit?  N	
   TLB miss? N
   PPN:      0x17

C. 0101 1111 1100 

D. CO:         0x0
   CI:         0xF
   CT:         0x17
   Cache hit?  N
   Cache byte: -

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PROBLEM 7.

A. 00 0011 1101 0100

B. VPN:      0xF
   TLBI:     0x3
   TLBT:     0x3
   TLB hit?  Y	
   TLB miss? N
   PPN:      0xD

C. 0011 0101 0100

D. CO:         0x0
   CI:         0x5
   CT:         0xD
   Cache hit?  Y
   Cache byte: 0x36

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PROBLEM 8.

Hex		unsigned	2's comp	floating point
0x00000000	0		0		0
0xFF800000	511 x 2^23	-2^23		-inf
0x0000002A	42		42		42 x 2^-149
0x80000000	2^31		-2^31		-0
0xFFFFFFFF	2^32 - 1 	-1		nan

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PROBLEM 9.

	separate	paired
A.	1/4		1/4
B. 	1/4		1/4
C. 	3/4		1/2


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PROBLEM 10. 

	c0  c1   c2   c3   c4   c5   c6   c7  c8  c9 c10  c11

li1	 f   d    e    m    w
li2	     f    d    e    m    w
addu3             f    d    e    m    w
lw4                    f    d    e    m    w
addu5                       f    d    d    e   m   w
sw6                              f    f    d   e   m   w



forwarding arcs: (inst:stage)

li1:m   -> addu3:e
li2:e   -> addu3:e
addu3:e -> lw4:e
lw4:m   -> addu5:e
addu5:e -> sw6:e
addu5:m -> sw6:m

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PROBLEM 11. 

A. B1 B2 A1 A2
B. A1 B1 A2 B2
C. sequentially inconsistent
C. A1 A2 B1 B2

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PROBLEM 12. 

A. 16 bytes
B. doubleword alignment
C. 16 bytes
D. doubleword alignment

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PROBLEM 13. 

A. min = 3, max = n, stride = 1

B. Dn-3, Cn-2, Bn-1, Dn-2
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