GATE Overflow Test Series | Computer Organization and Architecture | Test 1 | Question: 11

Question

A computer using expanding opcode technique has $32-bit$ instructions and $13-bit$ addresses. If there are $60$ two-address instructions, and $128$ zero-address instructions, approximately how many one-address instructions can it support?

• A. $32\;K$
• B. $20\;K$
• C. $64\;K$
• D. $16\;K$

Answer 1

No. of encoding for 32-bit instructions = $2^{32}$
    
We are using 13-bit addresses here and we can divide the entire possible encodings to 3 types of instructions as
    
$2^{26}\times 60 + 2^{13} \times n + 128 = 2^{32}$
    
$\implies 2^{13}\times n = 2^{32} - (2^{26} \times  60-2^7)$
    
$\implies n = \dfrac{2^{26}[2^6 - 60]  - 2^7}{2^{13}}$
    
$\implies n = \dfrac{2^{28}  - 2^7}{2^{13}}$   

$\implies n = \dfrac{2^{21}  - 1}{2^{6}} \approx 2^{15} \approx 32 K.$

Only option matching is option A.

Comments

@gatecse

I have 1 doubt.
Let x be the number of 1-address instruction,then,
((2^6-60)*2^13-x)*2^13>=128
(2^15-x)>=1/2^6
x<=(2^15-1/2^6)
x<=32767
And the largest number in the range of x is 20K.then why is the answer not 20K?

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how 20K and not 32K?

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x is the number  of 1-address instruction and the condition is that x<=32767
then

20K=20*2^10=20480
32K=2^15=32768
the value of 32K is greater than 32767(which is the largest value that x could be).
hence the largest number of 1-address instruction in the range of x is 20K

Answer 2

$[(4*2^{13})-n]*2^{13}=128$

solving this eqn,we’ll get,

$n=2^{15}-\frac{1}{2^{6}}$$\approx 2^{15}=32K$

Comments

got it.