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Student ID ________________
Semester / Year: Semester 2 2019
Faculty / Dept: Management and Marketing
Subject Code: MGMT20005
Subject Name: Business Decision Analysis
Writing Time: 2 hrs
Reading Time: 15 minutes
Open Book Status: No
Number of Pages (including this page): 6
Authorised Materials:
Scientific calculators permitted
English-other language translation dictionaries permitted
No lap top or note book computers
Instructions to Students:
This examination contributes 60% to the final subject mark.
This examination paper includes 6 questions with a total mark of 100. You must answer them all.
Use two decimal points for calculations where relevant.
Instructions to Invigilators:
Paper to be held by Library: Yes
Student may keep the paper: No
Student may annotate the paper during reading time: Yes
Extra Materials Required:
Note: Graph paper is attached at the end of this booklet.
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Question 1 [28 marks]
An appliance dealer must decide how many new microwave ovens to order for next month. The
ovens each cost $220 and sell for $300. Because the oven manufacturer is coming out with a new
product line in two months, any ovens not sold next month will have to be sold at the dealer’s half
price clearance sale. Additionally, the appliance dealer feels he suffers a loss of $25 for every oven
demanded when he is out of stock. On the basis of past months’ sales data, the dealer estimates
the probabilities of monthly demand (D) for 1, 2, or 3 ovens to be 0.5, 0.3, and 0.2, respectively.
Consistent with the demand scenarios, the dealer is looking to order 1, 2, or 3 ovens.
The dealer is considering conducting a telephone survey on the customers’ attitudes towards
microwave ovens. The results of the survey will either be favorable (F) or unfavorable (U). The
dealer’s probability estimates for the survey results based on the number of units demanded are:
P(F | D = 1) = 0.3 P(F | D = 2) = 0.6 P(F | D = 3) = 0.9
P(U | D = 1) = 0.7 P(U | D = 2) = 0.4 P(U | D = 3) = 0.1
a. Use a decision tree to recommend a decision based on EMV when the survey is not
conducted. [6 marks]
EV(order 1) = 62.5
EV(order 2) = 80
EV(order 3) = 45
Optimal decision based on EV is to order 2, and the maximum profit is 80.
b. Use EVPI to determine whether the dealer should attempt to obtain a better estimate of
demand. [4 marks]
In the presence of perfect information:
If D=1, then order 1, which gives 80
If D=2, then order 2, which gives 160
If D=3, then order 3, which gives 240
EVwPI = 80*0.5+160*0.3+240*0.2 = 136
EVPI = EVwPI – EvwoPI = 136 – 80 = 56
c. Use a decision tree to determine the optimal decision for the appliance dealer given that
the survey is conducted? [14 marks]
favourable
states of nature Prior conditional joint posterior
D=1 0.5 0.3 0.15 0.294
D=2 0.3 0.6 0.18 0.353
D=3 0.2 0.9 0.18 0.353
0.51
unfavourable
states of nature Prior conditional joint posterior
D=1 0.5 0.7 0.35 0.686
D=2 0.3 0.4 0.12 0.235
D=3 0.2 0.1 0.02 0.039
0.49
Optimal decision:
If favorable, then order 2 and the expected profit is 107.06;
Otherwise, order 1 and the expected profit is 69.02.
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d. What is the expected value of the information provided by the survey?
[4 marks]
EVwSI = 107.06*0.51+69.02*0.49=88.42
EVSI = EVwSI – EvwoSI = 88.42-80=8.42
Question 2 [12 marks]
Luke’s utility function is given by 𝑈𝑈(𝑥𝑥) = 100𝑥𝑥 − 𝑥𝑥2 for 0 ≤ 𝑥𝑥 ≤ 50 where 𝑥𝑥 represents total
wealth in dollars.
a. If Luke currently has $20, should he take a bet in which he will win $15 with probability 0.3
and lose $10 with probability 0.7, based on expected utility?
[6 marks]
NOT BET: U(20)=1600
BET: EU=0.3*U(35)+0.7*U(10)=1312.5
So, it is optimal not to bet based on the EU approach.
b. If Luke’s total assets are currently $25, what is the certainty equivalent if he takes the bet
in (a)? What is the risk premium he would pay for the bet? [6 marks]
EU=0.3*U(40)+0.7*U(15)=1612.5
Certainty equivalent is obtained by solving the equation 100*CE-CE^2=1612.5.
Thus, CE is 20.21.
EV=0.3*40+0.7*15=22.5
Risk Premium = EV – CE = 2.29
Question 3 [16 marks]
Consider the following linear program:
Max Z = 4 X1 + 5 X2 (total profit in dollars)
Subject to:
Constraint A: 2 X1 + 4 X2 <= 120 Constraint B: 4 X1 + 3 X2 <= 140 Constraint C: X1 + X2 >= 40
X1 ≥ 0, X2 ≥ 0
Use the graphical solution approach to solve the above linear programming model, and answer
the following questions (you need to use the graph paper at the end of this booklet to draw
graphs):
a. What are the optimal solutions (including the optimal values for decision variables and the
optimal profit)? [8 marks]
Optimal values for decision variables: x1=20, x2=20
Optimal objective value: 180
b. Suppose the unit profit for X1 is decreased from $4 to $3. Are the values for decision
variables in (a) still optimal? What is the optimal value of the objective function when this
unit profit is decreased to $3? [8 marks]
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The optimal decision variable values remain the same.
