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Furkat [3]
3 years ago
8

Assume a 8x1 multiplexer’s data inputs have the following present values: i0=0, i1=0, i2=0, i3=0, i4=0, i5=1, i6=0, i7=0. What s

hould be value of the select inputs s2, s1 and s0 for the value on the multiplexer’s output d to be 1?
s2=

s1=

s0
Computers and Technology
1 answer:
blagie [28]3 years ago
5 0

Answer: s2=1 s1=0 s0=1

Explanation:

Generally speaking, the multiplexer is a digital circuit , build with combinational logic, that acts like a switch, sending to the output the current value present at the input which order number (in decimal) is equal to the binary combination of the select inputs, expressed in decimal form.

If the multiplexer is 8x1, in order to be able to pass to the output any of the 8 inputs, the number of select inputs (n), must satisfy the following relationship:

M(number of inputs) = 2ⁿ

In this case, as the only input which present value  is "1" ,is the input i5, the value present at the input select must be the binary combination of s₀, s₁

and s₂, that yields the decimal 5, i.e.,  s₀ = 1  s₁ = 0  s₂ = 1.

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3 years ago
Illustrate that the system is in a safe state by demonstrating an order in which the threads may complete.If a request from thre
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Answer:

a. safe sequence is T2 , T3, T0, T1, T4.

b. As request(T4) = Available, so the request can be granted immediately

c. As request(T2) < Available, so the request can be granted immediately

d. As request(T3) < Available, so the request can be granted immediately.

Explanation:

It will require matrix

[i, j] = Max [i, j] – Allocation [i, j]

A B C D

T0 3 3 3 2

T1 2 1 3 0

T2 0 1 2 0

T3 2 2 2 2

T4 3 4 5 4

Available = (2 2 2 4)

1. Need(T2) < Available so, T2 can take all resources

Available = (2 2 2 4) + (2 4 1 3) (Allocation of T2) = (4 6 3 7)

2. Need(T3)<Available so, T3 will go next

Available = (4 6 3 7) + (4 1 1 0) = (8 7 4 7)

Like wise next T0, T1, T4 will get resources.

So safe sequence is T2 , T3, T0, T1, T4.

(Note, there may be more than one safe sequence).

Solution b.

Request from T4 is (2 2 2 4) and Available is (2 2 2 4)

As request(T4) = Available, so the request can be granted immediately.

Solution c.

Request from T2 is (0 1 1 0) and Available is (2 2 2 4)

As request(T2) < Available, so the request can be granted immediately.

Solution d.

Request from T3 is (2 2 1 2) and Available is (2 2 2 4)

As request(T3) < Available, so the request can be granted immediately.

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Answer:

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