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ADMS 3330

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ADMS 3330 Winter 2007 Final Exam Solutions
Question 1 Given the linear program
Min 3X – 4Y
S.T.
-X + 2Y ≤ 8 ___ (1)
X + 2Y ≥ 12 ___ (2)
2X + Y ≤ 16 ___ (3)
X, Y ≥ 0
A) Write the problem in standard form
B) Solve the problem using graphical solution procedure
C) Find the dual prices
D) What are the values of the three slack/surplus variables at the optimal solution?
E) If the second constraint becomes X + 2Y = 12, what is the feasible region (please mark it
properly)? What is the optimal solution?
ANSWERS:
a) Standard Form: Min 3X – 4Y + 0S1 + 0S2 + 0S3
S.T. -X + 2Y + S1 = 8
X + 2Y – S2 = 12
2X + Y + S3 = 16
X, Y, S1, S2, S3 ≥ 0
y
2x + y ≤ 16
9
8
-x + 2y ≤ 8
7
Optimal (4.8, 6.4)
x + 2y ≥ 12 Solution Z = - 11. 2
6
5
(2, 5) Feasible Region
Z = - 14
4
3 (6.67, 2.67)
Z = 9. 33
2
1
x
0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5
Page 1 of 11 ADMS 3330 Winter 2007 Final Exam Solutions
Optimal solution occurs at extreme point 1.
At Optimal solution X = 2, Y = 5 OFV = -14
c)
Dual price is the improvement in the OFV per unit increase in the right hand side of a
constraint.
Use at least one binding constraint in calculating dual price
Dual Price for constraint 1 Dual Price for constraint 2
(RHS of Constraint 1 increased by 1) (RHS of Constraint 2 increased by 1)
-X + 2Y = 9 ___ (1) -X + 2Y = 8 ___ (1)
X + 2Y = 12 ___ (2) X + 2Y = 13 ___ (2)
Adding (1) and (2) gives Adding (1) and (2) gives
4Y = 21 4Y = 21
Y = 5.25 Y = 5.25
Substituting in (2) gives Substituting in (2) gives
X + 2(5.25) = 12 X + 2(5.25) = 13
X = 12 – 10.5 = 1.5 X = 13 – 10.5 = 2.5
New OFV = 3(1.5) – 4(5.25) = -16.5 New OFV = 3(2.5) – 4(5.25) = -13.5
Dual Price = Old OFV – New OFV Dual Price = Old OFV – New OFV
Dual Price = -14 – (-16.5) = 2.5 Dual Price = -14 – (-13.5) = -0.5
Since constraint 3 is not binding (it did not touch the optimal solution point), its dual price is
0.
(Note that Dual Price = Old OFV –New OFV for minimization problems. For max, Dual Price = New
OFV – Old OFV)
d)
The values of slack/surplus for constraint (1) and (2) will be 0 at optimal solution (X = 2, Y = 5)
because they are binding:
Slack for Constraint 1 Surplus for Constraint 2 Slack for Constraint 3
-X + 2Y + S1 = 8 X + 2Y – S2 = 12 2X + Y + S3 = 16
-2 + 2(5) + S1 = 8 2 + 2(5) + S2 = 12 2(2) + 5 + S3 = 16
S1 = 8 – 8 = 0 S2 = 12 – 12 = 0 S3 = 16 – 9 = 7
The slack/surplus values are 0, 0, 7 respectively for constraints 1, 2, and 3.
Page 2 of 11 ADMS 3330 Winter 2007 Final Exam Solutions
e) The feasible region is the set of points on the line x+2y = 12 between extreme points and
The optimal solution remains the same as in part b)
X = 2, Y = 5 OFV = -14
Page 3 of 11 ADMS 3330 Winter 2007 Final Exam Solutions
Question 2
Max X + Y
ST
X – kY ≤ 1
X, Y ≥ 0
where k is a real number.
a) What value(s) of k makes the above LP have alternative optimal solutions?
b) What value(s) of k makes the above LP have a unique optimal solution?
ANSWERS:
a) The slope of the objective function =-1/1 = -1
(Note: The slope for aX + bY is –a/b)
The slope of the constraint X – kY ≤ 1 is -1/-k = 1/k
For the LP to have alternative optimal solutions, the slope of the objective function must equal
the slope of one of the constraints. Since there is only one constraint, alternative optimal
solutions occur when
1
−1 = 𝑘
−𝑘 = 1
Therefore 𝑘 = −1
b) In the absence of alternative optimal solution, we have a unique one.
Therefore a unique solution occurs when
𝑘 ≠ −1
Page 4 of 11 ADMS 3330 Winter 2007 Final Exam Solutions
Question 3
Molton Canadian is only making beer at 3 different locations. The capacity (in thousand bottles) or
3 plants in different locations: 200, 250 and 150. Once beer is made, it is shipped and stored in
either one, two warehouses. Table 1 gives the production and shipping cost per thousand bottles of
beer made at a particular location and shipped to a particular warehouse. Table 2 shows the cost of
shipping a thousand bottles of beer from each warehouse a customer. Customers would like to
receive 200 (thousand bottles for customer I), and 250 (thousand bottles for customer 2). Table 3
gives the annual fixed cost of operating each plant and warehouse. Molton Canadian is to minimize
the annual cost of marketing customer demands. Now we want to formulate this problem as a
mixed-integer linear program problem. Please answer the following questions.
Table I To
Warehouse Warehouse
1 2
Location 1 $900 $1,000
From Location 2 $1,100 $800
Location 3 $800 $700
Table 2
Customer 1 Customer 2
From Warehouse 1 $40 $80
Warehouse 2 $70 $40
Table 3
Fixed Annual
Cost
Location 1 $20,000
Location 2 $35,000
Location 3 $30,000
Warehouse 1 $40,000
Warehouse 2 $25,000
a) This is a combination of the transshipment problem and the fixed cost problem. Please
draw the network for the included transshipment problem.
b) Define your decision variable properly
c) Write down the objective function
d) Write down all constraints.
Page 5 of 11 ADMS 3330 Winter 2007 Final Exam Solutions
ANSWER:
a)
b) Let
Xij number of bottles (in thousands) shipped from node i to node j
for all i and j
Yi = 1 if node i is used, 0 otherwise for i = 1, 2, 3, 4, 5
c) Objective Function
Min 900X 14 + 1000X 15+ 1100X 24 + 800X25 + 800X 34+ 700X 35 + 40X46 + 80X47 + 70X56 +
40X 57+ 20,000Y 1 35,000Y +230,000Y +340,000Y + 45,000Y 5
d) Constraints
X14 + X15- 200Y1 ≤ 0 X14 + X24 + X34 – X46 – X47 = 0 X46 + X47 ≤ 450Y4
X24 + X25 - 250Y2 ≤ 0 X15 - X25 + X35 – X56 – X57 = 0 X56 + X57 ≤ 450Y5
X34 + X35 - 150Y3 ≤ 0 X46 + X56 = 200 Xij 0 for all i, j
X47 + X57 = 250 Yi = 0, 1 i = 1, 2, 3, 4, 5
Page 6 of 11 ADMS 3330 Winter 2007 Final Exam Solutions
Question 4
Dale Adams operates a delivery truck in South-Western Ontario. He has received 8 potential
shipments for the month of May. Dale realizes that he cannot accept all the shipments since their
combined total time exceeds 20 days, a limit on the number of days he is allowed to work in any
one month. The shipment duration in days and profit in dollars are given below.
Shipment 1 2 3 4 5 6 7 8
Profit 5 8 7 6 3 9 5 2
Duration 2 7 3 6 2 8

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