Question

John is given the task of collecting 100 rocks, 130 sticks, and 180 hay. He has the option of exploring the Farm area, the Forest area, or the Cave area for these resources. Each time he explores the Farm, he will find .5 rocks, 1 stick, and 3 hay. Exploring the farm will use up 2 stamina points. As for the forest, John will find 1.5 rocks, 4 sticks, and .5 hay for each attempt, with a 3 point stamina point reduction. For the cave, John will find 7 rocks, 1.5 sticks, and 1.3 hay, using up 4 stamina points per cave search. John can explore an area multiple times and as many times as he wants. How many times should John explore each area in order to have the smallest stamina point reduction while still fulfilling his task of collecting 100 rocks, 130 sticks, and 180 hay? (aka the least amount of stamina points John can lose while completing his task)

Answer 1

Say John explores Farm area xl times explores Forest area x2 times explores Cave area x3 times our objective is to minimize the stamina point reduction, so obiective function is and the task of collecting 100 rocks, 130 sticks, and 180 hay: 0.5x1 + 1.5x+7 x2 100 x + 4 x2 + 1.5 x3 = 130 3 1 +0.5x+ 1.32 180 and xi,x, r, 2 0; 12 3 As the constraint is of type''we should add artificial variable A After introducing artificial variable:s subject to 0.5x 1.5x, + 7 x3 + A = 1 00 130 + A3 180 3 1 +0.5x+ 1.3x2

Iteration-1 MinRatio 100 Mİ 100 0.5 14.2857 - 130 130 -= 86.6667 180 MI 180 0.5 1.3 - 138.4615 1.3 Z- 410M 4.5M 6M 9.8M MIMM Positive maximum Z,- C, is 9.8M- 4 and its column index is 3. So, the entering variable is x3 Minimum ratio is 14.2857 and its row index is 1. So, the leaving basis variable is A1 The pivot element is 7. Iteration-2 MinRatio 14.2857 0.2143 108.5714 4 14.2857 0.0714 0.2143 100 66.6667 M 108.5714 0.8929 (3.6786) 0 29.5146 3.6786 161.4286 M 161.4286 2.9071 0.2214 -729.0323 0.2214 Z- 270M 57.1429 3.8M 0.2857 3.9M 0.85714 M M -C 3.8M- 1.71433.9M-2.142910 00 Positive maximum Z, - C, is 3.9M- 2.1429 and its column index is 2. So, the entering variable is x Minimum ratio is 29.5146 and its row index is 2. So, the leaving basis variable is A2 The pivot element is 3.6786

Iteration-3 MinRatio 7.9612 4 7.9612 0.0194 0.0194 410 29.5146 3 29.5146 0.2427 .2427121.6 M 154.8932 0 01 154.8932 (2.8534) = 54.2838 → 2.853 Z- 154.8932M 120.3883 2.8534M+0.8058 34 M Z-C 2.8534M-1.19421 00 0 Positive maximum Z, - C is 2.8534M- 1.1942 and its column index is 1. So, the entering variable is x Minimum ratio is 54.2838 and its row index is 3. So, the leaving basis variable is A3 The pivot element is 2.8534 Iteration-4 x3 MinRatio 4 6.9071 16.3389 54.2838 x1 Z- 185.2127 Since all Z, - C. 0 Hence, optimal solution is arrived with value of variables as: XI = 54.2838, x,-16.3389, x3 = 6.9071 Min Z- 185.2127