CS计算机代考程序代写 decision tree Microsoft Word – MGMT20005_S1_2017_Solutions

Microsoft Word – MGMT20005_S1_2017_Solutions

Question 1

a. Use a decision tree to recommend a decision for Giant.

[6 marks]

Recommended Decision: Purchase Component (d2)

b. Compute EVPI to determine whether Giant should attempt to obtain a better estimate of

demand.

[4 marks]

EVPI = $12,000,000

c. Suppose Giant will conduct the above market study. Use a decision tree to recommend a

decision for Giant.

[14 marks]

If Favourable

State of nature

s
i

Prior

probabilities

P(s
j
)

Conditional

probabilities

P(F | s
j
)

Joint

probabilities

P(F ∩ s
j
)

Posterior

probabilities

P(s
j
| F)

s1 0.4 0.1 0.04 0.13

s2 0.4 0.3 0.12 0.4

s3 0.2 0.7 0.14 0.47

P(F) = 0.3 1.00

1

2

-30

40

90

s1

0.4

s2

0.4

s3

0.2

3

10

30

50

s1

0.4

s2

0.4

s3

0.2

d1

d2

If Unfavourable

State of nature

s
i

Prior

probabilities

P(s
j
)

Conditional

probabilities

P(U | s
j
)

Joint

probabilities

P(U ∩ s
j
)

Posterior

probabilities

P(s
j
| U)

s1 0.4 0.9 0.36 0.51

s2 0.4 0.7 0.28 0.4

s3 0.2 0.3 0.06 0.09

P(U) = 0.7 1.00

Decision strategy:

If F then d1

If U then d2

d. Determine whether Giant should conduct the market study if it costs 3.5 million dollars.

[4 marks]

Giant should conduct the market study.

Question 2:

Iteration 1 2 3

Random Number 0.51 0.97 0.16

Units sold 800 1000 600

Revenue 17600 22000 13200

Variable cost 4800 6000 3600

Fixed cost 8000 8000 8000

Net Profit 4800 8000 1600

Question 3:

a. What are the optimal solutions (including the optimal values for decision variables and the

optimal objective function value)? [8 marks]

The optimal decision is x1=2.4, x2=3.2, and the optimal profit is 3*2.4+4*3.2=20.

b. Suppose the unit profit on x1 is increased from $3 to $7. Are the values for decision

variables in (a) still optimal? What is the optimal value of the objective function when this

unit profit is increased to $7? [8 marks]

Yes. The optimal objective function value 29.6.

Question 4:

a. Fill out the above tables. Note that the unfilled cells are marked by “xxx”. [3 marks]

The first cell 318.

The second cell is “not binding”.

The third cell is 0.

b. How much storage space is left unused? How many display racks are left unused?[2 marks]

Zero units of storage space are left. There are 36 display racks left unused.

c. By how much can the total profit on women’s umbrellas increase before the optimal

solution would change? [4 marks]

12

d. Interpret the shadow price for the storage space constraint. [2 marks]

The optimal profit increases by 2 per unit increase in the storage space, given the available

storage space is within the interval [96, 168].

e. Sky-Umbrella considers stocking a new arrival umbrella, Travel Umbrella. It uses 3 units of

storage space and 2 special display racks. Sky-Umbrella makes a profit of $7 per unit of

Travel Umbrella. What is your recommendation on whether to stock this new arrival

umbrella or not? [5 marks]

Sky-Umbrella should produce.

Question 5:

Let Pij = number of toy i to produce in month j

Sij = surplus (inventory) of toy i at end of month j

Min 12P11 + 12P 12 + 12P13 + 14P21 + 14P22 + 14P23 + 1S11 + 1S12 + 1S13 + 1.2S21 + 1.2S22 + 1.2S23

s.t. P11 − S11 = 195

S11 + P12 − S12 = 350

S12 + P13 − S13 = 600

S13 = 150

P21 − S21 = 395

S21 + P22 − S22 = 700

S22 + P23 − S 23 = 520

S23 = 60

5P11 + 4P21 ≤ 3500

5P12 + 4P22 ≤ 5000

5P13 + 4P23 ≤ 4800

2P11 + 3P21 ≤ 2100

2P12 + 3P22 ≤ 3000

2P13 + 3P23 ≤ 2500

Pij, Sij ≥ 0

Question 6:

Let xij be the units of products to be shipped from warehouse i to boutique j;

yi = 1 if a warehouse is constructed in suburb i, 0 otherwise

Minimize the total cost or z

= (2×11 + 4×12 + 6×13 + 8×14 + 7×15 + 4×21 + 5×22 + 2×23 + 4×24 + 4×25

+ 3×31 + 3×32 + 3×33 + 5×34 + 5×35 + 5×41 + 6×42 + 3×43 + 2×44 + 4×45)

+ (27,000y1 + 45,000y2 + 35,000y3 + 25,000y4)

Subject to x11 + x12 + x13 + x14 + x15 – 24,000y1 ≤ 0

x21 + x22 + x23 + x24 + x25 – 29,000y2 ≤ 0

x31 + x32 + x33 + x34 + x35 – 21,000y3 ≤ 0

x41 + x42 + x43 + x44 + x45 – 26,000y4 ≤ 0

x11 + x21 + x31 + x41 = 10,000

x12 + x22 + x32 + x42 = 12,000

x13 + x23 + x33 + x43 = 15,000

x14 + x24 + x34 + x44 = 8,000

x15 + x25 + x35 + x45 = 6,000

xij ≥ 0 and integer; yi = 0 or 1 for all i and j