Question

17. A farmer has 100 acres of tillable land on which corn, tomatoes, beans, peas, and carrots can be planted. The labor requirements, plant costs, and gross income for I acre of each of these crops are as follows: Labor Gross (worker-hours) Costs (S) Income (S) Corn 95 1300 275 345 435 20 200 Tomatoes Beans Peas Carrots 120 25 35 40 75 nd The farmer has available up to 3600 hr of labor at $7.25/hr. However, the labor and plant costs must be paid before any income from the sale of the produce is realized. The farmer has $3000 in capital to invest in this year's planting and can borrow up to another $12,000 if desired. Any such loan would have a 9% annual interest rate but would be repaid within 4 months. Land unused for any of these vegetables must be maintained by planting ground cover. One acre of cover crops requires 2 hr of labor and costs $9. Determine a planting schedule that maximizes net income

Answer 1

Assume that given problem x1-C x2 =T x4 -D x7-L

Find solution using Simplex(BigM) method MAXz-40x1800x2 +145x3185x4250x5-9x6-7.25x7-0.03x8 subject to x1x2 +x3x4 x5x6 100 5x1 + 120x2 + 25x3 + 40x5 + 2x6-x7 = 0 20x 200x2 +55x340x4 75x5 +9x6 +3.25x7- 83000 x7 <ー3600 x8 12000 and x1.x2x3.x4,15.x6x1x8 >_ 0 Solution Problem is Max z = 40x1 + 800x2 + 145x3 + 18584 + 250x5-916-7.25x7-0.03x8 subject to =100 5 x 120x2 25x3 +40s+26 20x 200x2 55 x3 +40 x4 75*5 +9x6 +3.25x, - xg 3000 r7S3600 s 12000 and x,, x2.xz.x4.Xz, xx7,xg 2 0; The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate i. As the constraint l is of type···we should add artificial variable A 2. As the constraint 2 is of type we should add artificial variable A2 3. As the constraint 3 is of type ' we should add slack variable S 4. As the constraint 4 is of type we should add slack variable S 5. As the constraint 5 is of type'we should add slack variable S After introducing slack,artificial variables Maxz- 40xl + 800 x2 + 145x3 + 18584 + 250x5-9x6 . 7.25x7-0.03x8 subject to OSI + 0S2 + 0S3 . MAI . MA2 -100 + 40xs 2*6 - A2=0 5 x 120x2 25xj 20x' + 200x2 + 55x3 + 40x4 + 75x5 + 9x6 + 3.25x7·Kg+S, + 3000 3600 12000 Iteration-l 40 800 145 185 7.25 0.03 M100 (120) 25 40 0 3000 0 3600 0 120000 20 200 75 -6M -121M -26M -41M -3M 29 3 ctivate Winddws 100 Go to Settings to activate Negative minimum 2 is-121M-800 and its column index is 2. So, the entering variable is x2. Windows

Minimum ratio is 0 and its row index is 2. So, the leaving basis variable is Az + The pivot element is 120 Entering x2. Departing A. Key Element 120 R2(new) Rfold) 120 R (new) R (old-R(new) Rylnew) Rg(old)-200R(new) R(new)-R (old) R,(new) - Rs(old) Iteration-2 145 185 250 59 4, I-MI 100 24 120 2 800 0 24 24 120 35 3000 (40) 12 0 3600 0 12000 0 23M 100 24 3 23M 20 24 3 19M 500 24 3 19M 65 24 3 2M 800 59M 40M 20 60 3120 3 M 50 59M 67M7 60 3 120 12 Negative minimum Z,- C, is-M185 and its column index is 4. So, the entering variable is x4 Minimum ratio is 75 and s ow index is 3. So, the leaving basis variable is S The pivot element is 40. Entering- , Departing-Si, Key Element-40 R (new) - R fold 40 R new)R (old-R (new) R(new) R fold) R(new)-Rsfold) Iteration-3 145 185 101 120 -MI 25 0 800 0 0 24 24 120 Activaté Windows Go to SEttings toactivats9 Windo120 185 75 480

0 3600 0 12000 2M 2095 3 24 I1M 685 +185 24 3 11M 7325 - - 10M + | 0.11 46M + 13875 2.с. | 2M 1135 3 24 2420 24 M 1325 1OLM 1165 120 24 M 250 24 3 ↑ | 0.1146M 24 24 01M 1165 Negative minimum Z,-C is24 and its column index is 6. So, the entering variable is x- Minimum ratio is 0 and its row index is 2. So, the leaving basis variable is x :. The pivot element is Enteringg2, Key Element60 (new)R2(old) x 60 101 R3(new) = R3(old)-150R2(new) R4lne) R4(old) R,(new)-Rold) Iteration4 145 185 250 7.25 101 161 16 131 49 -MI 25 16 160 25 21 185 75 16 160 0 3600 0 12000 0 3M 545 101M 4225 161M 6055 16 23M 1185 10LM 5825 161M 837 16 131M 5325 124 = 13875 | | | 185 | |-9 | -0.30621| + 16 16 152 32 31M 7325 00.3062M 16 16 16 523 Negative minimum Z-C, is -0.3062M+ 32and its column index is 7. So, the entering variable is x7- Minimum ratio is 81.63 and its row index is 1. So, the leaving basis variable is A The p vot element is面 Enteringng1, Key Element - 160 Activate Windows Go to Settings to activate Windows 160 Ri(new) = Ri(old) x-19

Rlnew R (old)R(new) 160 R4(new) = R4(old)-R1(new) e-R (old) Iteration-5 145 185 250 9-7.25 0,03 230 49 8080 230 2620 49 49 49 49 330 49 1100 49 49 2900 49 172400 49 1149 49 8080 49 379 49 2620 49 83 185 | 0 230 230 0 12000 92080 49 79830 49 489500 49 18555 281045 6595 49 417 49 41553 4900 185 98 4635 241845 4565 49 41553 Negative minimum -C is 4900 and its column indcx is 8. So, the entering variable isxg Minimum ratio is 1000 and its row index is I. So, the leaving basis variable is x The pivot clement is Entering X Departing xKey Elemet49 49 49 R (new)-R (old(new) RnewR (oldRw) Iteration-6 800 145 185 250 7.25 0.03 115 49 -0.03 1000 2020 0 -655 0 (60) 0 20o to Sdttings to activato Windbws

25 23 185 100 -59 .19 115 805 49 0 11000 2020 655 89117 18520 11931 40 73507 38653 400 こ 18470 100 60972 4911020 13531 78507 41553 Negative minimum Z, C, is and its column index is 2. So, the entering variable isxz. Minimum ratio is 0 and its row index is 2. So, the leaving basis variable is The pivot element is 60. Entering xzDepartng Key Element R2(new)-R2(old) ÷ 60 Ri(new)R,(old)+2020R-(new) Ry(new) R (old)+59R2(new) R4(new) = R4(old) Rs(new) = Rs(old)-2020R2(new) -60 Iteration-7 145 185 250 0.03 x1 101 xg 0.03 1000 12 120 24 23 24 60 185 100 24 120 80 101 11000 12 12503 185 8393 40 6793 40 7789 20 2789 20 4856 25 5081 399 -18470 800 40 100 6703 181 Since all 7,-G 0 Hence, optimal solution is arrived with value of variables as: Max z-18470