Title: Chapter 10Solutions 1
1Chapter 10 Solutions 1
- 10.3.1) The owner of a chain of three grocery
stores has purchased 5 crates of fresh
strawberries. The estimated probability
distribution of potential sales of the
strawberries before spoilage differs among the
three stores. Therefore, the owner wants to know
how to allocate 5 crates to the three stores to
maximize expected profit. - For administrative reasons, the owner does not
wish to split crates between stores. However, he
is willing to distribute no crates to any of his
stores. The following table gives estimated
expected profit at each store when it is
allocated various numbers of crates. -
Soln
Let xn number crates allocated to a store
pn(xn) expected profit from allocating xn
crates to store n sn number of crates
remaining fn(sn) max pn(xn)
fn1(sn-xn)
2Chapter 10 Solutions 1
Soln
Let xn number crates allocated to a store
pn(xn) expected profit from allocating xn
crates to store n sn number of crates
remaining fn(sn) max pn(xn)
fn1(sn-xn)
3Chapter 10 Solutions 1
Soln
Let xn number crates allocated to a store
pn(xn) expected profit from allocating xn
crates to store n sn number of crates
remaining fn(sn) max pn(xn)
fn1(sn-xn)
Store 1 Store 2 Store
3
4Chapter 10 Solutions 1
Soln
Let xn number crates allocated to a store
pn(xn) expected profit from allocating xn
crates to store n sn number of crates
remaining fn(sn) max pn(xn)
fn1(sn-xn)
Store 1 Store 2 Store
3
5Chapter 10 Solutions 1
Store 1 Store 2 Store
3
6Chapter 10 Solutions 1
Store 1 Store 2 Store
3
Optimal Allocation
Optimal path
Optimal path from a given state
7Chapter 10 Solutions 1
- 10.3.2) A college student has 7 days remaining
before final examinations begin in her four
courses and she wants to allocate this study time
as effectively as possible. She needs at least 1
day on each course, and she likes to concentrate
on just one course each day, so she wants to
allocate 1, 2, 3, or 4 days to each course.
Having recently taken an O.R. course, she decides
to use dynamic programming to make the
allocations so as to maximize the total grade
points. -
Soln
Let xn number days allocated to a course
pn(xn) grade points earned from allocating xn
days to course n sn number of days
remaining fn(sn) max pn(xn)
fn1(sn-xn)
Course 1 Course 2 Course 3
6
5
5
4
4
3
3
2
8Chapter 10 Solutions 1
Soln
Let xn number days allocated to a course
pn(xn) grade points earned from allocating xn
days to course n sn number of days
remaining fn(sn) max pn(xn)
fn1(sn-xn)
Course 1 Course 2 Course 3
Course 4
6
5
5
4
4
4
3
3
3
2
2
9Chapter 10 Solutions 1
Soln
Let xn number days allocated to a course
pn(xn) grade points earned from allocating xn
days to course n sn number of days
remaining fn(sn) max pn(xn)
fn1(sn-xn)
Course 1 Course 2 Course 3
Course 4
6
5
5
4
4
4
3
3
3
2
2
10Chapter 10 Solutions 1
Soln
Let xn number days allocated to a course
pn(xn) grade points earned from allocating xn
days to course n sn number of days
remaining fn(sn) max pn(xn)
fn1(sn-xn)
Course 1 Course 2 Course 3
Course 4
6
5
5
4
4
4
3
3
3
2
2
11Chapter 10 Solutions 1
Soln
Let xn number days allocated to a course
pn(xn) grade points earned from allocating xn
days to course n sn number of days
remaining fn(sn) max pn(xn)
fn1(sn-xn)
Course 1 Course 2 Course 3
Course 4
6
5
5
4
4
4
3
3
3
2
2
12Chapter 10 Solutions 1
- 10.3.7) Consider an electronic system consisting
of four components, each of which must work for
the system to function. The reliability of the
system can be improved by installing several
parallel units in one or more components. The
following table gives the probability that the
respective components will function if they
consist of one, two, or three parallel units - The probability that the system will function is
the product of the probabilities that the
respective components will function. The cost
(in 100s) of installing 1, 2, or 3 parallel
units in the respective components is given by
the following table (total budget 1,000). -
Soln
Let xn number parallel units for component n
pn(xn) probability component will function
with xn parallel units cn(xn) cost of
installing xn units sn number of in
100s remaining fn(sn) max pn(xn)
fn1(sn-cn(xn)
13Chapter 10 Solutions 1
Soln
Let xn number parallel units for component n
pn(xn) probability component will function
with xn parallel units cn(xn) cost of
installing xn units sn number of in
100s remaining fn(sn) max pn(xn)
fn1(sn-cn(xn)
14Chapter 10 Solutions 1
Soln
15Chapter 10 Solutions 1
Soln
16Chapter 10 Solutions 1
Soln