Management ofBlood Component Preparation

Speaker: Chun-Cheng Lin National Taiwan University

Co-authors: Chang-Sung Yu, Yin-Yih Chang

2

Outline

• Introduction to

the blood component preparation problem (BCPP)

• A linear time algorithm for the BCPP

• Some variants of the BCPP

• Conclusion and future work

3

Introduction

• Transfusion therapy to transfuse the specific blood components

needed to replace particular deficits

for some medical purposes.

• Whole blood contain all blood elements a source for blood component production.

• Blood component preparation the indication for the use of unfractionated whole blood

almost does not exist now separating specific cell components from the whole blood a lot of different methods (processes) or equipment

different use, efficiency and quality

4

A Process of Separating Blood Components

• Implied value (ai): Consider both the revenue contributed by patients or insurance and the costs induced by blood of collection, testing, preparation, preservation, storage, processing time, etc.

• Demand limit (di).

Packed Red Blood CellsPacked Red Blood Cells

Platelet-Rich PlasmaPlatelet-Rich Plasma

Washed Red Blood CellsWashed Red Blood Cells

Fresh Frozen PlasmaFresh Frozen Plasma

Platelet ConcentratePlatelet Concentrate

Fresh PlasmaFresh Plasma

Frozen PlasmaFrozen Plasma

CryoprecipitateCryoprecipitate

soft-spin centrifugation

hard-spin centrifugationwashing

Whole BloodWhole Blood

freezing

thawing; centrifugation; freezing

5

Blood Component Preparation Problem

• Blood Component tree vertex vi = a blood component

with value ai and demand limit di

the amount xi of vi is derived from the amount of its parent

according to a given ratio ri;

• Initial assignment of { xi } x1 = Q; other xi = 0

• The Blood Component Preparation Problem (BCPP)Given an initial volume Q of the whole blood and

an n-vertex blood component tee T

(where demand limit di ; implied value ai),

determine the assignments of { xi }so that (1) the total value is maximized (2) while the demand limit of each component is satisfied

r2 r3

x1

a1 d1

x2

a2 d2

x3

a3d3

x4

a4 d4

x5

a5 d5

x6

a6 d6

x7

a7 d7

x8

a8 d8

x9

a9 d9

r8 r9

r4 r6r5r7

6

A linear programming approach• The BCPP problem can be solved by linear programming.•

2 1 1 2( ) 1 ( )( ) ( ) ( ) ( )

Maximize

such that

, for every vertex ;

( ( (...( ( ) )...) ) ),

for every leaf with ancestors;

i

k ki ii i i i

i iv V

i i i

i i p v p vp v p v p v p v

i

a x

x d v

x r r r r Q x x x x

v k

Comment:There exist a lot of software tools for the linear programming problem, users just need to describe the BCPP as a linear program and then use those tools to solve it without implement it.

r2 r3

x1

a1 d1

x2

a2 d2

x3

a3d3

x4

a4 d4

x5

a5 d5

x6

a6 d6

x7

a7 d7

x8

a8 d8

x9

a9 d9

r8 r9

r4 r6r5r7

7

Example• 1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9

4 1 2

5 1 2

7 1 3

8

Maximize 3 4 8 2 3 2 6

such that

10; 4; 5; 12; 7; 8; 9; 13; 6;

0.2(0.8(100 ) )

0.3(0.8(100 ) )

1.0(0.2(100 ) )

0.7(0.5

x x x x x x x x x

x x x x x x x x x

x x x

x x x

x x x

x

1 2 6

9 1 2 6

(0.8(100 ) ) )

0.3(0.5(0.8(100 ) ) )

x x x

x x x x

0.8 0.2x1

3 10

x2

4 4x3

8 5

x4

2 12x5

1 7x6

3 8x7

1 9

x8

2 13x9

6 6

0.7 0.3

0.2 0.50.3 1

82 1 6

flowinflowout

1 0.8(100 )

0.5 0.7

xx x x

Q = 100

8

Motivations• Linear programming (LP) problem

• 2 drawbacks to solve the BCPP by LP: The worst-case algorithm for the LP problem may not be executed

efficiently (its time complexity is nonlinear) It may not be convenient for users to directly describe the constraints

of the LP for a general derivatives tree.

result Reference

the well-known simplex algorithm Dantzig (1947)

the worst-case complexity of the simplex algorithm is exponential

Klee and Minty (1972)

the first polynomial-time algorithm Khachiyan (1979,1980)

the first algorithm performing well both in theory and in practice

Karmarkar (1984)

9

Main result• Main Theorem: There exists a linear time algorithm

for the BCPP in the size of vertices.

