I was wondering if you can help with a programing assignment in C. This is a weekly (now bi-weekly) assignment for operating systems class. These assignments are graded in an Ubuntu environment so it would be best to program in that, however I suspect OSX would work fine. (There will be threading issues in Windows.) I will paste the file description below and the attach the given file templates. This is due Tuesday 28FEB, I would need it by 9AM that day. Let me know if you would be able to do it! Thanks!!
I’ll pay $100.
Homework 6 – elevator scheduling
This is our first 10 point, 2 week homework. Compared to our previous (5 point) homeworks, there will be less emphasis on basic programming skills (which it is hoped you have acquired enough of by now), and more on algorithmic or system particulars. While the homework is due 2 weeks from now, you are encouraged to start early, so you can take full advantage of office hours.
This time, we continue our work with the elevator controller, based on the solution for hw5.
However, this time, the multithreaded/concurrent programming aspects are less of a concern (they are mostly taken care of by the hw5 solution), and scheduling is our focus. A few things have changed, to make things a little more interesting:
like in a normal building, passengers are more likely to travel to or from the bottom floor.
passengers take several trips on the elevator, separated by a short delay. Except for part 4, this delay grows linearly with the passenger number.
each run has a fixed duration, rather than a fixed number of trips
our goal is to maximize throughput (number of trips), under different sets of constraints.
The provided homework template does a reasonable job of maximizing “throughput”. However, it does this at the expense of fairness.
There are three test scenarios: make highload, make mediumload, and make lowload. Make sure your solution provides the desired performance on all three scenarios. The total number of trips should remain above 200, 450 and 1000 respectively.
HINT: for all of these, the complexity (as in run-time CPU consumption) of your solution is not of great importance: my solutions are quite inefficient, but still only use less than 15% of the CPU.
TRIP TIME FAIRNESS
With the template solution, some trip requests take a long time to complete (from request to exiting at the requested floor), and others go quickly. The difference between min and max is often more than 20x. Provide a solution where the max trip duration is no more than 100% over the mean trip duration, over all passengers.
HINT: a FIFO solution would probably work well here.
Turn-in requirement: put your entire solution in a file called hw6i1.c
PASSENGER SERVICE FAIRNESS
With the above solution, the more aggressive passengers (those with lower numbers), get more tripsand quicker trips. Provide a solution where the more aggressive passengers may get more trips, but on average need to wait longer for their trips (unless there are so many elevators that nobody needs to wait long!).
HINT: a round-robin scheduler solution would work well here.
Turn-in requirement: put your entire solution in a file called hw6i2.c
STATIC PASSENGER PRIORITIZATION
The passengers that wait the longest between elevator trips happen to be the most important passengers, and should get priority service. Here, you may assume that a higher-numbered passenger (as provided in passenger_request) is a higher-priority passenger. Provide a solution where higher-priority passengers reliably receive better service (shorter mean and max trip times) than lower-priority passengers, but make sure that no passenger completes less than 1/3 of the number of trips of the highest-priority passenger.
HINT: here, a weighted round-robin solution may be a good idea.
Turn-in requirement: put your entire solution in a file called hw6i3.c
DYNAMIC PASSENGER PRIORITIZATION
Here, do the same thing as in static passenger prioritization, but do not assume that the passenger number corresponds to their priority (we will use a modified main.c). Instead, have your solution base priority on inter-request time: passengers that issue frequent requests have lower priority than those that issue infrequent requests.
HINT: You’ll need to determine priority based on the timing of requests, not the passenger ID. Use gettimeofday() to get the current time. A scheduler like CFS from linux would be a good choice here.
Turn-in requirement: put your entire solution in a file called hw6i4.c
ASSIGNMENT DESCRIPTION.txt
HOMEWORK 6
This time, we continue our work with the elevator controller, based on the solution for hw5.
However, this time, the multithreaded/concurrent programming aspects are less of a concern (they are mostly taken care of by the hw5 solution), and scheduling is our focus. A few things have changed, to make things a little more interesting:
• like in a normal building, passengers are more likely to travel to or from the bottom floor.
