Presenter : Chang,Chun-Chih Authors : David Milne * , Ian H. Witten 2012, AI
Management of Blood Component Preparation Speaker: Chun-Cheng Lin National Taiwan University...
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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
18
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.
19
Thank you for your attention!