MA 114-004 Practice Exam 3
Justify your answers and show all relevant work. The exam paper will not be graded,
put all your work in the blue book provided.
Problem 1
A chicken rancher concludes that his flock of chickens needs at least 4000 pounds of
protein in their winter feed, at least 27,000 pounds of total digestible nutrients (TDN)
and at least 40,000 units (IUs) of Vitamin A. Each pound of seed provides 0.13 pounds
of protein, 0.48 pounds of TDN, and 2 IUs of Vitamin A. Each pound of corn supplies
0.065 pounds of protein, 0.96 pounds of TDN, and no Vitamin A. Seed costs $10 per
100-pound sack, while corn costs $15 per 100-pound sack.
(a) Let x and y be the number of pounds of seed and corn that the chicken rancher
must buy. Give the inequalities that x and y must satisfy.
(b) Find the objective function: Express the cost in dollars of buying x pounds of
seed and y pounds of corn that the chicken rancher would like to minimize. (Don’t
finish the problem!!!!)
Stop when you have listed the inequalities and objective function for this problem. Do
not finish the linear programming problem.
Solution:
(a) First we will find the constraint for the protein. Each pound of seed provides 0.13
pounds of protein and each pound of corn provides 0.065 pounds of protein, so
the function that expresses the total amount of protein in terms of x and y is
0.13(x) + 0.065(y). The total protein must be at least 4000 pounds, so the protein
constraint is
0.13x + 0.065y ≥ 4000
Next we find the constraint for the TDN. Each pound of seed provides 0.48 pounds
of TDN and each pound of corn provides 0.96 pounds of TDN, so the function
0.48(x) + 0.96(y) models the total amount of TDN in terms of x and y. We need
at least 27,000 pounds of TDN, so our second constraint is
0.48x + 0.96y ≥ 27, 000
Each pound of seed provides 2 IUs of Vitamin A and each pound of corn provides
no Vitamin A, of which we need 40,000 units. So we have
2(x) + 0(y) = 2x ≥ 40, 000
Note this last inequality required x ≥ 0 to hold, because it simplifies to x ≥ 20, 000
But we must also ensure y ≥ 0 because negative amounts of corn does not make
sense in the context of the problem. Our family of inequalities is


0.13x + 0.065y ≥ 4000



0.48x + 0.96y ≥ 27, 000

2x ≥ 40, 000



y ≥ 0
(b) Our objective function will be
(Number of pounds of seed) × ( price of seed per pound )+
(Number of pounds of corn) × ( price of corn per pound )
The price of seed per pound is $10/100 = $0.10 and the price per pound of corn
is $15/100 = $0.15. So our objective function is
0.1x + 0.15y
where cost is expressed in dollars.
Problem 2
The following picture represents a linear programming problem. Three of the boundary
lines for the feasible set have the value of their slopes labeled.
Find the vertex of the feasible set where the objective function 50x + 60y is maximized.
Solution: We want to find the form of the family of objective function lines first:
50x + 60y = c ⇐⇒ 60y = −50x + c ⇐⇒ y = −
50
c
5
c
x+
⇐⇒ y = − x +
60
60
6
60
The changes in slope at the vertices A, B, C and D give us the respecive intervals
1
7 1
7
− ,∞ ,
− ,−
,
−3, −
, (−∞, −3)
6
6 6
6
c
We want to maximize the cost c, which controls the y-intercept of the line y = − 65 x+ 60
.
We start with c large and, by letting c vary downward lower the line towards the feasible
set. It will touch a vertex with highest possible cost c when the slope is between the
changes in slope induced by the vertices (if it were equal to one of the slopes in the
picture, then it would intersect the feasible set along an entire edge). The slope − 65
is in the interval − 67 , − 61 which goes with the vertex B: The downward slope of the
cost line is steeper than the segment between A and B, but not as steep as the segment
between B and C. So the vertex B is the place on the feasible set where the objective
function is maximized.
Problem 3
Let A and B be transition matrices
between lily pads 1, 2, and 3.

