Revert "SERVER-35783 add integration test for plan cache oscillation"

This reverts commit 420d3a4c2f086374c00aada0e0e9ecbe724ef988.

1 files changed, 0 insertions, 120 deletions

diff --git a/jstests/core/plan_cache_replanning.js b/jstests/core/plan_cache_replanning.js
deleted file mode 100644
index 82fe85a8794..00000000000
--- a/jstests/core/plan_cache_replanning.js
+++ /dev/null
@@ -1,120 +0,0 @@
-/**
- * This test will attempt to create a scenario where the plan cache entry for a given query shape
- * oscillates. It achieves this by creating two indexes, A and B, on a collection, and interleaving
- * queries which are "ideal" for index A with queries that are "ideal" for index B.
-*/
-(function() {
-    "use strict"
-
-    load('jstests/libs/analyze_plan.js');  // For getPlanStage().
-    const coll = db.plan_cache_replanning;
-
-    function getPlansForCacheEntry(query) {
-        let key = {query: query, sort: {}, projection: {}};
-        let res = coll.runCommand("planCacheListPlans", key);
-        assert.commandWorked(res, `planCacheListPlans(${tojson(key)}) failed`);
-        assert(res.hasOwnProperty("plans"),
-               `plans missing from planCacheListPlans(${tojson(key)}) failed`);
-
-        return res;
-    }
-
-    function planHasIxScanStageForKey(planStats, keyPattern) {
-        const stage = getPlanStage(planStats, "IXSCAN");
-        if (stage === null) {
-            return false;
-        }
-
-        return bsonWoCompare(keyPattern, stage.keyPattern) == 0;
-    }
-
-    const queryShape = {a: 1, b: 1};
-
-    // Carefully construct a collection so that some queries will do well with an {a: 1} index
-    // and others with a {b: 1} index.
-    for (let i = 1000; i < 1100; i++) {
-        assert.commandWorked(coll.insert({a: 1, b: i}));
-    }
-
-    for (let i = 1000; i < 1100; i++) {
-        assert.commandWorked(coll.insert({a: i, b: 2}));
-    }
-
-    // This query will be quick with {a: 1} index, and far slower {b: 1} index. With the {a: 1}
-    // index, the server should only need to examine one document. Using {b: 1}, it will have to
-    // scan through each document which has 2 as the value of the 'b' field.
-    const aIndexQuery = {a: 1099, b: 2};
-    // Opposite of 'aIndexQuery'. Should be quick if the {b: 1} index is used, and slower if the
-    // {a: 1} index is used.
-    const bIndexQuery = {a: 1, b: 1099};
-
-    assert.commandWorked(coll.createIndex({a: 1}));
-    assert.commandWorked(coll.createIndex({b: 1}));
-
-    // Run a query where the {b: 1} index will easily win.
-    assert.eq(1, coll.find(bIndexQuery).itcount());
-
-    // The plan cache should now hold an inactive entry.
-    let entry = getPlansForCacheEntry(queryShape);
-    let entryWorks = entry.works;
-    assert.eq(entry.isActive, false);
-    assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true);
-
-    // Re-run the query. The inactive cache entry should be promoted to an active entry.
-    assert.eq(1, coll.find(bIndexQuery).itcount());
-    entry = getPlansForCacheEntry(queryShape);
-    assert.eq(entry.isActive, true);
-    assert.eq(entry.works, entryWorks);
-    assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true);
-
-    // Now we will attempt to oscillate the cache entry by interleaving queries which should use
-    // the {a:1} and {b:1} index. When the plan using the {b: 1} index is in the cache, running a
-    // query which should use the {a: 1} index will perform very poorly, and trigger
-    // replanning (and vice versa).
-
-    // The {b: 1} plan is currently in the cache. Run the query which should use the {a: 1}
-    // index. The current cache entry will be deactivated, and then the cache entry for the {a: 1}
-    // will overwrite it (as active).
-    assert.eq(1, coll.find(aIndexQuery).itcount());
-    entry = getPlansForCacheEntry(queryShape);
-    assert.eq(entry.isActive, true);
-    assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {a: 1}), true);
-
-    // Run the query which should use the {b: 1} index.
-    assert.eq(1, coll.find(bIndexQuery).itcount());
-    entry = getPlansForCacheEntry(queryShape);
-    assert.eq(entry.isActive, true);
-    assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true);
-
-    // The {b: 1} plan is again in the cache. Run the query which should use the {a: 1}
-    // index.
-    assert.eq(1, coll.find(aIndexQuery).itcount());
-    entry = getPlansForCacheEntry(queryShape);
-    assert.eq(entry.isActive, true);
-    assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {a: 1}), true);
-
-    // The {a: 1} plan is back in the cache. Run the query which would perform better on the plan
-    // using the {b: 1} index, and ensure that plan gets written to the cache.
-    assert.eq(1, coll.find(bIndexQuery).itcount());
-    entry = getPlansForCacheEntry(queryShape);
-    entryWorks = entry.works;
-    assert.eq(entry.isActive, true);
-    assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true);
-
-    // Now run a plan that will perform poorly with both indices (it will be required to scan 500
-    // documents). This will result in replanning (and the cache entry being deactivated). However,
-    // the new plan will have a very high works value, and will not replace the existing cache
-    // entry. It will only bump the existing cache entry's works value.
-    for (let i = 0; i < 500; i++) {
-        assert.commandWorked(coll.insert({a: 3, b: 3}));
-    }
-    assert.eq(500, coll.find({a: 3, b: 3}).itcount());
-
-    // The cache entry should have been deactivated.
-    entry = getPlansForCacheEntry(queryShape);
-    assert.eq(entry.isActive, false);
-    assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true);
-
-    // The works value should have doubled.
-    assert.eq(entry.works, entryWorks * 2);
-})();