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Chapter 10Solutions 1

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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 ... – PowerPoint PPT presentation

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Title: Chapter 10Solutions 1


1
Chapter 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)
2
Chapter 10 Solutions 1
  • 10.3.1)

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)
3
Chapter 10 Solutions 1
  • 10.3.1)

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
4
Chapter 10 Solutions 1
  • 10.3.1)

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
5
Chapter 10 Solutions 1
  • 10.3.1)

Store 1 Store 2 Store
3
6
Chapter 10 Solutions 1
  • 10.3.1)

Store 1 Store 2 Store
3
Optimal Allocation
Optimal path
Optimal path from a given state
7
Chapter 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
8
Chapter 10 Solutions 1
  • 10.3.2)

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
9
Chapter 10 Solutions 1
  • 10.3.2)

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
10
Chapter 10 Solutions 1
  • 10.3.2)

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
11
Chapter 10 Solutions 1
  • 10.3.2)

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
12
Chapter 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)
13
Chapter 10 Solutions 1
  • 10.3.7)

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)
14
Chapter 10 Solutions 1
  • 10.3.7)

Soln
15
Chapter 10 Solutions 1
  • 10.3.7)

Soln
16
Chapter 10 Solutions 1
  • 10.3.7)

Soln
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