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Solution: a. In increasing food production from 0 to 2 million units, production of clothing decreases from 16,000 to 15,000 units. Thus, the opportunity cost of producing the rst 2 million units of food is 1 thousand units of clothing. The opportunity cost of a second and third additional 2 million units is 2,000 and 3,000 units of clothing, respectively. b. The opportunity cost of increasing food production is increasing from 1,000 units of clothing to 2,000 to 3,000 units of clothing. c. Increasing clothing and food costs are re ected in a concave (outward-sloping) production-possibility frontier. Moving down the frontier from point A to points B, C, D, E, and F shows that to produce 2 million incremental units of food (the 2-million-unit-length horizontal dashed lines in Figure 1-4), we must give up more and more units of clothing (the vertical dashed lines of increasing length). Solved Problem 1.5 Explain how division of labor and specialization enhance production in an advanced society. Solution: Through the division of labor and specialization, the population within a given geographic region, instead of being self-suf cient and producing the full range of goods and services wanted, can concentrate its energies and time in the production of only a few goods and services in which its ef ciency is greatest. Thus, specialization and division of labor allow greater output. By then exchanging some of the goods and services so produced for different goods and services produced similarly within a different geographic region, the regions populations as a whole end up consuming a larger number and greater diversity of goods and services than would otherwise be the case.

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Object[] concatenate(Object[] a, Object[] b) // returns an array containing all of a[] followed by all of b[]

2.12 Write and test this method:

void shuffle(Object[] a) // randomly permutes the elements of a[]

2.13 Write and test this method:

2

int[] tally(String string) // returns an array a[] of 26 integers that count the frequencies // of the (case insensitive) letters in the given string

2.14 Write and test this method:

double innerProduct(double[] x, double[] y) // returns the algebraic inner product (the sum of the component// wise products) of the two given arrays as (algebraic) vectors

2.15 Write and test this method:

double[][] outerProduct(double[] x, double[] y) // returns the algebraic outer product of the two given arrays // as (algebraic) vectors: p[i][j] = a[i]*b[j]

In This :

2.16 Write and test this method:

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BI Object Metadata Aggregate Transformation rule InfoCube InfoObject InfoPackage InfoSet InfoSource DataStore Object Query (including variables, structures, restricted and calculated key figures) Reporting Agent scheduling package Reporting Agent settings Web item Web template Master data Characteristic value

double[][] product(double[][] a, double[][] b) // returns the matrix product of the two given arrays a matrix: // p[i][j] = Sum(a[i][k]*b[k][j]:k)

2.17 Write and test this method:

double[][] transpose(double[][] a) // returns the transpose ta of the specified array as a matrix: // ta[i][j] = a[j][i]

2.18 Write and test this method:

Demand Supply Equilibrium Price and Quantity Government and Price Determination Elasticity True or False Questions Solved Problems Demand

CHAP. 2]

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2.19 The Sieve of Eratosthenes is an array of boolean elements whose ith element is true if and only if i is a prime number. Use the following algorithm to compute and print a sieve of size 1000: (Precondition: p is an array of n bits.) (Postcondition: p[i] is true if and only if i is prime.) 1. Initialize p[0] and p[1] to be false, and all other p[i] to be true. 2. Repeat step 3 for each i from 3 to n, incrementing by 2. 3. If there is a prime the square root of i that divides i, set p[i] false. 2.20 Repeat Problem 2.19 using a java.util.Vector object. 2.21 Repeat Problem 2.19 using a java.util.BitSet object. 2.22 Define and test a Primes class with these methods:

public public public public public public public Use the BitSet static static static static static static static void setLast(int last) void setLast() void sizeSize(int size) void sizeSize() boolean isPrime(int n) int next() void printPrimes() // // // // // // // sets last sets last=1 sets size of bitset sets bitset size=1000 true if n is prime next prime after last prints sieve

12.1 The disadvantage of a binary search tree is that it may become very unbalanced, in which case searching degenerates into an O(n) algorithm. The advantage is the efficiency that a binary tree provides for insertions and deletions. The advantage of an AVL tree is that it is always balanced, guaranteeing the O(lgn) speed of the binary search algorithm. The disadvantages the complex rotations used by the insertion and removal algorithms needed to maintain the tree s balance.

The demand schedule for an individual speci es the units of a good or service that the individual is willing and able to purchase at alternative prices during a given period of time. The relationship between price and quantity demanded is inverse: more units are purchased at lower prices because of a substitution effect and an income effect. As a commodity s price falls, an individual normally purchases more of this good since he or she is like-

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