12. Dynamic Programming - Assembly Line Scheduling

Dynamic Programming

• Dynamic Programming was introduced by Richard Bellman in 1955.

• A dynamic programming algorithm solves every sub problem just once

• Saves its answer in a table (array)

• Avoids the work of re-computing the answer every time the sub problem is encountered.

• Dynamic programming is typically applied to optimization

problems.

• What is an optimization problem? • There can be many possible solutions. • Each solution has a value and

• If a problem has only one correct solution, then optimization is not required

• For example, there is only one sorted sequence

containing a given set of numbers.

• Optimization problems have many solutions.

• We want to compute an optimal solution e. g. with

minimal cost and maximal gain.

• There could be many solutions having optimal value

• Dynamic programming is very effective technique

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Elements of Dynamic Programming

• An optimization problem must have two elements for dynamic programming to be applicable

• Optimal Substructure

– A problem exhibits optimal substructure if an optimal solution to the problem contains within it optimal

solutions to the subproblems • Overlapping sub problems

– When a recursive algorithm revisits the same problem over and over again, then the optimization problem has

1. Characterize the structure of an optimal solution

2. Recursively define the value of an optimal

solution

3. Compute the value of an optimal solution in a

bottom-up fashion

4. Construct an optimal solution from computed

information

Note: Steps 1-3 form the basis of a dynamic

programming solution to a problem. Step 4 can be omitted only if the value of an optimal

solution is required.

• Dynamic programming, like divide and conquer method, solves problems by combining the

solutions to sub-problems.

• _{Divide and conquer algorithms:}

• partition the problem into independent sub-problem

• Solve the sub-problem recursively and

• Combine their solutions to solve the original problem

• In contrast, dynamic programming is applicable

when the sub-problems are not independent.

• Dynamic programming is typically applied to

optimization problems.

• There are two assembly lines each with n stations

• The jth station on line i is denoted by S_{i, j}

• The assembly time at that station is a_i,j.

• An auto enters factory, goes into line i taking time e_i

• After going through the jth station on a line i, the auto

goes on to the (j+1)st station on either line

• There is no transfer cost if it stays on the same line

• It takes time t_i,j to transfer to other line after station S_i,j

• After exiting the nth station on a line, it takes time x_i for

the completed auto to exit the factory.

• Problem is to determine which stations to choose from

lines 1 and 2 to minimize total time through the factory.

Stations S_i,j;

2 assembly lines, i = 1,2; n stations, j = 1,...,n.

a_i,j = assembly time at S_i,j;

t_i,j = transfer time from S_i,j (to S_i-1,j+1 OR S_i+1,j+1); e_i = entry time from line i;

x_i = exit time from line i .

In Out

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Total Computational Time

= possible ways to enter in stations at level n x one way Cost Possible ways to enter in stations at level 1 = 21

Possible ways to enter in stations at level 2 = 22 . . . Possible ways to enter in stations at level 2 = 2n

Total Computational Time = n.2n

Brute Force Solution

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• Let f_i[j] = fastest time from starting point station S_{i, j}

• f₁[n] = fastest time from starting point station S_{1 n}

• f₂[n] = fastest time from starting point station S_{2 n}

• l_i[j] = The line number, 1 or 2, whose station j-1 is used in a fastest way through station S_{i, j} .

• It is to be noted that l_i[1] is not required to be defined because there is no station before 1

• t_i[j-1] = transfer time from line i to station S_{i-1, j} or S_{i+1, j}

• Objective function = f* = min(f₁[n] + x₁, f₂[n] + x₂)

• _{l* = to be the line no. whose n}th station is used in a

fastest way.

S1 j-1

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In Out

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Notations: Finding Objective Function

S_i,j

a_i,j

t_1,j-1

f_i(j)

f₁[1] = e₁ + a_1,1; f₂[1] = e₂ + a_2,1.

f₁[j] = min (f₁[j-1] + a_1,j, f₂[j-1] + t_2,j-1 + a_1,j) for j ≥ 2; f₂[j] = min (f₂[j-1] + a_2,j, f₁[j-1] + t_1,j-1 + a_2,j) for j ≥ 2;

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Base Cases

• f₁[1] = e₁ + a_1,1

• f₂[1] = e₂ + a_2,1

Two possible ways of computing f₁[j]

