You have been asked to prove or disprove the sentence For al

You have been asked to prove or disprove the sentence: For all integers m, if m^2 - 2m is even, then m is even. Evaluate the following proofs and disproofs and select all those that you agree are correct. [At least one is correct!] The negation of the claim states: if m is odd, then m^- 2m is odd. Taking m = 3, we see that m^2 - 2m = 3, also odd. Since this example agrees with the negation of the claim, it is a counterexample to the original statement, which must be false. Assume (hoping for a contradiction) that for some m m^2 - 2m is even and m is odd. Then m = 2k + 1 for some integer k, leading to m^2 - 2m = 4k^2 - 1; but this is an odd number, so our assumption was impossible and the theorem must be true Suppose m is a particular but arbitrarily chosen even number: then m = 2p for some integer p. This leads to the calculation m^2 - 2m = 4p^2 - 4p = 2(2p^2 - 2p). Since 2p^2 - 2p is an integer, this means m^2 - 2m is even, and the theorem is true. Assume (hoping for a contradiction) that if m^2 - 2m is even, then m is odd. But in the example m = 2, m^2 - 2m = 0, we see that m^2 - 2m is even but m is also even. So our assumption is disproved and the theorem must be true.

Solution

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Step 1 Construct the transportation table entering the origin capacities ai, the destination requirement bj and the cost cij Step 2 Find an initial basic feasible solution by vogel’s method or by any of the given method. Step 3 For all the basic variables xij, solve the system of equations ui + vj = cij, for all i, j for which cell (i, j) is in the basis, starting initially with some ui = 0, calculate the values of ui and vj on the transportation table Step 4 Compute the cost differences dij = cij – ( ui + vj ) for all the non-basic cells Step 5 Apply optimality test by examining the sign of each dij If all dij 0, the current basic feasible solution is optimal If at least one dij < 0, select the variable xrs (most negative) to enter the basis. Solution under test is not optimal if any dij is negative and further improvement is required by repeating the above process. Step 6 Let the variable xrs enter the basis. Allocate an unknown quantity to the cell (r, s). Then construct a loop that starts and ends at the cell (r, s) and connects some of the basic cells. The amount is added to and subtracted from the transition cells of the loop in such a manner that the availabilities and requirements remain satisfied. Step 7 Assign the largest possible value to the in such a way that the value of at least one basic variable becomes zero and the other basic variables remain non-negative. The basic cell whose allocation has been made zero will leave the basis. Step 8 Now, return to step 3 and repeat the process until an optimal solution is obtained. 3.3 Worked Examples Example 1 Find an optimal solution W1 W2 W3 W4 Availability F1 19 30 50 10 7 F2 70 30 40 60 9 F3 40 8 70 20 18 Requirement 5 8 7 14 Solution 1. Applying vogel’s approximation method for finding the initial basic feasible solution W1 W2 W3 W4 Availability Penalty F1 5(19) (30) (50) 2(10) X X F2 (70) (30) 7(40) 2(60) X X F3 (40) 8(8) (70) 10(20) X X Requirement X X X X Penalty X X X X Minimum transportation cost is 5 (19) + 2 (10) + 7 (40) + 2 (60) + 8 (8) + 10 (20) = Rs. 779 2. Check for Non-degeneracy The initial basic feasible solution has m + n – 1 i.e. 3 + 4 – 1 = 6 allocations in independent positions. Hence optimality test is satisfied. 3. Calculation of ui and vj : - ui + vj = cij Assign a ‘u’ value to zero. (Convenient rule is to select the ui, which has the largest number of allocations in its row) Let u3 = 0, then u3 + v4= 20 which implies 0 + v4 = 20, so v4 = 20 u2 + v4= 60 which implies u2 + 20 = 60, so u2 = 40 u1 + v4= 10 which implies u1 + 20 = 10, so u1 = -10 u2 + v3= 40 which implies 40 + v3 = 40, so v3 = 0 u3 + v2= 8 which implies 0 + v2 = 8, so v2 = 8 u1 + v1= 19 which implies -10 + v1= 19, so v1 = 29 4. Calculation of cost differences for non basic cells dij = cij – ( ui + vj ) cij ui + vj (30) (50) -2 -10 (70) (30) 69 48 (40) (70) 29 0 dij = cij – ( ui + vj ) 32 60 1 -18 11 70 5. Optimality test ui (19) (10) u1= -10 (40) (60) u2 = 40 (8) (20) u3 = 0 vj v1 = 29 v2 = 8 v3 = 0 v4 = 20 dij < 0 i.e. d22 = -18 so x22 is entering the basis 6. Construction of loop and allocation of unknown quantity We allocate to the cell (2, 2). Reallocation is done by transferring the maximum possible amount in the marked cell. The value of is obtained by equating to zero to the corners of the closed loop. i.e. min(8-, 2-) = 0 which gives = 2. Therefore x24 is outgoing as it becomes zero. 5 (19) 2 (10) 2 (30) 7 (40) 6 (8) 12 (20) Minimum transportation cost is 5 (19) + 2 (10) + 2 (30) + 7 (40) + 6 (8) + 12 (20) = Rs. 743 7. Improved Solution cij ui + vj (30) (50) -2 8 (70) (60) 51 42 (40) (70) 29 18 dij = cij – ( ui + vj ) 32 42 19 18 11 52 Since dij > 0, an optimal solution is obtained with minimal cost Rs.743