cos / mirrors / cros / chromiumos / third_party / toolchain-utils / refs/heads/stabilize-13654.B / . / crosperf / machine_image_manager.py

# -*- coding: utf-8 -*- | |

# Copyright 2015 The Chromium OS Authors. All rights reserved. | |

# Use of this source code is governed by a BSD-style license that can be | |

# found in the LICENSE file. | |

"""MachineImageManager allocates images to duts.""" | |

from __future__ import print_function | |

import functools | |

class MachineImageManager(object): | |

"""Management of allocating images to duts. | |

* Data structure we have - | |

duts_ - list of duts, for each duts, we assume the following 2 properties | |

exist - label_ (the current label the duts_ carries or None, if it has an | |

alien image) and name (a string) | |

labels_ - a list of labels, for each label, we assume these properties | |

exist - remote (a set/vector/list of dut names (not dut object), to each | |

of which this image is compatible), remote could be none, which means | |

universal compatible. | |

label_duts_ - for each label, we maintain a list of duts, onto which the | |

label is imaged. Note this is an array of lists. Each element of each list | |

is an integer which is dut oridnal. We access this array using label | |

ordinal. | |

allocate_log_ - a list of allocation record. For example, if we allocate | |

l1 to d1, then l2 to d2, then allocate_log_ would be [(1, 1), (2, 2)]. | |

This is used for debug/log, etc. All tuples in the list are integer pairs | |

(label_ordinal, dut_ordinal). | |

n_duts_ - number of duts. | |

n_labels_ - number of labels. | |

dut_name_ordinal_ - mapping from dut name (a string) to an integer, | |

starting from 0. So that duts_[dut_name_ordinal_[a_dut.name]]= a_dut. | |

* Problem abstraction - | |

Assume we have the following matrix - label X machine (row X col). A 'X' | |

in (i, j) in the matrix means machine and lable is not compatible, or that | |

we cannot image li to Mj. | |

M1 M2 M3 | |

L1 X | |

L2 X | |

L3 X X | |

Now that we'll try to find a way to fill Ys in the matrix so that - | |

a) - each row at least get a Y, this ensures that each label get imaged | |

at least once, an apparent prerequiste. | |

b) - each column get at most N Ys. This make sure we can successfully | |

finish all tests by re-image each machine at most N times. That being | |

said, we could *OPTIONALLY* reimage some machines more than N times to | |

*accelerate* the test speed. | |

How to choose initial N for b) - | |

If number of duts (nd) is equal to or more than that of labels (nl), we | |

start from N == 1. Else we start from N = nl - nd + 1. | |

We will begin the search with pre-defined N, if we fail to find such a | |

solution for such N, we increase N by 1 and continue the search till we | |

get N == nl, at this case we fails. | |

Such a solution ensures minimal number of reimages. | |

* Solution representation | |

The solution will be placed inside the matrix, like below | |

M1 M2 M3 M4 | |

L1 X X Y | |

L2 Y X | |

L3 X Y X | |

* Allocation algorithm | |

When Mj asks for a image, we check column j, pick the first cell that | |

contains a 'Y', and mark the cell '_'. If no such 'Y' exists (like M4 in | |

the above solution matrix), we just pick an image that the minimal reimage | |

number. | |

After allocate for M3 | |

M1 M2 M3 M4 | |

L1 X X _ | |

L2 Y X | |

L3 X Y X | |

After allocate for M4 | |

M1 M2 M3 M4 | |

L1 X X _ | |

L2 Y X _ | |

L3 X Y X | |

After allocate for M2 | |

M1 M2 M3 M4 | |

L1 X X _ | |

L2 Y X _ | |

L3 X _ X | |

After allocate for M1 | |

M1 M2 M3 M4 | |

L1 X X _ | |

L2 _ X _ | |

L3 X _ X | |

After allocate for M2 | |

M1 M2 M3 M4 | |

L1 X X _ | |

L2 _ _ X _ | |

L3 X _ X | |

If we try to allocate for M1 or M2 or M3 again, we get None. | |

* Special / common case to handle seperately | |

We have only 1 dut or if we have only 1 label, that's simple enough. | |

""" | |

def __init__(self, labels, duts): | |

self.labels_ = labels | |

self.duts_ = duts | |

self.n_labels_ = len(labels) | |

self.n_duts_ = len(duts) | |

self.dut_name_ordinal_ = dict() | |

for idx, dut in enumerate(self.duts_): | |

self.dut_name_ordinal_[dut.name] = idx | |

# Generate initial matrix containg 'X' or ' '. | |

self.matrix_ = [['X' if l.remote else ' ' | |

for _ in range(self.n_duts_)] | |

for l in self.labels_] | |

for ol, l in enumerate(self.labels_): | |

if l.remote: | |

for r in l.remote: | |

self.matrix_[ol][self.dut_name_ordinal_[r]] = ' ' | |

self.label_duts_ = [[] for _ in range(self.n_labels_)] | |

self.allocate_log_ = [] | |

def compute_initial_allocation(self): | |