• passengers take several trips on the elevator, separated by a short delay. Except for part 4, this delay grows linearly with the passenger number.
• each run has a fixed duration, rather than a fixed number of trips
• our goal is to maximize throughput (number of trips), under different sets of constraints.
The provided homework template does a reasonable job of maximizing “throughput”. However, it does this at the expense of fairness.
There are three test scenarios: make highload, make mediumload, and make lowload. Make sure your solution provides the desired performance on all three scenarios. The total number of trips should remain above 200, 450 and 1000 respectively.
HINT: for all of these, the complexity (as in run-time CPU consumption) of your solution is not of great importance: my solutions are quite inefficient, but still only use less than 15% of the CPU.
TRIP TIME FAIRNESS
With the template solution, some trip requests take a long time to complete (from request to exiting at the requested floor), and others go quickly. The difference between min and max is often more than 20x. Provide a solution where the max trip duration is no more than 100% over the mean trip duration, over all passengers.
HINT: a FIFO solution would probably work well here.
Turn-in requirement: put your entire solution in a file called hw6i1.c
PASSENGER SERVICE FAIRNESS
With the above solution, the more aggressive passengers (those with lower numbers), get more tripsand quicker trips. Provide a solution where the more aggressive passengers may get more trips, but on average need to wait longer for their trips (unless there are so many elevators that nobody needs to wait long!).
HINT: a round-robin scheduler solution would work well here.
Turn-in requirement: put your entire solution in a file called hw6i2.c
STATIC PASSENGER PRIORITIZATION
The passengers that wait the longest between elevator trips happen to be the most important passengers, and should get priority service. Here, you may assume that a higher-numbered passenger (as provided in passenger_request) is a higher-priority passenger. Provide a solution where higher-priority passengers reliably receive better service (shorter mean and max trip times) than lower-priority passengers, but make sure that no passenger completes less than 1/3 of the number of trips of the highest-priority passenger.
HINT: here, a weighted round-robin solution may be a good idea.
Turn-in requirement: put your entire solution in a file called hw6i3.c
DYNAMIC PASSENGER PRIORITIZATION
Here, do the same thing as in static passenger prioritization, but do not assume that the passenger number corresponds to their priority (we will use a modified main.c). Instead, have your solution base priority on inter-request time: passengers that issue frequent requests have lower priority than those that issue infrequent requests.
HINT: You’ll need to determine priority based on the timing of requests, not the passenger ID. Use gettimeofday() to get the current time. A scheduler like CFS from linux would be a good choice here.
Turn-in requirement: put your entire solution in a file called hw6i4.c
THIS IS WHAT WE DID IN HW 5
Homework 5 – Concurrent Elevator Controller
A M -story highrise has N elevators in a single “elevator bank”, each able to serve every floor of the building. Contrary from most elevator designs, there are no buttons to choose a destination floor inside the elevator. Instead, the elevator lobby at each floor has M individual “destination floor” buttons, and passengers choose their destination by pushing the appropriate button. Once a button is pushed, a display near the buttons tells the passenger which elevator door to wait by. In this assignment, elevators serve only one passenger at a time.
Your assignment is to create an elevator controller that ensures that all passengers receive service, minimizing the mean and max time between button push and arrival at the destination floor.
THE CODE
The homework 5 template is a multithreaded program which simulates passengers and elevator travel. There is one thread per elevator, and one thread per passenger. There is a fixed number of passengers, that embark on a number of trips each. These threads interact with each other through a central elevator controller, which is your responsibility to implement.
The central controller responds to the following three function calls:
JUST IN CASE YOU NEED MORE INFO FROM HW5
Homework 5 – Concurrent Elevator Controller
A M -story highrise has N elevators in a single “elevator bank”, each able to serve every floor of the building. Contrary from most elevator designs, there are no buttons to choose a destination floor inside the elevator. Instead, the elevator lobby at each floor has M individual “destination floor” buttons, and passengers choose their destination by pushing the appropriate button. Once a button is pushed, a display near the buttons tells the passenger which elevator door to wait by. In this assignment, elevators serve only one passenger at a time.