• Characteristic value (vi) of vi

Compute the following formula in the bottom-up fashion

• e.g., (v8) = 2; (v9) = 6

(v6) = max( 3, 0.72 + 0.36 ) = 3.2

( )

( ) max , ( )j i

i i j jv Child v

v a r v0.8 0.2

x1

3 10

x2

4 4x3

8 5

x4

2 12 x5

1 7x6

3 8x7

1 9

x8

2 13 x9

6 6

0.7 0.3

0.2 0.50.3 1

10

Our linear time algorithm (1/4)• Step 1: vertex vi in top-down fashio

n of T, xi is assigned di, and then the remainin

g amount is forwarded to the next level; if not enough to satisfy any demand lim

it, then return false. for convenience, we express that xi = di

+ yi.

0.8 0.2100

3 10

0

4 4

0

8 5

02 12

01 7

03 8

0

1 9

02 13

06 6

0.7 0.3

0.2 0.50.3 1

0.8 0.210+03 10

4+0

4 4

5+0

8 5

12+1.6

2 12

7+13.4

1 7

8+0

3 8

9+4

1 9

13+5.2

2 13

6+1.8

6 6

0.7 0.3

0.2 0.50.3 1

100 10 90

.7 18.2

.3 7.8

.2 13.6

.3 20.4

.5 34 8 26

x1x2

x3 x7

x4

x5

x6

x8

x9

.8 72 4 68

.2 18 5 13

11

Our linear time algorithm (2/4)• Step 2: vertex vi in bottom-up fashio

n of T, If vi is a leaf, then yi

M = yi;

o.w., yiM = minvjChild(vi){ yj

M / rj },

yim = yj – rj yi

M for each vj Child(vi).

(In fact, yiM is the maximal possible am

ount of vi flowed from its descendents and yi

m is the amount of every descendent of vj of vi after yi

M is achieved) y1m

100

8

0

4

2

1.60

13.411

62

4

0

5.2

1

1.80

0.8 0.210+03 10

4+0

4 4

5+0

8 5

12+1.62 12

7+13.41 7

8+03 8

9+4

1 9

13+5.22 13

6+1.86 6

0.7 0.3

0.2 0.50.3 1

y1M

y2m

y2M

y3m

y3M

y7m

y7M

y6m

y6M

y4m

y4M

y8m

y8M

y9m

y9M

8 9

8 9

6 9 6

8 9

5 64 4

4 4 5 6

2 4

5.2 1.8eg., 6

0.7 0.3

is bottleneck 6

5.2 0.7 6 1; 1.8 0.3 6 0

eg., 8 is the minimum among { , , }.

is bottleneck

M M

M M

M

m m

M MM M

M M M M

y y

r r

v v y

y y

y yy y

r r r r

v v

2

4 5

6

8

1.6 0.2 8 0; 13.4 0.3 8 11

6 0.5 8 2

M

m m

m

y

y y

y

12

Our linear time algorithm (3/4)• Step 3:

Initially, all the leaves of T are marked.

For each internal vertex vi in the bottom-up fashion of T, compute (vi). If (vi) = ai, then vertex vi is marked.

v4, v5, v7, v8, v9 are leaves, and hence, marked.

(v6) = max(3, 0.72+0.36) = 3.2 > 3 = a6

v6 is unmarked.

(v3) = max(8, 11) = 8 = a3

v3 is marked.

(v2) = max(4, 0.22+0.31+0.53.2) = 4 = a2

v2 is marked.