0.3 0.5
A = 0.4 0.5
0.3 0
for two Markov processes modeling frogs jumping

0.2
0.7 ,
0.1


1 0.5 0.2
B = 0 0.5 0.7
0 0 0.1
Let d0 be the initial distribution matrix
 
.2
d0 = .3
.5 0
(a) Draw transition diagrams for each process A and B.
(b) Find the second distribution matrix d2 both under the action of Markov process
A and Markov Process B.
(c) Using the given initial distribution, what percent of frogs are on lily pad 2 after
running each process twice?
(d) Is A regular? Is B regular? Why or why not?
Solution:
(a) Here are the transition diagrams: The diagram for A is first, for B second.
(b) Under the action of the Markov Process represented by A, the second distribution
matrix d2 is
  
 

0.3 0.5 0.2 .2
0.3 0.5 0.2
d2 = A2 d0 = 0.4 0.5 0.7 0.4 0.5 0.7 .3 
0.3 0 0.1 .5 0
0.3 0 0.1


 


0.3 0.5 0.2 0.3(0.2) + 0.5(0.3) + 0.2(0.5)
0.3 0.5 0.2 0.31
0.4 0.5 0.7 0.4(0.2) + 0.5(0.3) + 0.7(0.5) = 0.4 0.5 0.7 0.58 =
0.3 0 0.1
0.3(0.2) + 0(0.3) + 0.1(0.5)
0.3 0 0.1 0.11 1

 

0.3(0.31) + 0.5(0.58) + 0.2(0.11)
0.405
0.(0.31)4 + 0.5(0.58) + 0.7(0.11) = 0.491
0.3(0.31 + 0(0.58) + 0.1(0.11)
0.104 2
Under the action of the Markov Process represented by B, the second distribution
matrix d2 is

 
  
1 0.5 0.2
1 0.5 0.2 .2
d2 = B 2 · d0 = 0 0.5 0.7 0 0.5 0.7 .3  =
0 0 0.1
0 0 0.1 .5 0


 


1 0.5 0.2 1(0.2) + 0.5(0.3) + 0.2(0.5)
1 0.5 0.2 0.45
0 0.5 0.7 0(0.2) + 0.5(0.3) + 0.7(0.5) = 0 0.5 0.7  0.5  =
0 0 0.1
0(0.2) + 0(0.3) + 0.1(0.5)
0 0 0.1 0.05 1

 

1(0.45) + 0.5(0.5) + 0.2(0.05)
0.71
0(0.45) + 0.5(0.5) + 0.7(0.05) = 0.285
0(0.45) + 0(0.5) + 0.1(0.05)
0.005 2
(c) For each Markov process, this information is given by the second entry of the
second distribution matrix d2 from part (b). For process A, 49.1% of the frogs are
on lily pad 2 after running the process twice. For process B, 28.5% of the frogs
are on lily pad 2 after running the process twice.
(d) Recall a stochastic matrix is regular if some power of the matrix has all nonzero
entries. We can check to see if the matrix A is regular by computing A2 ;



0.3 0.5 0.2 0.3 0.5 0.2
0.4 0.5 0.7 0.4 0.5 0.7 =
0.3 0 0.1 0.3 0 0.1


0.3(0.3) + 0.5(0.4) + 0.2(0.3) 0.3(0.5) + 0.5(0.5) + 0.2(0) 0.3(0.2) + 0.5(0.7) + 0.2(0.1)
0.4(0.3) + 0.5(0.4) + 0.7(0.3) 0.4(0.5) + 0.5(0.5) + 0.7(0) 0.4(0.2) + 0.5(0.7) + 0.7(0.1)
0.3(0.3) + 0(0.4) + 0.1(0.3)
0.3(0.5) + 0(0.5) + 0.1(0)
0.3(0.2) + 0(0.7) + 0.1(0.1)
so


0.35 0.4 0.43
A2 = 0.53 0.45 0.5 
0.12 0.15 0.07
Since A2 has no nonzero entries, A is regular.
We already know that B is not regular because B is an absorbing stochastic
matrix. The identity matrix in the upper left is the 1 × 1 identity matrix, just
like on Worksheet 3 Problem 4. So every power of B will have a 1 in the upper
left and a column of zeroes underneath; equivalently the first column of B will be
given by (1, 0, 0). So, every power of B has zero entries, and B is not regular.
Problem 4
Form the initial simplex tableau for the following LP problem: MInimize 5x+7y subject
to the constraints