Your assignment is to create an elevator controller that ensures that all passengers receive service, minimizing the mean and max time between button push and arrival at the destination floor.
THE CODE
The homework 5 template is a multithreaded program which simulates passengers and elevator travel. There is one thread per elevator, and one thread per passenger. There is a fixed number of passengers, that embark on a number of trips each. These threads interact with each other through a central elevator controller, which is your responsibility to implement.
The central controller responds to the following three function calls:
Homework 5 – Concurrent Elevator Controller
A M -story highrise has N elevators in a single “elevator bank”, each able to serve every floor of the building. Contrary from most elevator designs, there are no buttons to choose a destination floor inside the elevator. Instead, the elevator lobby at each floor has M individual “destination floor” buttons, and passengers choose their destination by pushing the appropriate button. Once a button is pushed, a display near the buttons tells the passenger which elevator door to wait by. In this assignment, elevators serve only one passenger at a time.
Your assignment is to create an elevator controller that ensures that all passengers receive service, minimizing the mean and max time between button push and arrival at the destination floor.
THE CODE
The homework 5 template is a multithreaded program which simulates passengers and elevator travel. There is one thread per elevator, and one thread per passenger. There is a fixed number of passengers, that embark on a number of trips each. These threads interact with each other through a central elevator controller, which is your responsibility to implement.
The central controller responds to the following three function calls:
1. /* called whenever a passenger pushes a button in the elevator lobby.
2. call enter / exit to move passengers into / out of elevator
3. return only after the passenger has been delivered to requested floor
4. */
5. void passenger_request(int passenger, int from_floor, int to_floor,
6. void (*enter)(int, int),
7. void(*exit)(int, int));
8.
9. /* called whenever the elevator needs to know what to do next.
10. call move_direction with direction -1 to descend, 1 to ascend.
11. must call door_open before letting passengers in, and door_close
12. before moving the elevator
13. */
14. void elevator_ready(int elevator, int at_floor,
15. void(*move_direction)(int, int),
16. void(*door_open)(int),
17. void(*door_close)(int));
PERFORMANCE EXPECTATIONS
The template code works correctly, but is terribly slow. In part, this is because passengers often “miss” the elevator (don’t get in in time), or miss their floor (stay on the elevator). In part, it is because the elevator stops at every floor, and we only use one elevator.
For a full score, have a fully working solution that uses all available elevators, and finishes the final case in “make test” in at most 20 seconds without using more than 10% of the CPU during that time.
NOTES
• Every passenger pushes a button.
• Passengers can’t change their minds, and will wait until they are told to enter the elevator, using the enter() callback function
• Passengers will stay in the elevator until they are told to leave, using the exit() callback function.
• Elevators can hold at most one passenger.
• The program finishes after all passengers have been served, and all functions have returned.
HINTS
• Start by making sure your passengers have time to get on and off the elevator. You can do this with condition variables (pthread_cond_t), or barriers (pthread_barrier_t). A barrier-based solution seems easier. Use one barrier to make the passenger wait for the door to open, and another to make the elevator wait for the passenger to enter.
• The template solution can a single set of global variables. You’ll probably want to have one set per elevator. Define an elevator struct that holds all your necessary state per elevator, and make an array of such structs.
• Handling multiple elevators should be the last item on your TODO list. An easy way to do it is to randomly decide, for each passenger, which elevator they should use, independent of everything else. Then you can treat each elevator+passengers group separately.
• You could use an extra set of mutexes around the whole passenger_request() function to make sure only one user’s request is handled by each elevator at one time.