(v1) = max(3, 0.84+0.28) = 4.8 > 3 = a1

v1 is unmarked.

y1m

100

8

0

4

2

1.60

13.411

62

4

0

5.2

1

1.80

0.8 0.210+03 10

4+0

4 4

5+0

8 5

12+1.62 12

7+13.41 7

8+03 8

9+4

1 9

13+5.22 13

6+1.86 6

0.7 0.3

0.2 0.50.3 1

y1M

y2m

y2M

y3m

y3M

y7m

y7M

y6m

y6M

y4m

y4M

y8m

y8M

y9M

y9M

13

0.3

Our linear time algorithm (4/4)• Step 4: Let U = V. vi U i

n top-down fashion of T, if vi is unmarked then output xi = di + 0; otherwise, do the following: Output xi = di + yi

M;

vj in the top-down fashion of the subtree Tvi rooted at vi, if vj is marked, then output xj = dj + yj

m; otherwise, for evry children vk of vj, yk

m y

km + rk yj

m, and then output xj = dj + 0 (i.e., yj

m = 0);

U U \ V(Tvi)

0.8 0.210+03 10

4+04 4

5+08 5

12+1.62 12

7+13.41 7

8+03 8

9+41 9

13+5.22 13

6+1.86 6

0.7 0.3

0.2 0.5 1

0.8 0.210+03 10

4+8

4 4

5+4

8 5

12+02 12

7+111 7

8+03 8

9+0

1 9

13+2.42 13

6+0.66 6

0.7 0.3

0.2 0.50.3 1

y1m

100

80

42

1.60

13.411

62

40

5.21

1.80

y1M

y2m

y2M

y3m

y3M

y7m

y7M

y6m

y6m

y4m

y4M

y8m

y8M

y9m

y9M

0

2.4 0.6

14

Observations of our outputs• Observation 1.

• Observation 2. In the output of our algorithm, For every vertex vi in R1, ai < (vi).

For every vertex vi in R2, ai = (vi).

• Observation 3. Any feasible solution of the BCPP is reachable from the output of our algorithm.

Mi iy y

0iy

mi iy y

Mi iy y on the band

above the band

below the band

R2

R3

R1

Output of Step 1 of our algorithm Output of our algorithm

0iy

15

Comparison of solutions

0iy

Mi iy y

0iy

mi iy y

Mi iy y on the band

above the band

below the band

R2

R3

R1

( )i

mi i i p v iy y rq q

band

0iy

…

… ( )ii i i p vy z r z

( )i

Mi i i p v iy y r z q

R11

R12

R13

R2

R3

0i iy z R1

Any feasible solution

Output of Step 1 of our algorithm Output of our algorithm

16

Blood Component Preparation Problem

• Theorem. The BCPP can be solved in O(n) time.

Proof. Skipped.

• Extension: How to choose multiple standardized processes so that the total value is maximized? The preparation and preservation of blood components are c

onsidered within a time frame.

In practice, we may execute more than one process simultaneously within a certain time frame.

Fortunately, for the standardization of executing processes, the number of alternative processes is fixed constant.

The extended problem can be solved roughly in polynomial time.

17

Modified Blood Component Preparation Problem

• The deriving operation may be nonreversible. require: the volume of the components at higher levels is more.

• The Modified Blood Component Preparation Problem (BCPP’)Given the initial volume Q of the whole blood and

an n-vertex blood component tree T (every vertex has its demand limit),

determine the assignments of { xi }so that (1) the volume of the components at higher levels

is remained as more as possible (2) while the demand limit of every component is satisfied.

• Algorithm:Steps 1 and 2 are the same as those of the previous.

Step 3: Output x1 = d1 + y1M ; xi = di + yi

m, vi V \ {v1}.

• Corollary. The BCPP’ can be solved in O(n) time.

mi i ix d y

1 1 1Mx d y

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Conclusion and future work• We have defined a new problem called the blood

component preparation problem (BCPP), and proposed not only a linear programming solution but also a linear time algorithm for the BCPP.

• Some variants are also given in this work.

• A line of future work includes: to investigate the tractability of the `derivatives graph

problem’ on some special cases of graphs; to take more factors (e.g., time and inventory) into

account in the BCPP; to evaluate the effectiveness of the BCPP applied in

practical environment; to investigate the sensitivity analysis and the critical

paths of the BCPP.

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Thank you for your attention!