• f₁[j] = f₂[j-1] + t_{2, j-1} + a_{1, j} OR f₁[j] = f₁[j-1] + a_{1, j} For j = 2, 3, . . ., n

f₁[j] = min (f₁[j-1] + a_{1, j}, f₂[j-1] + t_{2, j-1} + a_{1, j}) Symmetrically

For j = 2, 3, . . ., n

f₂[j] = min (f₂[j-1] + a_{2, j}, f₁[j-1] + t_{1, j-1} + a_{2, j})

Objective function = f* = min(f₁[n] + x₁, f₂[n] + x₂)

• _{f1[1] = e1 + a1,1 = 2 + 7 = 9} • _{f2[1] = e2 + a2,1 = 4 + 8 = 12}

f₁[j] = min (f₁[j-1] + a_{1, j}, f₂[j-1] + t_{2, j-1} + a_{1, j}) f₂[j] = min (f₂[j-1] + a_{2, j}, f₁[j-1] + t_{1, j-1} + a_{2, j})

j = 2

f₁[2] = min (f₁[1] + a_{1, 2}, f₂[1] + t_{2, 1} + a_{1, 2})

= min (9 + 9, 12 + 2 + 9) = min (18, 23) = 18, l₁[2] = 1

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In Out 2 4 2 2 3 1 1 2 3 2 4 1 3 2

• _{f1[1] = e1 + a1,1 = 2 + 7 = 9} • _{f2[1] = e2 + a2,1 = 4 + 8 = 12}

f₁[j] = min (f₁[j-1] + a_{1, j}, f₂[j-1] + t_{2, j-1} + a_{1, j}) f₂[j] = min (f₂[j-1] + a_{2, j}, f₁[j-1] + t_{1, j-1} + a_{2, j})

j = 2

f₂[2] = min (f₂[1] + a_{2, 2}, f₁[1] + t_{1, 1} + a_{2, 2})

= min (12 + 5, 9 + 2 + 5) = min (17, 16) = 16, l₂[2] = 1

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In Out 2 4 2 2 3 1 1 2 3 2 4 1 3 2

f₁[j] = min (f₁[j-1] + a_{1, j}, f₂[j-1] + t_{2, j-1} + a_{1, j}) f₂[j] = min (f₂[j-1] + a_{2, j}, f₁[j-1] + t_{1, j-1} + a_{2, j})

j = 3

f₁[3] = min (f₁[2] + a_{1, 3}, f₂[2] + t_{2, 2} + a_{1, 3}) = min (18 + 3, 16 + 1 + 3)

= min (21, 20) = 20, l₁[3] = 2

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f₁[j] = min (f₁[j-1] + a_{1, j}, f₂[j-1] + t_{2, j-1} + a_{1, j}) f₂[j] = min (f₂[j-1] + a_{2, j}, f₁[j-1] + t_{1, j-1} + a_{2, j})

j = 3

f₂[3] = min (f₂[2] + a_{2, 3}, f₁[2] + t_{1, 2} + a_{2, 3}) = min (16 + 6, 18 + 3 + 6)

= min (22, 27) = 22, l₂[3] = 2

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f₁[j] = min (f₁[j-1] + a_{1, j}, f₂[j-1] + t_{2, j-1} + a_{1, j}) f₂[j] = min (f₂[j-1] + a_{2, j}, f₁[j-1] + t_{1, j-1} + a_{2, j})

j = 4

f₁[4] = min (f₁[3] + a_{1, 4}, f₂[3] + t_{2, 3} + a_{1, 4}) = min (20 + 4, 22 + 1 + 4)

= min (24, 27) = 24, l₁[4] = 1

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f₁[j] = min (f₁[j-1] + a_{1, j}, f₂[j-1] + t_{2, j-1} + a_{1, j}) f₂[j] = min (f₂[j-1] + a_{2, j}, f₁[j-1] + t_{1, j-1} + a_{2, j})

j = 4

f₂[4] = min (f₂[3] + a_{2, 4}, f₁[3] + t_{1, 3} + a_{2, 4}) = min (22 + 4, 20 + 1 + 4)

