Solved problems for Semester test preparation
Conversion of LP problem formulation into standard form
for Simplex solving
Remember / important:
 For ≤  + Slack
 For ≥  - Surplus + Artificial
 For =  + Artificial
PPT sides example (Module 7 Part 2)
Minimize cost = R5X1+ R6X2
subject to
X1
X1, X2
+ X2
= 1,000 lb
X1
≤ 300 lb
X2
≥ 150 lb
≥0
Equalities for each constraint is different. Therefor the standard form will look like this:
Minimize cost = $5X1 + $6X2 + $0S1 + $0S2 + $MA1 + $MA2 (don’t forget to include S & A’s)
subject to:
1X1 + 1X2 + 0S1 + 0S2 + 1A1 + 0A2 = 1,000
1X1 + 0X2 + 1S1 + 0S2 + 0A1 + 0A2 = 300
0X1 + 1X2 + 0S1 – 1S2 + 0A1 + 1A2 = 150
X1, X2, S1, S2, A1, A2 ≥ 0
Now you can complete the first simplex tableau
Problem 1
Minimize cost = 50X1 + 10X2 + 75X3
Subject to:
X1 - X2 = 1,000
2X2 + 2X3 = 2,000
X1 ≤ 1,500
X1,X2,X3 ≥ 0
(a) Formulate the constraints and objective functions to standard form to use in the
simplex method.
Solution Problem 1
a)
Minimum cost = 50X1 + 10X2 + 75X3 + 0S1 + MA1 + MA2
Subject to:
1X1 – 1X2 + 0X3 + 0S1 + 1A1 + 0A2 = 1,000
0X1 + 2X2 + 2X3 + 0S1 + 0A1 + 1A2 = 2,000
1X1 + 0X2 + 0X3 + 1S1 + 0A1 + 0A2 = 1,500
Problem 2
Maximize earnings = R0.80 X1 + R0.40 X2 + R1.20 X3 - R0.10 X4
Subject to:
X1 + 2 X2 + X3 + 5 X4  150
X2 - 4 X3 + 8 X4 = 70
6 X1 + 7 X2 + 2 X3 - X4  120
X1, X2, X3, X4  0
Convert the LP problem into standard form for simplex solution
Solution Problem 2
Maximize earnings: 0.8X1 + 0.4X2 + 1.2X3 – 0.1X4 + 0S1 + 0S2 – MA1 – MA2
Subject to:
X1 + 2X2 + X3 + 5X4 + S1 = 150
X2 – 4X3 + 8X4 + A1 = 70
6X1 + 7X2 + 2X3 – X4 – S2 + A2 = 120
Minimization:
Let:
X1 = number of mattresses
X2 = number of box springs
Minimize cost = 20X1 + 24X2
subject to:
X1 + X2  30
X1 + 2X3  40
X1, X2  0
Initial tableau:
Cj

M
M
Solution
Mix
A1
A2
Zj
Cj – Z j
$20
X1
1
1
2M
–2M + 20
$24
X2
1
2
3M
–3M + 24
$0
S1
–1
0
–M
M
$0
S2
0
–1
–M
M
M
A1
1
0
M
0
M
A2
0
1
M
0
Quantity
30
40
70M
Second tableau:
Cj

M
Solution
Mix
A1
$20
X1
1
X2
$24
Zj
Cj – Z j
1
1

$24
X2
0
$0
S1
–1
$0
S2
1
2
1

0
2
24
M + 12
2
1 M + 12
2
0
–M
M
M
A1
1
1

2
1
2
M – 12
2
1 M + 12
2
M
A2
 12
0
0
0
Quantity
10
1
20
2
 1 2 M + 12
3
2
10M + 480
M – 12
Final tableau:
Cj