"""Compute the initial label-dut allocation. | |

This method finds the most efficient way that every label gets imaged at | |

least once. | |

Returns: | |

False, only if not all labels could be imaged to a certain machine, | |

otherwise True. | |

""" | |

if self.n_duts_ == 1: | |

for i, v in self.matrix_vertical_generator(0): | |

if v != 'X': | |

self.matrix_[i][0] = 'Y' | |

return | |

if self.n_labels_ == 1: | |

for j, v in self.matrix_horizontal_generator(0): | |

if v != 'X': | |

self.matrix_[0][j] = 'Y' | |

return | |

if self.n_duts_ >= self.n_labels_: | |

n = 1 | |

else: | |

n = self.n_labels_ - self.n_duts_ + 1 | |

while n <= self.n_labels_: | |

if self._compute_initial_allocation_internal(0, n): | |

break | |

n += 1 | |

return n <= self.n_labels_ | |

def _record_allocate_log(self, label_i, dut_j): | |

self.allocate_log_.append((label_i, dut_j)) | |

self.label_duts_[label_i].append(dut_j) | |

def allocate(self, dut, schedv2=None): | |

"""Allocate a label for dut. | |

Args: | |

dut: the dut that asks for a new image. | |

schedv2: the scheduling instance, we need the benchmark run | |

information with schedv2 for a better allocation. | |

Returns: | |

a label to image onto the dut or None if no more available images for | |

the dut. | |

""" | |

j = self.dut_name_ordinal_[dut.name] | |

# 'can_' prefix means candidate label's. | |

can_reimage_number = 999 | |

can_i = 999 | |

can_label = None | |

can_pending_br_num = 0 | |

for i, v in self.matrix_vertical_generator(j): | |

label = self.labels_[i] | |

# 2 optimizations here regarding allocating label to dut. | |

# Note schedv2 might be None in case we do not need this | |

# optimization or we are in testing mode. | |

if schedv2 is not None: | |

pending_br_num = len(schedv2.get_label_map()[label]) | |

if pending_br_num == 0: | |

# (A) - we have finished all br of this label, | |

# apparently, we do not want to reimaeg dut to | |

# this label. | |

continue | |

else: | |

# In case we do not have a schedv2 instance, mark | |

# pending_br_num as 0, so pending_br_num >= | |

# can_pending_br_num is always True. | |

pending_br_num = 0 | |

# For this time being, I just comment this out until we have a | |

# better estimation how long each benchmarkrun takes. | |

# if (pending_br_num <= 5 and | |

# len(self.label_duts_[i]) >= 1): | |

# # (B) this is heuristic - if there are just a few test cases | |

# # (say <5) left undone for this label, and there is at least | |

# # 1 other machine working on this lable, we probably not want | |

# # to bother to reimage this dut to help with these 5 test | |

# # cases | |

# continue | |

if v == 'Y': | |

self.matrix_[i][j] = '_' | |

self._record_allocate_log(i, j) | |

return label | |

if v == ' ': | |

label_reimage_number = len(self.label_duts_[i]) | |

if ((can_label is None) or | |

(label_reimage_number < can_reimage_number or | |

(label_reimage_number == can_reimage_number and | |

pending_br_num >= can_pending_br_num))): | |

can_reimage_number = label_reimage_number | |

can_i = i | |

can_label = label | |

can_pending_br_num = pending_br_num | |

# All labels are marked either '_' (already taken) or 'X' (not | |

# compatible), so return None to notify machine thread to quit. | |

if can_label is None: | |

return None | |

# At this point, we don't find any 'Y' for the machine, so we go the | |

# 'min' approach. | |

self.matrix_[can_i][j] = '_' | |

self._record_allocate_log(can_i, j) | |

return can_label | |

def matrix_vertical_generator(self, col): | |

"""Iterate matrix vertically at column 'col'. | |

Yield row number i and value at matrix_[i][col]. | |

""" | |

for i, _ in enumerate(self.labels_): | |

yield i, self.matrix_[i][col] | |

def matrix_horizontal_generator(self, row): | |

"""Iterate matrix horizontally at row 'row'. | |

Yield col number j and value at matrix_[row][j]. | |

""" | |

for j, _ in enumerate(self.duts_): | |

yield j, self.matrix_[row][j] | |

def _compute_initial_allocation_internal(self, level, N): | |

"""Search matrix for d with N.""" | |

if level == self.n_labels_: | |

return True | |

for j, v in self.matrix_horizontal_generator(level): | |

if v == ' ': | |

# Before we put a 'Y', we check how many Y column 'j' has. | |

# Note y[0] is row idx, y[1] is the cell value. | |

ny = functools.reduce(lambda x, y: x + 1 if (y[1] == 'Y') else x, | |

self.matrix_vertical_generator(j), 0) | |

if ny < N: | |

self.matrix_[level][j] = 'Y' | |

if self._compute_initial_allocation_internal(level + 1, N): | |

return True | |

self.matrix_[level][j] = ' ' | |

return False |