Question

d) Given the primal problem Max z= 8x/+3x2+xz Subject to: x;+6x,+8x3<118 X, + 5x+10x<240 X1, X2,X3, 20 Write down its problem (5 marks) dual

Answer 1

Criven Standard primal problem as- Max 2 = 8 + 3 x 2 + X3 conwert max Step I Fiorst we will function to min objective objective funcion there are Min Zw = 118w, + 240 W₂ As there only two constraint will be only two decision variable w, and we Subject to : Wat W2 28 6W y + 5w ₂ 3 3 + 10w, 21 8 Wa Min Zw = 118w, +240 w2 Sulject to W, + W, 28 bw, +562 23 8 w, +10021 هونه ر له

Given wiccnt is br - 24 V 6-22 222 341 {2.52 are 2 here Rz and RA parallel Their equivalent OxZ = 3=1.52 84 The wicient will be 62 6(V) iz 거 24 vi zar & 1.52 {2.52 a O a be O and at Let potential at node node 6 be v. Now at b, applying KCL we get ( + iz tiz=0 i, = V-29 ,iz k Now 2.5+1.5 9

- V-29 +y=0 + V b 4 V-24 + x = 0 V-24 + 3 v=0 7 40 = 24 += 6 volt -18 6 Now = U-24 = -3A (-ve sign shows that direction is opposite as shown in arcuit) 7 current through resistor IR -BA 1.5A Now 4 Current through R2 is 1.5 ISA Now zov 4. 4 i R = 1.5*1.5 = 2-25 V Current through (RS= ISA Voltage arrows 1.5.2 is so current through Rq = 225 - 225 - 1125 A

Current through R3 = 2.25 = -3751 6 16) Power dissipation in R4 iq=1.12SA = (1.125)² x 6 = 7.59375 W