$20
$24
Solution
Mix
X1
X2
Zj
Cj – Z j
X1 = 20,
X2 = 10,
Min Cost = R640
$20
X1
1
0
20
0
$24
X2
0
1
24
0
$0
S1
–2
1
–16
16
$0
S2
1
–1
–4
4
M
A1
2
–1
16
M – 16
M
A2
–1
1
4
M–4
Quantity
20
10
R640
Maximization:
Problem 1
A new furniture shop wants to optimize their profit in producing exclusive beds and
chairs. Each bed yields a profit of R10.00 and each chair yields a profit of R8.00. For
a production period of one week, 80 Labor hours are available for the abolstery
department and 50 hours are available for the final assembly department. Each bed
requires 4 hours in the abolstery department and 1 hour in the final assembly
department. Each chair requires 2 hours in the abolstery department and 2 hours in
the final assembly department.
Formulate the production mix of beds and chairs for the furniture shop as a LP problem
and then solve it using the Simplex algorithm to determine how many beds and chairs
should be produced each week to optimize their profits.
Let :
Beds = X1
Chairs = X2
1)
Formulate the above problem into a LP solution.
2)
Formulate / transform the above constraints and objective functions
into the standard simplex format
3)
Calculate the optimal solution by using the Simplex method, using the
transformed LP constraints and objective Function in question 2.
Solution – Problem 1
1)
Maximize profit : R10X1 + R8X2
Subject to :
4X1 + 2 X2  80
X1 + 2X2  50
(Abolstery department)
(Final assembly department)
X1, X2  0
2)
Maximize Profit = R10X1 + R8X2 + 0S1 +0S2
Subject to:
4X1 + 2X2 + 1S1 +0S2 = 80 
X1 + 2X2 + 0S1 +1S1 = 50 
X1, X2 ,S1 ,S2  0 
Initial tableau:
Cj

0
0
Solution
Mix
S1
S2
Zj
Cj – Z j
10
X1
4
1
0
10
8
X2
2
2
0
8
0
S1
1
0
0
0
0
S2
0
1
0
0
Quantity
80
50
0
Second tableau:
Cj

10
0
Solution
Mix
X1
S2
Zj
Cj – Z j
10
X1
1
0
10
0
8
X2
0.5
1.5
5
3
0
S1
0.25
–0.25
2.5
–2.5
0
S2
0
1
0
0
Quantity
Solution
Mix
X1
X2
Zj
10
X1
1
0
10
8
X2
0
1
8
0
S1
0.3333
–0.1667
2
0
S2
–0.3333
0.6667
2
Quantity
Cj – Zj
0
0
–2
–2
20
30
200
Third tableau:
Cj

10
8
10
20
260
Solution:
X1 = 10, X2=20, Profit = R260
Problem 2
The “Smurf company” produces tables and chairs for all the smurfs in there village. Each table yields a
profit of R9.00 and each chair yields a profit of R12.00. For a production period of one week, only 10
liters of white paint and 12 liters of blue paint are available. Each table requires 1 liter of white paint and
1 liter of blue paint. Each chair requires 1 liter of white paint and 2 liters of blue paint.
Formulate the production mix for the “Smurf company” as a LP problem and then solve it using the
Simplex algorithm to determine how many tables and chairs should be produced each week to optimize
their profits.
Solution Problem 2
a)
Maximize profit : 9X1 + 12X2
Subject to : X1 + X2  10
X1 + 2X2  12
X1, X2  0
b)
Initial tableau
Cj

R0
R0
Solution
Mix
S1
S2
Zj
Cj – Z j
R9
X1
1
1
0
9
R12
X2
1
2
0
12
R0
S1
1
0
0
0
R0
S2
0
1
0
0
Solution
Mix
S1
X2
Zj
R9
X1
0.5
0.5
6
R12
X2
0
1
12
R0
S1
1
0
0
R0
S2
-0.5
0.5
6
Cj – Zj
3
0
0
–6
Solution
Mix
X1
X2
Zj
Cj – Z j
R9
X1
1
0
9
0
R12
X2
0
1
12
0
R0
S1
2
–1
6
–6
R0
S2
–1
1
3
–3
Quantity
10
12
R0
Second tableau:
Cj

R0
R12
Quantity
4
6
R72
Final tableau:
Cj

R4
R12
c) The optimal solution is : X1 (Tables)= 8, X2 (Chairs)= 2 with a profit of R96.00.
Quantity
8
2
R96
DUAL
The primal LP formulation is as follows:
Minimize cost = 23X1 + 18X2
Subject to:
8X1 + 4X2  120
4X1 + 6X2  115
9X1 + 4X2  116
a) Formulate the dual of the primal LP problem.
Solution
Maximize Z = 120U1 + 115U2 + 116U3
Subject to:
8U1 + 4U2 + 9U3  23
4U1 + 6U2 + 4U3  18
U1, U2, U3  0
Please note that you also must be able to formulate the Dual back
to the primal LP formulation.