Ioe202 hw3 jahnavi - Operations Modeling Notes PDF

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Operations Modeling Notes...

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1.

Stage 4 Stage 1 Stage 2

Stage 3

As shown by the arrows, there are 4 stages in this diagram. 1. In stage 1, the minimal solution is from New York to Cleveland and the maximum solution is from New York to St. Louis. The possible solution for stage 1 can be written as x1(NY) = min{400+x1(CLE)} x1(NY) = min{800+x2(NASH)} x1(NY) = min{950+xz(STL)} 2. In stage 2, the possible solutions can be from Cleveland to Phoenix, Cleveland to Dallas, St. Louis to Phoenix, St. Louis to Dallas, Nashville to Dallas, or Nashville to Salt Lake City. This can be written as x2(CLE) = min{1800+x3(PHX)} = 2200 x2(CLE) = {900+x3(DALLAS) + 900+ x3(PHX)} = 2200 x2(STL) = min{1100+x3(PHX)} = 2050 x2(STL) = min{600+x3(DALLAS)} = 1550 x2(STL) = min{600+x3(DALLAS) + 900+x3(PHX)} = 2450 x2(NASH) = min{600+x3(DALLAS)} = 1400 x2(NASH) = {1200+x3(SLC)} = 2000 x2(NASH) = {600+x3(DALLAS)+ 1000+x3(SLC)} = 2400 3. In stage 3, the possible solutions can be Phoenix to LA, Dallas to Phoenix, Dallas to LA, Dallas to Salt Lake City, and Salt Lake City to LA. These can be written as From CLE: x3(PHX) = min{400+x4(LA)} = 2600 From CLE: x3(DALLAS + PHX) = min{400+x4(LA)} = 2600 From CLE: x3(DALLAS + SLC) = min{600+ x4(LA)} = 2900 From STL: x3(PHX) = min{400+x4(LA)} = 2450 From STL: x3(DALLAS) = min{1300+(LA)} = 2850 From STL: : x3(DALLAS + PHX) = min{400+x4(LA)} = 2850 From STL: x3(DALLAS + SLC) = min{600+ x4(LA)} = 3150 From NASH: x3(DALLAS) = min{1300+(LA)} = 2700 From NASH: x3(DALLAS + PHX) = min{400+x4(LA)} = 2700 From NASH: x3(DALLAS + SLC) = min{600+ x4(LA)} = 3000 From NASH: x3(SLC) = min{600+x4(LA)} = 2600 4. x4(LA) = 0

From looking at these calculations, going from New York to Cleveland has the smallest value of 400. However, from here the driver can either go to Dallas, Phoenix and then LA, Phoenix and then LA, or Dallas to Salt Lake City and then LA, but none of them are an optimal solution. Therefore, New York to Cleveland is not the best choice for the first stage. If the driver chooses St.Louis he can go to Phoenix and then LA, Dallas to Phoenix to LA, Dallas to LA, or Dallas to Salt Lake City to LA. Looking at the calculation going from Salt Lake City to Dallas seems like an optimal solution but then going from Dallas to LA is larger than from Phoenix to LA as seen in the calculations above. If you were to try Nashville, the possible routes could be from Nashville to Dallas and then LA, or Nashville to Salt Lake City and then LA, neither of which are optimal. Therefore, the optimal solution would be from New York to St.Louis, to Phoenix to LA for a total of 2450 (optimal solutions highlighted above). 2. x6 = 200 + 2000(10) = 20200 x5 = min{200 + 3000(10)+x6} = 50400 x5 = min{200+5000(10)+2000(0.05)} = 50300 x4 = min{200+2500(10)+x5} = 75500 x4 = min{200+5500(10)+3000(0.05)+x6} = 75550 x4 = min{200+7500(10)+7000(0.05)} = 75550 x3 = min{200+1000(10)+x4} = 85700 x3 = min{200+3500(10)+2500(0.05)+x5} = 85625 x3 = min{200+6500(10)+8500(0.05)+x6} = 85825 x3 = min{200+8500(10)+14500(0.05)} = 85925 x2 = min{200+3000(10)+x3} = 115825 x2 = min{200+4000(10)+1000(0.05)+x4} = 115750 x2 = min{200+6500(10) + 6000(0.05) +x5} = 115800 x2 = min{200+9500(10)+15000(0.05)+x6} = 116150 x2 = min{200+11500(10) + 23000(0.05)} = 116350 x1 = min{200+2000(10)+x2} = 135950 x1 = min{200+5000(10)+3000(0.05)+x3} = 135975 x1 = min{200+6000(10)+5000(0.05)+x4} = 135950 x1 = min{200+8500(10)+12500(0.05)+x5} = 136125 x1 = min{200+11500(10)+24500(0.05)+x6} = 136625 x1 = min{200+13500(10)+34500(0.05)} = 136925 Therefore, the optimal cost would be 135950

Silver-Meal Heuristics V(1) = 200 V(2) = (200+3000(0.05))/2 = 175 V(3) = (200+5000(0.05))/3 = 150 V(4) = (200+12500(0.05))/4 = 206.25 V(1) covers the production till V(4) After V(4) V(1) = 200 V(2) = (200+3000(0.05))/2 = 175 V(3) = (200+(7000(0.05))/3 = 183.33 This covers till quarter 5. This heuristic gives a total cost of 136000 which is close to what we got for the W-W Algorithm. 3. x8 = 0 x7 = {n7 + x8} = 0.6 x6 = {n67 + x7} = 1.2 x6 = {n68 + x8} = 0.85 x5 = {n56+x6} = 1.45 x5 = {n57+x7} = 1.45 x5 = {n58+x8} = 1.2 x4 = {n45+x5} = 2.25 x4 = {n46+x6} = 1.7 x4 = {n47+x7} = 1.8 x4 = {n48+x8} = 1.65 x3 = {n34+x4} = 2.25 x3 = {n35+x5} = 2.05 x3 = {n36+x6} = 2.05 x3 = {n37+x7} = 2.25 x3 = {n38+x8} = 2.25 x2 = {n23+x3} = 2.65 x2 = {n24+x4} = 2.5 x2 = {n25+x5} = 2.4 x2 = {n26+x6} = 2.5

x2 = {n27+x7} = 2.85 x2 = {n28+x8} = 2.95 x1 = {n12+x2} = 3.05 x1 = {n13+x3} = 2.9 x1 = {n14+x4} = 2.85 x1 = {n15+x5} = 2.85 x1 = {n16+x6} = 3.05 x1 = {n17+x7} = 3.55 x1 = {n18+x8} = 3.7 Therefore, the optimal policy would be to make a purchase at the beginning of year 1 and the beginning of year 5(or year 4) making the optimal cost $2.85 million. 4. Stages: the number of turns s1 – sc : from turn 1 to turn c States: xc represents the number of coins at each stage Action: ac represents the action at each stage pc = number of coins picked up by the other person Optimality equation: nc(sc) = max ac{nc=1(sc - ac – pc)} ∀ sc , c nc(sc) =

s =1 {01ifotherwise c

You are guaranteed to win if you go first when there is a multiple of the maximum number of coins plus 2 plus y (max coins + 2 + y) where y is between 1 and the maximum number of coins....