GATE Solved Paper 2020 - GATE 2020

56. Consider a paging system that uses a 1-level page table residing in main memory and a TLB for address translation. Each main memory access takes 100 ns. TLB lookup takes 20 ns. Each page transfer to/from the disk takes 5000 ns. Assume that the TLB hit ratio is 95%, page fault rate is 10%. Assume that for 20% of the total page faults, a dirty page has to be written back to disk before required page is read in from disk. TLB update time is negligible. The average memory access time is ns (round off to 1 decimal places) is ______.

  • Option : D
  • Explanation :
    Main memory access time, Tm = 100 ns
    TLB lookup, TTLB = 20ns
    Page transfer time, TPT = 5000 ns
    TLB hit ratio, x = 0.95 (95%)
    page fault rate, p = 0.10 (10%)
    Gate2020 cs
    We know,
    EMAT for multilevel paging,
    EMAT = x (Tc + Tm) + (1 – x) (Tc + (n + 1) Tm)
    EMAT, when there is a page fault, S → is service time
    EMAT = (1 – P) Tm + Ps
    Here, we are using TLB, and page fault occurs whenever there is a miss in TLB, So the required EMAT is ,
    EMAT = x(Ttlb + Tm) + (1 – x) [(1 – P) (Ttlb + Tm + Tm) + p(% dirty (Ttlb + Tm + 2TPT) + % clean (Ttlb + Tm + TPT)
    ∴ EMAT = 0.95 (20 + 100) + 0.05 (0.9 (20 + 100 + 100) + 0.1 (0.2 (20 + 100 + 2(5000)) + 0.8 (20 + 100 + 5000))
    = 154.5 ns

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57. Consider a graph G = (V, E), where V = {v1, v2, …, v100}, E = {(vi, vj)|1≤i≤j≤100}, and weight of the edge (vi, vj) is |i – j|. The weight of the minimum spanning tree of G is _____.

  • Option : A
  • Explanation :
    If we consider a small graph with 5 vertices, then the minimum spanning tree will have a weight 4.
    So, for n-vertices, MST weight would be (n – 1)
    As n = 100 (no. of vertices), So, minimum spanning tree weight = (100 – 1) = 99
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58. For n > 2, let a ∈ {0, 1}n be a non-zero vector. Suppose that x is chosen uniformly at random form (0,1)n. Then, the probability that Gate2020 cs is an number is _______ .

  • Option : A
  • Explanation :
    Let n = 3
    0 ϵ < 0,1]3
    Gate2020 cs
    x = 0 or 1
    Gate2020 cs
    = x1 = x3 (∵ a1 = 1, a2= 0, a3 = 1)
    = 0 or 1 or 1 or 2.
    Gate2020 cs
    Gate2020 cs
    = x3 (∵ a1 = a2 = 0)
    = 0 or 1
    Gate2020 cs
    Total number of cases = 2 × 3 + 4 × 3 = 18
    Now odd number of cases = 9
    P = 9/18 = 0.5
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59. Consider the array representation of a binary min-heap containing 1023 element. The minimum number of comparisons required to find the maximum in the heap is ______.

  • Option : A
  • Explanation :
    Min-heap contains 1023 elements.
    Min-heap means, parent should be minimum or equal to it’s children so, max children could be either left or right one.
    Following this logic, maximum can be definitely at leaf nodes.
    Gate2020 cs
    No. of elements in leaf = n/2 = 1023/2 = 512
    To find maximum among 512 elements, no. of comparisons needed is 511.
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