The reason for the answers to Exercises 55 and 56 are different.

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a) Suppose that we are carrying out the 1-phase simplex algorithm on a linear program in standard inequality form (with 3 variables and 4 constraints) and suppose that we have reached a point where we have obtained the following tableau. Apply one more pivot operation, indicating the highlighted row and column and the row operations you carry out. What can you conclude from your updated tableau? x1 x2 x3 81 82 83 84 81 -2 0 1 1 0 0 0 3 82 3 0 -2 0 1 2 0 6 12 1 1 -3 0 0 1 0 2 84 -3 0 2 0 0 -1 1 4 -2 -2 0 11 0 0-4 0 -8

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Chapter 6.2, Problem 57E

To determine

To explain:

The reason for the answers to Exercises 55 and 56 are different.

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a) Suppose that we are carrying out the 1-phase simplex algorithm on a linear program in standard inequality form (with 3 variables and 4 constraints) and suppose that we have reached a point where we have obtained the following tableau. Apply one more pivot operation, indicating the highlighted row and column and the row operations you carry out. What can you conclude from your updated tableau? x1 x2 x3 81 82 83 84 81 -2 0 1 1 0 0 0 3 82 3 0 -2 0 1 2 0 6 12 1 1 -3 0 0 1 0 2 84 -3 0 2 0 0 -1 1 4 -2 -2 0 11 0 0-4 0 -8

Microsoft Excel snapshot for random sampling: Also note the formula used for the last column 02 x✓ fx =INDEX(5852:58551, RANK(C2, $C$2:$C$51)) A B 1 No. States 2 1 ALABAMA Rand No. 0.925957526 3 2 ALASKA 0.372999976 4 3 ARIZONA 0.941323044 5 4 ARKANSAS 0.071266381 Random Sample CALIFORNIA NORTH CAROLINA ARKANSAS WASHINGTON G7 Microsoft Excel snapshot for systematic sampling: xfx INDEX(SD52:50551, F7) A B E F G 1 No. States Rand No. Random Sample population 50 2 1 ALABAMA 0.5296685 NEW HAMPSHIRE sample 10 3 2 ALASKA 0.4493186 OKLAHOMA k 5 4 3 ARIZONA 0.707914 KANSAS 5 4 ARKANSAS 0.4831379 NORTH DAKOTA 6 5 CALIFORNIA 0.7277162 INDIANA Random Sample Sample Name 7 6 COLORADO 0.5865002 MISSISSIPPI 8 7:ONNECTICU 0.7640596 ILLINOIS 9 8 DELAWARE 0.5783029 MISSOURI 525 10 15 INDIANA MARYLAND COLORADO

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