LOGGING
To add more logging output, try using the log() function instead of printf.
log(loglevel, format_string, parameters)
works just like fprintf(stderr,…) except it only prints if the configured log level is higher than the loglevel argument provided. To set the log level to, for example, 8 use the following flag to gcc -DLOGLEVEL=8
GOOD DOCUMENTS ON THREADED PROGRAMMING
http://www.ibm.com/developerworks/linux/library/l-posix1.html?S_TACT=105AGX03&S_CMP=EDU
http://www.ibm.com/developerworks/linux/library/l-posix2/?S_TACT=105AGX03&S_CMP=EDU
http://www.ibm.com/developerworks/linux/library/l-posix3/index.html?S_TACT=105AGX03&S_CMP=EDU
https://computing.llnl.gov/tutorials/pthreads/#ConVarSignal
http://pages.cs.wisc.edu/~travitch/pthreads_primer.html
__MACOSX/._ASSIGNMENT DESCRIPTION.txt
hw6/hw6.c
#include “hw6.h”
#include
#include
#include
#include
pthread_mutex_t lock;
pthread_barrier_t barrier;
int floor;
int direction;
int next;
int occupancy;
enum {ELEVATOR_ARRIVED, ELEVATOR_OPEN, ELEVATOR_CLOSED} state;
} elevators[ELEVATORS];
// global lock over all passengers and passenger_count
pthread_mutex_t passenger_lock=PTHREAD_MUTEX_INITIALIZER;
int passenger_count=0;
struct Passenger {
int elevator;
int from_floor;
int to_floor;
int finished;
pthread_mutex_t wakeup_lock;
pthread_cond_t wakeup;
} passengers[PASSENGERS];
void scheduler_init() {
memset(passengers,0,sizeof(passengers));
for(int i=0;i
passenger_count=passenger+1;
passengers[passenger].elevator=-1;
passengers[passenger].from_floor=from_floor;
passengers[passenger].to_floor=to_floor;
passengers[passenger].finished=0;
pthread_mutex_unlock(&passenger_lock);
// wait for elevator assignment
pthread_mutex_lock(&passengers[passenger].wakeup_lock);
pthread_cond_wait(&passengers[passenger].wakeup, &passengers[passenger].wakeup_lock);
pthread_mutex_unlock(&passengers[passenger].wakeup_lock);
int elevator=passengers[passenger].elevator;
log(3,”Passenger was %d assigned to elevator %d\n”,passenger,elevator);
log(3,”Passenger %d awaiting elevator %d floor arrival at %d, occupancy %d.\n”,passenger,elevator,from_floor,elevators[elevator].occupancy);
pthread_barrier_wait(&elevators[elevator].barrier);
log(3,”Passenger %d alerted to elevator %d floor arrival at %d, occupancy %d.\n”,passenger,elevator,from_floor,elevators[elevator].occupancy);
pthread_mutex_lock(&elevators[elevator].lock);
log(3,”Passenger %d alerted to elevator %d floor arrival at %d, occupancy %d (got lock).\n”,passenger,elevator,from_floor,elevators[elevator].occupancy);
enter_elevator(passenger, elevator);
elevators[elevator].occupancy++;
elevators[elevator].next = to_floor;
pthread_mutex_unlock(&elevators[elevator].lock);
pthread_barrier_wait(&elevators[elevator].barrier);
// get passenger out of elevator once it’s at destination floor
pthread_barrier_wait(&elevators[elevator].barrier);
pthread_mutex_lock(&elevators[elevator].lock);
log(3,”Passenger %d alerted to elevator %d floor delivery at %d, state %d.\n”,passenger,elevator,to_floor,elevators[elevator].state);
if(elevators[elevator].floor == to_floor && elevators[elevator].state == ELEVATOR_OPEN) {
exit_elevator(passenger, elevator);
elevators[elevator].next=-1;
// now schedule next trip
elevators[elevator].occupancy–;
}
pthread_mutex_unlock(&elevators[elevator].lock);
pthread_barrier_wait(&elevators[elevator].barrier);
}