2x + y ≥ 10




3x + 2y ≥ 18

x + 2y ≥ 10



x≥0




y≥0
Label the columns of the tableau with the variables they represent. Indicate which
variables are in Group I and Group II. Stop when you have set up the initial simplex
tableau. Do not finish solving the problem.
Solution:
To use the simplex tableau, we must convert the objective function to a maximization
problem and convert the constraints to standard form of [linear polynomial] ≤ [number].
So our new problem is
Maximize −5x − 7y subject to the constraints


−2x − y ≤ −10





−3x − 2y ≤ −18
−x − 2y ≤ −10



x≥0




y≥0
We obtain the initial tableau
u
v
w
M
x
y u v w M
−2 −1 1 0 0 0
−3 −2 0 1 0 0
−1 −2 0 0 1 0
5
7 0 0 0 1
−10
−18
−10
0
The Group I variables are x and y. The Group II variables are u, v, w and M (Note
that I labeled the rows of the tableau with the Group II variables).
Problem 5
A truck rental company in Raleigh only rents trucks on a daily basis. Rented trucks
must be returned to Offices 1, 2, or 3. The transition diagram below represents a
Markov process with regular time interval as ”a day” and the state set as Offices 1,
2, and 3. The probabilities are based on the percentages of trucks returned to each
location based on where they were picked up. We assume that all the truckss are rented
each day.
(a) Set up the stochastic matrix for this Markov process with rows and columns
indexed by the state set 1, 2, and 3.
(b) On Wednesday morning, 30% of the trucks are at Office 1, 50% of the trucks are
at Office 2, and 20% of the trucks are at Office 3. What percentage of the trucks
will be at Office 2 on Friday morning?
(c) What percentage of the trucks that are at Office 2 Wednesday morning will be at
Office 2 Thursday morning?
(d) In the long run, what can we expect the distribution of trucks to be between the
offices?
(a) The stochastic matrix has columns indexed by the current states, and rows indexed
by the next states:
1
2
3
1 0.6 0.3 0.2
2 0.1 0.6 0.1
3 0.3 0.1 0.7
(b) We interpret the statistics for Wednesday as the initial distribution matrix
 
0.3
d0 = 0.5
0.2 0
To find the percentage of trucks at Office 2 on Friday morning, we run the process
twice, or in other words we find the second distribution matrix d2 . So , d2 =
A2 · d0 =


 


0.6 0.3 0.2 0.6 0.3 0.2
0.3
0.383
0.1 0.6 0.1 0.1 0.6 0.1 0.5 = 0.275
0.3 0.1 0.7 0.3 0.1 0.7
0.2 0
0.342 2
The percentage of trucks at Office 2 is given by the 2nd entry of d2 , so there are
27.5% of the trucks at Office 2 on Friday morning.
(c) We need to track the flow of trucks from Office 2 to Office 2 for one iteration of
the process, and we are ignoring the flow of trucks between other states in this
sub-problem, so we start with an initial distribution matrix with 100% of the
trucks at Office 2:
 
0

d0 = 1
0 0
Thursday morning is one day later. We find the first distribution matrix:
 
 
 

0.3
0.6 0.3 0.2 0
0.6 0.3 0.2







d1 = A · d0 = 0.1 0.6 0.1 = 0.1 0.6 0.1 1 = 0.6
0.1 1
0.3 0.1 0.7 0 0
0.3 0.1 0.7
So, 60% of the trucks that started out at Office 2 on Wednesday morning are at
Office 2 Thursday morning.
(d) To answer this we use the stable distribution of the matrix representing this
Markov process: since the matrix A has no zero entries, we know it is regular
and this strategy is permissible. Let X = (x, y, z) be the stable distribution.
Then we can find X by solving the system of equations obtained from AX = X,
x + y + z = 1. The condition AX = X can be written as
0.6x +0.3y +0.2z = x
0.1x +0.6y +0.1z = y
0.3x +0.1y +0.7z = z
So we have the system
x
+y
+z =
0.6x +0.3y +0.2z =
0.1x +0.6y +0.1z =
0.3x +0.1y +0.7z =
1
x
y
z
We need to use GJE on this system, but we only know how to do that when the
last column of our augmented matrix has numbers in it. So, let’s rearrange the
last three equations by moving the variable on the right of the = sign in each
equation to the left hand side of the = sign, using subtraction:
x
+y
+z =
−0.4x +0.3y +0.2z =
0.1x −0.4y +0.1z =
0.3x +0.1y −0.3z =
1
0
0
0
Then I’ll rewrite it as an augmented coefficient matrix. While I’m at it, I’m going
to convert to fractions because I hate doing GJE on decimals for big matrices.


