Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: THEend8_
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