void elevator_ready(int elevator, int at_floor,
void(*move_direction)(int, int),
void(*door_open)(int), void(*door_close)(int)) {
pthread_mutex_lock(&elevators[elevator].lock);
if(elevators[elevator].state == ELEVATOR_ARRIVED && elevators[elevator].next == at_floor) {
door_open(elevator);
elevators[elevator].state=ELEVATOR_OPEN;
pthread_mutex_unlock(&elevators[elevator].lock);
log(3,”Elevator %d signaling arrival.\n”,elevator);
pthread_barrier_wait(&elevators[elevator].barrier);
// gives the passenger time to board
pthread_barrier_wait(&elevators[elevator].barrier);
return;
}
else if(elevators[elevator].state == ELEVATOR_OPEN) {
door_close(elevator);
elevators[elevator].state=ELEVATOR_CLOSED;
}
else {
// find an unserved passenger
if(elevators[elevator].next<0) {
log(7,"Elevator %d acquiring passenger scheduling lock.\n",elevator);
pthread_mutex_lock(&passenger_lock);
int closest_passenger = -1;
int closest_distance = 1000;
for(int i=0;i
if(at_floor < elevators[elevator].next) {
move_direction(elevator,1);
elevators[elevator].floor++;
}
else if(at_floor > elevators[elevator].next) {
move_direction(elevator,-1);
elevators[elevator].floor–;
}
}
elevators[elevator].state=ELEVATOR_ARRIVED;
}
pthread_mutex_unlock(&elevators[elevator].lock);
}
__MACOSX/hw6/._hw6.c
hw6/hw6.h
/* Change these experiment settings to try different scenarios. Parameters can
also be passed in using gcc flags, e.g. -DELEVATORS=5 */
#ifndef MAX_CAPACITY
#define MAX_CAPACITY 1
#endif
#ifndef ELEVATORS
#define ELEVATORS 4
#endif
#ifndef FLOORS
#define FLOORS 28
#endif
#ifndef PASSENGERS
#define PASSENGERS 50
#endif
#ifndef TRIPS_PER_PASSENGER
#define TRIPS_PER_PASSENGER 10
#endif
// these settings affect only the ‘looks’, will be tested at log level 1
#ifndef DELAY
#define DELAY 100000
#endif
#ifndef DURATION
#define DURATION 30
#endif
#ifndef LOG_LEVEL
#define LOG_LEVEL 9
#endif
#define log(level,format,…) do{ if(level<=LOG_LEVEL) fprintf(stderr,format,__VA_ARGS__); }while(0);
/* called once on initialization */
void scheduler_init();
/* called whenever a passenger pushes a button in the elevator lobby.
call enter / exit to move passengers into / out of elevator
return only when the passenger is delivered to requested floor
*/
void passenger_request(int passenger, int from_floor, int to_floor, void (*enter)(int, int), void(*exit)(int, int));
/* called whenever the doors are about to close.
call move_direction with direction -1 to descend, 1 to ascend.
must call door_open before letting passengers in, and door_close before moving the elevator
*/
void elevator_ready(int elevator, int at_floor, void(*move_direction)(int, int), void(*door_open)(int), void(*door_close)(int));
__MACOSX/hw6/._hw6.h
hw6/main.c
#include
#include
#include
#include
#include
#include
#include
#include”hw6.h”
// when stop == 1, all threads quit voluntarily
static int stop=0;
/* This is an internal struct used by the enforcement system
– there is no access to this from hw5 solution. */
static struct Elevator {
int seqno, last_action_seqno; // these two are used to enforce control rules
int floor;
int open;
int passengers;
int trips;
} elevators[ELEVATORS];
void elevator_check(int elevator) {
if(elevators[elevator].seqno == elevators[elevator].last_action_seqno) {
log(0,”VIOLATION: elevator %d make at most one action call per elevator_ready()\n”,elevator);
exit(1);
}