1
1
1
1


− 2
3
1
0
 5 10

5


 1

1
 10 − 25

0
10


1
10
3
10
3
− 10
0
First, pivot about a11 :

1 1
R2 +

0 7

10
1
R3 − 10 R1 , → 

0 − 12

3
R4 − 10 R1
0 − 15
1
2
5 R1 ,
Next, pivot about a22 :

1 1

0 1

10
R2 → 

7
0 − 21

0 − 51
1
6
7
0
− 35
1





1 
− 10 

4
7
3
− 10
0 0 0





13 

30 
4
7
0




1 
− 10 

3
5
2
5
0
− 35
3
− 10
1
7
3
7
6
7




13 

70 
3
7
0 0 0
3
7


1
R1 − R2 ,

0

1
R3 + 2 R2 , → 

0

R4 + 51 R2
0
We’ll go ahead an cancel the fourth row:

1 0

0 1

R4 + R3 → 

0 0

Now we pivot about a33 :

1 0 17

0 1 6

7
7
R3 → 

3
0 0 1

1
0
1
7
3
7

1
6
7
4
7
0
3
7
13
70







0 − 37
− 13
70

4
7
0

1 0 0

R1 − 71 R3 ,
0 1 0

→
6

R2 − 7 R3
0 0 1

0 0 0
11 
30

6 
30 

13 

30 
0
So the stable distribution is
 11 


0.367
  

1 

 5  ≈  0.2 
  

13
0.433
30
30
and in the long run, we expect 36.7% of the trucks to be at Office 1, 20% of the
trucks to be at office 2, and 43.3% of the trucks to be at Office 3.
Problem 6 Maximize 36x + 48y + 70z subject to the constraints


x≤4



y ≤ 6

4x + 3y + 2z ≤ 38



x ≥ 0, y ≥ 0, z ≥ 0
This is a maximization problem and all the inequalities are in standard form, so we
can immediately set up a simplex tableau. Note we are forced to use a simplex tableau
because there are three variables in the original problem: x, y and z.
The system of equations with slack variables that replaces our constraint family is
x
+u
y
4x
+3y +2z
−36x −48y −70z
= 4
= 6
= 38
= 0
+v
+w
+M
where x, y, z, u, v, w are all ≥ 0 and M is as large as possible.
We obtain the initial tableau
u
v
w
M
x
y
z
1
0
0
0
1
0
4
3
2
−36 −48 −70
u
1
0
0
0
v w M
0 0 0
1 0 0
0 1 0
0 0 1
4
6
38
0
There are no negative numbers in the upper right box that we need to get rid of, but
the solution obtained from the initial tableau is not a maximum because there are
negative numbers in the left side of the bottom row. We must pivot until the lower left
row has no negative numbers. To choose the pivot column, find the column with the
most negative entry in the bottom row. This is the -70 in row 3. Then, to choose the
pivot entry in column 3, compare positive ratios obtained by dividing entries in the last
column by the corresponding entry in column 3... but in this case, there are two zeros
in column 3 so our only positive ratio comes from 38/2 = 19. So we must pivot about
the 2 in column 3:
1
R3 →
2
x
y
z
1
0
0
0
1
0
3
2
1
2
−36 −48 −70
u
1
0
0
0
v w M
0 0 0
1 0 0
0 21 0
0 0 1
4
6
19
0
u
R4 + 70R3 → v
z
M
x
y
1
0
0
1
3
2
2
104 57
z
0
0
1
0
u
1
0
0
0
v w M
0 0 0
1 0 0
0 12
0
0 35 1
4
6
19
1330
Our new tableau has no negative entries in the left part of the bottom row, so this
tableau gives us a maximum and we’re done. To read off the solution, note that our
Group I variables are now x, y and w, set those equal to zero. The Group II variables
index the rows now (see the tableau) and we read their values from the last column.
We have x = 0, y = 0, z = 19, u = 4, v = 6 and M = 1330. So our objective function
36x + 48y + 70z reaches its maximum value of 1330 at the point (0, 0, 19).
Problem 7 The following transition diagram shows an absorbing stochastic process:
(a) Write down the stochastic matrix that represents this Markov process. Since this
is absorbing, the matrix should take the form
"
#
I S
A=
0 R
In this case, the choice of indexing the absorbing states first has been made for
you.
(b) Write down the sub-matrices S and R for your matrix from part (a).
(c) Find the fundamental matrix (I − R−1 ).
(d) Find the stable matrix
"
#
I S(I − R−1 )
0
of the absorbing stochastic process.
0
(a) We use the same indexing as the diagram gives us to write down the absorbing
stochastic matrix:
1
2
3
4
1
1
0
0
0
2
0
1
0
0
3
4
0.6 0
0.1 0
0.2 0.5
0.1 0.5
(b) We have
"
I
S
#
0 R
Therefore
=
1
0
0
0
0.6 0
S=
,
0.1 0
0
1
0
0
0.6 0
0.1 0
0.2 0.5
0.1 0.5
0.2 0.5
R=
0.1 0.5
(c) To find (I − R)−1 , we choose I to be the same dimensions as R so that the
subtraction is defined:
0.8 −0.5
0.2 0.5
1 0
=
−
I −R=
−0.1 0.5
0.1 0.5
0 1
We’ll use the formula for the inverse of a 2 × 2 matrix here. The determinant
D = (0.8)(0.5) − (−0.1)(−0.5) = 0.35 So the fundamental matrix is
1 0.5 0.5
1.4286 1.4286
−1
(I − R) =
=
0.2858 2.2858
0.35 0.1 0.8
(d) We simply use the formula as
0.6
−1
S(I − R) =
0.1
we have calculated all the ingredients:
0.8571 0.8571
0 1.4286 1.4286
=
0.1429 0.1429
0 0.2858 2.2858
and
"
#
I S(I − R−1 )
0
0

