<p> MATH 3170 Assignment #2 Solution (20 mark) </p><p>Questions marked: 3.8.10, 3.9.12, 3.10.9, 3.11.9</p><p>3.6.4 </p><p>Let Xi = fraction undertaken of investment I (I = 1, 2, …, 9).</p><p>The LP formulation is: max z = 14X1 + 17X2 + 17X3 + 15X4 + 40X5 + 12X6 + 14X7 + 10X8 + 12X9 s.t. 12X1 + 54X2 + 6X3 + 6X4 + 30X5 + 6X6 + 48X7 + 36X8 + 18X9<=50 (Year 1) 3X1 + 7X2 +6X3 + 2X4 + 35X5 + 6X6 + 4X7 + 3X8 + 3X9<=20 (Year 2) All variables>=0.</p><p>Note that the question does not mention that the cash not used in Year 1 can be used in Year 2.</p><p>3.7.1 </p><p>Graphically, we find the optimal solution to be x1 = 50, x2 = 100 with optimal value z = 2500. 3.8.10 (5 marks) </p><p>Let Mi = Tons of coal shipped from Mine i and Xij = Tons of coal shipped from Mine i to Customer j. Here is appropriate formulation in LINDO:</p><p>MIN 50 M1 + 55 M2 + 62 M3 + 4 X11 + 6 X12 + 8 X13 + 12 X14 + 9 X21 + 6 X22 + 7 X23 + 11 X24 + 8 X31 + 12 X32 + 3 X33 + 5 X34 SUBJECT TO 2) M1 <= 120 3) M2 <= 100 4) M3 <= 140 5) X11 + X21 + X31 = 80 6) X12 + X22 + X32 = 70 7) X13 + X23 + X33 = 60 8) X14 + X24 + X34 = 90 9) 0.08 M1 + 0.06 M2 + 0.04 M3 <= 15 10) 0.05 M1 + 0.04 M2 + 0.03 M3 <= 12 11) M1 - X11 - X12 - X13 - X14 = 0 12) M2 - X21 - X22 - X23 - X24 = 0 13) M3 - X31 - X32 - X33 - X34 = 0 END</p><p>Of course, all variables are nonnegative.</p><p>Constraint (9) comes from 0.08M1+0.06M2+0.04M3 <= 0.05(M1+M2+M3) but since we know M1+M2+M3 is simply the sum of all the Xij’s which is 80+70+60+90=300, the RHS is imply 0.05(300)=15. Establishing constraint (10) is similar.</p><p>You can do the above formulation without defining M1, M2 and M3 by simply replacing each Mi with the corresponding expression given in (11) to (13).</p><p>3.9.12 (5 marks) </p><p>Let A = hundreds of liters of A purchased and processed B = hundreds of liters of B produced from A and processed C = hundreds of liters of C from A D = hundreds of liters of D from processed C iS = hundreds of liters of i sold (i=B,C or D) CP = hundreds of liters of C processed iL = hundreds of liters of product i left unsold (i=B,C or D)</p><p>The correct LP formulation in LINDO is as follows: </p><p>MAX 12 BS + 16 CS + 26 DS - 9 A - CP SUBJECT TO 2) - CS - CP + C - CL = 0 3) - BS + B - BL = 0 4) - DS + D - DL = 0 5) BS <= 30 6) CS <= 60 7) DS <= 40 8) - 0.6 A - 0.4 CP + B = 0 9) - 0.4 A + C = 0 10) - 0.6 CP + D = 0 11) 3 A + CP <= 200 END </p><p>Of course, all variables are nonnegative. Note that there are many different ways to model this problem as an LP problem. The above formulation is just one of them.</p><p>3.10.9 (5 marks) </p><p>Let xi = number of workers who get quarter i off it = inventory of mixers at end of quarter t mt = mixers produced during quarter t</p><p>The LP formulation is: min z = 30(i1 + i2 + i3 + i4) + 30,000(x1 + x2 + x3 + x4) s.t. i1 = 600 + m1 - 4000, i2 = i1 + m2 - 2000 i3 = i2 + m3 - 3000, i4 = i3 + m4 - 10,000 m1500(x2 + x3 + x4), m2500(x1 + x3 + x4), m3500(x1 + x2 + x4), m4500(x1 + x2 + x3) All variables 0</p><p>3.11.9 (5 marks) </p><p>Let Pi = Fraction of project I invested in, Bt = Money borrowed at time t, Ct= ending cash position at time t max z = 5.5P1 +6P3-P2+1.03C2.5-1.035B2.5 s.t.</p><p>C0 = 2-3P1-2P2-2P3+B0 C.5=1.03C0-1.035B0-P1-.5P2-2P3+B.5 C1=1.03C.5-1.035B.5+1.8P1+1.5P2-1.8P3+B1 C1.5=1.03C1-1.035B1+1.4P1+1.5P2+P3+B1.5 C2=1.03C1.5-1.035B1.5+1.8P1+1.5P2+P3+B2 C2.5=1.03C2-1.035B2+1.8P1+.2P2+P3+B2.5 All Bt<=2 ALL VARIABLES >=0</p><p>The above formulation assumes that all debts with interest must be repaid in exactly 6 months.</p>