if(elevators[elevator].passengers > MAX_CAPACITY || elevators[elevator].passengers < 0) {
log(0,"VIOLATION: elevator %d over capacity, or negative passenger count %d!\n",elevator,elevators[elevator].passengers);
exit(1);
}
elevators[elevator].last_action_seqno=elevators[elevator].seqno;
}
void elevator_move_direction(int elevator, int direction) {
elevator_check(elevator);
log(8,"Moving elevator %d %s from %d\n",elevator,(direction==-1?"down":"up"),elevators[elevator].floor);
if(elevators[elevator].open) {
log(0,"VIOLATION: attempted to move elevator %d with door open.\n", elevator);
exit(1);
}
if(elevators[elevator].floor >= FLOORS || elevators[elevator].floor < 0) {
log(0,"VIOLATION: attempted to move elevator %d outside of building!\n", elevator);
exit(1);
}
sched_yield();
usleep(DELAY);
elevators[elevator].floor+=direction;
}
void elevator_open_door(int elevator) {
elevator_check(elevator);
log(9,"Opening elevator %d at floor %d\n",elevator,elevators[elevator].floor);
if(elevators[elevator].open) {
log(0,"VIOLATION: attempted to open elevator %d door when already open.\n", elevator);
exit(1);
}
elevators[elevator].open=1;
usleep(10*DELAY);
}
void elevator_close_door(int elevator) {
elevator_check(elevator);
log(9,"Closing elevator %d at floor %d\n",elevator,elevators[elevator].floor);
if(!elevators[elevator].open) {
log(0,"VIOLATION: attempted to close elevator %d door when already closed.\n", elevator);
exit(1);
}
sched_yield();
usleep(10*DELAY);
elevators[elevator].open=0;
}
void* start_elevator(void *arg) {
int elevator = (int)arg;
struct Elevator *e = &elevators[elevator];
e->last_action_seqno = 0;
e->seqno = 1;
e->passengers = 0;
e->trips = 0;
log(6,”Starting elevator %d\n”, elevator);
e->floor = 0;//elevator % (FLOORS-1);
while(!stop) {
e->seqno++;
elevator_ready(elevator,e->floor,elevator_move_direction,elevator_open_door,elevator_close_door);
sched_yield();
}
}
/* This is an internal struct used by the enforcement system
– there is no access to this from hw5 solution. */
static struct Passenger {
int id;
int from_floor;
int to_floor;
int in_elevator;
int trips;
int active;
struct timeval started_at;
double sum_duration;
double max_duration;
enum {WAITING, ENTERED, EXITED} state;
} passengers[PASSENGERS];
void passenger_enter(int passenger, int elevator) {
if(passengers[passenger].from_floor != elevators[elevator].floor) {
log(0,”VIOLATION: let passenger %d on on wrong floor %d!=%d.\n”, passengers[passenger].id, passengers[passenger].from_floor, elevators[elevator].floor);
exit(1);
}
if(!elevators[elevator].open) {
log(0,”VIOLATION: passenger %d walked into a closed door entering elevator %d.\n”, passengers[passenger].id, elevator);
exit(1);
}
if(elevators[elevator].passengers == MAX_CAPACITY) {
log(0,”VIOLATION: passenger %d attempted to board full elevator %d.\n”, passengers[passenger].id, elevator);
exit(1);
}
if(passengers[passenger].state!=WAITING) {
log(0,”VIOLATION: passenger %d told to board elevator %d, was not waiting.\n”, passengers[passenger].id, elevator);
exit(1);
}
log(6,”Passenger %d got on elevator %d at %d, requested %d\n”, passengers[passenger].id, elevator, passengers[passenger].from_floor, elevators[elevator].floor);
elevators[elevator].passengers++;
passengers[passenger].in_elevator = elevator;
passengers[passenger].state = ENTERED;
sched_yield();
usleep(DELAY);
}