1
0
=
0
0

0 0.8571 0.8571
1 0.1429 0.1429

0
0
0 
0
0
0
Problem 8
The victims of a disease are classified into three states: cured, dead, and sick. Once
cured, a person is permanently cured because they become immune. Each year 80%
of the sick are cured, 10% remain sick, and 10% die. We model this as an absorbing
stochastic process where the regular time interval is one year and the states are Cured,
Dead, and Sick.
(a) Make a transition diagram for this Markov process.
(b) Make the absorbing stochastic matrix
"
A=
I
S
#
0 R
for this Markov process. Be sure to put the columns indexing the absorbing states
first. Write down the sub-matrices S and R explicitly.
(c) Find the fundamental matrix F = (I − R)−1 . Use the fundamental matrix to
estimate how long someone who is sick will stay sick. Give your answer in years
rounded to two decimal places. Show your work and explain the answer in the
context of the problem. Recall that the ijth entry of F is the expected number of
times the process will be in nonabsorbing state i if if starts in nonabsorbing state
j.
(d) Find the long term-trend for the progression of the disease: in other words, what
will the long-term distribution of between the absorbing states of patients who
contract the disease? Hint: the formula for the stable matrix of an absorbing
stochastic process is
"
#
I S(I − R−1 )
0
0
(a) The transition diagram is shown below
(b) Note the absorbing states are Cured and Dead, which in our diagram are the
states with no strictly outbound arrows.
Cured Dead Sick
Cured
1
0
0.8
Dead
0
1
0.1
Sick
0
0
0.1
so that


1 0 0.8
A=
= 0 1 0.1
0 R
0 0 0.1
"
I
and
S
#
0.8
S=
,
0.1
R = [0.1]
(c) The fundamental matrix F = (I − R)−1 = (1 − 0.1)−1 = 1/(0.9) = 10/9. The
rows and columns of F are indexed by the nonabsorbing states, but there is only
one nonbsorbing state here, which is Sick. So, the single entry of the 1 × 1 matrix
F is telling us how long someone stays in nonabsorbing state sick once they enter
non-absorbing state sick, which is 10/9 years, approximately 1.11 years.
(d) The hint gave us the formula again, so we should use it: First
 
8
9
0.8 10
−1


S(I − R) =
=
0.1
1
9
9
and the stable matrix is
"


1 0 89


I S(I − R−1 )


