Bearbeite MS_SS_16 und vergleiche deine Lösungen. Aus dem Kurs Management Science an der Technische Universität München (TUM).
A chemical company wants to maximize the profit of its supply chain. In the supply chain the two glues L1 and L2 are produced. For the production, the two pre-products ammonia (A) and methanol (M) are required. Ammonia and methanol are produced on the same resource which has a total production capacity of 500.000 kg. To produce 1 kg of L1 0,5 kg of ammonia and 0,9 kg of methanol are needed. To produce 1 kg of L2 0,7 kg of ammonia and 0,4 kg of methanol are required. Ammonia and methanol which is not used for the production of L1 and L2 can be sold on the market. Due to supply chain contracts there are minimum quantities of L1 and L2 which have to be produced. Sales prices, production costs and minimum quantities are given in Table 1.
Products
Sales price (in monetary units)
Production costs (in monetary units)
Minimum quantities (in 1.000 kg)
A
M
L1
L2
10
20
30
35
5
8
14
16
0
0
5
7
Tabelle 1: Sales prices, production costs and minimum quantities
How many units shall the company produce in order to maximize profit? Formulate a linear program and explain the variables, the objective function value and the constraints.
Consider the following linear program (LP)
3.x1+0x2+2.x3=z
Min
subject to the constraints
2.x1+4.x2≥8
3.x1+3.x3≥2
2x1+5x2+1⋅x3=10
x1≤0
x2≥0
x3ЄR.
A chemical company wants to introduce a new product to the market. In case of a success (80% probability) a profit of 4 will be made. Otherwise, there will be a loss of 1.
Before introducing the product to the market, a market survey which costs x > 0 can be undertaken. In case the market survey is positive (80% probability) the chance of a market success is 98, 75%. If the market survey yields a negative result, the chance of a market success is 5%.
The decision tree provides the three alternatives “Market introduction”, “Market survey and No introduction to market” as well as the payoffs.
Consider the following integer program:
Max
z=6x1+5.x2
subject to
4x1+5x2<25
6x1+4x2≤30
x1≥0,x2≥0 and x1,x2=Z.
The LP relaxations of the subproblems are given in the following table:
Subproblem | Constraint added to P | Solution of relaxation |
---|---|---|
Pa | x1=3,75, x2=2,14, z=32,14 | |
Pb | x1≤3 | x1=3, x2=2,6, z=31 |
Pc | x1≤3, x2≥3 | x1=2,5, x2=3, z=30 |
Pd | x1≤3, x2≤2 | x1=3, x2=2, z=28 |
Pe | x1≤2, x2≤3 | x1=2, x2=3, z=27 |
Pf | x1≤4, x2≤1,5 | x1=4, x2=1,5, z=31,5 |
Pg | x1≤4, x2≥2 | infeasible |
Ph | x1≤4, x2≤1 | x1=4,33, x2=1, z=31 |
Pi | x1≤4, x2≥1 | x1=4, x2=1, z=29 |
Pj | x1≤5, x2≥1 | x1=5, x2=0, z=30 |
Determine the optimal solution with branch-and-bound. Generate the branch-and-bound tree according to LIFO with the larger character in the alphabet as tie breaker (i.e. when choosing between På and P₂ choose P₂ first). Draw the branch–and–bound tree and indicate in the tree which rule is used in order to fathom a subproblem. Indicate the order in which subproblems are considered.
At what stage would the branch-and-bound procedure have been terminated if the objective function value could have been not more than 5% below the optimal objective function value of the linear relaxation of the original problem?
opt,obj,Value=32.14
32.1432.14−31=0,035<0,05=5
Could have been stopped after Pb
Would the optimal solution have been found in less steps by applying the MUB-rule?
Provide an argument for your answer.
Yes because we would now we are going from Pa - Pf - Ph- P; - Pb
we are going from Pa-> Pf -> P -> P -> P -> P3 which is lower.
Consider the following maximization problem:
Max 1x1+1x2=N
subject to the constraints
1x2≤7−1x1
1x1+0x2≤4
0x1+1x2≤6
1x1+1x2≤7
x1,x2≥0
The start tableau is given by
x1,x2≥0
BV
Wert
X1
X2
X3
X4
X5
x6
X3
4
1
0
0
0
0
X4
6
0
0
1
0
0
X5
0
14
-1
0
0
0
x6
7
12
0
0
0
-2
0
1
1
0
0
0
0$
Which special case is given in the start tableau? What could be the effect of the special case when applying the Simplex algorithm?
Starting with the tableu given above perform the next iteration of the primal or dual Simplex in order to improve the objective function. Mark the pivot element in the start tableau. Write the resulting tableau in the table below.
Consider the following decision matrix for a minimization problem:
S1 | S2 | S3 | |
---|---|---|---|
a1 | X | 1 | X |
a2 | 2 | 1 | 2 |
a3 | 5 | 1 | -2 |
a) Determine the efficient alternatives depending on x with x in the interval [— 10, —9, . . ., 9, 10]. Provide an argument for your answer.
b) Determine the maximum value for x for which a decision maker applying the Laplace rule would choose alternative a₁. Provide a rationale for your answer.
Determine the efficient alternatives depending on x with x in the interval [— 10, —9, . . ., 9, 10]. Provide an argument for your answer.
Determine the maximum value for x for which a decision maker applying the Laplace rule would choose alternative a₁. Provide a rationale for your answer.