But the objective function value becomes 160.
Question 4 [16 marks]
A farmer in Iowa has a 100-acre farm on which to plant watermelons and cantaloupes. Every acre
planted with watermelons requires 50 gallons of water per day and must be prepared for planting
with 20 pounds of fertilizer. Every acre planted with cantaloupes requires 75 gallons of water per
day and must be prepared for planting with 15 pounds of fertilizer. The farmer estimates it will take
2 hours of labor to harvest each acre planted with watermelons and 2.5 hours of labor to harvest
each acre planted with cantaloupes. He believes that watermelons will sell for about $3 each, and
cantaloupes will sell for about $1 each. Every acre planted with watermelons is expected to yield
90 saleable units. Every acre planted with cantaloupes is expected to yield 300 saleable units. The
farmer can pump about 6,000 gallons of water per day for irrigation purposes from a shallow well.
He can buy as much fertilizer as he needs at a cost of $10 per 50-pound bag and can hire laborers
to harvest the fields at a rate of $5 per hour. Finally, the farmer decides that the land use for
watermelons cannot exceed that for cantaloupes.
If the farmer sells all the watermelons and cantaloupes he produces, how many acres of each crop
should the farmer plant to maximize the total profit?
The following table shows the partial sensitivity analysis report produced by Excel Solver.
a. Formulate an linear programming (LP) model for the problem.
[4 marks]
Let x1 be the number of acres of land for watermelon; x2 be the number of acres of land for
cantaloupes
Maximise total profit or
z = 3*90*x1 + 1*300*x2 – (2*x1+2.5*x2)*5 – (20*x1+15*x2)*10/50=256×1+284.5×2
Subject to x1 + x2 ≤ 100
50*x1 + 75*x2 ≤ 6000
X1 – x2 <= 0
x1 and x2 ≥ 0
b. For the optimal solution, how much land is used to grow each of the two crops, and what is
the maximum profit?
[2 marks]
x1=48 acres of land for watermelons
x2=48 acres of land for cantaloupes
Variable Cells
Final Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficient Increase Decrease
$B$16 ? 48 0 ? 1E+30 66.33333333
$C$16 ? 48 0 ? 99.5 540.5
Constraints
Final Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease
$B$21 land use LHS ? 0 ? 1E+30 4
$B$22 water LHS ? 4.324 ? 250 6000
$B$23 requirement LHS ? 39.8 ? 20 80
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The maximum profit is given by 25944
c. For the optimal solution, how much land is left unused and how much water is used?
[2 marks]
4 acres of land is left unused, i.e., 100 – 48 – 48 = 4.
No water is left used.
d. Would the optimal solution change if the price for watermelon is increased by $0.5?
[2 marks]
No. With $0.5 increase in the price of watermelon, there is $0.5*90=$45 increase in the
revenue and thus profit per acre, which is smaller than the allowable increase + infinity.
Thus, the optimal solution would not be changed.
e. What would be the optimal solution if the price for cantaloupe is increased by $0.5?
[2 marks]
The increase in the price of cantaloupe per acre is given by $.5*300=$150, which is greater
than the allowable increase 99.5. Thus, the optimal solution would change.
f. How much would the total profit increase if the water available is increased from 6,000
gallons to 6,200 gallons?
[2 marks]
The allowable increase for the RHS of the constraint is 250. The increase in the water is
200=6200-6000, smaller than the allowable increase 250. Thus, the shadow price $4.324 can
be used to examine the profit increase due to the increase in water. We obtain that the
profit increase is given by 200*$4.324=$864.8.
g. The farmer has an opportunity to sell 3 acres of land at a price of $1,000. Do you recommend
him make the sale?
[2 marks]
The allowable decrease for the RHS of the constraint is 4. The reduction in land is 3 acres,
which is smaller than the allowable decrease 4. Thus, the shadow price $0 can be used to
examine the revenue loss due to the reduction in land. We obtain that the profit loss is given
by 3*$0=$0, which is smaller than the price 1,000. Therefore, we recommend him make the
sale.
Question 5 [12 marks]
Johnson Bicycle makes a plastic tricycle that is composed of three major components: a
handlebar-front wheel-pedal assembly, a seat and frame unit, and rear wheels. The company has
orders for 12,000 of these tricycles. Current schedules yield the following information.
Requirements Cost to
Manufacture
Cost to
Purchase Component Plastic Time Space
Assembly 3 10 2 8 12
Seat/Frame 4 6 2 6 9
Rear wheel (each) 0.5 2 0.1 1 3
Available 50,000 160,000 30,000
The company obviously does not have the resources available to manufacture everything needed
for the completion of 12,000 tricycles so has gathered purchase information for each component.