= 0 1 19 


0
0
0 0 0
#
So in the long term, 98 of the patients who enter state Sick will end up in absorbing
state Cured, and 91 of the patients who enter state Sick will end up in absorbing
state Dead. So the long term prognosis of someone who becomes sick with the
disease is that they have a 89 chance of surviving.
Problem 9
A small craft shop makes toy stuffed bears and silk flower wreaths. Each bear requires
3 hours of preparation time, 6 hours of assembly time, and 2 hours of finishing time.
Each wreath requires 2 hours of preparation time, 5 hours of assembly time, and 3 hours
of finishing time. There are 36 hours of preparation time, 78 hours of assembly time,
and 42 hours of finishing time available each week. If each bear earns a profit of $10
and each wreath earns a profit of $, how many of each should be crafted to maximize
the profit?
Solution:
The table below summarizes the information in the problem:
Prep Time
Assembly
Finishing
Profit
Bears
Wreaths
Available
3 labor hours 2 labor hours 36 labor hours
6 labor hours 5 labor hours 78 labor hours
2 labor hours 3 labor hours 42 labor hours
$10
$8
Let x be the number of bears manufactured each day, let y be the number of wreaths
manufactured each day. We obtain 3 inequalities from the table: Prep : 3x + 2y ≤ 36,
Assembly 6x + 5y ≤ 78, Finishing 2x + 3y ≤ 42. Also, bears and wreaths can’t be
represented by negative numbers, so we require x ≥ 0, y ≥ 0. Our objective function is
10x + 8y. In standard form the inequalities are


y ≤ − 32 x + 18



y ≤ − 6 x + 78
5
5
2

y
≤
−
x
+
14

3


x ≥ 0, y ≥ 0
2
Let L1 be y = − 23 x + 18, L2 be y = − 65 x + 78
5 and L3 be y = − 3 x + 14. Then the
graph of the feasible set, with the boundary lines L1, L2, and L3 color coded, is shown
below.
The vertex A is the origin (0, 0), B is the y-intercept of L3 so is (0, 14), and E is the
x-intercept of L1, so is (12, 0). To find C, set L3 and L2 equal:
6
78
2
8
8
− x+
= − x + 14 ⇐⇒
x=
⇐⇒ x = 3
5
5
3
15
5
Then
78
6
= 12
− (3) +
5
5
So C has coordinates (3, 12). Then D is the intersection of L1 and L2:
6
78
3
3
12
− x+
= − x + 18 ⇐⇒
x=
⇐⇒ x = 8
5
5
2
10
5
Then
6
78
− (8) +
=6
5
5
and D has coordinates (8, 6). We check the value of each vertex in the objective function:
Vertex Profit = 10x + 8y
(0, 0)
10(0) + 8(0) = 0
(0, 14) 10(14) + 8(0) = 140
(3, 12) 10(3) + 8(12) = 126
(8, 6) 10(8) + 8(6) = 128
(12, 0) 10(12) + 8(0) = 120
The maximum value of $128 occurs at the vertex (8, 6). To maximize profit each week,
the craft shop should make 8 bears and 6 wreaths.
Problem 10
Form the initial simplex tableau corresponding to the following linear programming
problem: Maximize 2x + y − 3z subject to the constraints


x + y − 2z ≤ 10



2x − y + 3z ≤ 18
x + 3y + z ≤ 21



x ≥ 0, y ≥ 0, z ≥ 0
Stop when you have formed the simplex tableau. Do not solve the problem.
Solution:
Our simplex tableau will correspond to the system of equations
x
+y −2z +u
2x −6y +3z
+v
x
+3y +z
+w
−2x −y +3z
M
= 10
= 18
= 21
= 0
where x, y, z, u, v and w are all ≥ 0, and M is as large as posslble. The initial tableau
is
u
v
w
M
x
y
z u v w M
1
1 −2 1 0 0 0
2 −1 3 0 1 0 0
1
3
1 0 0 1 0
−2 −1 3 0 0 0 1
10
18
21
0
Problem 11
State the maximization problem corresponding to the following initial tableau, and
solve it using the simplex method.
u
v
M
x
6
15
−3
y
7
5
4
u
1
0
0
v M
0 0
1 0
0 1
120
195
0
Solution:
We observe there are no negative numbers in the upper right hand box that gives
values of the group II variables, we can take each row above the horizontal bar to
represent an inequality in standard form: [linear polynomial] ≤ [nonnegative constraint].
Furthermore we are simply told that the objective row represents a function we’d like
to maximize. So the problem is: Maximize 3x − 4y subject to


