test.js

function B1() {
  /* Benchmark.prototype.setup goes here {{{ */
  var arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
      seed_data = 3,
      result = 0;
  /* }}} */
  var start = new Date;
  while (...) { /* do N iterations of test body */
    /* body of the case goes here {{{ */
    for (var i = 0; i < arr1.length; i++) {
      result += (seed_data * i);
    }
    /* }}} */
  }
  var end = new Date;
  /* N / (end - start) gives ops per ms result */
}

function B2() {
  /* Benchmark.prototype.setup goes here {{{ */
  var arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
      seed_data = 3,
      result = 0;
  /* }}} */
  var start = new Date;
  while (...) { /* do N iterations of test body */
    /* body of the case goes here {{{ */
    var l = arr1.length;
    for (var i = 0; i < l; i++) {
      result += (seed_data * i);
    }
    /* }}} */
  }
  var end = new Date;
  /* N / (end - start) gives ops per ms result */
}

function B3() {
  /* Benchmark.prototype.setup goes here {{{ */
  var arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
      seed_data = 3,
      result = 0;
  /* }}} */
  var start = new Date;
  while (...) { /* do N iterations of test body */
    /* body of the case goes here {{{ */
    var arr2 = [1, 2, 3, 4, 5, 6, 7, 8, 9];

    for (var i = 0; i < arr2.length; i++) {
      result += (seed_data * i);

      if (i === 4) {
        arr2.push(10)
      }
    }
    /* }}} */
  }
  var end = new Date;
  /* N / (end - start) gives ops per ms result */
}

Questions to ask yourself here: does operation that we are measuring in B3 perform the same amount of work as in B1 and B2?

The key to the answer here is to perceive the whole case as an operation. Then the answer is obvious: no, in B3 we allocate one additional array per iteration and we alter the array.

You can try something more like this http://jsperf.com/cache-length-for/4 instead where you try to balance the amount of work you do in all cases (it should be equal).

Awesome! Thanks for help me to find the right case.