void passenger_exit(int passenger, int elevator) {
if(passengers[passenger].to_floor != elevators[elevator].floor) {
log(0,”VIOLATION: let passenger %d off on wrong floor %d!=%d.\n”, passengers[passenger].id, passengers[passenger].to_floor, elevators[elevator].floor);
exit(1);
}
if(!elevators[elevator].open) {
log(0,”VIOLATION: passenger %d walked into a closed door leaving elevator %d.\n”, passengers[passenger].id, elevator);
exit(1);
}
if(passengers[passenger].state!=ENTERED) {
log(0,”VIOLATION: passenger %d told to board elevator %d, was not waiting.\n”, passengers[passenger].id, elevator);
exit(1);
}
log(6,”Passenger %d got off elevator %d at %d, requested %d\n”, passengers[passenger].id, elevator, passengers[passenger].to_floor, elevators[elevator].floor);
elevators[elevator].passengers–;
elevators[elevator].trips++;
passengers[passenger].in_elevator = -1;
passengers[passenger].state = EXITED;
passengers[passenger].trips++;
struct timeval after;
gettimeofday(&after,0);
double duration = after.tv_sec – passengers[passenger].started_at.tv_sec + (after.tv_usec – passengers[passenger].started_at.tv_usec)/1000000.0;
passengers[passenger].sum_duration += duration;
if(duration > passengers[passenger].max_duration)
passengers[passenger].max_duration = duration;
sched_yield();
usleep(DELAY);
}
void* start_passenger(void *arg) {
int passenger=(int)arg;
int wait = passenger*10;
usleep(wait*DELAY);
struct Passenger *p = &passengers[passenger];
log(6,”Starting passenger %d\n”, passenger);
p->from_floor = random()%FLOORS;
p->in_elevator = -1;
p->trips = 0;
p->active = 1;
p->sum_duration = 0;
p->max_duration = 0;
p->id = passenger;
while(!stop) {
if(p->from_floor!=0 && (random()%2))
p->to_floor = 0;
else
p->to_floor = 1 + random() % (FLOORS-1);
log(6,”Passenger %d requesting %d->%d\n”,
passenger,p->from_floor,p->to_floor);
struct timeval before;
gettimeofday(&before,0);
passengers[passenger].state=WAITING;
gettimeofday(&passengers[passenger].started_at,0);
passenger_request(passenger, p->from_floor, p->to_floor, passenger_enter, passenger_exit);
struct timeval after;
gettimeofday(&after,0);
int ms = (after.tv_sec – before.tv_sec)*1000 + (after.tv_usec – before.tv_usec)/1000;
log(1,”Passenger %d trip duration %d ms, %d slots\n”,passenger,ms,ms*1000/DELAY);
p->from_floor = p->to_floor;
// usleep(DELAY); // sleep for some time
usleep(wait*DELAY);
}
}
void* draw_state(void *ptr) {
while(!stop) {
printf(“\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n”);
for(int floor=FLOORS-1;floor>=0;floor–) {
printf(“%d\t”,floor);
for(int el=0;el
}
printf(” “);
int align = 5+5*ELEVATORS-ELEVATORS*ELEVATORS/4;
for(int p=0;p
}
while(align–>0) printf(” “);
printf((floor%2?” / “:” \\ “));
for(int p=0;p
}
printf(“\n”);
}
usleep(DELAY);
}
}
int main(int argc, char** argv) {
scheduler_init();
pthread_t passenger_t[PASSENGERS];
for(int i=0;i
mediumload: clean $(SCHEDULER)
@gcc -DDURATION=30 -DDELAY=3000 -DLOG_LEVEL=9 -DPASSENGERS=50 -DFLOORS=27 -DELEVATORS=4 -g –std=c99 main.c $(SCHEDULER) -o hw6 $(LIBS) $(CCOPT)
@./hw6 2> /tmp/logs.tmp
lowload: clean $(SCHEDULER)
@gcc -DDURATION=30 -DDELAY=3000 -DLOG_LEVEL=9 -DPASSENGERS=50 -DFLOORS=27 -DELEVATORS=10 -g –std=c99 main.c $(SCHEDULER) -o hw6 $(LIBS) $(CCOPT)
@./hw6 2> /tmp/logs.tmp
clean:
@rm -rf *~ *.dSYM $(BINARIES)
__MACOSX/hw6/._Makefile
__MACOSX/._hw6
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