6x + 7y ≤ 120
15x + 5y ≤ 195


x ≥ 0, y ≥ 0
Note we are told to use the simplex method, so a graphical solution is not
acceptable in this case. If you are told to use the simplex method on a
two-variable problem on the exam and you use Chapter 3 Methods, you will
not get any points for that solution!!!
On to the tableau: we don’t need to get rid of any negative numbers in the upper
right-hand box of the tableau, so we can proceed using the methods of Section 4.2 .
First we examine the entries of the bottom row: since there is a negative number, −3,
in column 1, the initial tableau does not yield a maximum and we must pivot. We
choose the pivot column by choosing the column whose row 3 entry is most negative,
which is column 1.
To choose the pivot element in column 1, we compare the positive ratios obtained by
dividing elements of the last column by column 1 entries in the corresponding row. Now
195/15 = 13, and 120/6 = 20, so we pivot about the 15 in column 1, row 2:
1
R2 →
15
x y u
6
7 1
1
1
3 0
−3 4 0
v
0
1
15
0
M
0
0
1
120
13
0
R1 − 6R2
u
→
R3 + 3R2
x
M
x y u v
0
5 1 − 25
1
1
1
3 0
15
0
5 0 15
M
0
0
1
42
13
39
Now there are no negative entries in the bottom left box, so the current tableau is a
solution to the maximization problem. Set the group I variables equal to zero, and the
group II variables are written to the left of the tableau, whose values we get from the
last column: x = 13, y = 0, u = 42, v = 0, M = 39. So, the maximum value of 39
occurs at the point (13, 0).
Problem 12
A major coffee supplier has warehouses in Seattle and San José. The coffee supplier
receives orders from coffee retailers in Salt Lake City and Reno. The retailer in Salt
Lake City needs 400 pounds of coffee, and the retailer in Reno needs 350 pounds of
coffee. The Seattle warehouse has 700 pounds available, and the warehouse in San José
has 500 pounds available. The cost of shipping from Seattle to Salt Lake City is $2.50
per pound, from Seattle to Reno $3 per pound, from San José to Salt Lake City $ 4
per pound, and from San José to Reno $ 2 per pound. Find the number of pounds to
be shipped from each warehouse to each retailer such that the total shipping cost is
minimized.
Solution:
Let x be the number of pounds of coffee shipped San José (SJ) to Salt Lake City (SLC),
let y be the number of pounds of coffee shipped from SJ to Reno. Then, since SLC
requires 400 pounds of coffee, Seattle must ship 400 − x pounds to SLC, and since Reno
requires 350 pounds, Seattle must ship 350 − y to Reno. SJ only has 500 pounds to
ship, so x + y ≤ 500. Seattle only has 700 pounds to ship, so (400 − x) + (350 − y) ≤ 700.
Further, we require the amounts of coffee to be nonnegative, so we obtain the family of
constraints


x + y ≤ 500



(400 − x) + (350 − y) ≤ 700

x ≥ 0, y ≥ 0



400 − x ≥ 0, 350 − y ≥ 0
In standard form, these become


y ≤ −x + 500





y ≥ −x + 50
x ≤ 400



y ≤ 350




x ≥ 0, y ≥ 0
The objective function that we wish to minimize is the shipping cost, which is the sum
of four products, one from each shipping leg:
2.5(400 − x) + 4x + 2y + 3(350 − y) = 2050 + 1.5x − y
Here is a graph of the feasible set, which has six vertices:
where L1 is the line y = −x + 500, L2 is the line y = −x + 50, L3 is the line x = 400,
and L4 is the line y = 350. The coordinates of the vertices are very easy to solve for so
I will omit that discussion. We check the value of the objective at all six vertices:
Vertex
Cost = 2050 + 1.5x − y
A = (0, 50)
2050 + 1.5(0) − (50) = 2000
B = (0, 350)
2050 + 1.5(0) − (350) = 1700
C = (150, 350) 2050 + 1.5(150) − (350) = 1925
D = (400, 100) 2050 + 1.5(400) − (100) = 2550
E = (400, 0)
2050 + 1.5(400) − (0) = 2650
2050 + 1.5(50) − (0) = 2125
F = (50, 0)
So the cost is minimized at vertex B when x = 0 and y = 350. So, to minimize the
shipping cost, we ship 400 pounds of coffee from Seattle to SLC and 350 pounds from
SJ to Reno. (Note: in this case, since x = 0, two of the shipping routes are not used!!).
Problem 13 Worksheet 3 Solutions will be posted separately.