An instruction pipeline has five stages, namely, instruction fetch $$(IF),$$ instruction decode and register fetch $$(ID/RF)$$ instruction execution $$(EX),$$ memory access $$(MEM),$$ and register writeback $$(WB)$$ with stage latencies $$1$$ ns, $$2.2$$ $$ns,$$ $$2$$ $$ns,$$ $$1$$ $$ns,$$ and $$0.75$$ $$ns,$$ respectively ($$ns$$ stands for nanoseconds). To gain in terms of frequency, the designers have decided to split the $$ID/RF$$ stage into three stages $$(ID, RF1, RF2)$$ each of latency $$2.2/3$$ $$ns,$$ Also, the $$EX$$ stage is split into two stages $$(EX1, EX2)$$ each of latency $$1$$ ns. The new design has a total of eight pipeline stages. A program has $$20$$% branch instructions which execute in the $$EX$$ stage and produce the next instruction pointer at the end of the $$EX$$ stage in the old design and at the end of the $$EX2$$ stage in the new design. The IF stage stalls after fetching a branch instruction until the next instruction pointer is computed. All instructions other than the branch instruction have an average $$CPI$$ of one in both the designs. The execution times of this program on the old and the new design are $$P$$ and $$Q$$ nanoseconds, respectively. The value of $$P/Q$$ is _____________.

Consider the following rooted tree with the vertex labelled P as the root The order in which the nodes are visited during an in-order traversal of the tree is

Consider the pseudocode given below. The function Dosomething () takes as argument a pointer to the root of an arbitrary tree represented by the leftMostChild-rightSibling representation. Each node of the tree is of type treeNode.

typedef struct treeNode* treeptr; 
Struct treeNode 
{ 
    Treeptr leftMostchild, rightSibiling; 
}; 
Int Dosomething (treeptr tree) 
{ 
    int value =0; 
    if (tree ! = NULL) { 
      If (tree -> leftMostchild = = NULL) 
          value=1; 
    else 
        value = Dosomething (tree->leftMostchild); 
        value = value + Dosometing (tree->rightsibiling); 
    } 
    return (value); 
}

When the pointer to the root of a tree is passed as the argument to DoSomething, the value returned by the function corresponds to the

Consider the C function given below. Assume that the array listA contains n (> 0) elements, sored in ascending order.

int ProcessArray(int *listA, int x, int n) 
{ 
      int i, j, k; 
      i = 0; 
      j = n-1; 
      do 
      { 
         k = (i+j)/2; 
         if (x <= listA[k]) 
           j = k-1; 
         if (listA[k] <= x) 
           i = k+1; 
      } 
      while (i <= j); 
      if (listA[k] == x) 
           return(k); 
      else 
           return -1; 
}

Which one of the following statements about the function ProcessArray is CORRECT?