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Solved Problem 14.2 Refer to Figure 14-3, which contains the market demand curve facing a monopolist. a. What price should the monopolist charge without price discrimination if its best level of output (given by the point where MR = MC) is OB What would the TR be How much is the consumers surplus b. Suppose the monopolist sold OA units at price OF. In order to induce consumers to buy AB additional units, it lowers its price to OC only on AB units. How much would TR be now How much of the consumers surplus remains Solution: a. The highest price the monopolist can charge (without price discrimination) to sell OB units is OC. The TR would then equal the area of rectangle OCKB. Consumers surplus is CGK. b. TR is OFHA (for OA units) plus AJKB (for AB units). Note that price discrimination has increased TR by CFHJ (and this is the amount by which the consumers surplus declined). Consumers surplus is now only FGH plus HKJ.

public class TestSquares { public static void main(String[] args) { Primes.setSize(1000); for (int n = 1; n < 100; n++) { for (int i = n*n+1; i < (n+1)*(n+1); i++) { if (Primes.isPrime(i)) { System.out.printf("%d < %d < %d%n", n*n, i, (n+1)*(n+1)); break; } } } } } public class TestMersenne { public static void main(String[] args) { Primes.setSize(1000); for (int p = Primes.next(); p < 30; p = Primes.next()) { int n = (int)Math.round(Math.pow(2,p)) - 1; System.out.printf("%d\t2^%d-1%d", p, p, n); if (Primes.isPrime(n)) { System.out.println(" is prime "); } else { System.out.println(" is not prime "); } } } } boolean isPalindromic(int n) { if (n < 0) { return false; } int p10=1; // make p10 is the greatest power of 10 that is < n while (p10 < n) { p10 *= 10; } p10 /= 10; while (n > 9) { if (n/p10 != n%10) { return false; } n /= 10; // remove rightmost digit from n p10 /= 10; n %= p10; // remove leftmost digit from n } return true; // single digit integers are palindromic } }

Arrays work well for unordered sequences, and even for ordered sequences if they don t change much. But if you want to maintain an ordered list that allows quick insertions and deletions, you should use a linked data structure. This chapter shows how to do that. MAINTAINING AN ORDERED ARRAY 2 outlines how the binary search can find elements very quickly in an array that is sorted. This suggests that we should keep our arrays in sorted order. But inserting new elements in an ordered array is difficult. The main problem is that we have to shift all the larger elements forward to make room for the new element to be placed in its correct ordered position. This can be done by the insert() method shown in Example 3.1. EXAMPLE 3.1 Inserting into an Ordered Array

void insert(int[] a, int n, int x) { // preconditions: a[0] <= ... <= a[n-1], and n < a.length; 3 // postconditions: a[0] <= ... <= a[n], and x is among them; 4 int i = 0; 5 while (i < n && a[i] <= x) { 6 ++i; 7 } 8 System.arraycopy(a, i, a, i+1, n-i); // copies a[i..n) into a[i+1..n+1) 9 a[i] = x; 10 } The insert() method takes three arguments: the array a[], the number n of elements that are already sorted in the array, and the new element x to be inserted among them. The preconditions at line 2 specify that the first n elements of the array are in ascending order and that the array has room for at least one more element. The postconditions at line 3 specify that the array is still in ascending order and that x has

been successfully inserted among them. The code at lines 4 7 searches the array for the correct position for x to be inserted. It should be the smallest index i for which a[i] > x. For example, if x = 50 for the array shown in Figure 3.1, then the correct position for x is at index i = 1, because a[0] <= x < a[1]. After the correct position i has been located for x, the insert() method shifts the elements that are greater than x one position to the right. This is accomplished by the call

System.arraycopy(a, i, a, i+1, n-i);