Develop a linear programming model to tell the company how many of each component should
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be manufactured and how many should be purchased in order to provide 12,000 fully completed
tricycles at the minimum cost.
[12 marks]
Let xij denote the number of component i when manufactured (j=1) or purchased (j=2),
where i=1 corresponds to the assembly component, i=2 corresponds to the seat/frame
component, and i=3 corresponds to rear wheel.
Minimize the total cost or z=8 x11+6 x21+1 x31 + 12 x12 + 9 x22 + 3 x32
Subject to:
X11 + x12 = 12,000
X21 + x22 = 12,000
X31 + x32 = 24,000
3 x11 + 4 x21 + 0.5 x31 <=50,000
10 x11 + 6 x21 + 2 x31 <=160,000
2 x11 + 2 x21 + 0.1 x31 <= 30,000
Xij >= 0
Question 6 [16 marks]
The HotAir Company manufactures heaters that are sold to five different retail customers across
Australia. The company is evaluating its manufacturing and logistics strategies to ensure that it is
operating in the most efficient manner possible. The company can produce heaters at three plants
across the country and stock these units in any of four different warehouses. The cost of
manufacturing and shipping a unit between each plant and warehouse is summarized in the
following table along with the monthly production capacity and fixed cost of operating each plant.
Unit Shipping Cost to Warehouses
Plants Fixed Cost Warehouse 1 Warehouse 2 Warehouse 3 Warehouse 4 Capacity
Plant 1 5,000 9 4 3 7 700
Plant 2 2,000 12 6 2 6 200
Plant 3 4,000 4 10 11 5 500
Similarly, the per-unit cost of shipping units from each warehouse to each customer is given
below, along with the monthly fixed cost and stocking capacity of operating each warehouse as
well as the monthly demand at each customer.
Unit Shipping Cost to Customers
Warehouses Fixed Cost Customer 1 Customer 2 Customer 3 Customer 4 Customer 5 Capacity
Warehouse 1 1,000 5 7 8 7 4 1,000
Warehouse 2 1,500 10 8 6 8 2 400
Warehouse 3 2,000 9 4 3 4 3 600
Warehouse 4 1,400 4 10 11 3 6 400
Demand 200 300 200 150 250
HotAir would like to determine which plants and warehouses it should operate to meet demand
in the most cost-effective manner. Formulate an LP/IP model that helps HotAir make this decision.
[16 marks]
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Relabel all the nodes in the following manner:
Plant 1 (node 1), Plant 2 (node 2), Plant 3 (node 3), Warehouse 1 (node 4), Warehouse 2
(node 5), Warehouse 3 (node 6), Warehouse 4 (node 7), Customer 1 (node 8), etc.
Let yi indicate whether or not node i are used, where i=1,2,…, 7.
Let xij denote the shipping quantity from node i to node j.
Minimize the total cost or z=5000y1+2000y2+4000y3+1000y4+1500y5+2000y6+1400y7
+9×14+4×15+……+5×37
+5×48+7×49+……+6×712
Subject to
X14+x15+x16+x17<=700y1
X24+x25+x26+x27<=200y2
X34+x35+x36+x37<=500y2
X48+x49+x410+x411+x412<=1000y4
X58+x59+x510+x511+x512<=400y5
X68+x69+x610+x611+x612<=600y6
X78+x79+x710+x711+x712<=400y7
X48+x49+x410+x411+x412=x14+x24+x34
X58+x59+x510+x511+x512=x15+x25+x35
X68+x69+x610+x611+x612=x16+x26+x36
X78+x79+x710+x711+x712=x17+x27+x37
X48+x58+x68+x78=200
X49+x59+x69+x79=300
X410+x510+x610+x710=200
X411+x511+x611+x711=150
X412+x512+x612+x712=250
Xij>=0 for all ijs, and yi=0 or 1 for i=1,…,7.
Mathematical formula that may be used:
Bayes’ Theorem
𝑃𝑃�𝐵𝐵𝑗𝑗�𝐴𝐴𝑖𝑖� =
𝑃𝑃�𝐴𝐴𝑖𝑖�𝐵𝐵𝑗𝑗�𝑃𝑃(𝐵𝐵𝑗𝑗)
𝑃𝑃(𝐴𝐴𝑖𝑖|𝐵𝐵1)𝑃𝑃(𝐵𝐵1) + 𝑃𝑃(𝐴𝐴𝑖𝑖|𝐵𝐵2)𝑃𝑃(𝐵𝐵2) + ⋯+ 𝑃𝑃�𝐴𝐴𝑖𝑖�𝐵𝐵𝐽𝐽�𝑃𝑃(𝐵𝐵𝐽𝐽)
Roots of Quadratic Equations
The roots of the quadratic equation 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 are given by
𝑥𝑥 =
−𝑏𝑏 ∓ √𝑏𝑏2 − 4𝑎𝑎𝑐𝑐
2𝑎𝑎
,
where 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 are constants.
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END OF EXAMINATION PAPER
This examination contributes 60% to the final subject mark.