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Econ 7322
Assessment Q1 a. TC=FC+VC =18000+12000*0.9=$28800 TR=12000*$3.2=$38400 P=TR - TC= 38400-28800=$9600 b. TC=TR TFC+TVC=Price per cake*No.of cake $18000+$0.9*No.of cake=$3.2*No.of cake $1800=$2.3*No.of cake So, No.of cake =18000/2.3=7826 c. The revenue: 3.2x=y Cost: 18000+0.9x=y The x is number of cupcake and y is money So we can see the cross point is the break-even d. 7826/12000*100%=65.22% So the maximum capacity break-even is 65.22% e. $18000+$0.9*No.of cake=$2.75*No.of cake $1800=$1.85*No.of cake So, No.of cake =18000/1.85=9730 Q2. a. 1. Decision variables: x1 = necklaces x2 = bracelets 2. Objective: Maximize profits 3. Constraints: Gold used ≤ Gold available Platinum used ≤Platinum available All variables ≥ 0, x2≤4 Max Z =300x1 + 400x2 s.t. 3x1 + 2x2 ≤ 18 2x1 + 4x2 ≤ 200 x1 ≥ 0, 4≤x2 ≥ 0 b. c. If x2=4, so 3x1 + 2x2 ≤ 18 2x1 + 4x2 ≤ 200 x1=2 Plug in Z’: 300x1 + 400x2 = 300*2 + 400*4 Z’=2200 So, will not the maximum demand for bracelets be met, will miss 200 unit. d. If no bracelets being produced, so x2=0 The gold available is 18 units, platinum is 20 units One necklaces need 3 units gold and 2 units platinum, so 18/3=6 20/2=10 6/3=2 10/2=5 So the max produce is make 6 necklaces, so profit =2*$300=$600 e. So we can see, if increasing the profit on a bracelet from $400 to $600, the profit is $3000, and if changing the platinum requirement for a necklace from 2 units to 3 units, the profit is $2000. Q3. a. Decision variables: x1 = mine1 x2 = mine2 Objective: Minimize profits Constraints: mine1 coasts ≥ mine1 coasts available Mine2 coasts≥ mine2 coasts available All variables ≥ 0 Min Z =12x1 + 8x2+24x3 s.t. 6x1 + 2x2 ≥12 2x1 +2x2 ≥8 4x1 +12x2 ≥24 x1 ≥ 0, x2 ≥ 0 b. Q4. a. Decision variables: x1 = permanent x2 = temporary Objective: Minimize profits Constraints: permanent coasts ≥ permanent coasts available Temporary coasts≥ temporary coasts available All variables ≥ 0 Min Z =64x1 + 42x2 s.t. 16x1 + 12x2 ≥450 0.5x1 + 1.4x2 ≤ 25 x1 + x2 ≤ 40 x1 ≥ 0, x2 ≥ 0 b. c. We can see the data is decreasing. d. The data shows more decreasing on objective. e. f. Part3 Q1 a. Max g’y Subject to B’y ≤ d y ≥ 0 The LP problem: Min d’x Subject to Bx ≥ g X ≥ 0 b. Intuitive proof: Primal is Unbounded: If the original problem is unbounded, this means that the objective function can grow without bound while still satisfying all constraints. Maximize g′y subject to B′y≤d y≥0 Being unbounded means there exists a feasible y such that g′y=∞ No Corresponding Dual Feasibility: If the primal target is unbounded, then there is no limit in the dual to how far the primal target can grow. Therefore, the dual cannot find a feasible solution corresponding to the infinite growth of the primordial. minimize d′x subject to Bx≥g x≥0 If the dual is feasible, there exists x≥0 such that Bx≥g. Contradiction: Using the fact that g′y=∞, multiply the dual constraint by y: y′Bx≥y′g x′By≥g′y=∞ But since B′y≤d and x≥0, we have: x′B′y≤x′d c. Dual is Unbounded: If the dual problem is unbounded, this means that the dual objective function can be reduced without limit while still satisfying all constraints. minimize d′x subject to Bx≥g x≥0 Being unbounded means there exists a feasible x such that d′x=-∞ No Corresponding Primal Feasibility: This means that there is no limit in the primal object to how much the dual object's target can be reduced. Therefore, the primal cannot find a feasible solution corresponding to the dual infinite reduction. maximize g′y subject to B′y≤d y≥0
Contradiction: Using the fact that d′x=−∞, multiply the primal constraint by x: x′B′y≤x′d=−∞ But since Bx≥g and y≥0, we have: x′B′y≥x′g′ d. Primal Problem: maximize c′x subject to Ax≤b x≥0 Dual Problem: minimize b′y subject to A′y≥c y≥0 Proof: Primal Feasible and Dual Feasible: Since x0 and y0 are feasible for the primal and dual, we have that: Ax0≤b and x0≥0 A′y0≥c and y0≥0 Weak Duality: By weak duality, we know that for any feasible x and y: c′x ≤ b′y Given Condition: We are given that: b′y0=c′x0 Putting It All Together: By combining the given condition with weak duality, we have: c′x0=b′y0 Since c′x≤b′y for any feasible x and y, x0 and y0must be optimal.
Q2 a. b. c. We can see from the sensitivity form data, the shadow price show more higher, so can earn for each additional of that profit d. From question a we can see the process 2 is not produced in the optimal solution, need to increase the profit for that product by the absolute value of the reduced cost in order for it to be produced in the optimal solution. Q3 1. Decision variables: x1 = midnight -4:00am x2 =4:00am-8:00am X3=start 8:00am X4=start noon X5=start 4:00pm X6=8:00pm-midnught 2. Objective: Minimize profits 3. Constraints: midnight -4:00am ≥ 10 4:00am-8:00am ≥12 8:00am-noon ≥20 Noon-4:00pm ≥25 4:00pm -8:00pm ≥32 8:00pm-midnight ≥18 All variables ≥ 0 Min Z =80x1 + 500x2+420x3+300x4+270x5+210x6 s.t. x1≥10 x1 + x2 ≥ 12 X2 + x3 ≥20 X3+x4≥25 X4+x5≥32 X5+x6≥18 x123456 ≥ 0