Question

following problems, we will be performing sensitivity analysis on the following LPP: Maximize Z = 25x+30x, + 40x; subject to X; + 2xy + x2 < 40 8x, + X2 – 2x, 510 X-2x, +4x2 < 25 X,X23*, 20 and The final simplex tableau for this problem is given below. Basic Variable (0) (1) Z 1 0 X 0 3 /10 17/2 0 1 2150 Coefficient of: X2 X3 s 0 0 20 1 0 2/5 0 0 0 1 1/5 S2 S3 0 5 0 -1/10 1 1/2 01 /5 Right side 925 27/2 45/2 I 13 |0 17/2 (3) 1. Suppose the right-hand sides are changed to the values below: 540 b = 20 [20] Find the new solution. 2. Find the allowable range for the value of b, (which originally has a value of 25). 3. Find the new solution if we change the objective function to Z = 45x, +20x2 +22xz. 4. Find the allowable range for the value of cz (which originally has a value of 40). 5. Find the new solution if we change the problem to Maximize Z = 25x, +30x, +40x subject to 2x, + 2xy + xy S40 4 2x + x2 - 2x, 510 4x2 – 2x2 + 4x2 s 25 and X7, X7,20 (Note that this only changes the coefficients on the variable x,.)

Answer 1

D from the final table 3, 32 33 B- - 1/10 7 lo 1115 i o 10 l 15 ) new bis 40 b = then 20 Right hand side of final table 2 RHS = - B = 1 215 Vo 7 1 / 2 20 Y Y21 20t lo a w 4 tu then new solution. = 8, n = 0 2 = 8 and Max 7 = 8x30 + 8x40=560 & range of bez max {-XBi } = Abg - minh Xu Max Aba ming Bizo XB = (27/2 45/ 13] 1215 o 1/10 B's (Bij) = 1 o 1 1/2 Yoy

Max & s - 45/2 - - 13 Zz Abzt ming 1/10 L 1 /5 max f-us-658 s obze min { 135} -45 s obz s 125 b2 = 25 - 20 u bu tab 2 160 (3 it we change objective function 7 = 25x + 30% + 4083 to z = 450 trong + 2203 So final table will be 1 45 20 22 o o o (B xo a ez si s ses | RHS RHS/Yj 20 3110 o 25 o 1/10 | 27/2 45 O by 17/2 0 0 0 12 45/2 2.65 22 23 25 o I 15 o 1/5 13 o o 1-30.2 124 o 2.2 3-4 2o 45 2.0 ) 20 22 200 no 87 SRHS 1 215 -%7 216/17 00 217 17 45/17 22 m i vs 2 -cj o o o 12.4 3.55 4.17 635.94 so optimal solution is = us , = 216 85 17 17 n77 o 203 and max 7 = 635.94 TEST

nd the prima (4) Range of a му e , for - May - 100, -2s} & Azo - 25 S. Alo I : C = 40 then 4o -25 2 TA - 40 is u gaz L 40 it we use any value between (15, 40) then optimal solution will remain unchanged. because we change the coeficients of an si 8, 1 to (2 247 then values of 2 column in final table 1.215 کار o - 1/10 1/2 1 د B- 2 2 + ole d sin ۱ة then zi-9 = 35 So So there is no change por 22 = 27/7 = 13 C so in stulution still optimal optimal solution. PE