= min (26, 25) = 25, l₂[4] = 1

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f₁[j] = min (f₁[j-1] + a_{1, j}, f₂[j-1] + t_{2, j-1} + a_{1, j}) f₂[j] = min (f₂[j-1] + a_{2, j}, f₁[j-1] + t_{1, j-1} + a_{2, j})

j = 5

f₁[5] = min (f₁[4] + a_{1, 5}, f₂[4] + t_{2, 4} + a_{1, 5}) = min (24 + 8, 25 + 2 + 8)

= min (32, 35) = 32, l₁[5] = 1

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f₁[j] = min (f₁[j-1] + a_{1, j}, f₂[j-1] + t_{2, j-1} + a_{1, j}) f₂[j] = min (f₂[j-1] + a_{2, j}, f₁[j-1] + t_{1, j-1} + a_{2, j})

j = 5

f₂[5] = min (f₂[4] + a_{2, 5}, f₁[4] + t_{1, 4} + a_{2, 5}) = min (25 + 5, 24 + 3 + 5)

= min (30, 32) = 30, l₂[5] = 2

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In Out

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f₁[j] = min (f₁[j-1] + a_{1, j}, f₂[j-1] + t_{2, j-1} + a_{1, j}) f₂[j] = min (f₂[j-1] + a_{2, j}, f₁[j-1] + t_{1, j-1} + a_{2, j})

j = 6

f₁[6] = min (f₁[5] + a_{1, 6}, f₂[5] + t_{2, 5} + a_{1, 6}) = min (32 + 4, 30 + 1 + 4)

= min (36, 35) = 35, l₁[6] = 2

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f₁[j] = min (f₁[j-1] + a_{1, j}, f₂[j-1] + t_{2, j-1} + a_{1, j}) f₂[j] = min (f₂[j-1] + a_{2, j}, f₁[j-1] + t_{1, j-1} + a_{2, j})

j = 6

f₂[6] = min (f₂[5] + a_{2, 6}, f₁[5] + t_{1, 5} + a_{2, 6}) = min (30 + 7, 32 + 4 + 7)

= min (37, 43) = 37, l₂[6] = 2

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f* = min (f₁[6] + x₁, f₂[6] + x₂) = min (35 + 3, 37 + 2)

= min (38, 39) = 38 l* = 1

l* = 1 => Station S_{1, 6} l₁[6] = 2 => Station S_{2, 5} l₂[5] = 2 => Station S_{2, 4} l₂[4] = 1 => Station S_{1, 3} l₁[3] = 2 => Station S_{2, 2} l₂[2] = 1 => Station S_{1, 1}

Entire Solution Set: Assembly-Line Scheduling

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In Out 2 4 2 2 3 1 1 2 3 2 4 1 3 2

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j

f_i(j) 2 3 4 5 6

1 1 2 1 1 2

2 1 2 1 2 2

j l_i(j)

Fastest Way: Assembly-Line Scheduling

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l* = 1 => Station S_{1, 6}

FASTEST-WAY(a, t, e, x, n)

1 ƒ₁[1]  e₁ + a_1,1

2 ƒ₂[1]  e₂ + a_2,1

3 for j  2 to n

4 do if ƒ₁[ j - 1] + a_{1, j} ≤ ƒ₂[ j - 1] + t_{2, j - 1} + a_{1, j}

5 then ƒ₁[ j]  ƒ₁[ j - 1] + a_{1, j}

6 l₁[ j]  1

7 else ƒ₁[ j]  ƒ₂[ j - 1] + t_{2, j - 1} + a_{1, j}

8 l₁[ j]  2

9 if ƒ₂[ j - 1] + a_{2, j} ≤ ƒ₁[ j - 1] + t_{1, j - 1} + a _{2, j}

10. then ƒ₂[ j]  ƒ₂[ j - 1] + a_{2, j}

11. l₂[ j]  2

12. else ƒ₂[ j]  ƒ₁[ j - 1] + t_{1, j - 1} + a_{2, j}

13. l₂[ j]  1

14. if ƒ₁[n] + x₁ ≤ ƒ₂[n] + x₂

15. then ƒ*

= ƒ1[n] + x1 16. l* = 1

17. else ƒ*

= ƒ2[n] + x2 18. l* = 2

1. Print-Stations (l, n)

2. i  l*

3. print “line” i “, station” n

4. for j  n downto 2

5. do i  l_i[j]

6. print “line” i “, station” j - 1